Category: CBSE Class 9

In this page, we are providing Why Do we Fall Ill Class 9 Extra Questions and Answers Science Chapter 13 pdf download. NCERT Extra Questions for Class 9 Science Chapter 13 Why Do we Fall Ill with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 13 Extra Questions and Answers Why Do we Fall Ill

Extra Questions for Class 9 Science Chapter 13 Why Do we Fall Ill with Answers Solutions

Why Do we Fall Ill Class 9 Extra Questions Very Short Answer Type

Why Do We Fall Ill Class 9 Extra Questions Question 1.
What are infectious or communicable diseases?
Answer:
The diseases which are caused by infectious agents are called as infectious diseases as they can spread from one person to another through some medium or by direct contact.
Example: Pneumonia, common cold, tuberculosis, etc.

Why Do We Fall Ill Extra Questions Question 2.
What are congenital diseases?
Answer:
The diseases which are present in a person since birth are called congenital diseases.
Example: colour blindness.

Why Do We Fall Ill Class 9 Extra Questions Pdf Question 3.
Give one local and one general effect of the inflammation process.
Answer:
Swelling or pain is a local effect. Fever or headache is a general effect.

Class 9 Why Do We Fall Ill Extra Questions Question 4.
Name the organism causing the following diseases:
(a) Kala-azar
(b) Sleeping sickness
Answer:
Kala-azar is caused by Leishmania; Sleeping sickness is caused by Trypanosoma.

Class 9 Science Chapter 13 Extra Questions And Answers Question 5.
It was diagnosed that a patient has lost the power of fighting any infection.
(i) Name the disease the patient is suffering from.
(ii) Name the pathogen responsible for the disease.
Answer:
(i) AIDS.
(iii) HIV is the causative organism of AIDS. [HIV-Human immunodeficiency virus]

Extra Questions Of Why Do We Fall Ill Question 6.
What is immunity?
Answer:
The ability of the body of a person to fight against the disease causing organisms is called immunity. Name any disease that can be prevented by using vaccines.

Class 9 Science Ch 13 Extra Questions Question 7.
The diseases which can be prevented by using vaccines.
Answer:
The diseases which can be prevented by using vaccines are polio, small pox, diphtheria, tetanus, measles, etc.

Extra Questions Of Why Do We Fall Ill Class 9 Question 8.
What are principles of treatment of a disease?
Answer:
The principles of treatment are:

• reducing the effects of the disease.
• to kill the cause of the disease.

Extra Questions On Why Do We Fall Ill Question 9.
How do children in many parts of India get immune to hepatitis-A by the time they are five year old?
Answer:
The children become immune to hepatitis A virus as they are exposed to hepatitis A virus present in the water they drink.

Ch 13 Science Class 9 Extra Questions Question 10.
Name the causative organisms of tuberculosis and cholera.
Answer:
Tuberculosis: Caused by bacterium called as Mycobacterium tuberculosis.
Cholera is caused by a bacterium called as Vibrio cholera

Why Do we Fall Ill Class 9 Extra Questions Short Answer Type 1

Extra Questions Why Do We Fall Ill Question 1.
What are the differences between acute and chronic diseases?
Answer:
Acute disease:

• The disease which lasts for only very short periods of time is called acute disease.
• They get over soon and do not get much time to affect the health more adversely. Example: Common cold

Chronic disease:

• The disease which can last for a long time, even as much as a lifetime is called chronic disease.
• They have a long-term effect on the health of a person. Example: Elephantiasis

Class 9 Science Why Do We Fall Ill Extra Questions Question 2.
What is a pandemic disease? Give one example.
Answer:
The disease which affects the health of human population all over the world is called a pandemic disease. For example, AIDS.

Class 9 Biology Chapter 13 Extra Questions Question 3.
Mention the symptoms because of which you will visit a doctor and why?
Answer:
High fever, headache, tiredness, loose motion, cough and cold, loss of appetite and body weight are some of the symptoms for which we will visit a doctor because the doctor would be able to ascertain the disease caused on the basis of the symptoms produced and give medication accordingly.

Class 9 Biology Why Do We Fall Ill Extra Questions Question 4.
Why is DPT called triple antigen?
Answer:
DPT is called as triple antigen as it affects three antigens to produce the antibodies against them. The diseases are: Diphtheria, Pertussis and Tetanus.

Extra Questions From Why Do We Fall Ill Question 5.
What are the symptoms shown by a person if the
(i) lungs get infected?
(ii) stomach is infected?
Answer:
(i) Cough, breathlessness, tiredness are the symptoms produced if the lungs of a person get affected by a disease.

(ii) Loose motion, vomiting and stomachache are the symptoms produced when the stomach of a person gets affected by a disease.

Question 6.
“In our country majority of children are already immune to hepatitis A without giving its vaccine to them.” Justify this statement giving reasons.
Answer:
Majority of the children are already exposed to the hepatitis A virus by the time they are five years old because the water which the children generally drink contains the hepatitis A virus. The immune system thus helps to develop immunity against the virus.

Question 7.
List any four factors that must be taken care of by an individual for keeping good health.
Answer:
The four factors which must be taken care by an individual for keeping good health are:

• Proper nourishing balanced diet
• Clean and hygienic environment
• Good social environment
• Proper sanitation and cleanliness

Question 8.
Why are antibiotics effective against bacteria?
Answer:
Antibiotics block the biochemical pathways of the bacteria which inhibit their growth and kill them. For example: Penicillin blocks the pathway involved in the synthesis of cell wall which protects the bacteria. This inhibits their growth and kills them.

Question 9.
Influenza or common cold spreads faster and is difficult to control. Explain.
Answer:
Influenza and common cold are infectious diseases which are spread through air, so it becomes difficult to control them.

Question 10.
Why are overcrowded and poorly ventilated areas, major factors in the spread of air-borne diseases?
Answer:
In poorly ventilated and overcrowded areas, the little droplets containing air-borne microbes thrown by an infected person on coughing or sneezing, can be easily inhaled by a healthy person standing close by. This can start a new infection in the healthy person. Hence, such areas are major factors in the spread of air-borne diseases.

Question 11.
(a) Name the organ into which the malarial parasite enters after a mosquito bite.
(b) Give two examples of air-borne diseases.
Answer:
(a) The malarial parasite enters the liver of the healthy person after mosquito bite.
(b) Common cold and pneumonia are some air-borne diseases.

Question 12.
What are non-infectious diseases? Give two examples of non-infectious diseases.
Answer:
The diseases which do not spread from one person to the other person in the community are called as non-infectious diseases. They are caused due to internal factors.
Example: Cancer, high blood pressure, diabetes, etc.

Question 13.
Identify infectious and non-infectious diseases from the diseases given below: Tuberculosis, goitre, marasmus and typhoid.
Answer:

• Infectious diseases are tuberculosis and typhoid.
• Non-infectious diseases are goitre and marasmus.

Question 14.
Write the symptoms and diseases associated with the following:
(a) when microbes attack the lungs
(b) when the virus enters the liver
Answer:
(a) Tuberculosis may occur when microbes attack the lungs. The symptoms would be persistent cough, blood in sputum, prolonged low fever.

(b) Jaundice may occur when the virus enters the liver. The symptoms would be headache, high fever, loss of appetite and dark yellow coloured urine.

Question 15.
Why does a person suffering from the disease HIV-AIDS dies even due to small infection?
Answer:
The HIV damages the immune system of the person suffering from AIDS. Due to this the body of the person is unable to fight against even the small infections. The immune system of the person becomes very weak and ultimately leads to the death of the person due to several infections.

Why Do we Fall Ill Class 9 Extra Questions Short Answer Type 2

Question 1.
(a) Define disease.
(b) Explain briefly the two groups of diseases.
Or
Differentiate between infectious and non-infectious diseases.
Answer:
(a) Disease refers to any condition that disturbs the normal functioning of the living organism. It is a condition of disturbed ease when an individual feels uncomfortable.

(b) The two groups of diseases are infectious diseases and the non-infectious diseases.
Infectious diseases:

• These diseases are caused by infectious agents. These can spread from one person to another through some medium or by direct contact.
• Example: Pneumonia, common cold, tuberculosis, etc.

Non-infectious diseases:

• These do not spread form one person to the other person in the community. They are caused due to
• internal factors.
  Example: Cancer, high blood pressure, diabetes, etc.

Question 2.
What are acute and chronic diseases? Which one of the two is more harmful and why? Give an example in support of your answer.
Answer:
Acute disease:

• The disease which lasts for only very short periods of time is called acute disease.
• They get over soon and do not get much time to affect the health more adversely. Example: Common cold

Chronic disease:

• The disease which can last for a long time, even as much as a lifetime is called chronic disease.
• They have a long-term effect on the health of a person. Example: Elephantiasis

Chronic diseases are more harmful as they persist for a long time and thus adversely affect the general health of the person suffering from them.

Question 3.
Giving reasons, justify that it is difficult to prepare antiviral medicines than antibiotics.
Or
Why is it difficult to prepare anti-viral medicines than antibiotics?
Or
Explain why antibiotics do not work against viruses but work against many group of bacteria.
Answer:
Antibiotics are the chemicals which usually block the biochemical pathways of the bacteria. This inhibits their growth and kills them. But, the viruses do not have their own biochemical pathways as present in the bacteria. Viruses use the host machinery to multiply themselves. So, it becomes difficult to prepare anti-viral vaccines in order to inhibit their growth.

Question 4.
How do diseases spread through air? Name two such diseases.
Answer:
The diseases spread through air when little droplets thrown out by an infected person who sneezes or coughs are inhaled by a healthy person. This results in a new infection in the healthy person.
Example: Common cold, pneumonia and tuberculosis.

Question 5.
Why does a person who has suffered once from small pox cannot suffer with it again?
Answer:
The immune system of our body responds against the disease causing small pox virus when it encounters it for the first time. This interaction is slower and less vigorous, but the immune system remembers the virus specifically. Whenever the immune system encounters the virus after this first response, the interaction is more vigorous and the immune system completely eliminates the virus. Thus, the small pox virus is not able to affect the person who has suffered from it once.

Question 6.
How is the principle of immunisation implemented for eliminating polio?
Answer:
The polio vaccine which is administered to the children in the form of polio drops contains the weakened polio causing pathogen which is not able to cause the disease but enables the immune system of the child to produce antibodies against the polio virus. Subsequent encounter of the child to the polio virus activates the immune system which kills the polio virus and protects the immunised child from the disease. Thus, principle of immunisation helps in eliminating polio.

Question 7.
Give the cause and the remedy of:
(a) Hepatitis
(b) AIDS
(c) Malaria
Answer:

Question 8.
“Being disease free is not the same as being healthy.” Explain the statement by giving an example.
Answer:
Disease refers to any condition that disturbs the normal functioning of the living organism. It is a condition of disturbed ease when an individual feels uncomfortable. Being disease-free is a concept which applies to individual sufferer.

Even a disease free person can be considered as having a poor health if the person is unfit in terms of social and mental health. So, health is a concept which applies to societies and communities.

For example: a disease free person suffering from mental stress and tension cannot be considered as healthy.

Question 9.
A person was bitten by a stray dog. After some days his nature gets irritated, he started fearing water.
(a) Name the disease.
(b) Is there any vaccine available?
(c) Is there any plan of your local authority for the control of this disease?
Answer:
(a) The disease caused due to bite of a stray dog is rabies.
(b) Yes, anti-rabies vaccine is available.
(c) Local authorities have planned immunisation of the stray dogs against the disease.

Question 10.
It was diagnosed that the body of a patient had lost his power of fighting any infection. Name the disease the patient was suffering from. Which microbe is responsible for this disease? Give two ways by which it spreads from one person to another.
Answer:
The disease is called AIDS (Acquired immunodeficiency syndrome) and the microbe which is responsible for the disease is a virus called HIV (Human immunodeficiency virus). The ways to control the disease are:

• Creating awareness among people regarding the modes of spread of the disease.
• Use of disposable syringes in hospitals and clinics.
• Avoiding sexual contact with multiple partners.

Question 11.
Give two examples for each of the following:
(a) Acute diseases
(b) Chronic diseases
(c) Infectious diseases
(d) Non-infectious diseases
Answer:
(a) Acute diseases: Common cold, eye flu
(b) Chronic diseases: Tuberculosis (TB), asthma
(c) Infectious diseases: Chicken pox, small pox
(d) Non-infectious diseases: Goitre, Cancer

Question 12.
Name two diseases caused by Protozoans. What are their causal organisms.
Answer:
Malaria is caused by Plasmodium and Kala-azar caused by Leishmania Symptoms of malaria are: high fever and shaking chills that can range from moderate to severe, profuse sweating, headache, nausea, vomiting, etc. Symptoms of Kala-azar are breathing difficulty, recurrent fever and skin sores.

Question 13.
Which bacterium causes peptic ulcers? Who discovered the above pathogen for the first time?
Answer:
Heliobacterium pylori cause peptic ulcers. This pathogen was discovered for the first time by Marshall and Warren.

Question 14.
What is an antibiotic? Give two examples.
Answer:
The chemicals secreted by some microorganisms which can kill the bacteria and other disease causing pathogens are called antibiotic. Example: Penicillin, Streptomycin

Question 15.
Fill in the blanks
(a) Pneumonia is an example of ________ disease.
(b) Many skin diseases are caused by ________
(c) Antibiotics commonly block biochemical pathways important for the growth of ________
(d) Living organisms carrying the infecting agents from one person to another are called ________
Answer:
(a) Communicable
(b) Fungi
(c) Bacteria
(d) Vector

Question 16.
Name the target organs for the following diseases
(a) Hepatitis targets ________
(b) Fits or unconsciousness targets ________
(c) Pneumonia targets ________
(d) Fungal disease targets ________
Answer:
(a) Liver
(b) Brain
(c) Lungs
(d) Skin

Question 17.
Who discovered Vaccine’ for the first time? Name two diseases which can be prevented by using vaccines.
Answer:
Edward Jenner discovered vaccine for the first time. Polio, small pox, typhoid, etc. can be prevented by vaccines.

Question 18.
Fill in the blanks:
(a) ________ disease continues for many days and causes effect the on body.
(b) ________ disease continues for a few days and causes short term effect on the body.
(c) ________ is defined as physical, mental and social well-being and comfort.
(d) Common cold is an ________ disease.
(e) Many skin diseases are caused by ________
Answer:
(a) Chronic, long-term
(b) Acute
(c) Health
(d) Infectious
(e) Fungi

Question 19.
Classify the following diseases as infectious or non-infectious.
(a) AIDS
(b) Tuberculosis
(c) Cholera
(d) High blood pressure
(e) Heart disease
(f) Pneumonia
(g) Cancer
Answer:
(a) infectious
(b) infectious
(c) infectious
(d) non-infectious
(e) non-infectious
(f) infectious
(g) non-infectious

Question 20.
Name any two groups of microorganisms from which antibiotics can be extracted.
Answer:
Bacteria and fungi are the microorganisms from which antibiotics can be extracted.

Question 21.
Name any three diseases transmitted through vectors.
Answer:
Filariasis, Dengue, Chikungunya, Malaria, Kala-azar

Why Do we Fall Ill Class 9 Extra Questions Long Answer Type

Question 1.
(a) Give definition of health.
(b) State and explain in brief the four major factors which cause disease.
Answer:
(a) ‘Health’ is defined as a state of physical, emotional, mental and social well being of a person.
(b) The four factors whose non availability or absence can cause disease are:
(i) Proper nourishing food: If the person does not get proper nourishing food and a balanced diet, then disease may affect the person.

(ii) Proper hygienic conditions and cleanliness: If the surroundings are not hygienic and the public cleanliness is ignored in a society, then the individuals living there become more prone to diseases.

(iii) Good social conditions: The society in which the person lives should cater towards a healthy mind set of the members of the society. A bad social environment makes the individuals of the society develop an unhealthy attitude.

(iv) Good economic conditions: Poverty is a major cause of diseases and poor economic conditions increase the chances of spread of diseases in the society due to inadequate food and unhygienic conditions.

Question 2.
(a) Doctors diagnosed that Radha was suffering from HIV-AIDS. List any two methods by which she might have come in contact with the disease. Name the organ affected by this disease.
(b) Why antibiotics cannot be used for its treatment? Justify your answer.
Answer:
(a) AIDS can spread by four ways: Sexual contact, Transfusion of AIDS infected blood, use of infected syringe and from an infected mother to her baby during pregnancy or through breast feeding.

(b) Antibiotics block the biochemical pathways of the microorganism in order to inhibit their growth and kill them. But, the viruses do not employ such biochemical pathways of their own. They utilise the host machinery to multiply themselves. So, antibiotics cannot be used for the treatment of the viral diseases like AIDS.

Question 3.
(a) If a person is suffering from jaundice, name the mode of its transmission and the organ affected by the disease.
(b) List one general mode of prevention of jaundice.
(c) It has been observed that despite the availability of the vaccine of Hepatitis A in the market, it may not be necessary to be given to children by the time they are 5 years old. Why?
Answer:
(a) Mode of transmission of Jaundice: Contaminated water Organ affected by Jaundice: Liver

(b) Access to pure drinking water can prevent jaundice.

(c) Most of the children of India are already exposed to the hepatitis A virus through the drinking water and their immune system helps to develop immunity against the disease by the time they are five years old. So, it is not necessary to give them the hepatitis A vaccine.

Question 4.
Give an account of malaria, giving its causative agent, symptoms and control measures.
Answer:
Malaria is a protozoan disease caused by Plasmodium species. It gets transmitted by a vector called as the female Anopheles mosquito. The symptoms of malaria are high fever and body ache. The patient feels chills and shivering at regular time intervals. The disease can be controlled by eliminating the breeding grounds of mosquitoes, using mosquito nets or mosquito repellents at home, taking a dose of quinine medicine as prescribed by the doctor.

Question 5.
(a) Name two air-borne diseases. How does the disease causing microbes spread through air?
(b) How does HIV virus spread from a patient to a healthy person?
(c) How does the immune system of our body function?
Answer:
(a) Common cold, tuberculosis, pneumonia are some air-borne infections. They spread from one person to another when an infected person releases droplets into air while coughing or sneezing and these droplets get inhaled by a healthy person standing close by. The microbes start a new infection in the healthy person.

(b) HIV can spread by the following ways: Sexual contact, transfusion of AIDS infected blood, from an infected mother to her baby during pregnancy or through breast feeding, use of infected syringe.

(c) The immune system gets activated on encountering a disease causing microbe or on infection and recruits many cells to the affected tissue by a process called as inflammation. This results in pain, swelling, fever, etc., but ultimately it results in killing the disease causing microbe and prevent spread of infection.

Question 6.
(a) Discuss briefly the principle of immunisation.
(b) Mention any two diseases that can be prevented by immunisation.
Answer:
(a) The process by which immunity or resistance to a disease is developed in an individual by administration of biological preparations called vaccines is known as immunisation. In this process, a harmless or dead pathogen is introduced in the body of the organism by vaccination. The immune system of the organism gets activated and produces antibodies against the pathogen to kill it.

These antibodies remain in the memory of the immune system of the individual and in the case of any subsequent infection by the same microbe, these antibodies act quickly and more vigorously to kill the microbe. In this way, the microbe is not able to infect the person and the person gets immunised against the disease.

(b) Two diseases prevented by immunisation are: Measles and polio. Some other diseases which can be prevented by vaccination are diphtheria, pertusis, whooping cough, hepatitis, tetanus, etc.

Question 7.
Explain giving reasons
(а) Balanced diet is necessary for maintaining healthy body.
(b) Health of an organism depends upon the surrounding environmental conditions.
(c) Our surrounding area should be free of stagnant water.
(d) Social harmony and good economic conditions are necessary for good health.
Answer:
(a) Balanced diet provides raw materials and energy in appropriate amount needed for the proper growth and development of the body. It contains the nutrients like the protein, carbohydrates, fats, minerals etc., in the right amount which helps in the proper growth and functions. This ensures a good health of the individual.

(b) The physical, mental and the social well being of a person is affected by the surrounding environmental conditions. Since, these factors determine the health of a person, we can say that the health of a person depends on the environmental conditions. For example, Unhygienic surroundings as breeding grounds for disease causing organisms and their vectors can harm our health.

(c) Stagnant water is a source of many water-borne diseases and acts as breeding place for many insect vectors which can cause diseases in human beings. So, our surrounding area should be free of stagnant water.

(d) Social harmony ensures that the mental tensions and problems do not surround a person and friendly relations are maintained among the individuals of the society. They strive for the growth and well being of each other and thus better health is maintained in such environment and surroundings. Better economic conditions enable a person to invest more to attain good health by eating nutritive food, ensure protection from diseases, get better medical facilities, etc. So, health of a person is better maintained if economic conditions are good.

Question 8.
What is a disease? How many types of diseases have you studied? Give examples.
Answer:
Any condition that disturbs the normal functioning of the organism or its organs is called a disease. Diseases can be classified as acute, chronic, infectious, non-infectious, etc. The diseases which are for short duration are called acute whereas the diseases which persist for a long time are called chronic diseases, example common cold (acute) and tuberculosis (chronic).

The diseases which can be transmitted from one person to another are called infectious diseases whereas the ones which cannot be transmitted are called non-infectious diseases, example cancer and goitre respectively.

Question 9.
What do you mean by disease symptoms? Explain giving two examples.
Answer:
The physical disturbances or the visual changes which indicate the existence of a disease are called the symptoms of the disease.
Example:

• High fever and chills is a symptom of malaria.
• Appearance of lesions on the skin is a symptom of chickenpox.

Question 10.
Why is the immune system essential for our health?
Answer:
Immune system is essential as it helps our body to fight against disease causing microbes. It protects our body by secreting chemicals called antibodies or other substances which protect us from harmful microbes.

Question 11.
What precautions will you take to justify “prevention is better than cure”?
Answer:
The precautions which can be taken to protect ourselves from the disease are:

• Maintaining personal hygiene as well as clean surroundings.
• Being aware about diseases, their symptqms and the organism which causes then.
• Eating a balanced diet.
• Timely vaccination and regular medical, check-up.

Question 12.
Why do some children fall ill more frequently than others living in the same locality?
Answer:
The reasons due to which some children fall ill more frequently than others living in the same locality can be:

• Poor and inefficient immune system.
• Unhealthy food habits and not taking balanced diet.
• Not maintaining personal hygiene.

Question 13.
Why are antibiotics not effective for viral disease?
Answer:
The mode of action of the antibiotics is blocking the biosynthetic pathways of the bacteria and other microbes. However, viruses do not have such biochemical pathways or have very few biochemical mechanisms of their own, hence viruses are unaffected by antibiotics.

Question 14.
Becoming exposed to or infected with an infectious microbe does not necessarily mean developing noticeable disease. Explain.
Answer:
Becoming exposed to or infected with an infectious microbe does not necessarily mean developing noticeable disease because the immune system becomes active when it encounters a pathogen. If the immune system of the organism is able to ward off the disease causing microbe, then the disease will not occur and the person will remain disease-free. So, a strong immune system can ensure that we do not suffer from diseases even if we are exposed to infectious microbes.

Question 15.
Give any four factors necessary for a person to be healthy.
Answer:
The four factors necessary for a person to be healthy are:

• Proper nourishment and a balanced diet which help to provide a strong immune system.
• Maintaining personal hygiene to protect ourselves from infectious microbes.
• Timely vaccination against various diseases.
• Maintaining community hygiene and clean surroundings will to prevent the incidence of waterborne and the air-borne diseases.

Question 16.
Why is AIDS considered to be a ‘Syndrome’ and not a disease?
Answer:
The causative organism of AIDS is a virus called the HIV which gets transmitted through sexual contact, transfusion of contaminated blood, intravenous syringes used by drug addicts and from a mother suffering with AIDS to her child, spread to lymph nodes all over the body. The virus attacks the immune system of the organism and makes it so weak that it cannot fight against even the minor infections. Due to this even a small cold can develop into pneumonia or minor stomach infection may lead to severe diarrhoea and blood loss in stools.

Ultimately the immune system becomes very weak and leads to the death of the person suffering from AIDS. Many diseases simultaneously develop in such a person, so there is no specific disease symptom but a number of symptoms occur in the person. This group of symptoms are called as ‘syndrome’. So, AIDS is not a disease, it is a syndrome.

Why Do we Fall Ill Class 9 Extra Questions HOTS

Question 1.
The symptoms seen in a patient are high fever, constipation and stomach pain. A doctor wants to confirm whether the patient is suffering from Amoebiasis or typhoid. How can the doctor confirm it?
Answer:
The doctor can advise the patient to get a widal test done which helps to confirm the presence of typhoid causing microorganism in the body. If typhoid is ruled out then the patient is most likely to suffer from Amoebiasis.

Question 2.
Name the disease:
(a) that spreads by sexual contact but not by casual contact like handshake.
(b) that can be caused by virus, bacteria or protozoa.
(c) that can be caused due to bite of stray dog.
Answer:
(a) AIDS
(b) Diarrhoea
(c) Rabies

Question 3.
A mother who had suffered from chicken pox in her childhood, is now taking care of her child, who is suffering from the same disease. What are the chances of her mother having chicken pox? Explain.
Answer:
The mother will not suffer from chicken pox as she has become immune to chicken pox. During the first infection, her immune system develops antibodies against chicken pox. These antibodies respond with more vigour during the next infection, eliminate the microbe and thus prevent the person from the disease.

Question 4.
Name the disease:
(a) in which the liver of the person is the target.
(b) in which saliva of the infected animal spreads infection.
(c) against which BCG vaccine is given.
(d) for which widal test is done.
(e) in which sexual contact spreads the disease but not the physical contact in form of handshakes or hugging.
Answer:
(a) Jaundice and hepatitis
(b) Rabies
(c) Tuberculosis
(d) Typhoid
(e) AIDS

Question 5.
The immune system of a patient has been damaged by a virus. What is the probable disease which would have led to this effect on the immune system? What are the ways by which the pathogen would have been transferred into the individual?
Answer:
The disease is most likely to be AIDS caused by the HIV virus which damages the immune system of the person.

AIDS can be spread in four ways: Sexual contact, transfusion of AIDS infected blood, from an infected i mother to her baby during pregnancy or through breast feeding.

Question 6.
A patient bitten by a stray animal complained of excessive salivation, restlessness and a fear of water. What can be the probable disease and its cause? How can such disease be treated or controlled?
Answer:
The patient is probably suffering from rabies caused by the rabies virus. The disease can be controlled by ensuring proper vaccination of stray animals like dogs. The disease can be treated by timely administration of anti rabies vaccine.

Question 7.
A patient went to a doctor and complained of having a persistent cough, blood in sputum, breathlessness and loss of body weight. Name the disease and the causative microbe. How is the disease transmitted? Which vaccine is given to develop resistance against the disease?
Answer:
The disease from which the patient is suffering is Tuberculosis (T.B.) and the causative microbe is . a bacteria called Mycobacterium tuberculosis.. The disease gets transmitted when a healthy person inhales the droplets released by an infected person on coughing, sneezing or while talking. The vaccine i called BCG is given to develop resistance against the disease.

Question 8.
A child was suffering from loose motions, frequent vomiting, feeling of nausea and watery diarrhoea which led to an excessive loss of fluids from his body. Name the disease and its causative microbe. What measures, can be taken to treat this disease? How can one avoid getting infected by the disease?
Answer:
The person is infected with disease called cholera which is caused by a virus called Vibrio cholerae. The measures which can be taken for treatment of the disease are: Giving ORS-Oral rehydration solution to the patient and giving antibiotics prescribed by the doctor. The disease can be avoided by eating clean and hygienic food and timely immunisation against the disease.

Why Do we Fall Ill Class 9 Extra Questions Value Based (VBQs)

Question 1.
An HIV infected employee was dismissed from service by the employer. The colleagues of the employee in the office intervened and convinced the boss of the company to reinstate the services of the employee as HIV is not transmitted by mere physical contact. They told the employer that such person should not be made a victim of social stigma on contracting the disease.
(i) Name the disease which the employee was suffering from.
(ii) How does the disease get transmitted from one person to another?
(iii) What values are shown by the colleagues of the employee?
Answer:
(i) AIDS: Acquired immunodeficiency syndrome.

(ii) The disease gets transmitted by four ways:
(a) Sex with multiple partners.
(b) Transfusion of HIV contaminated blood.
(c) Use of HIV contaminated syringe.
(d) From infected mother to her child at the time of delivery of the baby.

(iii) The values shown by the colleague of the employee are: awareness, helpfulness, empathy and knowledge about their rights.

Question 2.
Rohit and’ Mayank are roommates in the hostel. Mayank is an avid reader of science magazines and articles. Mayank immediately took Rohit to a doctor when Rohit complained of tiredness, nausea, vomiting and pale eyes. What is the probable disease Rohit was suffering from? Name the causative microbe. What values are shown by Mayank?
Answer:
The symptoms of Rohit show that he was suffering from Jaundice. The microbe which causes the disease is virus. The values shown by Mayank are awareness, scientific temper, concern for others and helpful nature.

Question 3.
The parents of a new bom child were afraid to take their child for vaccination as someone told them that the vaccine contains harmful germs. However, when they consulted Dr. Shweta, she listened to them patiently and then cleared their doubts about vaccines. She told them that the vaccines only contain dead germs or inactivated germs which do not cause the disease. How do the dead germs in vaccines protect from diseases? What are the values shown by Dr. Shweta?
Answer:
The dead or the inactivated germs in the vaccines stimulate our immune system to develop antibodies against the disease causing microbes present in the vaccine and store the antibodies for action during any subsequent infection. In this way the vaccines provide immunity to the person against the disease. The values shown by Dr. Shweta are concern for others, helpfulness and patience.

Question 4.
Megha’s told her that he had been feeling tired, breathless, had persistent cough and blood was coming out in his sputum. She told him that these are symptoms of an infectious disease and advised him to consult a doctor. She also took him for diagnostic tests prescribed by the doctor. The helps was diagnosed as suffering from an infectious disease. He took the medicines prescribed by the doctor and got cured of the disease.
(i) Name the most likely disease and microbe from which the help suffered?
(ii) How does the disease get transmitted?
(iii) What are the values depicted by Megha?
Answer:
(i) The disease is most likely Tuberculosis caused by bacteria.

(ii) The disease gets transmitted when a healthy person inhales the droplets released into air by an infected person while coughing or sneezing.

(iii) Megha shows a helpful nature, scientific attitude, concern for others and sympathetic nature.

Question 5.
The municipality of an area advised the residents of the area to eliminate the breeding grounds of mosquitoes and use mosquito repellents in homes to avoid a disease which results in chills and recurring fever in patients. The children of the area formed teams which inspected the surrounding areas and put few drops of oil or petrol on stagnant water pools of the area.
(i) Name the disease, causative organism and the vector of the disease which was spreading in the area.
(ii) How do the drops of petrol or oil on stagnant water prevent the spread of the disease?
(iii) What are the values shown by the students?
Answer:
(i) The disease is Malaria caused by Plasmodium and transmitted by the vector female Anopheles mosquito.

(ii) The drops of oil or petrol in the stagnant water pools kill the larvae of the mosquito as they will not be able to breathe.

(iii) The values shown by the children are: public good, helpful attitude, scientific temper and work for social good.

In this page, we are providing Improvement in Food Resources Class 9 Extra Questions and Answers Science Chapter 15 pdf download. NCERT Extra Questions for Class 9 Science Chapter 15 Improvement in Food Resources with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 15 Extra Questions and Answers Improvement in Food Resources

Extra Questions for Class 9 Science Chapter 15 Improvement in Food Resources with Answers Solutions

Improvement in Food Resources Class 9 Extra Questions Very Short Answer Type

Improvement In Food Resources Class 9 Extra Questions And Answers Question 1.
What is the advantage of selecting seeds of crops with wider adaptability for agriculture?
Answer:
Wider adaptability helps in stabilizing crop production under different environmental conditions.

Class 9 Science Chapter 15 Extra Questions And Answers Question 2.
Name the type of nutrient that we get from mustard seeds and linseed.
Answer:
Mustard seeds and linseed are oilseed crops that provide fats.

Improvement In Food Resources Class 9 Notes Extra Questions And Answers Question 3.
Mention any two abiotic factors that affect crop production.
Answer:
Drought, salinity, waterlogging, heat, cold and frost are the abiotic factors that affect crop production.

Improvement In Food Resources Class 9 Important Questions Question 4.
Students were asked to select one that is not a source of starch amongst the following rice, wheat, sunflower seeds, and potato tuber.
Answer:
Sunflower seeds are not a source of starch. They are a source of fats.

Improvement In Food Resources Class 9 Questions And Answers Pdf Question 5.
Improved varieties can be produced in both animals and plants. How?
Answer:
Improved varieties can be produced in both animals and plants by hybridization and genetic modification.

Ch 15 Science Class 9 Extra Questions Question 6.
Name two protein-containing Rabi crops.
Answer:
Protein containing Rabi crops are gram and peas.

Improvement In Food Resources Class 9 Questions And Answers Question 7.
Identify two crops from the following which provide us carbohydrates for energy requirement. Black gram, wheat, lentil, and rice.
Answer:
Wheat and rice provide energy.

Improvement In Food Resources Class 9 Notes Questions And Answers Question 8.
Name two plants which are used as biopesticide in organic farming.
Answer:
Turmeric and leaves of Neem plant are used as biopesticide in organic farming.

Improvement In Food Resources Question Answer Question 9.
Name the two vitamins which are added in the poultry feed.
Answer:
Vitamin A and K are the vitamins that are added in the poultry feed.

Extra Questions From Chapter Improvement In Food Resources Question 10.
Name the major nutrient which we get from fish.
Answer:
The major nutrient which we get from fish is protein.

Class 9 Improvement In Food Resources Extra Questions Question 11.
How does Catla differ from Mrigal?
Answer:
Callas are surface feeders while Mrigals are bottom feeders.

Class 9 Science Ch 15 Extra Questions Question 12.
What is mariculture?
Or
What do you mean by mariculture?
Answer:
Mariculture is a practice in which marine fishes like mullets, bhetki, and pearl spots are cultured in coastal waters on a commercial scale.

Class 9 Chapter 15 Extra Questions Question 13.
Give one example of local variety and foreign variety of bees.
Answer:
Local variety: Apis cerana Indica is commonly known as the Indian bee.
Foreign variety: Apis mellifera is the Italian variety of bee.

Important Questions Of Improvement In Food Resources Question 14.
State one factor which affects the quality of honey produced.
Answer:
The quality of honey depends upon the pasturage or the flowers available to the bees for nectar and pollen collection.

Question 15.
Give two examples of exotic breeds of cows that are selected for a long lactation period.
Answer:
Exotic or foreign breeds like Jersey and Brown Swiss are selected for long lactation periods.

Question 16.
What is meant by beekeeping?
Answer:
Beekeeping is the practice of rearing the honeybees for obtaining honey and bee wax.

Question 17.
Which species of the honeybee is commonly used for commercial honey production throughout the country?
Answer:
Apis mellifera, the Italian variety of bee is commonly used for commercial honey production throughout the country.

Improvement in Food Resources Class 9 Extra Questions Short Answer Type 1

Question 1.
Mention the names of four marine fish of high economic value.
Answer:
Fishes like mullets, bhetki, pearl spots and prawns are of high economic value.

Question 2.
Give two examples of shellfishes.
Answer:
Shellfish include prawns, mussels and oysters.

Question 3.
Name two desirable traits for variety improvement in poultry farming.
Answer:
The two desirable traits for variety improvement in poultry farming are:

• number and quality of chicks;
• dwarf broiler parent for commercial chick production.

Question 4.
Which method is commonly used for improving cattle breeds and why?
Answer:
Crossbreeding between the indigenous and exotic breeds is commonly used for improving cattle breeds. This is done as it helps to incorporate the desirable qualities like a long lactation period of exotic breeds with the disease resistance of indigenous breeds in the progeny.

Question 5.
What are ‘Sahiwal’ and ‘Jersey’ breeds?
Answer:
Sahiwal is an indigenous breed of cow whereas Jersey is the exotic breed of cow.

Question 6.
State the food requirements of dairy animals.
Answer:
The food requirements of dairy animals are of two types:

• Maintenance requirement, which is the food required to support the animal to live a healthy life.
• Milk producing a requirement, which is the type of food required during the lactation period.

Question 7.
What is mixed cropping? How does it help a farmer?
Answer:
Mixed cropping is the practice of growing two or more crops simultaneously on the same piece of land. For example, wheat and gram, or wheat and mustard, or groundnut and sunflower. It helps the farmer as it reduces the risk and gives some insurance against the failure of one of the crops.

Question 8.
State two advantages of fertilizers over manure.
Answer:
Fertilizers are more advantageous than manure as:

• Fertilizers are nutrient specific and provide the specific nutrients like N, P, K to the soil.
• They are not bulky, so are easier to transport.

Question 9.
What do you mean by vermicompost?
Answer:
The compost prepared by using earthworms to hasten the process of decomposition of plant and animal refuse is called vermicompost.

Question 10.
What are the advantages of fish farming?
Answer:
The advantages of fish farming are:

• It helps to get a large amount of desired fishes from a small area.
• It enables a variety of improvement in a better way.

Question 11.
What is a GM crop? Name anyone such crop which is grown in India.
Answer:
GM crops are the crops obtained by introducing a gene that would provide the desired characteristic to the crop plant. Bt Cotton is a genetically modified (GM) crop which has been made insect-resistant by introducing a gene from bacteria.

Question 12.
Give any two differences between macronutrients and micronutrients.
Answer:
Macronutrient:

1. Macronutrients are generally present and utilized by plant tissues in large amounts.
2. Apart from carbon, hydrogen, oxygen, the macronutrients include six elements- nitrogen, phosphorous, sulfur, potassium, calcium, and magnesium.

Micronutrient

1. The nutrients which are needed by plant tissues in small amounts are called micronutrients.
2. These include seven elements: iron, manganese, copper, zinc, molybdenum, boron, and chlorine.

Question 13.
What is a green revolution?
Answer:
An increase in crop production (especially wheat and rice) due to the use of plant breeding techniques and better farming practices, is called the green revolution.

Question 14.
What are the weeds? How do they harm crop plants?
Or
How do they prevent the growth of crops?
Answer:
The unwanted plants which grow along with the crop in the cultivated field are called weeds. Weeds compete for food, space, and light with the crop plant and reduce the growth of the crop.

Question 15.

1. What are many animals?
2. Give two examples of such animals.

Answer:

1. Milk: producing females of cattle are called many animals (dairy animals).
2. Examples of such animals are: cows, buffaloes, goats etc.

Question 16.
Mention two advantages of animal husbandry.
Answer:
Animal husbandry is advantageous to get

1. Higher milk production from cattle and higher egg production from poultry.
2. Utilization of animal wastes in a beneficial way.

Question 17.
State two characteristics of a healthy animal.
Answer:
The two characteristics of a healthy animal are:

1. A healthy animal feeds regularly.
2. Healthy animals have a normal posture.

Question 18.
Why would a cattle breeder choose to cross-breed a Jersey cow with a Red Sindhi? State two reasons.
Answer:
Jersey cows have a longer lactation period while Red Sindhi are resistant to diseases. So, the two are cross-bred to get animals with

• Long lactation period
• Resistance to diseases.

Question 19.
How are new varieties of poultry birds with desired traits produced?
Answer:
Poultry birds with desired traits are produced by crossbreeding the indigenous variety like Aseel with the exotic variety like Leghorn.

Question 20.
What desirable traits are focussed to develop by cross-breeding indigenous and exotic breeds of fowl?
Or
Mention any four desirable traits for which new varieties are produced?
Answer:
Desirable traits of poultry are:

• number and quality of chicks;
• dwarf broiler parent for commercial chick production;
• summer adaptation capacity/tolerance to high temperature;
• low maintenance requirements;
• reduction in the size of the egg-laying bird with the ability to utilize more fibrous cheaper diets formulated using agricultural by-products.

Question 21.
1. State one demerit with a composite fish culture system.
2. How can this problem be overcome?
Answer:
1. The demerit of composite fish culture system is:
(a) Many of the fishes used in the composite culture system breed only during monsoon.
(b) Lack of availability of good quality seed.
2. The problem is overcome by breeding these fish in ponds using hormonal stimulation. This ensures the supply of pure fish seed in desired quantities.

Improvement in Food Resources Class 9 Extra Questions Short Answer Type 2

Question 1.
(a) Suppose you are in charge of a grain store. How will you find out the presence of pests? Mention any two indicators.
(b) Which method is most effective for destroying insects in stored food grains, spraying or fumigation?
Answer:
(a) The indicators for the presence of pests in grain store are:
(i) Damaged or broken grains having perforations or holes.
(ii) Discoloration of grains, degraded quality of food grains.
(b) Fumigation is a better method than spraying of chemicals as the fumigants completely fill the area with gaseous particles that suffocate the pests and kill them.

Question 2.
Write a short note on marine fisheries.
Answer:
Marine fishery resources include 7500 km coastline and the deep seas beyond it. Some marine fish varieties are pomfret, mackerel, tuna, sardines, and Bombay duck. Fishes like mullets, bhetki, and pearl spots; shellfish such as prawns, mussels, and oysters as well as seaweed are of high economic value.

Question 3.
Differentiate between:
1. Inland fishery and marine fishery
2. Apiculture and aquaculture
Or
Distinguish between
1. Inland fishery and marine fishery
2. Culture fishery and capture fishery
3. Apiculture and aquaculture
Answer:
Inland Fishery:

• Inland fishery resources include canals, ponds, reservoirs, rivers and brackish water resources like estuaries and lagoons.
• Most fish production from these resources is through aquaculture.

Marine Fishery:

• Marine fishery resources include 7500 km coastline and the deep seas beyond it.
• Most fish production from these resources is through mariculture.

Apiculture:
The practice of rearing and maintenance of honeybees for obtaining honey and bee wax is called as apiculture.

Aquaculture:
The practice of breeding, rearing, and harvesting of plants and animals in all types of water environments including ponds, rivers, lakes, and the ocean.

Capture fishery:
Fish are obtained from natural resources in capture fishery.

Culture fishery:
Fish farming done in ponds or tanks is called culture fishery.

Question 4.
Differentiate between layers and broilers. What type of food should be given to broilers?
Answer:
Layers:

• The egg-laying birds which are managed for egg production are called layers.
• The layers start laying eggs at the age of 20 weeks. So, they are kept for longer periods of, around 500 days, called laying period.
• They require enough space, proper light, and hygienic conditions.
• Their feed consists of vitamins, minerals, and certain micronutrients that affect the hatchability of the eggs.

Broilers:

• The birds maintained for meat production are called broilers.
• The broilers are raised in poultry farms up to 6-7 weeks and usually weigh around 700 g to 1.5 kg.
• More stress is given to maintain the feathering, carcass quality, and low levels of mortality.
• The requirement of the broilers is protein and fat-rich food. The level of vitamin A and vitamin K is kept high in their feed.

Question 5.
What are the advantages of bee-keeping?
Answer:
The advantages of bee-keeping are:

• It helps to obtain honey and bee wax which are used in medicines and many cosmetic preparations respectively.
• It requires a very low amount of investment.
•  It is less labor-intensive.
• It helps to increase the pollination of flowers.

Question 6.
What is honey? What does the quality of honey depend upon?
Answer:
It is a sugary substance produced and stored by insects like bees in the bee-hives. It has a very high nutritional and medicinal value. The value or quality of honey depends upon the pasturage, or the flowers available to the bees for nectar and pollen collection.

Question 7.
Differentiate between the following:

1. Capture fishery and culture fishery
2. Beekeeping and Poultry farming

Answer:

1. Capture fishery: Fish are obtained from natural resources in capture fishery.
   Culture fishery: Fish farming is called culture fishery.
2. Beekeeping is the practice of rearing honeybees for obtaining honey and bee wax.
   Poultry farming is undertaken to raise domestic fowl called layers for egg production and the broilers for chicken meat.

Question 8.
Discuss various methods of weed control.
Answer:
The various methods of weed control are:
Mechanical removal, a spray of chemicals called weedicides, and preventive methods like proper seedbed preparation, timely sowing of crops, intercropping, and crop rotation.

question 9.
Discuss the role of hybridization in crop improvement.
Answer:
Hybridization is one of the methods of crop production which ensures high yield. Hybridization refers to crossing between two genetically dissimilar plants each of which possesses a particular desired character. The two varieties are cross-bred during the process to incorporate both the desirable characteristics in a single variety. This method of hybridization improves crops with respect to yield, disease resistance, pest resistance, etc.

Question 10.
What are the macro and micronutrients of plants? Name two of each kind.
Answer:
Macronutrients: The nutrients which are required in large quantities. They are six.
Example: Nitrogen, Phosphorus.
Micronutrients: The nutrients which are required in small quantities. They are seven.
Example: Iron, Copper.

Question 11.
Name any three processes in plants that are affected by deficiency in nutrients.
Answer:
Deficiency of nutrients affects the following things in plants:

• reproduction
• growth
•  susceptibility to diseases.

Question 12.
Distinguish between intercropping and mixed cropping. List any two advantages of intercropping over mixed cropping.
Answer:
Mixed cropping Growing two or more crops simultaneously on the same piece of land.
Inter-cropping: Growing two or more crops simultaneously on the same field in a definite pattern.
Intercropping is better than mixed cropping because:

• The application of pesticides to individual crop is easier and done as per the need of the crop.
• Different crops can be separately harvested and threshed.

Question 13.
What is hybridization? How is it done?
Answer:
Hybridization refers to crossing between two genetically dissimilar plants each of which possesses a particular desired character. The two varieties are cross-bred during the process to incorporate both the desirable characteristics in a single variety. This method of hybridization improves crops with respect to yield, disease resistance, pest resistance, etc.

Question 14.
How are cultivation practices and crop yield related to weather?
Answer:
1. Agriculture in India is largely dependent on the monsoon. The success of crops in most areas is dependent on timely monsoons and sufficient rainfall spread through most of the growing season. Hence, poor monsoons cause crop failure.

2. The cultivation of different varieties of crops is dependent on abiotic stresses like droughts and floods situation in an area.

Question 15.
Name any three methods of irrigation and briefly describe them.
Or
(a) Describe any two irrigation systems adopted in India to supply water to agricultural lands.
(b) Write two advantages of building check dams.
Answer:
(a) The three different types of irrigation systems are:
Wells: There are two types of wells: dug wells and tube wells. In a dug well, water is collected from water-bearing strata. Tube wells can tap water from the deeper strata. From these wells, water is lifted by pumps for irrigation.

Canals: In this system canals receive water from one or more reservoirs or from rivers. The main canal is divided into branch canals having further distributaries to irrigate fields.

River Lift Systems: In this system, water is directly drawn from the rivers for supplementing irrigation in areas close to rivers.

The advantage of check dams are-

• Leads to an increase in groundwater levels.
• The check-dams stop the rainwater from flowing away and reduce soil erosion.

Question 16.
What factors may be responsible for losses of grains during storage? Give one example of both. Give one method/preventive measure to control such losses.
Or
What preventive and control measures should be taken before the storage of grains?
Answer:
Factors responsible for losses of grains during storage are:

• Biotic: insects, rodents, fungi, mites, and bacteria
• Abiotic: inappropriate moisture and temperatures in the place of storage.

Ways to reduce loss during storage of grains:

1. Proper treatment and systematic management of warehouses.
2. They include strict cleaning of the produce before storage.
3. Proper drying of the product first in sunlight and then in shade.
4. Fumigation using chemicals that can kill pests.

Question 17.
List six facilities that must be provided to cattle to ensure their good health and production of clean milk.
Answer:
Six facilities that must be provided to cattle to ensure their good health and production of clean milk are:

• The cattle should be housed in well-ventilated sheds,
• The floor of the cattle shed needs to be sloping so as to stay dry and to facilitate cleaning.
• The feeding of cattle should be carried out in a scientific manner. The feed should have:

(a) Roughage, which is largely fiber and
(b) Concentrates, which are low in fiber and have high levels of proteins and other nutrients.

• Leads to an increase in groundwater levels.
• The check-dams stop the rainwater from flowing away and reduce soil erosion.

Question 18.
What is the lactation period? Name two breeds of cattle that are selected for their lactation period. Why are they crossed with local breeds?
Answer:
The period of milk production after the birth of a calf is called the lactation period. Milk production can be increased by increasing the lactation period. Exotic or foreign breeds like Jersey and Brown Swiss are selected for long lactation periods. They (exotic breeds) are crossed with local breeds to get animals with the desired qualities of both the animals.

Question 19.
Which variety of honeybee is advantageous – Apis cerana indica or Apis mellifera and why?
Answer:
Apis mellifera, the Italian variety of bee is better than Apis cerana indica. It has the following advantages:

• They have high honey collection capacity.
• They sting somewhat less.
• They stay in a given beehive for long periods and breed very well.

Question 20.
Match the column A with column B

Answer:
(a) (ii) Surface feeders
(b) (iii) Middle-zone feeders
(c) (i) Bottom feeders
(d) (iv) Culture fishery

Question 21.
Fill in the blanks
(a) Pigeon pea is a good source of ………..
(b) Berseem is an important ………… crop.
(c) The crops which are grown in the rainy season are called ………… crops.
(d) ………… are rich in vitamins.
(e) ………… crop grows in the winter season.
Answer:
(a) protein
(b) fodder
(c) Kharif
(d) Vegetables
(e) Rabi

Question 22.
What is a GM crop? Name anyone such crop which is grown in India.
Answer:
GM crops are the crops obtained by introducing a gene that would provide the desired characteristic to the crop plant. Bt Cotton is a genetically modified (GM) crop which has been made insect-resistant by introducing a gene from bacteria.

Question 23.
List out some useful traits in the improved crop.
Answer:
Some useful traits of improved crops are:
(a) higher yield
(b) improved nutritional quality
(c) resistance to biotic and abiotic stresses
(d) change in maturity duration
(e) wide range of adaptability
(f) desired agronomic characteristics.

Question 24.
Why is organic matter important for crop production?
Answer:
Organic matter is important for crops production because:
(a) it helps in improving soil structure.
(b) it helps in increasing water holding capacity of sandy soil.
(c) It helps in drainage and in avoiding water logging in clayey soils.

Question 25.
Why is excess use of fertilizers detrimental for the environment?
Answer:
Excess use of fertilizers is detrimental for the environment as their residual and unused amounts become pollutants for air, water, and soil. The fertilizers get washed away in the nearby water bodies and cause excessive growth of algae and aquatic plants which can adversely affect the aquatic life and drain the oxygen from water bodies.

Question 26.
Give one word for the following:
(a) Farming without the use of chemicals as fertilizers, herbicides and pesticides is known as ……….
(b) Growing of wheat and groundnut on the same field is called as ……….
(c) Planting soybean and maize in alternate rows in the same field is called as ……….
(d) Growing different crops on a piece of land in pre-planned succession is known as ……….
(e) Xanthium and Parthenium are commonly known as ……….
(f) The causal organism of any disease is called as ……….
Answer:
(a) organic farming
(b) mixed cropping
(c) intercropping
(d) crop-rotation
(e) weeds
(f) pathogen

Question 27.
Match the following A and B

Answer:
(a) → (iii) Drought animals
(b) → (v) Aseel
(c) → (iv) The local breed of cattle
(d) → (i) Milk producing female
(e) → (ii) Broiler

Question 28.
If there is low rainfall in a village throughout the year, what measures will you suggest to the farmers for better cropping?
Answer:
If there is low rainfall in a village throughout the year, then the farmers will be suggested to:
(a) Cultivate crops varieties that are drought resistant and early maturing.
(b) Increase the humus content of the soil in order to increase its water holding capacity.

Question 29.
Group the following and tabulate them as energy-yielding, protein yielding, oil yielding, and fodder crop. Wheat, rice, berseem, maize, gram, oat, pigeon gram, sudangrass, lentil, soybean, groundnut, castor, and mustard.
Answer:

1. Energy yielding: wheat, rice, maize
2. Protein yielding: gram, pigeon gram, lentil, soybean
3. Oil yielding: groundnut, castor, mustard, soybean
4. Fodder crops: berseem, oat, sudangrass

Question 30.
Define the term hybridization and photoperiod.
Answer:
Hybridization is one of the methods of crop production which involves crossing between two genetically dissimilar plants each of which possesses a particular desired character. The photoperiod is the duration of sunlight available to the plant. It affects the growth, flowering, and maturation of crops.

Question 31.
Fill in the blanks
(a) Photoperiod affects the ………..
(b) Kharif crops are cultivated from ……….. to ………..
(c) Rabi crops are cultivated from ……….. to ………..
(d) Paddy, maize, green gram and black gram are ……….. crops.
(e) Wheat, gram, pea, mustard are ……….. crops.
Answer:
(a) Flowering of plants
(b) June to October
(c) November to April
(d) Khalil
(e) Rabi

Question 32.
Cultivation practices and crop yield are related to environmental conditions. Explain.
Answer:
Different crops and cultivation practices require different climatic conditions, temperature, photoperiod for their growth and completion of life cycle. Depending upon these conditions some crops called Kharif crops are grown in rainy season while others called as the Rabi crops are grown during winter season.

Question 33.
Fill in the blanks
(a) A total of ……….. nutrients are essential to plants.
(b) ……….. and ……….. are supplied by air to p1ants.
(c) ……….. is supplied by water to plants.
(d) Soil supply ……….. nutrients to plants.
(e) ……….. nutrients are required in large quantity and called.
(f) ……….. nutrients are needed in small quantity for plants and are called.
Answer:
(a) 16 (Sixteen)
(b) Carbon, oxygen
(c) Hydrogen
(d) 13 (Thirteen)
(e) Six, macro
(f) Seven, micro

Question 34.
Differentiate between compost and vermicompost.
Answer:
1. Compost: Compost is prepared by decomposition of the farm waste materials like livestock excreta (cow dung, etc.), vegetable waste, animal refuse, domestic waste, sewage waste, straw, eradicated weeds, etc., in pits.
2. Vermicompost: The compost is called as vermicompost if it is prepared by using earthworms to hasten the process of decomposition of plant and animal refuse.

Question 35.
Arrange these statements in the correct sequence of preparation of green manure.
(a) Green plants are decomposed in soil.
(b) Green plants are cultivated for preparing manure or crop plant parts are used.
(c) Plants are plowed and mixed into the soil.
(d) After decomposition, it becomes green manure.
Answer:
(b)→(c)→(a)→(d)

Question 36.
An Italian bee variety A. mellifera has been introduced in India for honey production. Write about its merits over other varieties.
Answer:
The Italian bee variety A. mellifera has the following advantages:

• They have high honey collection capacity.
• They sting somewhat less.
• They stay in a given beehive for long periods and breed very well.

Question 37.
In agricultural practices, higher input gives a higher yield. Discuss how.
Answer:
The financial conditions of the farmers allow them to take up different farming practices and technologies. The cropping system and production practices of the farmer are decided by the farmer’s purchasing capacity for input. A higher money input helps to raise the yield. So, we can say that in agricultural practices, higher input gives a higher yield.

Improvement in Food Resources Class 9 Extra Questions Long Answer Type

Question 1.
Mention the modern initiatives undertaken in India to supply water to the fields.
Answer:
Indian agriculture is mainly dependent on the monsoons. The irregular or scarcity of rainfall often results in crop failure. To overcome the problem, different types of irrigation systems are in practice in India for the supply of water in agricultural fields. Wells, canals, river lift systems, tanks, etc. are used for irrigation. Some new initiatives like rainwater harvesting and watershed management are being used.

For this small check-dams are constructed to stop the rainwater from flowing and lead to an increase in groundwater levels. The different types of irrigation systems are:

1. Wells: There are two types of wells – dug wells and tube wells. In a dug well, water is collected from water-bearing strata. Tube wells can tap water from the deeper strata. From these wells, water is lifted by pumps for irrigation.

2. Canals: In this system canals receive water from one or more reservoirs or from rivers. The main canal is divided into branch canals having further distributed to irrigate fields.

3. River Lift Systems: In this system, water is directly drawn from the rivers for supplementing irrigation in areas close to rivers.

4. Tanks: These are small storage reservoirs, which intercept and store the run-off of smaller catchment areas.

Question 2.
What do you understand by composite fish culture? Describe in detail with advantages and disadvantages. What are the advantages of composite fish culture?
Answer:
A combination of five or six fish species is used in a single fish pond in the composite fish culture system. The selected species do not compete for food among them as they have different types of food habits.

The types of fishes used are:
Callas are surface feeders, Rohus feed in the middle-zone of the pond, Mrigals, and Common Carps are bottom feeders, and Grass Carps feed on the weeds. As a result, the food available in all the parts of the pond is used.

Advantages of Composite fish culture:

• The species of fishes in the pond utilize all the food available in the pond.
• The species do not compete with each other for food as they have different food habits.
• The yield of fish is increased by such a culture system.

The disadvantage of Composite fish culture:
A major problem in fish farming is the lack of availability of good quality fish seeds.

Question 3.
How do plants get their nutrients?
Or
List the nutrients supplied by air, water, and soil.
Answer:

Question 4.
Define manures. What are its different kinds? State two limitations of manures.
Answer:
Manure is prepared by the decomposition of animal excreta and plant waste and contains a lot of organic matter which helps in enriching the soil with nutrients and increasing soil fertility. Manure is classified on the basis of kind of biological material used as:

1. Compost: Compost is prepared by decomposition of the farm waste materials like livestock excreta (cow dung etc.), vegetable waste, animal refuse, domestic waste, sewage waste, straw, eradicated weeds, etc., in pits.

2. Vermicompost: The compost is called vermicompost if it is prepared by using earthworms to hasten the proœss of decomposition of plant and animal refuse.

3. Green manure: Green plants like sun hemp or guar are grown and then mulched by plowing them into the soil prior to the sowing of the crop seeds to enrich the soil in nitrogen and phosphorus.

The limitations of manure are:

1. They do not supply a specific nutrient.
2. They are not readily soluble in water so their absorption by plants is slower.
3. They are bulky in nature and difficult to be transported.
4. Only small amounts of nutrients are supplied by the manure.

Question 5.
Differentiate between manures and fertilizers.
Or
Compare manure and fertilizer in three points.
Manure:

1. Manure is prepared by the decomposition of animal excreta and plant waste.
2. They are bulky and difficult to be transported.
3. They are not nutrient-specific in nature.
4. They are not readily soluble in water and thus absorbed slowly by plants.
5. Manure contains a lot of organic matter.
6. They are environment friendly in nature.
7. Difficult to store.

Fertilizer:

1. Fertilizers are commercially produced plant nutrients.
2. They are not bulky, so easier to transport.
3. They are nutrient-specific in nature.
4. They are easily absorbed by the plants as they are soluble in water.
5. They do not contain organic matter.
6. The comparatively less environmentally friendly.
7. Easier to store.

Question 6.
Mention the management practices that are common between dairy and poultry farming.
Answer:
Management practices which are common between dairy and poultry fanning are:

• Should be housed in the well-ventilated shelter.
• Proper cleanliness should be maintained.
• Regular visits by a doctor and timely vaccination.
• Balanced food should be provided with additives.
• Regular inspections, with proper record keeping.

Question 7.
What are the differences between broilers and layers with respect to their purpose of breeding and daily requirement? What necessary steps have to be taken to prevent the occurrence of infectious diseases in poultry farms?
Answer:
The meat-producing birds are called broilers and the egg-laying birds are called layers. The housing, nutritional and environmental requirements of the broilers are different from the layers. The broilers are raised in poultry farms up to 6-7 weeks and usually weigh around 700 g to 1.5 kg.

The requirement of the broilers is protein and fat-rich food. The level of vitamin A and vitamin K is kept high in their feed. Care is taken to avoid mortality and to maintain the feathering and carcass quality. The layers start laying eggs at the age of 20 weeks. So, they are kept for longer periods of around 500 days, called laying period. They require enough space, proper light, and hygienic conditions.

heir feed consists of vitamins, minerals, and certain micronutrients that affect the hatchability of the eggs. Necessary steps for prevention of infectious diseases are:
1. Proper cleaning, sanitation, and spraying of disinfectants at regular intervals.
2. Appropriate vaccination to prevent the occurrence of infectious diseases.

Question 8.
Discuss the role of hybridization in crop improvement.
Answer:
Hybridization is one of the methods of crop production which ensures high yield. Hybridization refers to crossing between two genetically dissimilar plants each of which possesses a particular desired character. The two varieties are cross-bred during the process to incorporate both the desirable characteristics in a single variety. This method of hybridization improves crops with respect to yield, disease resistance, pest resistance, etc.

Question 9.
Define
1. Vermicompost
2. Green manure
3. Biofertilizer
Answer:
1. Vermicompost: The compost prepared by using earthworms to hasten the process of decomposition of plant and animal refuse.
2. Green manure: The manure which is prepared by decomposing green plants in field itself is called green manure. For example: sun hemp is grown in fields, mulched by plowing and allowed to
decompose infield for the preparation of green manure. It helps in enriching the soil in nitrogen and phosphorus.
3. Biofertiliser: Living organisms which are used as fertilizer to supply the nutrients to plants, are called as biofertilizers. For example, blue-green algae which fix nitrogen in soil, rice fields, are called biofertilizers.

Question 10.
Discuss various methods of weed control.
Answer:
The various methods of weed control are mechanical removal, a spray of chemicals called weedicides and preventive methods like proper seedbed preparation, timely sowing of crops, intercropping and crop rotation.

Question 11.
Differentiate between the following

1. Capture fishery and Culture fishery
2. Mixed cropping and Inter-cropping
3. Beekeeping and Poultry farming

Answer:

1. Capture fishery: Fish are obtained from natural resources in capture fishery. Culture fishery: Fish farming is called culture fishery.
2. Mixed cropping: Growing two or more crops simultaneously on the same piece of land. Inter-cropping: Growing two or more crops simultaneously on the same field in a definite pattern.
3. Beekeeping is the practice to rear honeybee for obtaining honey. Poultry farming is undertaken to raise domestic fowl called layers for egg production and the broilers for chicken meat.

Question 12.
Give the merits and demerits of fish culture.
Answer:
The merits and demerits of fish culture are:
Merits:

• It helps to get a large amount of desired fishes from a small area.
• It enables a variety of improvements in a better way.

Demerits:

• It is a threat to bio-diversity as only selected varieties are cultured on a large scale.
• It involves the culture of only economically important and valued fishes.

Question 13.
What do you understand by composite fish culture?
Answer:
A combination of five or six fish species is used in a single fish pond in the composite fish culture system. The selected species do not compete for food among them as they have different types of food habits.

The types of fishes used are:
Callas are surface feeders, Rohus feed in the middle-zone of the pond, Mrigals and Common Carps are bottom feeders, and Grass Carps feed on the weeds. As a result, the food available in all the parts of the pond is used.

Question 14.
Why should beekeeping be done in good pasturage?
Answer:
Beekeeping should be done in good pasturage as it:

• Helps in increasing the honey yield.
• Helps in increasing pollination efficiency.
• Provides better quality and quantity of nectar for the honey bees.

Question 15.
Write the modes by which insects affect the crop yield.
Answer:
The modes by which insects affect the crop yield are

• they cut the root, stem and leaf,
• they suck the cell sap from various parts of the plant, and
• they bore into stem and fruits.

Question 16.
Discuss why pesticides are used in very accurate concentration and in a very appropriate manner.
Answer:
Pesticides should be used in very accurate concentration and in a very appropriate manner because it
may have the following adverse effects:
(a) They harm the soil and causes loss of fertility
(b) They act as a check on the replenishment of organic matter
(c) They kill the microorganism of soil and destroy the soil structure
(d) They cause air, water and soil pollution.

Question 17.
Name two types of animal feed and write their functions.
Answer:
The two types of animal feed are:

• Roughage, which is largely fiber.
• Concentrates, which are low in fiber and have high levels of proteins and other nutrients.

Question 18.
What would happen if poultry birds are larger in size and have no summer adaptation capacity? In order to get small sized poultry birds, having summer adaptability, what method will be employed?
Answer:
Maintenance of temperature is needed for better egg production by poultry birds. The egg production is declined due to larger size (increase in surface area of the body) and no adaptability of summer in poultry birds. So, cross-breeding of poultry birds is done to obtain smaller size and higher summer adaptability in them. Small size is ais? needed for better housing and lower feed requirements.

Question 19.
Suggest some preventive measures for the diseases of poultry birds.
Answer:
Some preventive measures of poultry bird diseases are
(a) cleaning of poultry farms
(b) proper sanitation of poultry farms
(c) spraying of disinfectants at regular intervals
(d) appropriate vaccination of birds.

Question 20.
The figure shows the two crop fields [Plots A and BI have been treated by manures and chemical fertilizers respectively, keeping other environmental factors same. Observe the graph and answer the following questions.

1. Why does plot B show a sudden increase and then a gradual decrease in yield?
2. Why is the highest peak in plot A graph slightly delayed?
3. What is the reason for the different patterns of the two graphs?

Answer:
1. With the addition of chemical fertilizer, there is a sudden increase in yield due to the release of nutrients Nitrogen, Phosphorus, Potassium (NPK), etc. in high quantity. The gradual decline in the graph may be due to continuous use and a high quantity of chemicals that kill microbes useful for replenishing the organic matter in soil. This decreases soil fertility.

2. Manures supply small quantities of nutrients to the soil slowly as it contains large amounts of organic matter. It enriches soil with nutrients thereby increasing soil fertility continuously.

3. The difference in the two graphs indicates that use of manure is beneficial for a long duration in cropping as the yield tends to remain high when the number of manure increases.

4. In the case of Plot B the chemical fertilizers may cause various problems when used continuously for long time. Loss of microbial activity reduces the decomposition of organic matter and as a result, the soil fertility is lost that affects the yield.

Question 21.
Complete the crossword puzzle

Across:
1. Oil yielding plant (9)
3. Crop grown in winter season (4)
5. Fixed by Rhizobium (8)
9. Common honey bee (4)

Down:
2. Animal feed (6)
4. A micronutrient (5)
6. Unwanted plant in crop fields (4)
7. An exotic breed of chicken(7)
8. Bottom feeders in fish pond(7)
10. A marine fish (4)
Answer:

Improvement in Food Resources Class 9 Extra Questions HOTS

Question 1.
Irrigation systems are in practice in India for the supply of water in agricultural fields. Why?
Answer:
Most agriculture in India is rain-fed and the success of crops in most areas is dependent on timely monsoons and sufficient rainfall. Poor monsoons can cause crop failure. Hence, in order to ensure that the crops get water at the right stages during their growing season, irrigation systems are in practice in India for the supply of water to agricultural fields.

Question 2.
Nitrogenous fertilizers are usually not required by the leguminous crops. Why?
Answer:
The leguminous crops have root nodules that harbor nitrogen-fixing bacteria like Rhizobium in them. These bacteria convert the atmospheric nitrogen to nitrates for their utilization by plants. So, nitrogenous fertilizers are usually not required by the leguminous crops.

Question 3.
Why are crops like guar grown by some farmers before sowing the seeds of a crop?
Answer:
The crops like guar are grown by farmers before sowing the seeds of a crop as the guar crop is mulched by plowing. It turns into green manure which helps in enriching the soil in nitrogen and phosphorus.

Question 4.
Why is the enhancement of food production becoming a major necessity? Which group of activities are required for improving the crop yields?
Answer:
Enhancement of food production has become a major necessity because:
1. The population of the world is increasing at a faster rate, so it is needed to fulfill the food requirements of the population.
2. The land available for cultivation is limited, so the only option available is to increase crop production.

Major groups of activities for improving crop yields:

• Crop variety improvement
• Crop production improvement
• Crop protection management

Question 5.
List some dairy farming practices needed to increase the milk yield potential of cattle.
Answer:
The dairy farming practices needed to increase the milk yield potential of cattle are:

1. The cattle should be housed well and have adequate water.
2. The cattle should be kept disease-free.
3. The feeding of cattle should be carried out in a scientific manner with special emphasis on the quality and quantity of fodder.
4. Stringent cleanliness and hygiene should be maintained.
5. Regular inspections of cattle should be done with proper record keeping.
6. Regular visits by a veterinary doctor should be ensured.

Improvement in Food Resources Class 9 Extra Questions Value Based (VBQs)

Question 1.
Aniket took admission in a dairy research institute and took an active interest in his studies. During the course, he got an opportunity to visit a village. He met the farmers there and advised them about dairy farm practices which can help to increase the milk yield of their cattle. The farmers were able to get an increased yield due to his advice.

1. Name two species of cattle in India.
2. List two measures that would have been suggested by Aniket to increase milk yield.
3. What values are shown by Aniket?

Answer:
1. Indian cattle belong to two different species, Bos indicus, cows, and Bos bubble.s, buffaloes.
2. The measures which would have been suggested by Aniket to increase milk yield are:
(a) The cattle should be housed well and have adequate water.
(b) The cattle should be kept disease-free.
(c) The feeding of cattle should be carried out in a scientific manner with special emphasis on the quality and quantity of fodder.
3. The values shown by Aniket are a concern for others, caring nature, scientific approach, and knowledge.

Question 2.
Sonam got admission in a reputed institute engaged in agricultural research. After the completion of course she decided to go and serve the people of her village. Her village was suffering from a reduction in infertility of the soil and high salinity in soil. She called a meeting of farmers and told them that the use of fertilizers has increased the salinity of the soil. She suggested to them the use of manures and cropping patterns to get better yields.

1. What are the different types of manures which can be used?
2. Name the different cropping patterns which she would have suggested.
3. What values are shown by Sonam?

Answer:

1. The types of manures which can be used are: Compost, vermicompost, and green manure.
2. The different cropping patterns can be inter-cropping, mixed cropping, and crop rotation.
3. The values shown by Sonam are sincerity towards work, scientific aptitude, and concern for others.

In this page, we are providing Motion Class 9 Extra Questions and Answers Science Chapter 8 pdf download. NCERT Extra Questions for Class 9 Science Chapter 8 Motion with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 8 Extra Questions and Answers Motion

Extra Questions for Class 9 Science Chapter 8 Motion with Answers Solutions

Motion Class 9 Extra Questions Very Short Answer Type

Motion Class 9 Extra Questions Question 1.
Define the following terms:
(a) Distance
(b) Displacement
(c) Speed
(d) Velocity
(e) Acceleration
(f) Uniform motion
(g) Uniform circular motion
(h) Scalar quantity
(i) Vector quantity
Answer:
(a) Distance: The total path length traveled by a body in a given interval of time is called distance.
(b) Displacement: The shortest distance measured from the initial to the final position of an object is known as displacement.
(c) Speed: The speed of a body is defined as the distance traveled per unit time.
(d) Velocity: Velocity is defined as displacement per unit time.
(e) Acceleration: The rate of change of velocity with respect to time.
(f) Uniform motion: A body moving in a straight line has a uniform motion if it travels the equal distance in equal intervals of tune.
(g) Non-uniform motion: A body has a non-uniform motion if it travels unequal distances in equal intervals of time.
(h) Scalar quantity: A physical quantity which is described completely by its magnitude only, is called a scalar quantity.
(i) Vector quantity: A physical quantity that has magnitude as well as direction and obeys the vector addition is called a vector quantity.

Motion Chapter Class 9 Extra Questions And Answers Question 2.
Draw position-time graph of a body that started from a position other than origin and moving with uniform speed.
Answer:

Class 9 Motion Extra Questions Question 3.
Give one example where the displacement is zero but the distance traveled is not zero.
Answer:
When a body completes one rotation in a circular path its initial and final positions are the same and hence its displacement is zero.

Class 9 Motion Numericals Extra Questions Question 4.
The area under the velocity-time curve represents which physical quantity?
Answer:
Distance and displacement.

Motion Extra Questions Class 9 Question 5.
Under what conditions, distance, and magnitude of the displacement are equal?
Answer:
Distance and displacement are equal fan object move along a straight line in one direction.

Class 9 Science Chapter 8 Extra Questions And Answers Question 6.
What is the numerical ratio of average velocity to the average speed of an object when it is moving in a straight path?
Answer:
In this case, both are equal, so the ratio is 1.

Motion Chapter Class 9 Extra Questions And Answers Numericals Question 7.
What does the speedometer of an automobile measure?
Answer:
The speedometer of a vehicle measures its instantaneous speed.

Extra Questions Of Motion Class 9 Question 8.
Classify the following quantities on the basis of scalar quantities and vector quantities.
(a) Distance
(b) Displacement
(c) Speed
(d) Velocity
(e) Acceleration
Answer:
(a) Distance: Scalar quantity
(b) Displacement: vector quantity
(c) Speed: Scalar quantity
(d) Velocity: Vector quantity
(e) Acceleration: Vector quantity

Extra Questions On Motion Class 9 Question 9.
An object starts with initial velocity y and attains a final velocity y. The velocity is changing at a uniform rate. What is the formula for calculating average speed in this situation?
Answer:

Extra Questions For Class 9 Physics Motion Question 10.
What will you say about the motion of an object if its distance-time graph is a straight line with having a constant angle with a time axis?
Answer:
The motion of the body is uniform motion (constant speed).

Class 9 Science Motion Extra Questions Question 11.
What will you say about the motion of a body if its velocity-time graph is a straight line parallel to the time axis?
Answer:
The velocity of the body is constant.

Class 9 Science Ch 8 Extra Questions Question 12.
A physical quantity measured is -20 ms^-1 It is speed or velocity?
Answer:
It is velocity because velocity can be positive or negative while speed cannot be negative.

Class 9 Physics Motion Extra Questions Question 13.
A motorcycle drives from A to B with a uniform speed of 30 kmh¹ and returns hack with a speed of 20 kmh¹. Find its average speed.
Answer:
Let distance AB = x km.
Time taken from going A to B; t₁ = \(\frac {x}{30}\) h
Time taken from goingBto A; t₂ = \(\frac {x}{20}\) h
Total time taken,t = t₁ + t₂ = \(\frac {x}{30}\) + \(\frac {x}{20}\) = \(\frac {x}{50}\) h
total distance = x + x = 2x km

∴ Average speed = 24 km/h

Ch 8 Science Class 9 Extra Questions Question 14.
A man swims in a pool of width 300 m. He covers 600 m in 10 minutes by swimming from on end to the other, and back along the same path. Find the average speed and average velocity of the man for the entire journey.
Answer:
Total distance covered by the man in 10 minutes = OA + AB = 600 m
Displacement of man in 10 minutes = 0 m

Average speed = 1 m/s

Average velocity = 0 m/s

Extra Questions For Class 9 Science Chapter Motion With Answers Question 15.
What is the acceleration of a body moving with uniform velocity along a straight line?
Answer:
Acceleration of the body is zero as the velocity is constant.

Question 16.
Give an example of a uniform circular motion.
Answer:
An athlete running on a circular track with constant speed.

Question 17.
Convert speed 54 km in meter per second.
Answer:
Speed = 54 km/h =54 x \(\frac{1000 m}{3600}\) = 15 m/s

Question 18.
What ¡s the nature of the distance-time graph for accelerated motion?
Answer:

Motion Class 9 Extra Questions Short Answer Type 1

Question 1.
Distinguish between speed and velocity.
Answer:
Speed:

• Speed is the ratio of distance and time.
• Speed is always positive
• Speed is a scalar quantity.

Velocity:

• Velocity is the ratio of displacement and time.
• Velocity may be negative or positive.
• Velocity is a vector quantity.

Question 2.
Distinguish between distance and displacement.
Answer:
Distance

• It is the actual length of the path covered by a moving body.
• It is always positive or zero.
• It is a scalar quantity.

Displacement:

• It is the shortest distance measured between the initial and final positions.
• It may be positive, negative, or zero.
• it is a vector quantity.

Question 3.
Write down the SI unit of the following quantities:
(a) Displacement
(b) Speed
(c) Velocity
(d) Acceleration
Answer:
(a) m
(b) m/s
(c) m/s
(d) m/s²

Question 4.
Distinguish between uniform motion and non-uniform motion.
Answer:
1. Uniform motion:
A body moving in a straight line has a uniform motion if it travels the equal distance in equal intervals of time

2. Non.uniform motion:
A body has a non-uniform motion if it travels the unequal distance in equal intervals of time

Question 5.
Distinguish speed at any instant and average speed.
Answer:
1. Instantaneous speed:
The speed at any particular instant is known as instantaneous speed.

2. Average speed:
Average speed is the ratio of total distance traveled by a body and time taken to travel that distance.

Question 6.
Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Answer:
velocity-time graph

Question 7.
What is uniform circular motion? How is uniform circular motion regarded as an acceleration motion? Explain.
Answer:
When an object is moving in a circular path with a constant speed, the motion of an object is said to be uniform circular motion. When a body has a uniform circular motion, its velocity changes due to the continuous change in the direction of its motion. Hence, the motion of the body is accelerated motion.

Motion Class 9 Extra Questions Numericals

Question 8.
A person travels a distance of 4.0 m towards the east, then turns left and travels 3.0 m towards the north.
1. What is the total distance traveled by the person? North
2. What is his displacement?
Answer:
1. Total distance = OA + AB
= 4m + 3m
Total distance = 7m

2. Total displacement = OB = \(\sqrt{(O A)^{2}+(A B)^{2}}\)
=\(\sqrt{(4)^{2}+(3)^{2}}\) = \(\sqrt{25}\) = 5
Displacement = 5 m

Question 9.
A person travels on a semi-circular track of radius 50 m during a morning walk. If he starts from one end of the track and reaches the other end, calculates the distance traveled and displacement of the person.
Answer:
Let the person start moving from A and reach B via O.
The distance travelled by the person
= Length of track = πr
= \(\frac{22}{7}\) x 50 m = 157.14m
Distance = 157.14 m
The displacement is equal to the diameter of the semi-circular track joining A to B via O.
= 2r = 2 x 50 m = 100m
∴ Displacement = 100 m

Question 10.
Dinesh takes 20 minutes to cover a distance of 5 km on a bicycle. Calculate his average speed in
1. m/s
2. km/h
Answer:
1. Average speed in metre/second
Distance travelled by Dmesh = 5 km = 5 x 1000 m = 5000 m
time taken (t) = 20 min = 20 x 60 seconds = 1200 seconds

2. Average speed in kilometre/hour
Average speed = \(\frac{5 \mathrm{km}}{\frac{20}{60} \mathrm{h}}\) = 15 km/h
∴  Average speed = 15 km/h

Question 11.
A train starting from a railway station and moving with uniform acceleration attains a speed of 20 m/s in 10 seconds. Find its acceleration.
Answer:
Given,
initial velocity, u = 0 m/s
Final velocity, υ = 20 m/s
time taken, t = 10 s
Acceleration, a = \(\frac{v-u}{t}=\frac{20-0}{10}\)
∴ a = 2 m/s²

Question 12.
A car moving with a speed of 72 km/h is brought to rest in 10 seconds by applying brakes. Find the magnitude of average retardation due to brakes and distance traveled by car after applying the brakes.
Answer:
Given, initial velocity = 72 km/h = 72 x \(\frac{5}{18}\) = 20 m/s
Final velocity = 0 m/s
Time taken = 10s.

1. Retardation
Acceleration, a = \(\frac{υ – u}{t}\)
a = \(\frac{0 – 20}{10}\) = -2m/s²
∴ Retardation = 2 m/s²

2. Distance travelled using equation, υ² – u² = 2as
(0)² – (20)² = 2 (-2)s
∴ s = 100 m

Question 13.
The velocity of a particle moving along a straight line in a certain time interval is shown below. What is the distance traveled during acceleration?

Answer:
Here particle accelerates for 2 seconds.
Distance traveled during acceleration = area under υ – t graph for 2 sec.
\(\frac{1}{2}\) x Base x Height = \(\frac{1}{2}\) x 2 x 20 = 20m

Question 14.
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement—
The time graph is shown in the figure. Plot a velocity-time graph for the same.
Answer:
Velocity-time graph

Question 15.
Obtain a relation for the distance traveled by an object moving with uniform acceleration in the time interval between the 4th and 5th seconds.
Answer:
Using equation, s = ut + \(\frac{1}{2}\) at²
Distance travelled in 5 s; s₁ = u x 5 + \(\frac{1}{2}\) a(5)²
s₁ = 5u + \(\frac{25a}{2}\)

Distance travelled in 4 s; s₂ = u x 4 + \(\frac{1}{2}\) a(4)²
s₂ = 4u x 8a
Distance travelled in the interval between 4^th and 5^th sec.
= s₁ – s₁ = (5u + \(\frac{25a}{2}\)) – (4u + 8a)
∴ Distance travelled = u + \(\frac{9a}{2}\)

Motion Class 9 Extra Questions Short Answer Type 2

Question 1.
Draw a position-time graph of for
(a) rest
(b) uniform motion
(c) Non-uniform motion
Answer:
(a) Position – time graph for rest

(b) Position – time graph for uniform motion

(c) Position – time graph for non-uniform motion

Question 2.
A draw velocity-time graph for
(a) Uniform motion
(b) Uniform acceleration
(c) Uniform retardation
Answer:
(a) Velocity – time graph for uniform motion (Const. speed)

(b) Velocity – time graph for uniform acceleration (Const. acceleration)

(c) Velocity – time graph for uniform retardation

Question 3.
Explain the difference between the two graphs.

Answer:
(a) In the first graph, the body is starting its motion from origin but in the second graph, the body is starting its motion from any other point.
(b) In the first graph, the body has zero initial velocity but in the second graph the body has non-zero initial velocity.

Question 4.
Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of \(\frac{1}{2}\)
(Assume upward acceleration is – g and downward acceleration to be +g).
Answer:
The maximum height reached by the stone can be calculated by the formula
υ² – u² = 2as
Here, υ = 0 and a= – g
Applying this equation for the first ball

Motion Class 9 Extra Questions Numericals

Question 5.
Study the velocity-time graph of a traveling metro train as shown in the figure. What is the acceleration of the train
(a) during the first two seconds?
(b) between the second and tenth seconds?
(c) during the last two seconds?

Answer:
(a) Acceleration, a = \(\frac {v – u}{t}\)
Here, u = 0, u = 20m/s and t = 2s
a = \(\frac {20 – 0}{2}\) = 10 m/s²

(b) Here, u = 20m/s, υ = 20m/s and t = 8s
a = \(\frac {20 – 0}{2}\) = 0 m/s²

(c) Here, u = 20 m/s, u = 0 m/s and t = 2s
a = 0 – 20 = – 10 m/s²

Question 6.
An electron moving with a velocity of 5 x 10 ms¹ enters into a uniform electric field and acquires a uniform acceleration of 10 ms² in the direction of its initial motion.
1. Calculate the time in which the electron would acquire a velocity double its initial velocity.
2. How much distance would the electron cover in this time?
Answer:
Given, u =5 x 10 m/s²
a = 10⁴ m/s²

1. From υ = u + at
take υ = 2u
2u = u + at
u = at
5 x 10⁴ = 10⁴ x t
∴ t = 5s

2. From s = ut + \(\frac{1}{2}\) at²
s = 5 x 10⁴ x 5 + \(\frac{1}{2}\) x 10⁴ x (5)²
∴ s = 37.5 x 10⁴ m

Question 7.
The motion of a train is described by the velocity-time graph as shown in the figure given below.

Find distance travelled by car between
1. 0 to 20 s
2. 20 s to 50 s
Answer:
Distance travelled = Area of velocity – time curve
1. For (0 to 20 s)
Distance = Area of ΔOAC = \(\frac{1}{2}\) x OC x AC = \(\frac{1}{2}\) x 20 x 10 = 100 m

2. For (20 s to 50 s)
Distance = Area of rectangle ABDC + Area of ΔBED
Distance = AC x CD + \(\frac{1}{2}\) DE x DB = 10 x (40 – 20) + \(\frac{1}{2}\) x 10 x 5 = 225m

Motion Class 9 Extra Questions Long Answer Type

Question 1.
Derive an expression for three equations of motion for uniform accelerated motion graphically.
Answer:
Equation of motion by graphical method
Let us consider a body is moving with acceleration where u is initial velocity and u is final velocity, s is the displacement of object and t is a time interval.

1. υ = u + at
We know that slope of υ – t graph gives acceleration so slope
= a = \(\frac{v-u}{t-0}\)
a = \(\frac{v-u}{t}\)
∴ υ = u + at

2. s = ut + \(\frac{1}{2}\) at²
We know that area under u – t graph gives the displacement.
Area = s = area of triangle CDE + area of rectangle ABCE
s = ut + \(\frac{1}{2}\) x t x (υ – u) from (υ – u = at)
Putting the value of υ – u
s = ut + \(\frac{1}{2}\) at²

3. υ² – u² = 2as
We know that area under υ – t graph gives displacement
Area = s = area of trapezium ABDE
s = \(\frac{1}{2}\) x (υ+u) x t From I (t = \(\frac{υ – u}{a}\))
Putting the value of t.
υ² – u² = 2as

Motion Class 9 Extra Questions HOTS

Question 1.
The distance-time graph of two trains is given below. The trains start simultaneously in the
same direction.
1. How much ahead of A is B when the motion starts?
2. What is the speed of B ?
3. When and where will A catch B?
4. What is the difference between speeds of A and B?

5. Is the speed of both the trains uniform or non-uniform? Justify your answer.

Answer:

1. 100km
2. Speed of B, υ = \(\frac{x_{2}-x_{1}}{t}\) = \(\frac{150-100}{2-0}\) = 25 km/h
3. After 2 hours, A catches B at 2. i.e.,150 km from origin 0.
4. The slope of the distance-time graph of A is greater than that of B. Therefore, the speed of A is greater than
   speed B. Speed of A = Slope of OQ = \(\frac{150}{2}\) = 75 km/h
   Difference between speed of A and B =75 km/h – 25 km/h = 50 km/h
5. The speed of both trains is uniform.

Question 2.
A car starts from rest and moves along the x-axis with constant acceleration 5 ms2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from rest?
Answer:
Given,
u = 0, a = 5ms^-2, t₁ = 8s and t₂ = 12s.
Total distance = distance travelled during acceleration + distance travelled during constant velocity
s = s₁ + s₂

For S₁
s₁ = u₁t₁ + \(\frac{1}{2}\) \(a t_{1}^{2}\)
s₁ = \(\frac{1}{2}\) x 5 x (8)²
s₁ = 160 m

For S₁
Applying υ = u + at
υ = 0 + (5 x 8)
u = 40 m/s
s₁ = ut₂
s₂ = 40 x 12
s₂ = 480 m
Total distance, s = s₁ + s₂
s = 160 + 480 = 640m

Question 3.
An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2s if both the objects drop with the same acceleration? How does the difference in heights vary with time?
Answer:
At t = 0, difference in heights of two stones A and B
= 150 – 100 = 50 m
In t = 2s, distance through which each stone falls is –
s = ut + \(\frac{1}{2}\) at² = 0 x \(\frac{1}{2}\) x 10 x (2)² = 20m
∴ Difference in heights of two stones A and B
= (150 – 20) – (100 – 20) = 50
The difference in heights will remain the same at all times during their fall.

Question 4.
An object starting from rest travels 20 m in the first 2s and 160 m in the next 4s. What will be the velocity after 7s from the start?
Answer:
Given, the object travels 20 m first 2 s and 160 m in the next 4s.
From s = ut + \(\frac{1}{2}\) at²
In first 2 s, s = 20 m
20 = 0 + \(\frac{1}{2}\) a(2)²
20 = 2 a
∴ a = 10 m/s²
Velocity at the end of 2s is υ = u + at
υ = 0 + 10 x 2 = 20 m/s

In next 4 s,
s = ut + \(\frac{1}{2}\) a’t²
160 = 20 x 4 + \(\frac{1}{2}\) a’(4)²
80 = 8 a’
a’ = 10 m/s²

Motion Class 9 Extra Questions Value Based (VBQs)

Question 1.
Sumeet and Jaikishan are close friends. Sumeet is a science graduate and Jaikishan is a commerce graduate. Sumeet finds that while driving on a clear highway, Jaikishan often exceeds the speed limit and argues that there is no harm in doing so when the road is clear. Sumeet does not agree with him and tells him that with an increase in speed, the stopping distance of the car would increase and he would not be able to manage if someone appears suddenly on the way.
Read the above passage and answer the following questions:

1. is Sumeet right in his statement?
2. What values are displayed by Sumeet in his statement?
3. How is stopping distance related to the speed of the vehicle?

Answer:

1. Yes, Sumeet is absolutely right.
2. Sumeet is an intelligent boy and he can apply his knowledge of science in daily life.
3. Stopping distance (velocity)²

Question 2.
Raju and Raman are close friends. One day they went for a picnic. Raju was driving the car. When Raman saw the speedometer, he found that the speed of the car was more than the allowed speed limit. Raman advised Raju to reduce speed to avoid the punishment of Overspeed.
Answer the following questions:

1. What is the function of the speedometer?
2. What values are displayed by Raman?

Answer:

1. Speedometer provides instantaneous speed. B
2. Raman is a careful and law-abiding person.

Question 3.
Three friends Ram, Mohan, and Roy have to reach point B from point A. The path taken by them is shown in the figure.
Answer the following questions:

1. Distinguish between distance and displacement.
2. If all of them travel at the same speed, who will reach first at point B and why?
   

Answer:
1. Distance:

1. It is the actual length of the path covered by a moving body.
2. It is always positive or zero.
3. It is a scalar quantity.

Displacement:

1. It is the shortest measured distance between the initial and final positions.
2. It may be positive, negative, or zero.
3. It is a vector quantity.

2.  Mohan will reach first as he is traveling the shortest path.

In this page, we are providing Force and Laws of Motion Class 9 Extra Questions and Answers Science Chapter 9 pdf download. NCERT Extra Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 9 Extra Questions and Answers Force and Laws of Motion

Extra Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers Solutions

Force and Laws of Motion Class 9 Extra Questions Very Short Answer Type

Force And Laws Of Motion Class 9 Numericals Extra Questions Question 1.
Define the following terms

1. Force
2. Balanced forces
3. Unbalanced forces
4. Inertia
5. Momentum
6. Conservation of momentum.

Answer:

1. Force: It is an entity which when applied on a body changes or tends to change its state of motion and shape.
2. Balanced forces: When a number of forces acting simultaneously on a body do not bring about any change in the state of rest or of uniform motion along a straight line, then the forces acting on the body are said to be balanced.
3. Unbalanced forces: When a number of forces acting simultaneously on a body bring about a change in its state of rest or of uniform motion along a straight line, then these forces acting on the body are said to be unbalanced forces.
4. Inertia: Inertia is the natural tendency of an object to resist a change in its state of motion or rest.
5. Momentum: Momentum of a body is the product of its mass and velocity.
6. Conservation of momentum: In an isolated system, the total momentum remains conserved.

Force And Laws Of Motion Class 9 Extra Questions Question 2.
Name the physical quantity that measures inertia.
Answer:
Mass of body measures its inertia.

Force And Laws Of Motion Extra Questions Numericals Question 3.
There are three solids made up of aluminum, steel, and wood, of the same shape and same volume. Which of them would have the highest inertia?
Answer:
Steel ball would have the highest inertia as steel is denser than the other two.

Force And Laws Of Motion Class 9 Extra Numericals Question 4.
Name the factors on which the momentum of a body depends.
Answer:

• Mass and
• The velocity of the body.

Class 9 Force And Laws Of Motion Extra Questions Question 5.
Define one newton of force.
Answer:
Force is said to be 1 newton if it produces an acceleration of 1 rn/s² in a body of 1 kg.

Force And Laws Of Motion Extra Questions Question 6.
An object is thrown vertically upwards. What is its momentum at the highest point?
Answer:
Since the velocity at the highest point is zero, so the momentum of the object is zero at the highest point.

Extra Numericals Of Force And Laws Of Motion Class 9 Question 7.
State action and reaction when a bullet is fired from the gun.
Answer:
Action: Force exerted by a spring on the bullet.
Reaction: Force exerted on the gun.

Class 9 Science Chapter 9 Extra Questions And Answers Question 8.
An athlete runs some distance before taking a jump. Why?
Answer:
To gain momentum so that he may jump higher.

Class 9 Physics Chapter 2 Extra Questions Question 9.
Mention any two effects of force.
Answer:

• It changes the state of rest or motion of a body.
• It changes the shape of the body.

Class 9 Force And Laws Of Motion Extra Numericals Question 10.
What happens to the momentum of a body whose velocity is halved?
Answer:
The momentum of the body becomes half.

Class 9 Science Ch 9 Extra Questions Question 11.
Give the magnitude and the direction of the net force acting on the cork of mass 10 g floating on water.
Answer:
Zero.

Class 9 Science Chapter 9 Extra Numericals Question 12.
A car of mass 1000 kg is moving with velocity 5 m/s. Calculate the momentum of the car.
Answer:
Given, mass, m = 1000 kg
Velocity, u = 5 m/s
Momentum, P = mυ
P= 1000 x 5 = 5000kg m/s
P= 5000 kg m/s

Class 9 Physics Force And Laws Of Motion Extra Questions Question 13.
A meteorite burns in the atmosphere before it reaches the Earth’s surface. What happens to the linear momentum?
Answer:
Meteorite burns due to heat produced by frictional force. The linear momentum of the meteorite decreases as a frictional force acts on it.

Class 9th Science Chapter 9 Extra Questions Question 14.
Find the acceleration produced by a force of 2000 N acting on a car of mass 800 kg.
Answer:
Given,
Mass, m =800kg
Force, F = 2000 N
Using F = ma
a = \(\frac{F}{m}\)
a = \(\frac{2000}{800}\) = 2.5 m/s²

Force and Laws of Motion Class 9 Extra Questions Short Answer Type 1

Question 1.
Write down SI unit of (i) force (ii) momentum.
Answer:

• newton (N)
• kg rn/s.

Question 2.
When a person hits a stone, his foot is injured. Why?
Answer:
When a person hits a stone, the stone exerts an equal force on his foot. Due to this force, his foot gets injured.

Question 3.
Why no force is required to move an object with a constant velocity?
Answer;
We know, F = ma
When, velocity is constant, then acceleration, a = 0. Hence, F = 0.

Question 4.
Why is it easier to catch a table tennis ball than a cricket ball, even both are moving with the same velocity?
Answer:
It is easier to catch a table tennis ball because the table tennis ball has less mass (inertia).

Question 5.
Write down the expression for the recoil velocity of the gun.
Answer:
Recoil velocity of gun is given by,
\(\mathrm{V}_{\mathrm{G}}=\frac{m_{b} v_{b}}{\mathrm{M}_{\mathrm{G}}}\)
m_b = mass of bullet
m_G = mass of gun
υ_b = velocity of bullet
V_G = recoil velocity of gun.

Force and Laws of Motion Class 9 Extra Questions Numericals

Question 6.
A car of mass 1000 kg moving with a velocity of 54 km/h hits a wall and comes to rest in 5 seconds. Find the force exerted by the car on the wall.
Answer:
Given, Mass = 1000 kg
Time = 5 s
Initial velocity, u = 54 km/h = 15 m/s
Final velocity, υ = 0 m/s
Time, t = 5s
using, F = m a
F = m\(\left(\frac{v-u}{t}\right)\) = 1000 \(\left(\frac{0-15}{5}\right)\) = -3000N

Question 7.
A body of mass 100g is at rest on a smooth surface. A force of 0.1-newton act on it for 5 seconds. Calculate the distance traveled by the body.
Answer:
Given, Mass of body = 100 g = 0.1 kg
Force, F = 0.1N
Time, t = 5s
Initial velocity, u = 0 m/s
Using formula, F = ma
⇒ a = \(\frac{F}{m}=\frac{0.1}{0.1}\) = 1 m/s²

Using formula, s = ut + at²
s = ut + \(\frac {1}{2}\)at²
⇒ s = 0 + \(\frac {1}{2}\) x 1 x (5)²
∴ s = 12.5 m

Question 8.
A bullet of mass 200 g is fired from a gun of mass 10 kg with a velocity of 100 m/s. Calculate the velocity of recoil.
Answer:
Given, Mass of bullet, m = 200 g = 0.2 kg
Massofgun, M = 10 kg
Velocity of bullet, V_b = 100 m/.s
Recoil velocity of gun, V_G = \(\frac{m v_{b}}{\mathrm{M}}\)
V_G = – \(\frac{0.2 \times 100}{10}\) = 2m/s
Recoil velocity of gun, V_G = 2 m/s

Question 9.
Two spheres of masses 20 g and 40 g moving in a straight line in the same direction with velocities of 3 mIs and 2 m/s respectively. They collide with each other and after the collision, the sphere of mass 20 g moves with a velocity of 2.5 miles. Find the velocity of the second ball after confusion.
Answer:
Given, m₁ = 20 g = 20 x 10^-3 kg
m₁ = 40g = 40g x 10^-3kg
u₁ = 3 m/s
u₂ = 2 m/s
υ₁ = 2.5 m/s
Applying conservation of linear momentum,
m₁u₁ + m₂u₂ = m₁υ₁ + m₂υ₂
20 x 10^-3 x 3 + 40 x 10^-3 x 2 = 20 x 10^-3 x 2.5 + 40 x 10^-3 x υ²
υ₂ = 2.25m/s.

Force and Laws of Motion Class 9 Extra Questions Short Answer Type 2

Question 1.
Newton’s first, second, and third law of motion.
Answer:
Newton’s first law of motion: An object remains in a state of rest or of uniform motion along a straight line unless compelled to change that state by an applied force.

Newton’s second law of motion: The second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Newton’s third law of motion: According to the third law of motion to every action, there is an equal opposite reaction and they act on two different bodies.

Question 2.
What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically.
(a) momentum versus velocity when mass is fixed.
(b) momentum versus mass when velocity is constant.
Answer:
Momentum gives an idea about the quantity of motion contained in a body.
The momentum of (P) of an object is defined as the product of its mass (m) and velocity (υ).
P = mυ
SI unit of momentum is kg ms^-1
Graphs
(a) If m = constant
P ∝ υ

(b) If v = constant
P ∝ m

Question 3.
“Action and reaction are equal and opposite but even they do not cancel each other.” Explain, why?
Answer:
Two equal and opposite forces can cancel each other if they act on the same body. But action and reaction do not act on the same body. They act on two different bodies. Hence cannot cancel each other.

Question 4.
Why are the wheels of vehicles provided with mudguards?
Answer:
The rotating wheels of a vehicle throw out mud sticking to it tangentially due to the inertia of direction. The mudguards stop this mud to fall on another vehicle just moving behind the vehicle.

Question 5.
A car weighing 1600 kg moving with a velocity of 30 mIs retards uniformly coming to rest in 20 seconds. Calculate the
1. Initial and final momentum of the car.
2. Rate of change of linear momentum of the car.
3. Acceleration of the car.
Answer:
Given, mass, m = 1600 kg
Initial velocity, u = 30 m/s
Final velocity, υ = 0 m/s
Time, t = 20s
1. Initial momentum, P_i = mu = 1600 x 30
P_i = 4800 kg m/s
Final momentum, P_f = mυ = 1600 x 0
P_f = 0 kg m/s

2. Rate of change of linear momentum
= \(\frac{P_{f}-P_{i}}{t}\) = \(\frac{0-4800}{20}\) = – 240N
a = \(\frac{v-u}{t}\)

3.  Acceleration,
\(\frac{0 – 30}{20}\)
a = – 1.5 m/s².

Question 6.
Using the second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sand-bag at a speed of 10³ ms² and gets embedded after traveling 5 cm. Calculate
1. the resistive force exerted by the sand on the bullet.
2. the time is taken by the bullet to come to rest.
Answer:
According to the second law of motion,
The rate of change of momentum is directly proportional to the force applied.
F ∝ \(\frac{m(v-u)}{t}\)
F = \(\frac{km(v-u)}{t}\)
∴ F = kma
Take k = 1 in SI system
∴ F = ma

1. Given, mass of bullet, m = 10g = 10 x 10^-3kg
Initial velocity of the bullet, u = 10³m/s
Distance travelled by the bullet, s = 5 cm = 5 x 10^-2 m
Final velocity of the bullet = 0
Using the formula, υ^-2 – u^-2 = 2as
⇒ 0 – (10³)² = 2 x a x 5 x 10^-2

f = m x a = 10 x 10^-3 x (10⁷) = – 10⁵ N

2. From the formula,
υ = u + at
⇒  t = \(\frac {υ – u}{a}\) = \(\frac{0-10^{3}}{-10^{7}}\) = 10^-4 s = 0.0001 s

Question 7.
An object is placed on a rough surface. The external force of 20 N is acting on the body and 10 N frictional force is acting on the body.
Find
1. The net force on an object
2. Acceleration of an object if the mass of an object is 2 kg
3. The velocity of the object after 2 seconds if the object is started from rest.
Answer:
Given,

∴ υ = 10 m/s

Force and Laws of Motion Class 9 Extra Questions Long Answer Type

Question 1.
Explain why?
(a) A cricket player lowers his hands while catching the ball.
(b) The vehicles are fitted with shockers.
(c) A karate player breaks the pile of tiles or bricks with a single blow.
(d) In a high jump athletic event, the athletes are allowed to fall either on a sand bed or cushioned bed.
(e) In a moving car, the drivers and other passengers are advised to wear seat belts.
(f) China and glassware are packed with soft materials.
(g) Athletes are advised to come to a stop slowly after finishing a fast race.
Answer:
(a) if a player does not lower his hands while catching the ball, the time to stop the ball is very small. So a large force has to be applied to reduce the velocity of the ball to zero or to change the momentum of the ball. When a player lowers his hands, the time to stop the ball is increased and hence less force has to be applied to cause the same change in the momentum of the ball. Therefore, the hands of the player are not injured.

(b) The vehicles are fitted with shockers (i.e., springs)
The floor of a vehicle is cónnected to the lower part of the vehicle by springs or shockers. When the vehicle moves over a rough road, the force due to jerks is transmitted to the floor of the vehicle through the shockers. The shockers increase the time of transmission of the force of jerk to reach the floor of the vehicle. Hence less force is experienced by the passengers in the vehicles.

(c) A karate player can break a pile of tiles with a single blow of his hand because he strikes the pile of tiles with his hand very fast, during which the entire linear momentum of the fast-moving hand is reduced to zero in a very short interval of time. This exerts a very large force on the pile of tiles which is sufficient to break them, by a single blow of his hand.

(d) In a high jump athletic event., the athletes are allowed to fall either on a sand bed or cushioned bed: This is done to increase the time of athletes fall to stop after making the high jump, which decreases the rate of change of linear momentum and decreases the impact.

(e) In a moving car, the drivers and other passengers are advised to wear seat belts: When brakes are applied sudden1y, the passengers of the car fall forward due to the inertia of motion. The seat belt worn by passengers of the car prevents them from falling forward suddenly. This enables the entire linear momentum of the passengers to reduce to zero over a long interval of time, hence it prevents injury.

(f) China and glassware are packed with soft material: China and glassware are wrapped in paper before packing to avoid breakage while transporting. During transportation, there may be collisions due to ta jerks of the packed wares. Soft material like paper slows down their rate of change of linear momentum. As a result, the impact is reduced and items are not broken.

(g) Athletes are advised to come to stop slowly after finishing a fast race: By doing so, he decreases the rate of change of linear momentum by increasing the time interval and hence, reducing the impact, which reduces injury.

Question 2.
Explain
(a) How do we swim?
(b) Why does a gun recoil?
(c) It is difficult to walk on sand or ice.
(d) The motion of rocket.
(e) Why does a fireman struggle to hold a hose-pipe?
(f) Rowing of a boat.
(g) When a man jumps out from a boat, the boat moves backward.
(h) Walking of a person.
Answer:
(a) While swimming, a swimmer pushes the water backward with his hands (i.e., he applies force in the backward direction, which is known as action.) The reaction offered by the water to the swimmer pushes him forward.

(b) Recoiling of a gun: When a bullet is fired from a gun, it exerts a forward force on the bullet and the bullet exerts an equal (in magnitude) and opposite (in direction) force on the gun. Due to the high mass of the gun, it moves a little distance backward and gives a backward jerk to the shoulder of the gunman.

(c) It is difficult to walk on sand or ice: When our feet press the sandy ground in the backward direction, the sand gets pushed away and as a result, we get only a small reaction (forward) from the sandy ground making it difficult to walk.

(d) Rocket propulsion Before firing the rocket, the total linear momentum of the system is zero because the rocket is in the state of rest. When it is fired, chemical fuels inside the rocket are burnt and the hot gases are passed through a nozzle with great speed. According to the law of conservation of linear momentum, the total linear momentum after firing must be equal to zero. As the hot gases gain linear momentum to the rear on leaving the rocket, the rocket acquires equal linear momentum in the upward i.e., opposite direction.

(c) A fireman has to make a great effort to hold a hose-pipe to throw a stream of water on the fire to extinguish it. This is because the stream of water rushing through the hose-pipe in the forward direction with a large speed exerts a large force on the hose-pipe in the backward direction which is known as the reaction force. This reaction force tends to move the hose-pipe in the backward direction. Therefore, a fireman struggles to hold the hose-pipe strongly to keep it at rest.

(f) Rowing of a boat: The boatman during the rowing of a boat pushes the water backward with oars (this is an action of the boatman). According to the third law of motion, water exerts an equal (in magnitude) and opposite (in direction) push on the boat which moves it forward (this is a reaction by water). Thus, harder the boatman pushes back the water with oars (i.e., greater is the action), greater is the reaction force exerted by water, and faster the boat moves in the forward direction.

(g) When a man jumps out from a boat, the boat moves backward: When the passengers start jumping out of a rowing boat, they push the boat backward with their feet. The boat exerts an equal (in magnitude) and opposite (in direction) force on passengers in the forward direction which enables them to move forward.

(h) Walking of a person: When a person walks on the ground, he pushes the ground with his foot in the backward direction by pressing the ground. This push is known as the action, According to Newton’s third law of motion, an equal and opposite reaction acts on the foot of the person by the ground. This reaction (force) of the ground on the person pushes him forward.

Force and Laws of Motion Class 9 Extra Questions HOTS

Question 1.
Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed? Give reasons for your answer
Answer:
Yes, both the balls will start rolling in the direction opposite to the motion of the train. The speed of two balls will be different as the inertia of the two balls are different.

Question 2.
Two identical bullets are fired on by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulders more and why?
Answer:
The light rifles will hurt more as the recoil velocity of the light rifle will be greater.

Question 3.
Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown in the figure. Calculate the acceleration and frictional force of the floor on the ball.

Answer:
Given, Mass of ball, m = 50 g = 50 x 10^-3 kg
Acceleration can be calculated by υ – t graph,

= \(\frac{80 \mathrm{m} / \mathrm{s}}{8 \mathrm{s}}\) = – 10 m/s²
a = – 10 m/s²
Friction force, F = ma
F = 50 x 10^-3 x 10
F = 0.5 N

Question 4.
A truck of mass M is moved under a force F. if the truck is then loaded with an object equal to the mass of the truck and the driving force is halved, then how does the acceleration change?
Answer:
Given, Initially
M₁ = M and F₁ = F
Finally
M₂ = 2M and F₂=\(\frac {F}{2}\)
Hence, a₁ = \(\frac{F_{1}}{M_{1}}\)
a₁ = \(\frac{F}{M}\)

and a₂ = \(\frac{F_{2}}{M_{2}}\)
a₂ = \(\frac{\mathrm{F}}{2 \times 2 \mathrm{M}}\) = \(\frac{F}{4M}\)
\(\frac{a_{1}}{a_{2}}=\frac{F / M}{F / 4 M}=4\)
a₂ = \(\frac{a_{1}}{4}\)
Acceleration will become one fourth.

Question 5.
Derive the unit of force using the second law of motion. A force of 5 N produces an acceleration of 8 ms^-2 on a mass m₁ and an acceleration of 24 ms^-2 on a mass m₂. What acceleration would the same force provide if both the masses are tied together?
Answer:
Given, Force, F = 5N
a₁ = 8 m/s²
and a₁ = 24 m/s²
From formula
m₁ = \(\frac{F}{a_{1}}\) = \(\frac{5}{8}\) kg
and m₂ = \(\frac{F}{a_{2}}\) = \(\frac{5}{24}\) kg
When both the masses tied together
a =\(\frac{\mathrm{F}}{m_{1}+m_{2}}\)
a = \(\frac{5}{\frac{5}{8}+\frac{5}{24}}\) = 6m/s²
a = 6m/s²

Force and Laws of Motion Class 9 Extra Questions Value Based (VBQs)

Question 1.
During a cricket match, a new player Ayush injured his hand while catching the ball. His friend Rudra advised him to catch the ball by lowering his hands backward. When Ayush got another chance to catch the ball, he successfully caught the ball without injuring his hands.
Answer the following questions:
(a) A cricket player lowers his hands while catching the ball. Explain why.
(b) Write down the values shown by Rudra.
Answer:
(a) If a player does not lower his hands while catching the ball, the time to stop the ball will be very small. So a large force has to applied to reduce the velocity of the ball to zero or to change the momentum of the ball. When a player lowers his hands, the time to stop the ball is increased and hence less force has to be applied to cause the same change in momentum of the ball. Therefore, the hands of the player are not injured.

(b) Rudra is a good person as he helped his friend Ayush. He is a knowledgeable person.

Question 2.
During servicing his bike Aman advised the mechanic to oil the shockers for its proper functioning.
Answer the following questions:
(a) The vehicles are fitted with shockers. Explain why.
(b) Write down the values shown by Aman.
Answer:
(a) The floor of a vehicle is connected to the lower part of the vehicle by springs or shockers. When a vehicle moves over a rough road, the force due to jerks is transmitted to the floor of the vehicle through the shockers. The shockers increase the time of transmission of the force of the jerk to reach the floor of the vehicle. Hence less force is experienced by the passengers in the vehicles.
(b) Aman is an intelligent and careful person.

Question 3.
Ranjan advised his son Ayush to wear a seat belt while driving the car.
Answer the following questions
(a) While driving the car, the drivers and other passengers are advised to wear seat belts. Explain why.
(b) Write down the values shown by Ranjan.
Answer:
(a) When brakes are applied suddenly, the passengers of the car fall forward due to the inertia of motion. The seat belt worn by passengers of the car prevents them from falling forward suddenly. This enables the entire linear momentum of the passengers to reduce to zero over a long interval of time, hence it prevents injury.
(b) Ranjan is a noble, intelligent, and careful person.

In this page, we are providing Structure of the Atom Class 9 Extra Questions and Answers Science Chapter 4 pdf download. NCERT Extra Questions for Class 9 Science Chapter 4 Structure of the Atom with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 4 Extra Questions and Answers Structure of the Atom

Extra Questions for Class 9 Science Chapter 4 Structure of the Atom with Answers Solutions

Structure of the Atom Class 9 Extra Questions Very Short Answer Type

Structure Of Atom Class 9 Extra Questions Question 1.
Is it possible for the atom of an element to have one electron, one proton and no neutron? If so, name the element. [NCERT Exemplar]
Answer:
Yes, it is true for hydrogen atom which is represented as \(_{ 1 }^{ 1 }{ H }\)

Class 9 Science Chapter 4 Extra Questions Question 2.
Will ³⁵Cl and ³⁵Cl have different valencies? Justify your answer. [NCERT Exemplar]
Answer:
No, ³⁵Cl and ³⁵Cl are isotopes of an element.

Class 9 Structure Of Atom Extra Questions Question 3.
Why did Rutherford select a gold foil in his a-ray scattering experiment? [NCERT Exemplar]
Answer:
Rutherford selected a gold foil in his a-ray scattering experiment because gold has high malleability.

Structure Of Atom Class 9 Important Questions With Answers Pdf Question 4.
One electron is present in the outermost shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outermost shell? [NCERT Exemplar]
Answer:
+ 1.

Extra Questions On Structure Of Atom Class 9 Question 5.
Write any two observations which support the fact that atoms are divisible. [NCERT Exemplar]
Answer:
Discovery of electrons and protons.

Structure Of Atom Extra Questions Question 6.
Write down the electron distribution of chlorine atom. How many electrons are there in a L-shell? (Atomic number of chlorine is 17). [NCERT Exemplar]
Answer:
The electron distribution of chlorine atom is 2, 8, 7 and the L-shell has 8 electrons.

Class 9 Science Chapter 4 Extra Questions And Answers Question 7.
In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed? [NCERT Exemplar]
Answer:
– 2.

Extra Questions Of Structure Of Atom Class 9 Question 8.
Helium atom has 2 electrons in its valence shell but its valency is not 2, Explain.
Answer:
Helium atom has 2 electrons in its outermost shell and its duplet is complete. Hence the valency is zero.

Structure Of Atom Extra Questions Class 9 Question 9.
An element X has a mass number 4 and atomic number 2. Write the valency of this element.
Answer:
Valency is zero, as K-shell is completely filled.

Class 9 Science Ch 4 Extra Questions Question 10.
Who identified the sub-atomic particle electron?
Answer:
J.J. Thomson.

Class 9 Chapter 4 Science Extra Questions Question 11.
Name the rays discovered by E. Goldstein.
Answer:
Canal rays.

Structure Of An Atom Class 9 Extra Questions Question 12.
Give two examples from everyday life where we use cathode ray tubes.
Answer:
Television picture tube and Fluorescent light tubes.

Structure Of Atom Class 9 Extra Questions And Answers Question 13.
What are the actual values of charge and mass of the electron?
Answer:
Charge = 1.60 x 10^-19 C (one unit negative charge)
Mass = 9.11 x 10^-31 kg (1/1840th of the mass of hydrogen atom).

Chapter 4 Science Class 9 Extra Questions Question 14.
Compare the radius of the nucleus with that of the atom.
Answer:
Radius of the atom is of the order of 10^-10 m while that of the nucleus is of the order of 10^-15 m. Thus, nucleus is 10^-15/10^-10 i.e., 1/100,000^th of the size of the atom.

Ch 4 Science Class 9 Extra Questions Question 15.
Why are Bohr’s orbits called stationary states?
Answer:
According to Bohr, the orbits in which the electrons revolve have fixed energies. Hence, they are called stationary states (stationary means fixed).

Question 16.
Who discovered neutron?
Answer:
Chadwick.

Question 17.
What are nucleons? What is their number called?
Answer:
Protons and neutrons present in the nucleus are collectively called nucleons. Their total number is called ‘mass number’.

Question 18.
What do you mean by ‘Atomic number’?
Answer:
Atomic number of an element is the number of protons present in the nucleus of the atom of that element.

Question 19.
Who discovered the nucleus of the atom?
Answer:
Rutherford.

Question 20.
What is the charge on alpha particle?
Answer:
+ 2.

Question 21.
The mass number of an element is 18. It contains 7 electrons. What is the number of protons and neutrons in it?
Answer:
Number of protons = 7 Number of neutrons = 11 ,

Question 22.
What is the maximum number of electrons that can be present in M-shell?
Answer:
M-shell means 3rd shell for which n = 3. Hence, maximum number of electrons that can be present in M-shell = 2n² = 2 x 3² = 18.

Question 23.
Identify the isotopes out of A, B, C and D? ³³A₁₇, ⁴⁰B₂₀, ³⁷C₁₇, ³⁸D₁₉.
Answer:
³³A₁₄ and ³⁷C₁₇? are isotopes.

Question 24.
Represent the three isotopes of hydrogen and give their names.
Answer:
\(_{ 1 }^{ 1 }{ H }\) (Protium), \(_{ 2 }^{ 1 }{ H }\) (Deuterium), \(_{ 3 }^{ 1 }{ H }\) (Tritium).

Question 25.
How many electrons are present in the species He²⁺ ion? Suggest another name for it.
Answer:
He²⁺ ion has no electrons. It is also called alpha (α) particle.

Question 26.
What is the reason for the identical chemical properties of all the isotopes of an element?
Answer:
Isotopes have identical electronic configuration.

Question 27.
Give one similarity and one difference between a pair of isotopes.
Answer:
They have same number of protons but different number of neutrons in their nuclei.

Question 28.
Give the number of protons and neutrons in an atom of uranium (U-235) used in a nuclear reactor.
Answer:
In nuclear reactor U-235 is used. Atomic number of Uranium is 92.
∴ Number of protons = Atomic number = 92
Number of neutrons = Mass number – Atomic number = 235 – 92 = 143.

Question 29.
Out of proton and neutron, which is heavier?
Answer:
Neutron is slightly heavier (1.675 x 10^-27 kg) than proton (1.67 x 10^-27 kg).

Question 30.
If K and L shells of an atom are completely filled what will be its name?
Answer:
The atom will belong to the noble gas element neon (Ne).

Question 31.
Do isobars have also identical chemical characteristic like isotopes?
Answer:
No, these are not identical because the isobars have different atomic number as well as different electronic configuration.

Question 32.
What is the number of electrons in the valence shell of chlorine (Z = 17)?
Answer:
The electronic distribution of the element is:
K(2), L(8), M(7). This means that the valence shell of chlorine has 7 electrons.

Question 33.
Which radioisotope is used for the treatment of cancer?
Answer:
Radioisotope Co-60 is used for the treatment of cancer.

Question 34.
Out of C-12 and C-14 isotopes of carbon, which is of radioactive nature?
Answer:
C-14 isotope is of radioactive nature.

Question 35.
The electron configuration of an element is: 2(K), 8(L), 5(M). Predict its valency.
Answer:
The valency of the element is 3. It is calculated as: 8 – 5 = 3.

Question 36.
Will \(_{ 12 }^{ 6 }{ C }\) and \(_{ 14 }^{ 6 }{ C }\) have different valencies?
Answer:
Both the species are the isotopes of the same element i.e., the carbon. Since their atomic numbers Eire same, their electronic configurations as well as valencies will also be the same.

Question 37.
Do the elements \(_{ 3 }^{ 1 }{ X }\) and \(_{ 3 }^{ 2 }{ Y }\) represent pair of isotopes?
Answer:
No, they do not because the two elements differ in their atomic numbers. Isotopes have the same atomic numbers.

Structure of the Atom Class 9 Extra Questions Short Answer Type 1

Question 1.
In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer. [NCERT Exemplar]
Answer:
No, the statement is incorrect. In an atom the number of protons and electrons is always equal.

Question 2.
Calculate the number of neutrons present in the nucleus of an element X which is represented as \(_{ 31 }^{ 15 }{ X }\). [NCERT Exemplar]
Answer:
Mass number = No. of protons + No. of neutrons = 31
∴ Number of neutrons = 31 – number of protons
= 31 – 15
= 16

Question 3.
The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements? [NCERT Exemplar]
Answer:
∴ They are isobars because isobars have different atomic numbers but same mass number.

Question 4.
Why do Helium, Neon and Argon have a zero valency?
Answer:
Helium has two electrons in its only energy shell, while Argon and Neon have 8 electrons in their valence shells. As these have maximum number of electrons in their valence shells, they do not have any tendency to combine with other elements. Hence, they have a valency equal to zero.

Question 5.
What are isobars? Give one example.
Answer:
Isobars are the atoms of different elements which have different atomic numbers but same mass number.
Example is \(_{ 40 }^{ 18 }{ Ar }\), \(_{ 40 }^{ 20 }{ Ca }\) .

Question 6.
What is the number of valence electron in:
(i) Sodium ion (Na⁺)
(ii) Oxide ion (O^2-)?
Answer:
(i) Sodium ion (Na⁺)
No. of electrons: (11 – 1); Electronic configuration = 2, 8.
∴ Na⁺ ion has 8 valence electrons.

(ii) Oxide ion (O^2-)
No. of electrons: (8 + 2) = 10; Electronic configuration = 2, 8.
∴ O^2- ion has 8 valence electrons.

Question 7.
The number of electrons in the outermost ‘L’ shell of an atom is 5.
(a) Write its electronic configuration.
(b) What is its valency and why?
Answer:
(a) 2, 5
(b) 3, because it needs three more electrons to complete its octet.

Question 8.
Write two difference between isobars and isotopes.
Answer:
Isobars

• Same mass number but different atomic numbers.
• Different number of protons.

Isotopes:

• Same atomic number but different mass numbers.
• Same number of protons.

Question 9.
Mention the postulates Neils Bohr put forth to overcome the objections raised against Rutherford’s atomic model.
Answer:
Bohr put forward the following postulates in order to overcome the objections raised against Rutherford s atomic method: .

Only certain special orbits known as discrete orbits of electrons are allowed inside the atom.
While revolving in discrete orbits the electrons do not radiate energy. These orbits or shells are called energy levels.

Question 10.
Given that natural sample of iron has isotopes \(_{ 54 }^{ 28 }{ Fe }\), \(_{ 58 }^{ 28 }{ Fe }\) and \(_{ 57 }^{ 28 }{ Fe }\) in the ratio of 5%, 90% and 5% respectively. What will be the average atomic mass of iron (Fe)?
Answer:
Average atomic mass
= \(\frac { 54 × 5 + 56 × 90 + 57 × 5 }{ 100 }\)
= \(\frac { 270+5040+285 }{ 100 }\)
= 55.95 u.

Question 11.
(a) What are canal rays? Who discovered them?
What is the charge and mass of canal ray?
(b) How are the canal rays different from electron in terms of charge and mass?
Answer:
(a) New radiations in a gas discharge tube which are positively charged. E. Goldstein discovered them. Charge on canal rays is positive and mass is one unit.

(b) Electrons are negatively charged and their mass is approximately 1/2000 of that of canal rays.

Question 12.
The average atomic mass of a sample of an element y is 35.5 u. What are the percentages of isotopes \(_{ 37 }^{ 17 }{ y }\) and \(_{ 35 }^{ 17 }{ y }\), in the sample?
Answer:
Let the % composition of \(_{ 37 }^{ 17 }{ y }\) be x.
Then the % composition of \(_{ 37 }^{ 17 }{ y }\) will be 100 – x
Now,

Question 13.
(i) Write the postulates of Bohr model of atom.
(ii) Draw a sketch of Bohr model of an atom with atomic number 15.
Answer:
(i) Only certain special orbits-discrete orbits of electrons are allowed.

(ii) While revolving in discrete orbits, the electrons do not radiate energy.

Question 14.
List the main differences between an atom and an ion.
Answer:
Atom:

• It is electrically natural.
• The valence shell of atom may or may not have 8 electrons.
• The atoms are less stable.
• It may or may not exist freely in solution.

Ion:

• It has positive or negative charge.
• The valence shell of an ion has 8 electrons.
• The ions are more stable.
• It can exist freely in solution.

Question 15.
An ion has one neutron more than the number of protons but has one electron less than the number of protons. Which two ions are possible which satisfy this condition? Explain.
Answer:
(i) Na⁺ ion, because it has atomic number = 11 and mass number = 23. Thus, no. of protons = 11, no. of neutrons = 23 – 11 = 12 and number of electrons = 11 – 1 = 10.

(ii) K⁺ ion, because it has atomic number = 19 and mass number = 39. Thus, no. of protons = 19, No. of neutrons = 39 – 19 = 20 and no. of electrons = 19 – 1 = 18.

Question 16.
Give the main points of Dalton’s atomic theory.
Answer:

• All matter is made up of a large number of extremely small indivisible particles called atoms.
• Atoms of the different elements are different.
• Atom is the smallest unit of matter which takes part in a chemical reaction.
• Atoms can neither be created nor be destroyed.

Question 17.
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign proper symbol of the species.
Answer:
Number of electrons present = 18
Number of neutrons present = 16
Atomic number of the element = 16
The element with atomic number 16 is sulphur (S). Since it has 18 electrons and not 16, it is therefore, an anion (S^2- ion).

Question 18.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. [NCERT Exemplar]
Answer:
We know
Mass number = No. of protons + No. of neutrons = 81
i.e., p + n = 81
Let number of protons = x

Structure of the Atom Class 9 Extra Questions Short Answer Type 2

Question 1.
Show diagrammatically the electron distributions in a sodium atom and a sodium ion and also given their atomic number. [NCERT Exemplar]
Answer:

Since the atomic number of sodium atom is 11, it has 11 electrons. A positively charged sodium ion (Na⁺) is formed by the removal of one electron from a sodium atom. So, a sodium ion has 11 – 1 = 10 electrons. Thus, electronic distribution of sodium ion will be 2, 8. The atomic number of an element is equal to the number of protons in its atom. Since, sodium atom and sodium ion contain the same number of protons, therefore, the atomic number of both is 11.

Question 2.
The ratio of the radii of hydrogen atom and its nucleus is ~ 10⁵. Assuming the atom and the nucleus to be spherical. [NCERT Exemplar]
(i) What will be the ratio of their sizes?
(ii) If atom is represented by planet Earth ‘R_e’ = 6.4 x 10⁵ m, estimate the size of the nucleus.
Answer:
(i) Volume of the sphere = \(\frac { 4 }{ 3 }\) πr³
Let R be the radius of the atom and r be that of the nucleus.

(ii) If the atom is represented by the planet earth (R_e = 6.4 x 10⁶ m) then the radius of the nucleus
would be r_n = \(\frac{\mathrm{R}_{e}}{10^{5}}\)
r_n = \(\frac{6.4 \times 10^{6} \mathrm{m}}{10^{5}}\)
= 6.4 x 10 m = 64 m.

Question 3.
The number of protons, neutrons and electrons in particles from A to E are given below:

(i) Which one is a cation?
(ii) Which one is an anion?
(iii) Which represent pair of isotopes?
Answer:
(i) B is a monovalent cation (B⁺)
(ii) E is a monovalent anion (E^–)
(iii) A and D represent pair of isotopes.

Question 4.
Explain Bohr Bury rules for distribution of electrons into different shells. Write the distribution of electrons in sodium atom (Z = 11).
Answer:
Bohr Bury Rules:

• The maximum no. of electrons present in a shell is given by the formula 2n² (where n is shell number).
• The maximum number of electrons that can be accommodated in the outermost orbit is 8.
• Electrons are not accommodated in a given shell, unless the inner shells are filled.
• Electronic configuration of Na = 2, 8, 1.

Question 5.
Atom A has a mass number 238 and atomic number 92 and atom B has mass number 235 and atomic number 92.
(i) How many protons, atoms A and B have?
(ii) How many neutrons, atoms A and B have?
(iii) Are atoms A and B isotopes of the same element? How?
Answer:
(i) Protons in atoms A and B = 92
(ii) Neutrons in atoms A and B = 146 and 143
(iii) Yes, atom A and B are isotopes of the element since they have the same atomic number or same electronic configuration.

Question 6.
State three features of the nuclear model of an atom put forward by Rutherford.
Answer:

1. There is a positively charged centre in an atom called the nucleus. Nearly, all the mass of an atom resides in the nucleus.
2. The electrons revolve around the nucleus in well-defined orbits.
3. The size of the nucleus is very small as compared to the size of the atom.

Question 7.
The nucleus of an atom is found to have a total mass of nearly 20.088 x 10^-27 kg and a total charge of 9.612 x 10^-19 coulombs. Calculate the atomic number and mass number of the atom. Name the element.
Answer:
Mass of a nucleon (proton or neutron)
= 1.674 x 10^-27 kg
Total mass of nucleus = 20.088 x 10^-27 kg
∴ No. of nucleons (protons + neutrons) present in the nucleus
= \(\frac{20.088 \times 10^{-27}}{1.674 \times 10^{-27}}\) = 12
∴ Mass number = 12u.
Charge on one proton = 1.602 x 10^-19 coulombs; Total charge on the nuclear = 9.612 x 10^-19
∴ No.of protons in the nucleus
= \(\frac{9.612 \times 10^{-19}}{1.602 \times 10^{-19}}\) = 6
Atomic number = 6. Thus, the element is carbon.

Question 8.
Give reasons why?
(а) Atom is electrically neutral.
(b) Atom as a whole is an empty space.
(c) Rutherford model of atom could not provide stability to the nucleus.
Answer:
(a) An atom is electrically neutral because the number of protons and number of electrons in it are equal.

(b) According to Rutherford’s experiment, the size of the nucleus is very small as compared to the size of an atom, therefore, atom as a whole is an empty space.

(c) According to Rutherford, the protons are present inside the nucleus and electrons revolve around the nucleus. According to the elecromagnetic theory, a charged particle moving in a circular path continuously loses energy in the form of electromagnetic radiations and finally falls into the nucleus.

Question 9.
Write the complete symbol for
(i) the nucleus with atomic number 56 and mass number 138.
(ii) the nucleus with atomic number 26 and mass number 55.
(iii) the nucleus with atomic number 4 and mass number 9.
Answer:
(i) The element with atomic number 56 is Ba. Its symbol is \(_{ 138 }^{ 56 }{ Ba }\)

(ii) The element with atomic number 26 is Fe. Its symbol is \(_{ 56 }^{ 26 }{ Fe }\)

(iii) The element with atomic number 4 is Be. Its symbol for \(_{ 9 }^{ 4 }{ Be }\)

Question 10.
How many protons, electrons and neutrons are there in the following nuclei?
(i) \(_{ 17 }^{ 8 }{ O }\)
(ii) \(_{ 25 }^{ 12 }{ Mg }\)
(iii) \(_{ 80 }^{ 35 }{ Br }\)
Answer:
(i) \(_{ 35 }^{ 79 }{ Br }\)
Atomic number, Z = 8
Mass number, A = 17
No. of protons = No. of electrons = Z = 8
No. of neutrons + No. of protons = A
No. of neutrons + 8 = 17
or No. of neutrons = 17 – 8 = 9

(ii) \(_{ 25 }^{ 12 }{ Mg }\)
Atomic number, Z = 12
Mass number, A = 25
No. of protons = No. of electrons = Z = 12
No. of neutrons = A – No. of protons = 25 – 12 = 13.

(iii) \(_{ 80 }^{ 35 }{ Br }\)
Atomic number, Z = 35
Mass number, A = 80
No. of protons = No. of electrons = Z = 35
No. of neutrons = A – No. of protons = 80 – 35 = 45.

Question 11.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol for the ion.
Answer:
Since the iron carries 3 units of positive charge, it will have 3 electrons less than the number of protons. Let number of electrons = x
No. of protons = x + 3
No. of neutrons = x + \(\frac { x × 30.4 }{ 100 }\)
= x + 0.304x
= 1.304x
Now, No. of protons + No. of neutrons = 56
x + 3 + 1.304 x = 56
2.304 x = 53
x = \(\frac { 53 }{ 2.304 }\) = 23
No. of electrons = 23
No. of protons = 23 + 3 = 26
Symbol = \(_{ 56 }^{ 26 }{ Fe }\)³⁺.

Question 12.
The nuclear radius is of the order of 10^-13 cm while atomic radius is of the order 10^-8 cm. Assuming the nucleus and the atom to be spherical, what fraction of the atomic volume is occupied by the nucleus?
Answer:
The volume of a sphere = 4 πR^3/3 where R is the radius of the sphere.
∴ Volume of the nucleus = 4 πr^3/3 = 4π (10^-13)^3/3 cm³
Similarly,
Volume of the atom = 4 πR^3/3 = 4π (10^-8)^3/3 cm³
∴ Fraction of the volume of atom occupied by the nucleus
= \(\frac{4 \pi\left(10^{-13}\right)^{3} / 3 \mathrm{cm}^{3}}{4 \pi\left(10^{-8}\right)^{3} / 3 \mathrm{cm}^{3}}\)
= 10^-15.

Question 13.
Calculate the number of electrons, protons and neutrons in the following species:
(i) Phosphorus atom
(ii) Phosphide ion (P^3-)
(iii) Magnesium ion (Mg²⁺)
Mass numbers: P = 31; Mg = 24
Atomic numbers: P = 15, Mg = 12.
Answer:
(i) Phosphorus atom
Number of electrons = Atomic number = 15
Number of protons = Atomic number = 15
Number of neutrons = Mass number – Atomic number
= 31 – 15 = 16.

(ii) Phosphide ion (P^3-)
Phosphide ion (P^3-)
= Phosphorus atom + 3 electrons
P^3- = P + 3e^–
Thus, phosphide ion has same number of protons and neutrons as phosphorus atom but has three electrons more.
Number of electrons = 15 + 3 = 18
Number of protons = 15
Number of neutrons = 31 – 15 = 16

(iii) Magnesium ion (Mg²⁺)
Mg²⁺ ion is formed by loss of two electrons by Mg atom. Therefore, it has two electrons less than number of electrons in Mg atom.
Mg²⁺ = Mg – 2e^–
Number of electrons = 12 – 2 = 10
Number of protons = 12
Number of neutrons = (24 – 12) = 12

Structure of the Atom Class 9 Extra Questions Long Answer Type

Question 1.
(a) Enlist the conclusions drawn by Rutherford from his a-ray scattering experiment.
(b) In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?
Answer:
(a) Rutherford concluded from the a-particle scattering experiment that:

(i) Most of the space inside the atom is empty because the a-particles passed through the gold foil without getting deflected.

(ii) Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.

(iii) A very small fraction of a-particles were deflected by 180°, indicating that all the positive charges and mass of the gold atom were concentrated in a very small volume within the atom.

From the data he also calculated that the radius of the nucleus is about 105 times less than the radius of the atom.

(b) Rutherford proposed a model in which electrons revolve around the nucleus in well-defined orbits. There is a positively charged center in an atom called the nucleus. He also proposed that the size of the nucleus is very small as compared to the size of the atom and nearly all the mass of an atom is centred in the nucleus.

Whereas, Thomson proposed the model of an atom to be similar to a Christmas pudding. The electrons are studded like currants in a positively charged sphere like Christmas pudding and the mass of the atom was supposed to be uniformly distributed.

Question 2.
(a) What were the drawbacks of Rutherford’s model of an atom?
(b) What are the postulates of Bohr’s model of an atom?
Answer:
(a) The orbital revolution of the electron is not excepted to be stable. Any particle in a circular orbit would undergo an acceleration and the charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know.

(b) The postulates put forth by Neils Bohr’s about the model of an atom:
(i) Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.

(ii) While revolving in discrete orbits the electrons do not radiate energy.
These orbits are called energy levels. Energy levels in Em atom are shown by circles.
These orbits are represented by the letters K, L, M, N, …………….. or the numbers, n = 1, 2, 3, 4, ………………

Question 3.
An atom of an element has three electrons in the third shell which is the outermost shell. Write
(a) the electronic configuration
(b) the atomic number
(c) number of protons
(d) valency
(e) the name of the element
(f) its nature whether metal or non-metal.
Answer:
The third shell is M-shell. If the atom of the element has three electrons in the third shell, this means that K and L shells are already filled.
(a) Electronic configuration: 2, 8, 3
(b) Atomic number = No. of electrons = 13
(c) Number of protons = No. of electrons = 13
(d) Valency of the element = 3
(e) The element with Z = 13 is aluminium (Al)
(f) It is a metal.

Question 4.
Answer the following in one line or two:
(a) What is the maximum number of electrons that can be accommodated in the outermost energy shell in an atom?
(b) On the basis of Thomson’s model of an atom, explain how an atom is neutral as a whole.
(c) How many neutrons are present in hydrogen atom?
(d) Do isobars belong to the same element?
(e) An element has five electrons in the M shell which is the outermost shell. Write its electronic configuration.
Answer:
(a) The outermost energy shell in an atom can have a maximum of eight electrons.

(b) According to Thomson’s model of an atom, all the protons in an atom are present in the positively charged sphere while negatively charged electrons are studded in this sphere. Since the electrons and protons are equal in number each carrying one unit charge, the atom as a whole is electrically neutral.

(c) Hydrogen atom has no neutron.

(d) No, isolars belong to different elements since they differ in their atomic numbers.

(e) The electronic configuration of the element is K(2), L(8), M(5).

Question 5.
Explain why did Rutherford select a gold foil in his alpha particle scattering experiments? What observations in a-scattering experiment led Rutherford to make the following observations:
(i) Most of the space in an atom is empty.
(ii) Nucleus is positively charged.
Mention any two drawbacks of Rutherford’s model.
Answer:
Rutherford selected a gold foil because he wanted a very thin layer as possible. This gold foil was about 1000 atoms thick.

(i) As most of the alpha particles passed through the foil undeviated, it means that they did not come across any obstruction in their part. Thus, most of the space in an atom was thought to be empty.

(ii) Very few particles deviated by small angles from their path which suggested that nucleus is positively charged.

The revolution of the electron in any circular orbit is not expected to be stable. Any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy. Thus, the revolving electrons would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and matter would not exist in the form that we know. We know that atoms are quite stable.

Question 6.
(a) Why are anode rays called canal rays?
(6) Mention two postulates of J.J. Thomson’s model.
(c) Compare the properties of protons and electrons.
Answer:
(a) The anode rays produced at the anode of the discharge tube are called canal rays because they pass through the holes of the cathode.

(b)

• Atom consists of positively charged sphere and electrons are embedded in it.
• The negative and positive charges are equal in magnitude.

(c) Protons:

• Positively charged.
• Mass of 1 proton is equal to mass of H atom.

Electrons:

• Negatively charged.
• Mass of electron is 1/1840 times that of a proton.

Question 7.
(a) What are isobars?

(b) Atomic number of an element Y is 17.
(i) Write its electronic configuration.
(ii) What is the number of valence electron in Y?
(iii) How many electrons are needed to complete the octet of Y?
(iv) Is it a metal or non-metal?

(c) The valency of Na is 1 and not 7. Give reason.
Answer:
(a) Atoms of different elements with different atomic numbers which have same mass number are called isobars.

(b) (i) 2,8, 7
(ii) 7
(iii) 1
(iv) Non-metal

(c) Electronic configuration of Na is 2, 8,1. It can lose one electron to attain the electronic configuration of neon. Therefore, the valency of Na is 1. The valency of Na may be 7 when it gains 7 electrons in its valence shell, but gain ring 7 electrons is difficult. Therefore, the valency of Na is not 7.

Question 8.
An atom of an element has 2 electrons in the M-shell i.e., third shell. What will be the atomic number of this element? Name this element. Find the valency of this element. Also, find the number of neutrons in the atom of this element.
Answer:
(i) Atomic number is 12

(ii) Element is magnesium.
(iii) Valency is 2 (⁺²)
(iv) No. of neutrons = Atomic mass – No. of protons = 24 – 12 = 12.

Structure of the Atom Class 9 Extra Questions HOTS

Question 1.
What information do you get from the figure given below about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form. [NCERT Exemplar]

Answer:

Question 2.
In the Gold foil experiment of Geiger and Marsdein, that paved the way for Rutherford’s model of an atom, ~ 1.00% of the a-particles were found to deflect at angles > 50°. If one mole of α – particles were bombarded on the gold foil, compute the number of α – particles that would deflect at angles less than 50°. [NCERT Exemplar]
Answer:
% of α – particles deflected more than 50° = 1% of α – particles
% of α – particles deflected less than 50° = 100 – 1 = 99%
Number of α – particles bombarded = 1 mole = 6.022 x 10²³ particles
Number of particles that deflected at an angle less than 50°
= \(\frac { 99 }{ 100 }\) x 6.022 x 10²³
= \(\frac { 596.178 }{ 100 }\) x 10²³
= 5.96 x 10²³.

Question 3.
An ion Y^3- contains 18 electrons and 16 neutrons. Calculate the atomic number and mass number of the element Y. Name the element Y.
Answer:
Number of electrons in Y^3- ion = 18
Since negative charge is formed by gain of electrons by the neutral atom and the number of electrons gained is equal to the number of units of negative charge on the ion.
∴ Number of electrons in the neutral atom = 18-3 =15.
Now, for a neutral atom, Atomic number = Number of protons = Number of electrons.
∴ Atomic number of element Y = 15
Mass number of the element = Number of protons + Number of neutrons = 15 + 16 = 31
The given element Y with atomic number 15 is phosphorus.

Question 4.
The following data represent the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D.

Answer the following questions:
(a) Give the electronic distribution of element B.
(b) The valency of element A.
(c) The atomic number of element B.
(d) The mass number of element D.
Answer:
(а) Electronic distribution of elements B : 2, 8, 6.
(b) Valency of A is 1(-1).
(c) Atomic number of element B is 16.
(d) Mass number of element D is 39.

Question 5.
Give a reason to explain why:
(a) isotopes of an element show identical chemical properties?
(b) the atomic masses of elements are in fractions?
(c) atoms combine with other atoms?
Answer:
(a) Since all the isotopes of an element have identical electronic configuration containing the same number of valence electrons. Therefore all the isotopes of an element show identical chemical properties.

(b) The fractional atomic masses of elements are due to their isotopes having different masses.

(c) The atoms combine with other atoms to achieve the electronic configuration of the nearest noble gas and thus, become more stable.

Question 6.
(a) Describe the main features of Bohr’s model of an atom. Draw a neat and labelled diagram of energy levels.
(b) Which of the following pairs are isotopes and which are isobars?
(i) \(_{ 58 }^{ 26 }{ A }\), \(_{ 58 }^{ 28 }{ B }\)
(ii) \(_{ 79 }^{ 35 }{ X }\), \(_{ 80 }^{ 35 }{ Y }\)
Give reasons for your choice.
(c) Elements A and B have atomic numbers 18 and 16 respectively. Which of these two would be more reactive and why?
Answer:
(a) Features of Bohr’s Models:
(i) Electrons revolve in certain permitted orbits which are associated with fixed amount of energy. So they are called energy levels (K, L, M, N) or sub-shells.
(ii) As long as electron revolves in the same energy level, they do not lose energy.
(iii) The energy of orbit closest to the nucleus is lowest and farthest away is highest.
(iv) When we supply energy to an electron, it goes to a higher energy level and when it comes back to a lower level it radiates energy.

(b) \(_{ 79 }^{ 35 }{ X }\), \(_{ 80 }^{ 35 }{ Y }\) are isotopes because their atomic no. is same but mass no. is different. \(_{ 58 }^{ 26 }{ A }\), \(_{ 58 }^{ 28 }{ B }\) are iso¬bars, as mass no. is same but atomic no. is different.

(c) B would be more reactive as its electronic configuration is 2, 8, 6 and it requires two more electrons to complete its octet.

Question 7.
Choose the noble gases from the elements shown in the table below:

Answer:

A, B and D have completely filled valence shells, i.e., either 2 or 8 electrons in their valence shells.
∴ A, B and D are noble gases.

Question 8.
The mass number and electronic configuration of five elements A, B, C, D and E are as follows:

(а) Name the elements which has 22 neutrons in nucleus.
(b) What is atomic number of C?
(c) Name the elements which will form most stable ionic bond.
(d) Give the formation of the compound between B and D.
(e) Name the elements which will not take part in chemical combination.
Answer:
(a) E
(b) 11
(c) B and C
(d) DB₄
(e) E

Question 9.
Write the electronic configurations of the following elements and write the number of valence electrons present in it.
(a) \(_{ 14 }^{ 7 }{ N }\)
(b) \(_{ 28 }^{ 14 }{ Si }\)
(c) \(_{ 40 }^{ 20 }{ Ca }\)
(d) \(_{ 40 }^{ 18 }{ Ar }\)
(e) \(_{ 9 }^{ 4 }{ Be }\)
Answer:

Question 10.
Complete the following table.

Answer:
For F atom
Atomic no. = 9 (Give)
No. of electrons = 9
No. of protons = 9
No. of neutrons = 10 (Given)
Mass no. = Z + n = 9 + 10 = 19

For Mg²⁺ ion
No. of protons = 12 (Given)
Atomic no. = 12
No. of electrons = 12 – 2 = 10
Mass no. = 24 (Give)
No. of neutrons = A – Z = 24 – 12 = 12

For S atom
No. of protons = 16 (Given)
Atomic no. = 16
No. of protons = 16
Mass no. = 32 (Given)
No. of neutrons = A – Z = 32 – 16 = 16

For P^3- ion
No. of electrons = 18 (Given)
No. of protons = 18 – 3 = 15
Atomic no. = 15
No. of neutrons = 16 (Given)
Mass no. = Z + n = 15 + 16 = 31

The complete table is as follows:

Note: The value of positive charge on cation shows that the number of electrons in it are less than the number of protons by the same value. Similarly, the value of negative charge on anion shows that the number of electrons in it are more than the number of protons by the same value.

In this page, we are providing Diversity in Living Organisms Class 9 Extra Questions and Answers Science Chapter 7 pdf download. NCERT Extra Questions for Class 9 Science Chapter 7 Diversity in Living Organisms with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 7 Extra Questions and Answers Diversity in Living Organisms

Extra Questions for Class 9 Science Chapter 7 Diversity in Living Organisms with Answers Solutions

Diversity in Living Organisms Class 9 Extra Questions Very Short Answer Type

Class 9 Science Chapter 7 Extra Question Answer Question 1.
Who proposed the two kingdom classification?
Answer:
Carolus Linnaeus.

Diversity In Living Organisms Class 9 Extra Questions And Answers Question 2.
What is biodiversity?
Answer:
The variety of life forms including plants, animals and microscopic organisms which inhabit this earth constitute biodiversity.

Diversity In Living Organisms Class 9 Questions Question 3.
What is classification?
Answer:
The grouping of organisms on the basis of their similarities and differences is called classification.

Diversity In Living Organisms Class 9 Extra Questions Question 4.
What is the taxonomy?
Answer:
The branch of science which deals with the classification of organisms is called taxonomy.

Diversity In Living Organisms Extra Questions Question 5.
Who proposed the five kingdom system of classification?
Answer:
R.H. Whittaker.

Diversity In Living Organisms Class 9 Questions And Answers Question 6.
Name the fundamental unit of classification.
Or
Which is the lowermost category in the hierarchy of classification?
Answer:
Species.

Diversity In Living Organisms Class 9 Extra Questions And Answers Pdf Question 7.
Give one point of difference between gymnosperms and angiosperms.
Answer:
The seeds of gymnosperms are naked whereas the seeds of angiosperms are enclosed within a fruit.

Diversity In Living Organisms Class 9 Important Questions Question 8.
Name the largest phylum of kingdom Animalia.
Answer:
Arthropoda.

Class 9 Science Ch 7 Extra Questions Question 9.
Name the class to which the sea horse belongs.
Answer:
Sea horse belongs to class Pisces.

Questions On Diversity In Living Organisms Class 9 Question 10.
Which group of plants is referred to as vascular cryptogams?
Answer:
Pteridophytes.

Class 9 Diversity In Living Organisms Extra Questions Question 11.
Which phylum in animial kingdom consists of pseudocoelomate organisms?
Answer:
Nematoda

Class 9 Science Chapter 7 Extra Questions Answers Question 12.
Which group of organisms are called as the ‘Amphibians of plant kingdom*?
Answer:
Bryophytes

Diversity In Living Organisms Class 9 Fill In The Blanks Question 13.
Name the two classes of angiosperms with one example of each.
Answer:
The two classes are –

• Monocots: Wheat
• Dicots: Pea

Diversity In Living Organisms Important Questions Question 14.
Name the division of cryptogams to which algae belong.
Answer:
Thallophyta

Class 9 Diversity In Living Organisms Questions Question 15.
Which is the highest unit of classification?
Answer:
Kingdom

Question 16.
What is the characteristic feature due to which echinoderms are named?
Answer:
Spiny skin (Echino- spiny; derm-skin)

Question 17.
Name the group which comprises of bacteria and blue green algae.
Answer:
Monera

Question 18.
Name an organism which is called saprophyte. Why is it called so?
Answer:
Yeast. It is called so as it feeds on dead and decaying matter to obtain its nutrition.

Question 19.
Identify the kingdom in which the organisms do not have a well-defined nucleus and are not able to show multicellular designs.
Answer:
Monera

Question 20.
Give reason, why blue green algae are classified along with bacteria and placed in the kingdom Monera.
Answer:
As the blue green algae are unicellular prokaryotes like bacteria.

Question 21.
Mention any one characteristic feature of saprophytes.
Answer:
The saprophytes feed on dead and decaying organic matter.

Question 22.
Write one point of difference between monocotyledonous and dicotyledonous plants.
Answer:
Monocots are plants which bear single cotyledon in their seeds. Dicots are the plants which bear two cotyledons in their seeds.

Question 23.
Poriferans have hole or pores all over the body that lead to a system that helps in circulating water to bring in food and oxygen. Name the system.
Answer:
The system is called as the canal system.

Question 24.
State the phylum to which liver fluke and Planaria belong.
Answer:
Phylum Platyhelminthes

Question 25.
Write the type of body cavity and symmetry possessed by nematodes.
Answer:
Body cavity is pseudocoelom and symmetry is bilateral symmetry.

Question 26.
Name one mammal that lays eggs.
Answer:
Platypus

Question 27.
Name the substance which Coelomic cavity of arthropods is filled with. What type of symmetry do they have?
Answer:
Coelomic cavity of arthropods is filled with blood. They show bilateral symmetry.

Question 28.
Name the phylum to which centipede and prawn belong.
Answer:
Phylum Arthropoda

Question 29.
Echinoderms are marine animals. What is their skeleton made up of?
Answer:
Their skeleton is made up of calcium carbonate.

Question 30.
Name one reptile with a four chambered heart.
Answer:
Crocodile

Question 31.
Shyam knew the correct scientific name of mango but did not follow the convention while writing it and wrote it as Mangifera Indica. Rewrite the scientific name as per the convention.
Answer:
Mangifera indica

Question 32.
Rewrite the scientific names correctly:
(a) Panthera tigris
(b) Periplaneta Americana
Answer:
(a) Panthera tigris
(b) Periplaneta americana

Diversity in Living Organisms Class 9 Extra Questions Short Answer Type 1

Question 1.
How can we say that classification of organisms is closely related to their evolution?
Answer:
We can say that classification of organisms is closely related to their evolution because the simple organisms have a primitive body design as they appeared earlier whereas the complex organisms have more advanced body designs as they are more recent. This shows that during the course of evolution more complex body designs were formed from simpler ones.

Question 2.
What is the difference between algae and fungi?
Answer:
Algae:

• Have chlorophyll.
• Autotrophic mode of nutrition.
• Cell wall made of cellulose.
• Food stored in the form of starch.
• Examples: Chlamydomonas, Spirogyra, Ulo-thrix

Fungi:

• Lack chlorophyll.
• Heterotrophic mode of nutrition.
• Cell wall made of chitin.
• Food stored in the form of glycogen.
• Examples: Rhizopus, Agaricus, Yeast

Question 3.
Pick the odd one out and justify your choice by giving reasons:
(a) Moss, Fern, Pinus, Spirogyra
(b) Sea cucumber, Octopus, Feather star, Star fish
Answer:
(a) The odd one out in this case is Pinus as it is a phanerogams having covered reproductive parts whereas the other three are cryptogams which bear hidden reproductive organs.

(b) The odd one out in this case is Octopus as it belongs to phylum Mollusca while others are the members of phylum Echinodermata.

Question 4.
Why were fungi and bacteria considered as plants even though they do not have chlorophyll?
Answer:
Fungi and bacteria were considered as plants as they have cell wall which is a characteristic feature of the plants. So, some earlier classification systems included them under plants.

Question 5.
Why do bryophytes and pteridophytes grow in moist and shady places?
Answer:
Bryophytes and pteridophytes grow in moist and shady places as they need water for their reproduction. Male gametes are carried towards the female gamete by water in order to bring about fertilisation in them.

Question 6.
Which divisions of the plant kingdom are called cryptogams? Why are they called so?
Answer:
Thallophyta, Bryophyla and Pteridophyta are considered as cryptogams. They are called so because they bear hidden and inconspicuous reproductive orgAnswer:

Question 7.
Characteristics of some organisms are given. Identify their group and give one example of each.
(а) Single celled, eukaryotic and photosynthetic
(b) The body is divided into segments, may be unisexual or hermaphrodite
Answer:
(a) Protista: Euglena
(b) Annelida: Earthworm

Question 8.
How do saprophytes get their food? Give one example of saprophyte.
Answer:
Saprophytes derive their nutrition from the dead and decaying materials. For example, Agaricus (Mushroom), Rhizopus (Bread mould), Yeast etc.

Question 9.
What are phanerogams? How are they classified?
Answer:
The phanerogams are the plants which produce seeds and have a well differentiated body with true roots, stem and leaves. They are advanced members of kingdom Plantae. It includes gymnosperms and angiosperms.

Question 10.
What are gymnosperms? Give two examples.
Answer:
The plants which bear naked seeds which are not enclosed in fruit are called gymnosperms. For example, Cycas, Pinus, etc.

Question 11.
Write two peculiar characters of sponges.
Answer:
The two peculiar features of sponges are:
(i) They have pores called ostia all over their body and a single large opening at the top, called osculum which helps to develop a canal system for water movement.

(ii) They have a skeleton made up of calcareous or siliceous spicules or spongin fibres which gives strength and support.

Question 12.
Classify the following in their respective phylum/class: Jellyfish, earthworm, cockroach, rat
Answer:

• Jellyfish: Phylum Coelenterata
• Earthworm: Phylum Annelida
• Cockroach: Phylum Arthropoda
• Rat: Phylum Vertebrata, Class: Mammalia

Question 13.
What are the two peculiar features of phylum Echinodermata?
Answer:
The phylum Echinodermata has organisms which have

• spiny skin
• water vascular system

Question 14.
What are the three germ layers present in the organisms? What are the two groups of organisms on the basis of germ layers?
Answer:
The three germ layers are ectoderm, mesoderm and endoderm. The groups are diploblastic and triploblastic.

Question 15.
How are vertebrates different from the other chordates?
Answer:
The notochord is present at any stage of their life cycle in the case of chordates. In the vertebrates the notochord gets replaced by the vertebral column.

Question 16.
How are pteridophytes are different form Phanerogams? Give one example for each group.
Answer:
Pteridophytes have hidden, inconspicuous reproductive organs example ferns.
Phanerogams have well differentiated reproductive organs which are not hidden, example rose, apple.

Question 17.
(a) Given below are few plant species. Identify the divisions to which they belong and write the major characteristic of each division.
(i) Spirogyra
(ii) Deodar
(iii) Moss
(b) What is the mode of nutrition for all of them?
Answer:
(a) (i) Thallophyta: Plant body is not well differentiated
(ii) Gymnosperms: have naked seeds
(iii) Bryophyta: have rhizoids for absorption of water, have stem-like and leaf-like structures.

(b) All of them are autotrophic organisms.

Question 18.
What do you understand by the term ‘naked embryo’? Name any two divisions in kingdom Plantae that have naked embryo. Give one example of each division.
Answer:
Naked embryo is the term which refers to an embryo which is not borne inside the seed. The pteridophytes and gymnosperms bear naked embryo. For example, Ferns and horsetail are pteridophytes. Pinus and Cycas are gymnosperms.

Question 19.
Write the difference between Gymnosperms and Angiosperms giving example of each type.
Or
What are gymnosperms? Give two characteristics.
Answer:
Gymnosperm:

• Bear naked seeds.
• Are woody, evergreen, perennials.
• Examples: Pinus and Cycas

Angiosperm:

• Bear seeds enclosed in fruit.
• Can be woody, non-woody annual, biennial or perennials.
• Examples: Mango, Neem

Question 20.
Thallophyta, bryophyta and pteridophyta are classified as cryptogams whereas gymnosperms and angiosperms are classified as phanerogams, why?
Answer:
Due to the presence of hidden and inconspicuous reproductive organs, thallophyta, bryophyta and pteridophyta are called as cryptogams. Gymnosperms and angiosperms are phanerogams as they have well developed and distinct reproductive organs, flowers, fruits and seeds.

Question 21.
State reasons for the following:
(а) Platyhelminthes are called so.
(b) Birds have hollow bones.
Answer:
(a) Platyhelminthes are called so because they have a dorsoventrally flattened body.
(b) Presence of hollow bones is an adaptation in birds which helps them to keep low body weight which is helpful in flight.

Question 22.
Identify the phylum of animals by the given characteristics and give an example of each.
(a) The coelomic cavity is blood-filled and the animals have jointed legs.
(b) The animals are called as flatworms and are either free living or parasitic.
Answer:
(a) Phylum arthropoda: eg., cockroach, butterfly, spider, etc.
(b) Phylum Platyhelminthes: Planaria, liver fluke, tapeworm, etc.

Question 23.
Write one point of difference between the following:
(а) Bilateral symmetry and radial symmetry
(b) Annelids and Nematodes
Answer:
(a) Bilateral symmetry – Body can be divided into two exact halves from one plane only.
Radial symmetry – Body can be divided into equal halves from any plane.

(b) Annelids – Have true coelom
Nematodes – Have pseudocoelom.

Question 24.
Give two examples of each:
(а) Egg laying mammals
(b) Organisms with open circulatory system.
Answer:
(а) Duck-billed platypus and Echidna are the egg laying mammals.
(b) Cockroach and Octopus have open circulatory system.

Question 25.
Select the odd one out with respect to classification. Also give reason for your choice: prawn, scorpion, octopus, butterfly.
Answer:
Octopus is the odd one out as it belongs to phylum Mollusca whereas others belong to phylum Arthropoda.

Question 26.
Given below are the two groups of organisms belonging to kingdom Animalia. Write the names of the phylum to which they belong.
(a) Octopus, Pila, Unio
(b) Centipede, prawn, scorpion
Answer:
(а) Octopus, Pila, Unio belong to Phylum Mollusca.
(b) Centipede, prawn, scorpion belong to phylum Arthropoda.

Question 27.
What is binomial nomenclature? Who introduced it?
Answer:
The system of scientific naming of the organism which consists of a generic name and a specific epithet is called as binomial nomenclature. It was introduced by Carolus Linnaeus.

Question 28.
Give two differences between bony fish and cartilaginous fish. Give one example of each.
Answer:
Bony Fish:

• Endoskeleton is made up of bones.
• Operculum covers the gill slits.
• Terminal mouth Example: Rohu, sea horse

Cartilaginous Fish:

• Endoskeleton is made up of cartilage.
• Operculum absent.
• Ventral mouth Example: Shark, saw fish

Question 29.
State any two characteristics of Mammalia. Name two egg laying mammals.
Answer:
Two characteristics of Mammalia are:

• Presence of mammary glands, four chambered heart and are warm-blooded
• Skin has hairs, sweat glands and oil glands Egg laying mammals are platypus and Echidna.

Question 30.
Write appropriate terms for the following.
(a) Animals which are able to maintain a certain body temperature over a wide range of temperature in the environment.
(b) Animals which have pseudocoelom.
Answer:
Kingdom Mammalia: Warm-blooded animals
Phylum Nematoda: Its members have pseudocoelom.

Diversity in Living Organisms Class 9 Extra Questions Short Answer Type 2

Question 1.
Explain the three basic features for grouping all organisms into five major kingdoms.
Answer:
The three basic features for grouping the organisms into five kingdoms are
(i) Cell structure: On the basis of this the two groups are prokaryotes and the eukaryotes which are distinguished on the basis of absence or presence of well defined nuclear membrane.

(ii) Thallus organisation: The organisms are grouped as unicellular or multicellular organisms on the basis of their being composed of a single cell or of many cells respectively.

(iii) Mode of nutrition: The organisms are grouped as autotrophs or heterotrophs on the basis of their ability to synthesise their own food or being dependent on other organisms for their food.

Question 2.
What are the steps in building a hierarchy of classification?
Answer:
The characteristics chosen for developing a hierarchy of classification should start with the characteristic which forms the broadest division and then the next characteristic chosen should be dependent on the previous characteristic and division, besides having its own new characteristic features.

This process should be continued for the next levels in order to build a hierarchy in classification. For example, the classification of organisms into two broad categories prokaryotes and eukaryotes forms the basis of further characteristics on which their classification is based.

Question 3.
Differentiate between Bryophyta and Pteridophyta. Give example of each group.
Or
Write four main features of pteridophyta and give two examples.
Answer:
Bryophyta:

• They are called the ‘amphibians of the plant kingdom’.
• They lack vascular tissues.
• Body is not well-differentiated into true root, stem or leaves.
• The dominant phase or the main plant body is gametophyte (haploid).
• Sporophyte depends upon gametophyte for its support and nutrition.
• Spores are formed in capsule of sporophyte.
• Examples: Liverworts, Mosses

Pteridophyta:

• They are the first land plants.
• They have vascular tissues xylem and phloem.
• Body is well-differentiated into true roots, stem and leaves.
• The dominant phase or the main plant body is sporophyte (diploid).
• Sporophyte and gametophyte are independent structures in them.
• Spores are produced inside the sporangia borne on leaves or cones.
• Examples: Ferns, Horsetail, Marsilea

Question 4.
Name three groups which are placed under Cryptogamae. State and explain two characteristics which are exhibited by each category of these plant bodies.
Answer:
The three groups placed under Cryptogamae are: Thallophyta, Bryophyta and Pteridophyta.

The two characteristic features of cryptogams are:

• They have inconspicuous and hidden reproductive organs.
• They produce naked embryos called spores.

Question 5.
Define the following: (a) Radial symmetry (b) Bilateral symmetry
Answer:
When any plane passing through the central axis can divide the organism into identical halves, it is called radial symmetry, example coelenterates and the adults of Echinoderms.

When only one plane can divide the body of the organism into identical right and left halves, the symmetry is called as bilateral symmetry, example annelids, arthropods and humans.

Question 6.
Give main features of phylum Chordata.
Answer:
The main features of phylum Chordata are presence of:

• Notochord
• Dorsal Central nervous system
• Pharynx perforated by gill slits
• Ventral heart
• A post-anal tail

Question 7.
(a) Write any two features that are present in all chordates.
(b) Write one difference between pseudocoelom and true coelom.
Answer:
(a) The features present in all chordates are:

• Dorsal central nervous system and dorsal nerve cord.
• Triploblastic, coelomate with organ system level of organisation.

(b) If the body cavity of an organism is not lined by mesoderm and the space is filled with vacuolated cells, then the body cavity called pseudocoelom.

Question 8.
Define classification. Give any two of its significance.
Answer:
Grouping of organisms on the basis of their similarities and differences is called classification. Classification is important as it helps to make the study of vast variety of organisms easier and also helps us to understand the inter-relationships which occur among the organisms.

Question 9.
Classify the following plants into different plant divisions. Spirogyra, Fern, Funaria, Pinus, Apple tree, Mustard plant
Answer:
Spirogyra: Thallophyta, Fern: Pteridophyta, Funaria: Bryophyta, Pinus: Gymnosperm, Apple tree: Angiosperm, Mustard plant: Angiosperm

Question 10.
To which group do the following organisms belong and give one reason for each.
(a) Cyanobacteria
(b) Euglena
(c) Ulothrix
Answer:
(a) Cyanobacteria: Kingdom Monera; does not have a well-defined nucleus and lacks membrane bound cell organelles, prokaryotes.

(b) Euglena: Kingdom Protista; are aquatic, unicellular eukaryotes with well-defined nucleus and membrane bound cell organelles.

(c) Ulothrix: Kingdom Plantae, Thallophyta; Have a thalloid body, photosynthetic and eukaryotic.

Question 11.
Why are bryophytes called as ‘Amphibians of plant kingdom9?
Answer:
Bryophytes are called as ‘amphibians of plant kingdom’ because:

• They live in moist, damp places in order to get water from soil either directly or with the help of their rhizoids.
• They require water for the transfer of their gametes and are dependent on water for sexual reproduction.

Question 12.
(а) What is coelom?
(b) Presence of ‘coelom’ in an animal is considered advantageous. Why?
Answer:
(a) Coelom is a body cavity lined by mesodermal cells and lies between the body wall and alimentary canal (gut) of the organism.

(b) The coelom is considered advantageous as it helps to accommodate the various organs of the body in a proper way and gives a greater flexibility to the body of the organism.

Question 13.
(а) Write one characteristic each of amphibia and aves.
(b) Write the name of the class to which following belong:
(i) Sea-horse
(ii) King cobra
Answer:
(a) Amphibia: Can live both on land and in water, slimy to touch, cold blooded with three chambered heart.
Aves: Have forelimbs modified into wings for flight. Body covered with feathers, are warm-blooded with four-chambered heart

(b) Sea-horse: Class Pisces;
King cobra: Class Reptilia

Question 14.
Give the three characteristic features of Class Mammalia.
Answer:
The characteristic features of Class Mammalia are:

• Presence of Mammary glands.
• The skin has hairs.
• Their body has sweat glands and oil glands.
• They have a four-chambered heart, are warm-blooded.

Question 15.
Distinguish between monocots and dicots.
Answer:
Monocots:

• Possess a single cotyledon.
• Have fibrous roots.
• Have parallel venation. Examples: Wheat, Maize

Dicots:

• Possess two cotyledons.
• Have tap root.
• Have reticulate venation. Examples: Pea, gram

Question 16.
(a) What is symbiotic relationship?
(b) Name a symbiotic life form that grows on the bark of tree as large coloured patches.
Answer:
(a) The mutually dependent relationship between two organisms where both are benefitted is called symbiotic relationship.

(b) Lichen is a symbiotic life form that grows on the bark of trees as large coloured patches. It is symbiotic association between fungi and algae.

Question 17.
State the bases of classifying plants and animals into different categories.
Answer:
The bases for classifying animals and plants into different categories are:

• Mode of nutrition – Plants are autotrophic whereas animals heterotrophic.
• Presence or absences of cell wall – Plant cells have cell wall whereas animal cells do not have a cell wall.

Question 18.
(a) Differentiate between Fungi and Plantae.
(b) Mention the basis of classification among plants to different levels.
Answer:
Fungi:

• They are non-chlorophyllous and heterotrophic.
• Their cell wall is made of chitin.
• They are mostly saprophytes or parasites.

Plantae:

• They have chlorophyll and are autotrophic.
• Their cell wall is made of cellulose.
• They are autotrophs.

(b) The basis of classification of plants to different levels is –

• Presence or absence of well-differentiated body tissues
• Presence or absence of a well developed transport system for water, minerals and organic substances.
• Seed bearing ability and presence of seeds enclosed in fruits or not.

Question 19.
Name the group which is called as Amphibian of the plant kingdom. Cite an example of this group, also mention one important feature of the same group.
Answer:
Bryophytes are called amphibians of the plant kingdom, example Marchantia, Funaria.

Characteristic feature: Plant body is thalloid, not well-differentiated and has root-like, stem-like and leaf-like structures.

Question 20.
Identify the following:
(a) Amphibians of the plant kingdom.
(b) Plants with hidden reproductive organs
(c) Mutually benefitted relationship between two organisms.
Answer:
(a) Bryophytes
(b) Phanerogams comprising of Algae, Bryophytes and Pteridophytes.
(c) Symbiotic relationship as shown by algae and fungi in Lichen.

Question 21.
Describe the following:
(a) Lichens
(b) Cryptogams
(c) Phanerogams
Answer:
(a) Lichen is a symbiotic relationship between an algae and fungi.
(b) Cryptogams have hidden or inconspicuous reproductive organs.
(c) Phanerogams have well differentiated reproductive tissues which result in formation of seed.

Question 22.
(а) Which group of plants is known as ‘flowering plants’?
(b) On the basis of seed how is a maize plant different from a pea plant?
Answer:
(a) Angiosperms are called the flowering plants.
(b) Maize is a monocot plant as it bears one cotyledon in its seed whereas pea is a dicot as it has two cotyledons in its seed.

Question 23.
Name the largest group of animals. Write the salient features of this group. Give two examples.
Answer:
Phylum Arthropoda is the largest group of animals. They have jointed legs, bilateral symmetry and a body cavity filled with fluid. For example, Cockroach, spider, butterfly, etc.

Question 24.
What is binomial nomenclature? Who gave it? Write its advantage.
Answer:
The system of scientific naming of organisms which consists of a generic name and a specific epithet is called binomial nomenclature. It was given by Carolus Linnaeus. Its advantage is that it helps in the systematic study and understanding of the living organisms.

Question 25.
List the conventions used for writing a scientific name. What is the importance of scientific name?
Answer:
Convention for writing the scientific names:

• The name of the genus begins with a capital letter.
• The name of the species begins with a small letter.
• When printed, the scientific name is given in italics.
• When written by hand, the genus name and the species name have to be underlined separately.

Common names cannot be used in the same way by the scientist world over and can often result in confusion. To avoid this, a system of scientific names has been proposed.

Question 26.
Write true (T) or false (F)
(а) Whittaker proposed five kingdom classification.
(b) Monera is divided into Archaebacteria and Eubacteria.
(c) Starting from Class, Species comes before the Genus.
(d) Anabaena belongs to the kingdom Monera.
(e) Blue-green algae belongs to the kingdom Protista.
(f) All prokaryotes are classified under Monera.
Answer:
(a) True
(b) True
(c) False
(d) True
(e) False
(f) True

Question 27.
Fill in the blanks:
(a) Fungi shows _______ mode of nutrition.
(b) Cell wall of fungi is made up of _______
(c) Association between blue-green algae and fungi is called as _______
(d) Chemical nature of chitin is _______
(e) _______ has smallest number of organisms with maximum number of similar characters.
(f) Plants without well differentiated stem, root and leaf are kept in _______
(g) _______ are called the amphibians of the plant kingdom.
Answer:
(a) saprophytic
(b) chitin
(c) lichens
(d) Carbohydrate
(e) Species
(f) Thallophyta
(g) Bryophytes

Question 28.
You are provided with the seeds of gram, wheat, rice, pumpkin, maize and pea. Classify them whether they are monocot or dicot.
Answer:
Gram – dicot
Wheat – monocot
Rice – monocot
Pumpkin – dicot
Maize – monocot
Pea -dicot

Question 29.
Match items of column (A) with items of column (B).

Answer:
(a) (ii)
(b) (i)
(c) (iv)
(d) (iii)
(e) (vi)
(f) (u)
(g) (vii)

Question 30.
Match items of column (A) with items of column (B).

Answer:
(a) (iii)
(b) (ii)
(c) (vi)
(d) (i)
(e) (v)
(f) (iv)

Question 31.
Classify the following organisms based on the absence/presence of true coelom (i.e., acoelomate, pseudocoelomate and coelomate) Spongilla, Sea anemone, Planaria, Liver fluke, Wuchereria, Ascaris, Nereis, Earthworm, Scorpion, Birds, Fishes, Horse.
Answer:

• Spongilla: Acoelomate
• Sea anemone: Acoelomate
• Planaria: Acoelomate
• Liver fluke: Acoelomate
• Wuchereria: Pseudocoelomate
• Ascaris: Pseudocoelomate
• Nereis: Coelomate
• Scorpion: Coelomate
• Earthworm: Coelomate
• Birds, Fishes and Horse: Coelomate

Question 32.
Endoskeleton of fishes are made up of cartilage and bone; classify the following fishes as cartilagenous or bony:
Torpedo, Sting ray, Dog fish, Rohu, Angler fish, Exocoetus.
Answer:

• Torpedo: cartilagenous
• Sting ray: cartilagenous
• Dog fish: cartilagenous
• Rohu: bony
• Angler fish: cartilagenous
• Exocoetus: bony

Question 33.
Classify the following based on number of chambers in their heart.
Rohu, Scoliodon, Frog, Salamander, Flying lizard, King Cobra, Crocodile, Ostrich, Pigeon, Bat, Whale
Answer:

• Rohu, Scoliodon: 2 chambered,
• Frog, Salamander, Flying lizard, King Cobra: 3 chambered,
• Crocodile, Ostrich, Pigeon, Bat, Whale: 4 chambered

Question 34.
Classify Rohu, Scolidon, Flying lizard, King Cobra, Frog, Salamander, Ostrich, Pigeon, Bat, Crocodile and Whale into the cold-blooded/warm-blooded animals.
Answer:

• Cold-blooded: Rohu, Scolidon, Flying lizard, King Cobra, Frog, Salamander Crocodile
• Warm-blooded: Ostrich, Pigeon, Bat, Whale

Question 35.
Name two egg laying mammals.
Answer:

• Platypus
• Echidna.

Question 36.
Fill in the blanks:
(а) Five kingdom classification of living organisms is given by _______
(b) Basic smallest unit of classification is _______
(c) Prokaryotes are grouped in Kingdom _______
(d) Paramecium is a Protista because it is an _______
(e) Fungi do not contain _______
(f) A fungus _______ can be seen without microscope.
(g) Common fungi used in preparing bread is _______
(h) Algae and fungi form symbiotic association called _______
Answer:
(a) Robert Whittaker
(b) species
(c) Monera
(d) eukaryotic unicellular organism
(e) chlorophyll
(f) mushrooms
(g) yeast
(h) lichens

Question 37.
Give True (T) and False (F).
(а) Gymnosperms differ from Angiosperms in having covered seed.
(b) Non flowering plants are called Cryptogamae.
(c) Bryophytes have conducting tissue.
(d) Funaria is a moss.
(e) Compound leaves are found in many ferns.
(f) Seeds contain embryo.
Answer:
(a) False
(b) True
(c) False
(d) True
(e) True
(f) True

Question 38.
Give examples for the following.
(a) Bilateral, dorsiventral symmetry is found in _______
(b) Worms causing the disease elephantiasis is _______
(c) Open circulatory system is found in _______ where coelomic cavity is filled with blood.
(d) _______ are known to have pseudocoelom.
Answer:
(a) Liver Fluke
(b) Filarial worms
(c) Arthropods
(d) Nematodes

Question 39.
Label a, b, c and d, given in the figure below. Give the function of (b).
Answer:

(a) Dorsal fin
(b) Caudal fin
(c) Pelvic fin
(d) Pectoral fin
Function of Caudal fins: Caudal fins help in streamlined movement in water.

Question 40.
Fill in the boxes given in figure with appropriate characteristics/plant groups.

(a) Thallophyta
(b) Without specialized vascular tissue
(c) Pteridophyta
(d) Phanerogams
(e) Bear naked seeds
(f) Angiosperm
(g) Have seeds Thallophyta with two cotyledons
(h) Monocots

Diversity in Living Organisms Class 9 Extra Questions Long Answer Type

Question 1.
Explain the various categories of taxonomical hierarchy.
Answer:
The categories in taxonomical hierarchy are

• Kingdom: The highest category of classification
• Phylum (for animals) / Division (for plants): Group of related classes
• Class: Group of related orders.
• Order: Group of related families.
• Family: Group of related genus.
• Genus: Group of related species.
• Species: Group of organisms which can interbreed among themselves to produce a fertile offspring.

Question 2.
(a) Draw labelled diagrams of three protozoa.
(b) Euglena is a dual organism. Why?
Answer:
(a)

(b) Euglena is a dual organism as:

• It does not have a cell wall and behaves as a heterotrophic organism in the absence of sunlight.
• It contains chloroplast and can perform photosynthesis and behaves as autotroph in the presence of sunlight.

Question 3.
Mention the class to which they belong and give one characteristic feature of each. Frog, fish, lizard, pigeon, bat.
Answer:
(i) Fish: Class Pisces; Aquatic, respiration with gills, moist scales present on the body, two-chambered heart, cold-blooded

(ii) Frog: Class Amphibia; Can live both on land and in water, cold-blooded, three-chambered heart, slimy to touch

(iii) Lizard: Class Reptilia; Cold-blooded, three-chambered heart, hard covering present on eggs to prevent from desiccation

(iv) Pigeon: Class Aves; Forelimbs modified into wings for flight and have feathers, have beak, warm-blooded, four-chambered heart

(v) Bat: Class Mammalia; has mammary glands, warm-blooded, four-chambered heart.

Question 4.
(i) Draw a neat labelled diagram of Hydra.
(ii) Label mesoglea and gastro-vascular cavity.
(iii) Name the group of animals it belongs to.
(iv) Name one species of this group that lives in colonies.
Answer:
(i) and (ii)-See figure

(iii) Phylum Coelenterata
(iv) Corals live in colonies.

Question 5.
Write one difference for each of the following pairs:
(i) Thallophyta and Bryophyta
(ii) Nematoda and Annelida
(iii) Amphibians and Reptiles
Answer:
(i)

• Thallophyta Bryophyta – Plant body is not well differentiated into root, stem and leaves. It is a thalloid example – Spirogyra.
• Bryophyta – Plant body slightly more differentiated than Thallophyta with root-like, stem-like and leaf-like structures, called as Amphibians of the plant kingdom.

(ii)

• Nematoda – Called as roundworms, bilaterally symmetrical having pseudocoelom, example Ascaris
• Annelida – Have a metamerically segmented body and a true coelom, example Earthworm

(iii) Amphibians

• Can live both in land as well as water.
• Lack scales on body and have slimy skin.
• Lay eggs in water.
• Eggs are devoid of tough covering.

Reptiles:

• Most of them are terrestrial animals.
• Have dry scales on body.
• Lay their eggs mostly on land.
• Eggs have a tough covering to protect from drying.

Question 6.
(a) In which two ways are amphibians different from fishes?
(b) Identify the phylum of organisms having the characteristics:
(i) Pore bearing animals and radial symmetry
(ii) Body spiny and radial symmetry
(c) Why do gymnosperms not require water for fertilisation?
Answer:
(a) Amphibians can live both on land as well as water, have three-chambered heart and have lungs for respiration in adult stage.
Fishes are aquatic, have two-chambered heart and have gills for respiration.

(b) (i) Phylum Porifera
(ii) Phylum Echinodermata

(c) Gymnosperms do not require water for fertilisation as they bear pollen grains which are carried away by agents like wind to cause pollination.

Question 7.
Give four features of phylum Coelenterata. Give two examples.
Answer:
The four distinguishing features of phylum Coelenterata are:

• They have a radially symmetrical body.
• The body is made of two layers of cells, so called diploblastic.
• Has tentacles which surround their mouth. Tentacles bear cnidoblasts, stinging cells.
• They have a characteristic cavity called as coelenteron.
• They are solitary like Hydra and can be colonial like Obelia. example Hydra, Obelia, Jelly fish.

Question 8.
List the convention that is followed while writing the scientific names. Give scientific name of Mango and Tiger.
Answer:
The conventions followed while writing the scientific names are:

• The name of the genus begins with a capital letter.
• The name of the species begins with a small letter.
• When printed, the scientific name is given in italics.
• When written by hand, the genus name and the species name have to be underlined separately.

Scientific name of Mango is Mangifera indica and of Tiger is Panthera tigris.

Question 9.
Enlist the features of organisms placed in Protista. Give two examples.
Answer:
The members of the kingdom Protista have the following features:

• All of them are unicellular and eukaryotic.
• Their mode of nutrition is either autotrophic or heterotrophic.
• They are usually aquatic but some of them are parasitic.
• Have cilia, flagella or pseudopodia for movement. example Amoeba, Plasmodium

Question 10.
Give the general characteristics of fungi. Give two examples.
Answer:
The general characteristics of fungi are:

• They are heterotrophic organisms which can occur as saprophytes, parasites or symbionts.
• Their cell wall is made up of chitin and lack chlorophyll.
• Food is stored in the form of glycogen in them instead of starch as stored in plants.
• They are eukaryotes as they have membrane bound nucleus.
• For example: Yeast, Rhizopus, Agaricus, Penicillium, etc.

Question 11.
What are the five kingdoms of Whittaker? Give the most important characteristic feature of each kingdom.
Answer:
The five kingdoms proposed by R. H. Whittaker comprises of – Monera, Protista, Fungi, Plantae and Animalia.

• Kingdom Monera: Unicellular, prokaryotes
• Kingdom Protista: Unicellular eukaryotes
• Kingdom Fungi: Non-photosynthetic, chlorophyll absent, heterotrophic nutrition
• Kingdom Plantae: Chlorophyll bearing, photosynthetic autotrophs
• Kingdom Animalia: Heterotrophic nutrition, cell wall absent

Question 12.
(a) To which group do algae belong? Write one characteristic of the division. Give two examples.
(b) Name the group:
(i) Which includes unicellular eukaryotic organisms
(ii) In which mode of nutrition is saprophytic.
(iii) In which seeds are not enclosed in fruits.
(c) Classify flowering plants on the basis of number of cotyledons present in the seed.
Answer:
(a) Algae are members of division Thallophyta. Characteristic – Aquatic, chlorophyll bearing autotrophs.

(b) (i) Protista
(ii) Fungi
(iii) Gymnosperms

(c) The flowering plants are classified as monocots and dicots on the basis of presence of one cotyledon or two cotyledon respectively.

Question 13.
List three groups of plants and tell which plants are referred to as vascular plants? Also mention out of these which group is further classified on the basis of number of cotyledon? State its two characteristics.
Answer:
The pteridophytes, gymnosperms and angiosperms are the vascular plants. Out of these three, angiosperms are classified on the basis of number of cotyledon as – monocots having one cotyledon and dicots having two cotyledons.

Characteristic features of angiosperms are:

• Their seeds are enclosed within fruit.
• Embryo bears the cotyledons also called seed leaves which provide nutrition to the young plant/seedling on the germination of the seed.

Question 14.
(a) On what basis does the embryo of cryptogam differ from that of phanerogams?
(b) Describe the feature that divides the angiosperms into two groups.
(c) State the two sub-groups of angiosperms.
Answer:
(a) Cryptogams bear naked embryos whereas the phanerogams have embryos which are enclosed in seed.
(b) The feature that divides the angiosperms into two groups is the number of cotyledons present in their embryo.
(c) The two sub-groups of angiosperms are monocots (bear single cotyledon) and dicots (have two cotyledons).

Question 15.
Draw a neat diagram of Spirogyra and label the following parts:
(a) Outermost layer of the cell
(b) Organelle that performs the function of photosynthesis
(c) Jelly-like substance in the cell where all organelles are suspended.
(d) Dark coloured and dot-like structure generally present in the centre of the cell.
Answer:

(а) Outermost layer of the cell – Cell wall
(b) Organelle that performs the function of photosynthesis – Chloroplast
(c) Jelly-like substance in the cell where all organelles are suspended – Cytoplasm
(d) Dark coloured and dot-like structure generally present in the centre of the cell – Nucleus

Question 16.
Thallophyta, bryophyta and pteridophyta are called as ‘Cryptogams’. Gymnosperms and Angiosperms are called as ‘phanerogams’. Discuss why. Draw one example of Gymnosperm.
Answer:
Thallophyta, bryophyta and pteridophyta are called as ‘Cryptogams’ because they have hidden or inconspicuous reproductive organs. Spores are formed in them instead of seeds. Gymnosperms and Angiosperms are called as ‘phanerogams’ as they have well differentiated reproductive tissue/organs. Seed harbours the embryo and provides it nourishment too.

Question 17.
Define the terms and give one example of each
(a) Bilateral symmetry
(b) Coelom
(c) Triploblastic
Answer:
(a) If the organism can be divided exactly into two halves from one median plane only, the symmetry is called bilateral symmetry, example liver fluke.

(b) The internal body cavity present between visceral organs and body wall in which well developed organs can be accommodated is called as coelom, example butterfly.

(c) The organisms who have three embryonic layers are called as triploblastic organisms example star fish.

Question 18.
You are given leech, Nereis, Scolopendra, prawn and scorpion; and all have segmented body organisation. Will you classify them in one group? If no, give the important characters based on which you will separate these organisms into different groups.
Answer:
No, all the organisms given in the question do not belong to one group.

Leech and Nereis belong to phylum Annelida because they have metamerically segmented body i.e., body is divided into many segments internally by septa. Body segments are lined up one after the other from head to tail.

The characteristic identifying feature of Scolopendra, prawn and scorpion are the jointed legs and open circulating system due to which they are placed in phylum Arthropoda.

Question 19.
Which organism is more complex and evolved among Bacteria, Mushroom and Mango tree? Give reasons.
Answer:
Mango tree is more complex and evolved because, it is eukaryotic, autotrophic, terrestrial sporophyte with covered seed. Bacteria is unicellular prokaryote and fungi are the heterotrophic, simple thallophyte with no tissue systems.

Question 20.
Differentiate between flying lizard and bird. Draw the diagram.
Answer:
Flying lizard belongs to the group reptiles and is characterised as cold-blooded, body covered with scales and have three-chambered heart, while birds belong to group aves and are characterised as warm¬blooded, having feather covered body, forelimbs modified as wings and having four-chambered heart.

Question 21.
List out some common features in cat, rat and bat.
Answer:
Bat, rat and cat belong to the class Mammalia and have following common features:
(а) All have notochord at some stage of their life cycle.
(b) All are warm-blooded:
(c) All have four-chambered heart.
(d) All have skin covered with hair and with sweat and oil glands.

Question 22.
Why do we keep both snake and turtle in the same class?
Answer:
Because both are –

• Cold-blooded
• Have scales
• Breathe through lungs
• Have three-chambered heart
• They lay eggs with tough covering.

Diversity in Living Organisms Class 9 Extra Questions HOTS

Question 1.
Which group among the Amphibia and Pisces is more advanced and why?
Answer:
The amphibians are more advanced than the Pisces as

• Amphibians can survive both on land as well as in water whereas the members of Pisces can survive only in water.
• Amphibians have a three-chambered heart compared to the two-chambered heart present in the Pisces.
• Amphibians have the ability to respire through lungs but fishes respire through gills.

Question 2.
The protozoans have been included in Protista and not in kingdom Animalia. Give reason.
Answer:
Protozoans are unicellular, eukaryotes so they have been kept in the Kingdom Protista comprising of only the unicellular, eukaryotic organisms. Kingdom Animalia consists of multicellular eukaryotes.

Question 3.
Three types of animal specimen were collected by Rajeev and labelled as A, B and C. The specimen A had slimy skin, respired through lungs, B had dry scales with eggs having tough covering and the specimen C had moist scales with terminal mouth. Identify the class to which the specimens belong.
Answer:
A: Amphibia
B: Reptilia
C: Pisces

Question 4.
While playing near a pond Anmol experienced a pain on his feet. He saw a black coloured organism with metamerically segmented body was clinging to his foot and trying to suck blood from his foot. On the basis of this identify the organism and the phylum to which it may belong to. Name two other members of the phylum to which this organism belongs.
Answer:

• Organism: Leech
• Phylum: Annelida
• Other members: Nereis, Earthworm

Question 5.
Shobha went for a school trip and was shown some organisms which were told by their teachers as the members of the second largest phylum of kingdom Animalia. The organisms had a shell on their body and moved around with their muscular foot. Name the phylum to which the mentioned organism will belong. Give one more characteristic feature and two examples of such organisms.
Answer:
Phylum : Mollusca
Feature : Reduced Coelomic cavity
Examples : Snails and mussels

Question 6.
Three types of plant specimen were observed by Renu and labelled as A, B and C. The specimen A had green colour with undifferentiated thalloid body, B had well differentiated body which formed spores and had a vascular system and the specimen C had cones which had seeds but no fruits were formed in them. Identify the class to which the specimens belong.
Answer:
A: Thallophyta
B: Pteridophyta
C: Gymnosperms

Diversity in Living Organisms Class 9 Extra Questions Value Based (VBQs)

Question 1.
Reena read an article in newspaper regarding the problems caused by the chemical fertilisers. She surfed through the internet and came to know that certain blue-green algae are a good source of fertilisers. She advised the farmers of the area to use them and helped them to stop using chemical fertilisers.
(a) Which group comprises of the blue-green algae?
(b) How do these blue-green algae increase soil fertility?
(c) What are the characteristic features of the group to which they belong?
(d) What values are shown by Reena by her work?
Answer:
(a) Kingdom Monera
(b) They fix atmospheric nitrogen and increase soil fertility
(c) They are unicellular prokaryotes which do not have a well defined nucleus.
(d) Values shown by Reena are concern for environment, helpfulness, eco-friendliness and scientific attitude.

Question 2.
During their trip to a hill station, Shiksha observed some thalloid plants having little differentiation of body which were growing on the moist and damp surface in the form of dense mats. She asked her teacher about them and their role. Her teacher told her which group of organism they were and said that these dense mats were helpful in preventing soil erosion and need water for fertilisation.
(a) What is the probable group of organisms which her teacher would have told her?
(b) What is the peculiar feature of this group of organisms?
(c) Give two examples of members of the group.
(d) What are the values shown by Shiksha?
Answer:
(a) Bryophytes
(b) Can live both on land and in water
(c) Marchantia, Funaria
(d) The values shown by Shiksha are keen observation and scientific attitude.

In this page, we are providing Tissues Class 9 Extra Questions and Answers Science Chapter 6 pdf download. NCERT Extra Questions for Class 9 Science Chapter 6 Tissues with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 6 Extra Questions and Answers Tissues

Extra Questions for Class 9 Science Chapter 6 Tissues with Answers Solutions

Tissues Class 9 Extra Questions Very Short Answer Type

Class 9 Science Chapter 6 Tissue Extra Question Answer Question 1.
Name any two types of simple permanent plant tissues.
Answer:
The simple permanent tissues of plants are: Parenchyma, collenchyma and sclerenchyma.

Tissues Class 9 Extra Questions With Answers Question 2.
What are blood platelets?
Answer:
Blood platelets are the cell fragments present in the plasma of blood which help in the clotting of blood.

Tissue Class 9 Extra Questions And Answers Question 3.
Name the connective tissue that is found between skin and muscles.
Answer:
Areolar tissue.

Tissues Extra Questions And Answers Question 4.
Name the tissue present in the brain.
Answer:
Nervous tissue which comprises of its basic unit called neurons.

Class 9 Tissue Extra Questions Question 5.
In which of the simple plant tissue, deposition of lignin is found?
Answer:
Sclerenchyma.

Tissues Class 9 Extra Questions Question 6.
Name the basic packing tissue of plant.
Answer:
Parenchyma.

Tissue Extra Questions Class 9 Question 7.
Name the tissue which is present in the veins of leaves.
Answer:
Sclerenchyma.

Class 9 Science Chapter 6 Extra Questions And Answers Question 8.
Why is cork impervious to gases and water?
Answer:
Due to presence of a chemical substance called suberin.

Class 9 Tissues Extra Questions Question 9.
What is the function of phloem?
Answer:
Phloem helps in the transport of food from leaves to the various parts of the plant.

Extra Questions On Tissues Class 9 Question 10.
Which body cell provides resistance against infections?
Answer:
White blood cells (WBC) provide resistance against infections.

Extra Questions Of Tissue Class 9 Question 11.
Which biochemicals compose the solid matrix of cartilage?
Answer:
Proteins and sugars make up the solid matrix of cartilage.

Tissue Class 9 Extra Questions Question 12.
Name the connective tissue which helps in the repair of tissues.
Answer:
Areolar connective tissue helps in the repair of tissue.

Class 9 Science Ch 6 Extra Questions Question 13.
Which connective tissue is specialised for fat storage and acts as heat insulator?
Answer:
Adipose tissue helps in storage of fats and acts as heat insulator.

Tissue Chapter Class 9 Extra Questions Question 14.
Which muscle has spindle-shaped cells?
Answer:
Smooth muscle cells have spindle shaped cells.

Ch 6 Science Class 9 Extra Questions Question 15.
Which meristem is present at growing tips of stems and roots?
Answer:
Apical meristem is present at the growing tips of the stem and roots.

Tissues Class 9 Extra Questions Short Answer Type 1

Question 1.
List any four salient features of meristematic tissue.
Answer:
The salient features of meristematic tissue are:

• This tissue consists of cells which continuously divide to produce new cells.
• The cells of this tissue lack vacuoles.
• The cells of this tissue have dense cytoplasm.
• The cells of this tissue have thin cellulosic cell walls and prominent nuclei.

Question 2.
Write the four elements of xylem.
Answer:
The four elements of xylem are tracheids, vessels, xylem parenchyma and xylem fibres.

Question 3.
How is ligament different from tendons?
Or
Differentiate between tendon and ligament.
Answer:
Ligament is a connective tissue which joins bone to bone and is elastic in nature.

Tendons join bone to muscles and are less elastic as compared to the ligaments.

Question 4.
Write a short note on the different types of meristematic tissue with their location and functions in the plants.
Answer:
The meristematic tissues are classified as apical, lateral and intercalary meristematic tissue based on the region where they are present.

Apical meristem – It is present at the growing tips of stem and roots and results in increase in the length of the stem and the root. Lateral meristem (cambium): It is present on the lateral sides of stem and roots. It helps to increase the girth of the stem or root.

Intercalary meristem – It is present at the base of the leaves or internodes. It helps in the longitudinal growth of plants.

Question 5.
Show the diagrammatic representation of location of lateral meristem and intercalary meristem in plant body.
Answer:

Question 6.
Differentiate between chlorenchyma and aerenchyma.
Or
Write the difference between aerenchyma and chlorenchyma.
Answer:
Chlorenchyma:

• It is a type of parenchyma which contains chlorophyll.
• It helps to perform photosynthesis.
• It is present in green parts of plants like the leaves.

Aerenchyma:

• This type of parenchyma that has large air cavities in it.
• It helps to provide buoyancy to the plants.
• It is present in the aquatic plants, example in their floating leaves.

Question 7.
Write the functions of collenchyma in plants.
Answer:
Collenchyma allows easy bending in various parts of a plant (leaf, stem) without breaking. It also provides mechanical support to plants like in the leaf stalks below the epidermis.

Question 8.
What are the roles of epidermis in plants?
Answer:
The functions of epidermis are:

• Epidermis is usually made up of a single layer of cells and gives protection.
• The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.
• The epidermis of leaves have small pores called as stomata which help in gaseous exchange and transpiration.
• The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

Question 9.
Answer the following:
(i) How is the epidermis of the plants living in very dry habitats adapted?
(ii) Write functions of guard cells of stomata in the leaf.
Answer:
(i) The epidermis of plants living in dry habitats may be thicker or often secrete a waxy, water-resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.

(ii) The guard cells of stomata in the leaf help in gaseous exchange and transpiration.

Question 10.
What is the function of areolar tissues?
Answer:
The functions of areolar tissues are:

• It fills the space inside the organs
• It supports internal organs.
• It helps in repair of tissues.

Question 11.
Determine the location of the following tissues:

1. Unstriated muscle fibres
2. Cuboidal epithelium
3. Adipose tissue
4. Striated muscle fibres

Answer:

1. Unstriated muscle fibres: Present in iris of the eye, ureters, blood vessels, alimentary canal and bronchi of lungs.
2. Cuboidal epithelium: Present in lining of kidney tubules and ducts of salivary glands.
3. Adipose tissue: It is found below the skin and between internal orgAnswer:
4. Striated muscle fibres: It is present in muscles of our limbs

Tissues Class 9 Extra Questions Short Answer Type 2

Question 1.
Explain how the bark of a tree is formed. How does it act as a protective tissue?
Answer:
In the older stem, a strip of secondary meristem replaces the epidermis. The secondary meristem cuts off cells towards outside to form a several-layer thick tissue; This is called the cork or the bark of the tree.

Cells of cork or bark are dead, compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them impervious to gases and water. In this way it acts as a protective tissue.

Question 2.
Draw a diagrammatic labelled sketch of stem tip to show location of meristematic tissue. Mention the functions of different types of meristematic tissue.
Answer:

The types of meristematic tissue are:
(i) Apical meristem: It is present at the growing tips of stems and roots and results in increase in the length of the stem and the root.

(ii) Lateral meristem (cambium): It is present on the lateral sides of stems and roots. It helps to increase the girth of the stem or root.

(iii) Intercalary meristem: It is present at the base of the leaves or internodes. It helps in the longitudinal growth of plants.

Question 3.
What are the two main components of blood? Why is blood considered a type of connective tissue?
Answer:
Blood is a special connective tissue consisting of a fluid matrix, plasma, and formed elements. The formed elements are red blood cells (RBCs), white blood cells (WBCs) and blood platelets. Blood is considered as a type of connective tissue as they have the same origin as other types of connective tissue and helps to connect the different parts of the body to facilitate exchange of various components like nutrients and gases.

Question 4.
Give one function of each of the following.
(i) Stomata
(ii) Root nodules
(iii) Cardiac muscle fibres
Answer:
(i) Stomata: Help in exchange of gases in the plants.

(ii) Root nodules: In leguminous plants, the root nodules harbour nitrogen fixing bacteria which convert atmospheric nitrogen into nitrates.

(iii) Cardiac muscle fibres: They help in rhythmic contraction and relaxation of the heart.

Question 5.
Differentiate between bone and cartilage.
Or
Differentiate between bone and cartilage with respect to structure, function and location.
Answer:
Bone:

• Bones have a hard and non-pliable ground substance.
• Its matrix is rich in calcium salts and collagen fibres.
• It is the main tissue that provides structural frame to the body.
• The bone cells (Osteocytes) are present in the spaces called lacunae.
• Bones are present in the limbs and form main skeletal framework of the body.

Cartilage:

• Cartilage is pliable, flexible and resist compression.
• Its matrix is rich in protein called chondrin and sugars.
• It is present in bones of the vertebral column, limbs and hands in adults.
• Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix secreted by them.
• Cartilage is present in the tip of nose, outer ear joints, between adjacent bones of the vertebral column.

Question 6.
Explain the basic criteria for classification of permanent tissue in plants.
Answer:
The permanent tissues are classified on the basis of the following criteria:

1. Simple (made of one type of cell) or complex (made of more than one type of cells)
2. Cell wall: Thin or thick
3. Type of cell: living or dead
4. Type of function the tissue performs: epidermis is protective, parenchyma is packing or supportive tissue and sclerenchyma makes up conducting tissue.

Question 7.
Identify the given two slides A and B as a parenchyma or sclerenchyma. Sclerenchyma can be identified by which characteristic?

Answer:
Slide A is parenchyma and Slide B is sclerenchyma.
Sclerenchyma can be identified by the type of cells which are long and narrow as the walls are thickened due to presence of lignin.

Question 8.
(i) Identify the given figures.

(ii) Give any two major differences between the structures identified.
(iii) Describe the role performed by these two in the plant body.
Answer:
(i) Structure (A) is a tracheid and structure (B) is a vessel.

(ii) Tracheid:

• Tracheids are elongated or tube-like cells with thick and lignified walls and tapering ends.
• They are in the form of single cells.

Vessel:

• Vessel is a long cylindrical tube-like structure made up of many cells called vessel members.
• They are composed of a number of cells fused together.

(iii) Tracheids and vessels help in vertical transport of water and minerals in the plants. They also help to provide mechanical strength to the plants.

Question 9.
Draw a well labelled diagram of cardiac muscle found in the human body. Write two differences between striated and smooth muscles.
Answer:

Question 10.
Draw a labelled diagram of unstriated muscle tissue and mention its occurrence, features and functions.
Answer:

(i) The cells are long and spindle-shaped.
(ii) They do not have striations.
(iii) Involuntary in nature as they are not under control of our will.
(iv) The cells of smooth muscles are uninucleate.
(v) Smooth muscle fibres are present in iris of the eye, ureters, blood vessels, alimentary canal and bronchi of lungs.

Question 11.
Name the kinds of muscles found in your limbs and lungs. How do they differ from each other structurally and functionally?
Answer:
Striated muscle fibres are found in limbs whereas smooth muscle fibres are present in lungs. The differences in their structure are:
(i) Striated muscle fibres have alternate light and dark bands which are not present in the smooth muscle fibres.

(ii) Striated muscle fibres are cylindrical and multinucleate whereas the smooth muscle fibres are spindle-shaped and uninucleate.

(iii) Striated muscles are voluntary in nature (under control of our will) whereas the smooth muscle fibres are involuntary in nature (not under control of our will).

Question 12.
What are neurons? Where are they found in the body? What function do they perform in the body of an organism?
Answer:
The cells of nervous tissue are called nerve cells or neurons. Neurons are the structural and functional unit of the nervous system. They are found in the brain, spinal cord and nerves.

Their functions are:

• They are highly specialised for transmitting the stimulus from one place to another within the body on being stimulated.
• They help to coordinate the various functions of the body.

Question 13.
Animals of colder regions and fishes of cold water have thicker layer of subcutaneous fat. Describe why?
Answer:
The thick layer of subcutaneous fat acts as insulator and prevents the heat of the body to escape out. The layer of fat acts as a subcutaneous insulation of body for thermoregulation.

Question 14.
Match the column (A) with the column (B).

Answer:
(a) (v)
(b) (iv)
(c) (iii)
(d) (i)
(e) (ii)
(f) (vi)

Question 15.
Match the column (A) with the column (B).

Answer:
(a) (i)
(b) (c)
(d) (iii)
(e) (iv)

Question 16.
If a potted plant is covered with a glass jar, water vapours appear on the wall of glass jar. Explain why.
Answer:
The water is lost by the plant in the form of water vapour due to the process of transpiration. These water vapours appear on the wall of the glass jar.

Question 17.
Name the different components of xylem and draw a living component.
Answer:
Xylem consists of four elements which are:
(a) tracheids
(b) vessels
(c) xylem parenchyma
(d) xylem fibres
The only living component of xylem is xylem parenchyma whose basic structure is shown below:

Question 18.
Draw and identify different elements of phloem.
Answer:
Phloem has four elements called sieve tubes, companion cells, phloem fibres and the phloem parenchyma.

Question 19.
Write true (T) or false (F).
(a) Epithelial tissue is protective tissue in animal body.
(b) The lining of blood vessels, lung alveoli and kidney tubules are all made up of epithelial tissue.
(c) Epithelial cells have a lot of intercellular spaces.
(d) Epithelial layer is permeable layer.
(e) Epithelial layer does not allow regulation of materials between body and external environment.
Answer:
(a) True
(b) True
(c) False
(d) True
(e) False

Question 20.
Differentiate between voluntary and involuntary muscles. Give one example of each type.
Answer:
Voluntary muscles are present in our limbs as skeletal muscles and can be moved by our conscious will whenever we want. Involuntary muscles cannot function on their own. They cannot be controlled by our will or desire. The cardiac muscle and the smooth muscles are involuntary in nature.

Question 21.
Differentiate the following activities on the basis of voluntary (V) or involuntary (IV) muscles.
(a) Jumping of frog
(b) Pumping of the heart
(c) Writing with hand
(d) Movement of chocolate in your intestine
Answer:
(a) (V)
(b) (IV)
(c) (V)
(d) (IV)

Question 22.
Fill in the blanks.
(a) Lining of blood vessels is made up of _______
(b) Lining of small intestine is made up of _______
(c) Lining of kidney tubules is made up of _______
(d) Epithelial cells with cilia are found in _______ of our body.
Answer:
(a) Squamous epithelium
(b) Columnar epithelium
(c) Cuboidal epithelium
(d) Respiratory tract

Question 23.
Water hyacinth floats on water surface. Explain.
Answer:
The parenchyma present in the swollen petiole of water hyacinth is called aerenchyma which has large cavities to provide buoyancy and help them float on the water surface.

Question 24.
Which structure protects the plant body against the invasion of parasites?
Answer:
The epidermis of plants has thick cuticle and waxy substances to prevent the invasion of parasites.

Question 25.
Fill in the blanks.
(а) Cork cells possesses _______on their walls that makes it impervious to gases and water.
(b) _______ have tubular cells with perforated walls and are living in nature.
(c) Bone possesses a hard matrix composed of and _______ and _______
Answer:
(a) suberin
(b) sieve tubes
(c) calcium and phosphorus

Question 26.
Why is epidermis important for the plants?
Answer:
The outermost layer of cells covering an organism is called epidermis. It is usually made up of a single layer of cells and gives protection.

The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin (chemical substance with waterproof quality) to prevent water loss.

The stomata present on the epidermis of leaves helps in gaseous exchange and the loss of water vapour by transpiration.

The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

Question 27.
Fill in the blanks.
(a) _______ are forms of complex tissue.
(b) _______ have guard cells.
(c) cells of cork contain a chemical called _______
(d) Husk of coconut is made of _______ tissue.
(e) _______ gives flexibility in plants.
(f) _______ and _______ are both conducting tissues.
(g) Xylem transports and _______ and _______ from soil.
(h) Phloem transport from _______ and _______ to other parts of the plant.
Answer:
(a) Xylem and phloem
(b) Stomata
(c) suberin
(d) sclerenchyma
(e) Collenchyma
(g) water; minerals
(h) Food; leaf

Tissues Class 9 Extra Questions Long Answer Type

Question 1.
Differentiate between
(i) Xylem and phloem
(ii) Vessel and sieve tube
(iii) Tracheid and vessel
Answer:
(i) Xylem and phloem –
Xylem:

• Xylem consists of tracheids, vessels, xylem parenchyma and xylem fibres.
• All the cells of xylem except the xylem parenchyma are dead.
• Xylem helps to transport water and minerals.
• The transport is unidirectional through xylem.

Phloem:

• Phloem has four elements called sieve tubes, companion cells, phloem fibres and the phloem parenchyma.
• All cells of phloem are living except the phloem fibres.
• Phloem transports food from leaves to other parts of the plant.
• The transport is bidirectional through the phloem.

(ii) Vessel and sieve tube –
Vessel:

• They are tubular structures having a hollow lumen and composed of dead cells.
• Vessel helps to conduct water and minerals in plants.
• The walls of vessels are lignified.
• They also provide mechanical strength to the plants.
• Their end walls are completely dissolved.

Sieve Tube:

• They are tubular structures having vacuolated cytoplasm and composed of living cells.
• They help to transport food from leaves to other parts of the plant.
• Their walls are not lignified.
• They do not provide mechanical strength to the plants.
• Their end walls have perforations in form of sieve plate.

(iii) Tracheid and vessel _
Tracheid:

• Tracheids are elongated or tube-like cells with thick and lignified walls and tapering ends.
• They are in the form of single cells.
• The inner layers of the cell walls are more thickened.
• They have narrow lumen.
• They have pointed ends.

Vessel:

• Vessel is a long cylindrical tube-like structure made up of many cells called vessel members.
• They are composed of a number of cells fused together.
• Their walls are less thickened.
• They have wide lumen.
• They have blunt ends.

Question 2.
Differentiate between striated, unstriated and cardiac muscle fibres.
Answer:

Question 3.
(i) What is nervous tissue?
(ii) Draw a well labelled diagram of neuron. (Label any 4 parts)
Answer:
Nervous tissue is a tissue made of neurons. It is divided into two parts: the central nervous system (CNS) consisting of the brain and spinal cord; and the peripheral nervous system (PNS) which regulates and controls the various functions and activities of the body.

Question 4.
Write the differences between animal tissue and plant tissue.
Answer:
Plant Tissue:

• The tissue is well differentiated into meristematic tissue and permanent tissue.
• The tissue can grow throughout life due to activity of meristematic tissue.
• They are autotrophic in nature.
• The tissue has more amount of dead tissue which provides mechanical strength to the plants.
• The tissue organisation is comparatively simple.

Animal Tissue:

• The tissue is not much differentiated like the plant tissue.
• The tissue does not show growth throughout life.
• They are heterotrophic in nature.
• The tissue has more amount of living tissue than dead tissue.
• The tissue is complex as it is organised into organs and organ systems.

Question 5.
Write a note on the protective tissue in plants. (Give appropriate diagram also)
Answer:
The protective tissues in plants are epidermis and the cork.
(i) Epidermis: The outermost layer of cells covering an organism is called epidermis. It is usually made up of a single layer of cells and gives protection.

The epidermis may be thicker in some plants living in dry habitats or often secrete a waxy, water- resistant layer on their outer surface called cutin to prevent water loss.

The epidermis of leaves have small pores called stomata which are enclosed by two kidney-shaped cells called guard cells. Stomata help in gaseous exchange and transpiration.

The epidermal cells of roots bear root hairs that greatly increase the total absorptive surface area of the roots for absorption of water.

(ii) Cork: A strip of secondary meristem replaces the epidermis of the older stem and cuts off cells towards compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them impervious to gases and water.

Question 6.
Explain the significance of the following:
(i) Hair-like structures on epidermal cells.
(ii) Epidermis has thick waxy coating of cutin in desert plants.
(iii) Small pores in epidermis of leaf.
(iv) Numerous layers of epidermis in cactus.
(v) Presence of a chemical suberin in cork cells.
Answer:
(i) To increase the total absorptive surface area for absorption of water.
(ii) To prevent water loss by transpiration and protection from pathogens.
(iii) To help in gaseous exchange and transpiration.
(iv) To prevent water loss by transpiration.
(v) To make tissue impervious to gases and water.

Question 7.
Differentiate between sclerenchyma and parenchyma tissues. Draw well labelled diagram.
Answer:
Parenchyma:

• Cells are thin walled and thickened with cellulose.
• It is made up of living cells.
• Cells are usually loosely pac ked with large intercellular spaces.
• Helps to store nutrients and water in stem and roots.
• It is called chlorenchyma if it contains chlorophyll and performs photos ynthesis. The parenchyma of
• aquatic plants have large cavities to provide buoyancy to the plants to help them float, it is then called aerenchyma.

Sclerenchyma:

• Cells are thick and thickened with lignin.
• This tissue is made up of dead cells.
• There are no intercellular spaces between the cells.
• Provides strength to the various parts of the plant.
• The cells are long and narrow, make the plant hard and stiff. This tissue provides strength to the plants and is present in stems, around vascular bundles, in the veins of leaves and in the hard covering of seeds and nuts.

Question 8.
Describe the structure and function of different types of epithelial tissues. Draw diagram of each type of epithelial tissue.
Answer:
Epithelial tissues are the covering or protective tissues and cover most organs and cavities in the animal body. These cells are tightly packed, form a continuous sheet and are almost without any intercellular spaces between them. E.g., skin, the lining of the mouth, the lining of blood vessels, lung alveoli and kidney tubules are all made of epithelial tissue.

All epithelium is usually separated from the underlying tissue by an extracellular fibrous basement membrane. The types of epithelium on the basis of their structure and functions are:

(a) Squamous epithelium: Consists of flattened cells. Present in oesophagus and lining of the mouth. Skin epithelial cells are arranged in many layers to prevent wear and tear and are called stratified squamous epithelium.

(b) Columnar epithelium: Has tall or ‘pillar-like’ cells. It forms the inner lining of the intestine.

(c) Cuboidal epithelium: Has cube-shaped cells. It forms the lining of kidney tubules and ducts of salivary glands, where it provides mechanical support.

(d) Ciliated epithelium: Have cilia on the outer surfaces of epithelial cells. The cilia can move and their movement pushes the mucus in the respiratory tract forward to clear it.

(e) Glandular epithelium: Has gland cells which secrete substances at the epithelial surface.

Question 9.
Give reasons for
(a) Meristematic cells have a prominent nucleus and dense cytoplasm but they lack vacuole.
(b) Intercellular spaces are absent in sclerenchymatous tissues.
(c) We get a crunchy and granular feeling, when we chew pear fruit.
(d) Branches of a tree move and bend freely in high wind velocity.
(e) It is difficult to pull out the husk of a coconut tree.
Answer:
(a) Because the meristematic cells are actively dividing cells and there is no need of storage.
(b) Because they have a thick deposition of lignin in them.
(c) Due to the presence of stone cells (sclerenchyma) in the pear fruit.
(d) Due to the presence of collenchyma which provides flexibility to the various parts of the plant.
(e) Due to the sclerenchyma present in the husk of the coconut.

Question 10.
List the characteristics of cork. How are they formed? Mention their role.
Answer:
The characteristics of cork are:

• Cells of cork are dead at maturity.
• These cells are compactly arranged.
• Cells do not possess intercellular spaces.
• Cells possess a chemical substance suberin in their walls.
• They are several layers thick.

A strip of secondary meristem replaces the epidermis of the older stem and cuts off the outside cells to form a several-layer thick cork or the bark of the tree. Cells of cork are dead, compactly arranged without intercellular spaces and have a chemical called suberin in their walls that makes them protective in function and impervious to gases and water.

Question 11.
(a) Differentiate between meristematic and permanent tissues in plants.
(b) Define the process of differentiation.
(c) Name any two simple and two complex permanent tissues in plants.
Answer:
(a) Meristematic tissue:

• This tissue consists of cells which continuously divide to produce new cells.
• They are located at specific regions of the plant, i.e., apical, lateral and intercalary.
• The cells of this tissue are very active, lack vacuoles, have dense cytoplasm, thin cell walls and prominent nuclei.

Permanent tissue:

• Consists of cells which have taken up a specific role and lost the ability to divide.
• They are distributed throughout the plant body.
• They are vacuolated, vary in shape and size. Their cell wall may be thick.

(b) The process of taking up a permanent shape, size, and a function by the cells is called differentiation.

(c) Simple: Parenchyma/collenchyma/sclerenchyma Complex: Phloem/xylem

Tissues Class 9 Extra Questions HOTS

Question 1.
The walls of the sclerenchymatous cells are thickened and have narrow lumen. Which substance thickens it and what is its role?
Answer:
The walls of the sclerenchymatous cells are thickened due to presence of lignin. It helps in providing mechanical strength to the various parts of the plant.

Question 2.
Which type of muscle fibres will contract to remove your hands instantly when you touch a hot object?
Answer:
Striated muscle fibres will contract to remove our hands instantly when we touch a hot object.

Question 3.
Which tissue helps the leaves of lotus plant to float on water? Why?
Answer:
Aerenchyma helps the leaves of lotus plant to float on water. Aerenchyma has large cavities to provide buoyancy to the parts of aquatic plants.

Question 4.
A tissue present in plants helps in storing food and sideways conduction of water. Identify the type of tissue.
Answer:
The tissue is xylem parenchyma.

Question 5.
Which tissue enables the heart to pump blood to various parts of the body? Why?
Answer:
The cardiac muscles help the heart to pump blood to various parts of the body as they show rhythmic contraction and relaxation throughout their life.

Question 6.
What will be the consequence of
(i) removal of blood platelets from blood?
(ii) removal of cutin from the layer of epidermis?
Answer:
(i) Removal of blood platelets from blood will inhibit clotting of blood if an injury occurs and the person may bleed to death.

(ii) Removal of cutin would increase the amount of water loss from the leaves of the plants.

Question 7.
Some actions of our body are under our control but many of them are not under our control. Why is it so?
Answer:
The actions of our body are controlled by our muscles. The voluntary actions are under the control of our will and are caused by the activity of striated muscles, eg., movement of our limbs. The involuntary . actions are not under the control of our will and are performed by the smooth muscles, e.g., the activity of bronchi of lungs. Even the activity of cardiac muscles which helps in the rhythmic contraction and relaxation of heart are involuntary in nature.

Question 8.
Which kind of meristem can help grasses to regenerate parts removed by the grazing herbivores?
Answer:
Intercalary meristem can help grasses to regenerate parts removed by the grazing herbivores.

Question 9.
Name the tissue which replaces the epidermal tissue in older stem and is rich in suberin. What is the function of suberin?
Answer:
Cork is the tissue which replaces the epidermal tissue in older stem and is rich in suberin. Suberin present in the walls of cork cells makes them impervious to gases and water.

Question 10.
The process of transpiration does not occur properly when the leaves are covered by a layer of oily substance. Why? Which other functions will get affected due to this covering?
Answer:
The layer of oily substance will close the stomata present in leaves and this would decrease the rate of transpiration. The rate of exchange of gases decreases and consequently the rate of photosynthesis would also decrease.

Tissues Class 9 Extra Questions Value Based (VBQs)

Question 1.
Raman got injured while playing football. His injured leg started bleeding and his friends immediately rushed to take him to the doctor to give him first aid. The blood flowing from the wound stopped after some time and the doctor applied antiseptic on the wound.
(i) Why did the blood stop flowing after some time from the wound?
(ii) What kind of tissue is blood? Why?
(iii) What values are shown by Raman’s friends?
Answer:
(i) The blood stopped flowing after some time from the wound as the blood platelets present in blood helped in clotting of blood.

(ii) Blood is a connective tissue. Blood is considered as a type of connective tissue as they have the same origin as other types of
connective tissue and helps to connect the different parts of the body.

(iii) The values shown by Raman’s friends are presence of mind, helpful and a caring nature.

Question 2.
During a sports event, Shivani suffered a sprain due to which she was not able to run. Her teacher gave her support and told her that it was due to a ligament tear. She also called the doctor to give treatment to Shivani.
(i) What is a ligament? What kind of tissue is it?
(ii) Which type of fibrous tissue has great strength, limited flexibility and is similar to ligament?
(iii) What values are shown by Shivani’s teacher?
Answer:
(i) Ligament is the connective tissue which connects two bones. It is a kind of connective tissue.

(ii) Tendon is a fibrous tissue that has great strength, limited flexibility and is similar to ligament.

(iii) The values shown by her teacher are knowledge, scientific approach and a caring nature.

Question 3.
Rishi brought an aquatic plant which was floating on the surface to the science laboratory of water. He cut a section of the leaf of the plant and saw a tissue with lot of air cavities in it. He went to his teacher and discussed about the role of the air cavities in the leaves of the aquatic plant.
(i) Which type of tissue present in plants has air cavities?
(ii) What is the role of large air cavities in the leaves of such plants?
(iii) What values are shown by Rishi?
Answer:
(i) Aerenchyma is the tissue present in the plants and has large air cavities.

(ii) The large air cavities in the leaves of such plants help in providing buoyancy to the leaves to help them float on water.

(iii) Rishi shows a scientific attitude, inquisitive nature and empirical approach.

Here we are providing Probability Class 9 Extra Questions Maths Chapter 15 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Probability with Answers Solutions

Extra Questions for Class 9 Maths Chapter 15 Probability with Solutions Answers

Probability Class 9 Extra Questions Very Short Answer Type

Probability Class 9 Extra Questions Question 1.
The blood groups of some students of Class IX were surveyed and recorded as below :

If a student is chosen at random, find the probability that he/she has blood group A or AB.
Solution:
Here,
total number of students = 19 + 6 + 13 + 12 = 50
Number of students has blood group A or AB = 19 + 13 = 32
Required probability = \(\frac{38}{50}\) = \(\frac{16}{25}\)

Class 9 Probability Extra Questions Question 2.
A group of 80 students of Class X are selected and asked for their choice of subject to be
taken in Class XI, which is recorded as below :

If a student is chosen at random, find the probability that he/she is a student of either commerce or humanities stream.
Solution:
Here, total number of students = 80
Total number of students of Commerce or Humanities stream = 33
Required probability = \(\frac{33}{80}\)

Probability Extra Questions Class 9 Question 3.
A box contains 50 bolts and 150 nuts. On checking the box, it was found that half of the bolts and half of the nuts are rusted. If one item is chosen at random, find the probability that it is rusted.
Solution:
Total number of nuts and bolts in the box = 150 + 50
= 200
Number of nuts and bolts rusted = \(\frac{1}{2}\) × 200 = 100
P(a rusted nut or bolt) = \(\frac{100}{200}\) = \(\frac{1}{2}\)

Class 9 Maths Chapter 15 Extra Questions Question 4.
A dice is rolled number of times and its outcomes are recorded as below:

Find the probability of getting an odd number.
Solution:
Total number of outcomes = 250
Total number of outcomes of getting odd numbers = 35 + 50 + 53 = 138
∴ P(getting an odd number) = \(\frac{138}{250}\) = \(\frac{69}{125}\)

Extra Questions Of Probability Class 9 Question 5.
The probability of guessing the correct answer to a certain question is \(\frac{x}{2}\) If probability of
not guessing the correct answer is \(\frac{2}{3}\), then find x.
Solution:
Here, probability of guessing the correct answer = \(\frac{x}{2}\)
And probability of not guessing the correct answer = \(\frac{x}{2}\)
Now, \(\frac{x}{2}\) + \(\frac{2}{3}\) = 1
⇒ 3x + 4 = 6
⇒ 3x = 2
⇒ x = \(\frac{2}{3}\)

Extra Questions On Probability Class 9 Question 6.
A bag contains x white, y red and z blue balls. A ball is drawn at the random, then what is the probability of drawing a blue ball.
Solution:
Number of blue balls = Z
Total balls = x + y + Z
∴ P(a blue ball) = \(\frac{z}{x+y+z}\)

Probability Class 9 Extra Questions Short Answer Type 1

Probability Questions For Class 9 Question 1.
750 families with 3 children were selected randomly and the following data recorded:

If a family member is chosen at random, compute the probability that it has :
(i) no boy child
(ii) no girl child
Solution:
(i) P(no boy child) = \(\frac{100}{750}\) = \(\frac{2}{15}\)
and P (no girl child) = \(\frac{120}{750}\) = \(\frac{4}{25}\)

Probability Questions Class 9 Question 2.
If the probability of winning a race of an athlete is \(\frac{1}{6}\) less than the twice the probability of losing the race. Find the probability of winning the race.
Solution:
Let probability of winning the race be p
∴ Probability of losing the race = 1 – p
According to the statement of question, we have
p = 2 (1 – p) – \(\frac{1}{6}\)
⇒ 6p = 12 – 12p – 1
⇒ 18p = 11
⇒ p = \(\frac{11}{18}\) .
Hence, probability of winning the race is \(\frac{11}{18}\).

Questions On Probability Class 9 Question 3.
Three coins are tossed simultaneously 150 times with the following frequencies of different outcomes :

Compute the probability of getting :
(i) At least 2 tails
(ii) Exactly one tail
Solution:
Here, total number of chances = 150
(i) Total number of chances having at least 2 tails = 32 + 63 = 95
∴ Required probability = \(\frac{95}{150}\) = \(\frac{19}{30}\)
(ii) Total number of chances having exactly one tail = 30
∴ Required probability = \(\frac{30}{150}\) = \(\frac{1}{5}\)

Probability Class 9 Extra Questions Short Answer Type 2

Probability Class 9 Extra Questions Pdf Question 1.
The table shows the marks obtained by a student in unit tests out of 50 :

Find the probability that the student get 70% or more in the next unit test. Also, the probability that student get less than 70%.
Solution:
Here, the marks are out of 50, so we first find its percentage (i.e., out of 100)

Total number of outcomes = 5
Probability of getting 70% or more marks = \(\frac{3}{5}\)
Probability of getting less than 70% = \(\frac{2}{5}\)

Class 9 Maths Probability Extra Questions Question 2.
Books are packed in piles each containing 20 books. Thirty five piles were examined for defective books and the results are given in the following table :

One pile was selected at random. What is the probability that it has :
(i) no defective books ?
(ii) more than 0 but less than 4 defective books ?
(iii) more than 4 defective books ?
Solution:
Total number of books = 700
(i) P(no defective books) = \(\frac{400}{700}\) = \(\frac{4}{7}\)
(ii) P(more than 0 but less than 4 defective books) = \(\frac{269}{700}\)
13 (iii) P(more than 4 defective books) = \(\frac{13}{700}\)

Probability Class 9 Extra Questions Short Answer Type 1 and 2

Ch 15 Maths Class 9 Extra Questions Question 1.
Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and number of times ‘1 Tail’ appeared is double the number of times ‘No Tail’ appeared. Find the probability of getting ‘Two tails’.
Solution:
Total number of outcomes = 360
Let the number of times ‘No Tail’ appeared be x
Then, number of times 2 Tails’ appeared = 3x
Number of times ‘1 Tail’ appeared = 2x
Now, x + 2x + 3x = 360
⇒ 6x = 360
⇒ x = 60
P(of getting two tails) = \(\frac{3 \times 60}{360}\) = \(\frac{1}{2}\)

Probability Class 9 Important Questions Question 2.
A die was rolled 100 times and the number of times, 6 came up was noted. If the experimental probability calculated from this information is \(\frac{2}{5}\), then how many times 6 came up ? Justify your answer.
Solution:
Here, total number of trials = 100
Let x be the number of times occuring 6.
We know, probability of an event
= \(\frac { Frequency\quad of\quad the\quad event\quad occuring }{ Total\quad number\quad of\quad trails }\)
⇒ \(\frac{x}{100}\) = \(\frac{2}{5}\) [∵ Probability is given)
⇒ x = 40

Probability Class 9 Extra Questions Long Answer Type

Probability Class 9 Extra Questions With Answers Question 1.
Three coins are tossed simultaneously 250 times. The distribution of various outcomes is listed below :
(i) Three tails : 30,
(ii) Two tails : 70,
(iii) One tail : 90,
(iv) No tail : 60
Find the respective probability of each event and check that the sum of all probabilities is 1.
Solution:
Here, the total number of chances = 250
Total number of three tails = 30

Probability Extra Questions For Class 9 Question 2.
A travel company has 100 drivers for driving buses to various tourist destination. Given
below is a table showing the resting time of the drivers after covering a certain distance (in km).

What is the probability that the driver was chosen at random :
(a) takes a halt after covering 80 km?
(b) takes a halt after covering 115 km?
(c) takes a halt after covering 155 km?
(d) takes a halt after crossing 200 km?
Solution:
Total number of drivers = 100
(a) P (takes a halt after covering 80 km) = \(\frac{13}{100}\)
(b) P (takes a halt after covering 115 km) = \(\frac{60}{100}\) = \(\frac{3}{5}\)
(c) P (takes a halt after covering 155 km) = \(\frac{90}{100}\) = \(\frac{9}{10}\)
(d) P (takes a halt after crossing 200 km) = \(\frac{10}{100}\) = \(\frac{1}{10}\)

Probability Class 9 Extra Questions With Solutions Question 3.
A company selected 2300 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below :

If a family is chosen at random, find the probability that the family is :
(i) earning ₹7000 – ₹13000 per month and owning exactly 1 vehicle.
(ii) owning not more than one vehicle. (iii) earning more than ₹13000 and owning 2 or more than 2 vehicles. (iv) owning no vehicle
Solution:
Here, we have a total number of families = 2300
(i) Number of families earning ₹7000 to ₹13000 per month and owning exactly 1 vehicle = 295 + 525 = 820

Class 9 Maths Ch 15 Extra Questions Question 4.
A survey of 2000 people of different age groups was conducted to find out their preference
in watching different types of movies :
Type I + Family Type II → Comedy and Family
Type III → Romantic, Comedy, and Family 242.
Type IV → Action, Romantic, Comedy and Family

Find the probability that a person chosen at random is :
(a) in 18-29 years of age and likes type II movies
(b) above 50 years of age and likes all types of movies
(c) in 30-50 years and likes type I movies. :
Solution:
(a) Let E₁ be the event, between the age group (18 – 29) years and liking type II movies
Favourable outcomes to event E₁ = 160
∴ P(E₁) = \(\frac{160}{2000}\) = \(\frac{160}{2000}\)
(b) Let E₂ be the event, of age group above 50 years and like all types of movies
Favourable outcomes to event E₂ = 9
∴ P(E₂) = \(\frac{9}{2000}\)
(c) Let E₃ be the event, between age group (30 – 50) years and liking type I movies
Favourable outcomes to event E₃ = 505
∴ P(E₃) = \(\frac{505}{2000}\) = \(\frac{101}{400}\)

Probability Class 9 Extra Questions HOTS

Question 1.
In a kitchen, there are 108 utensils, consisting of bowls, plates, and glasses. The ratio of bowls, plates the glasses is 4:2:3. A utensil is picked at random. Find the probability that :
(i) it is a plate.
(ii) it is not a bowl.
Solution:
Total utensils in the kitchen = 108
Let number of bowls be 4x, number of plates be 2x and number of glasses be 3x
∴ 4x + 2x + 3x = 108
9x = 108
x = \(\frac{108}{9}\) = 12
Thus, number of bowls = 4 × 12 = 48
Number of plates = 2 × 12 = 24
Number of glasses = 3 × 12 =
(i)P (a plate) = \(\frac{24}{108}\) = \(\frac{2}{9}\)
(ii) P (not a bowl) = \(\frac{24+36}{108}\) = \(\frac{60}{108}\) = \(\frac{5}{9}\)

Question 2.
A bag contains 20 balls out of which x are white.
(a) If one ball is drawn at random, find the probability that it is white.
(b) If 10 more white balls are put in the bag, the probability of drawing a white ball will be double that in part (a), find x.
Solution:
Here, total no. of balls = 20
No. of white balls = x
∴ P(white ball) = \(\frac{x}{20}\)
Now, 10 more white balls are added
∴ Total no. of balls = 20 + 10 = 30
Total no. of white balls = x + 10
According to statement of question, we have
\(\frac{x+10}{30}\) = 2 × \(\frac{x}{20}\)
⇒ \(\frac{x+10}{3}\) = x
⇒ x + 10 = 3x
⇒ 2x = 10
⇒ x = 5

Question 3.
Here is an extract from a mortality table.

(i) Based on this information, what is the probability of a person aged 60′ of dying within a year ?
(ii) What is the probability that a person aged 61′ will live for 4 years ?
Solution:
(i) Clearly 16090 persons aged 60, (16090 – 11490), i.e., 4600 died before reaching their 61st birthday.
Therefore, P (a person aged 60 die within a year) = \(\frac{4600}{16090}\) = \(\frac{460}{1609}\)
(ii) Number of persons aged 61 years = 11490
Number of persons surviving for 4 years = 2320
P (a person aged 61 will live for 4 years) = \(\frac{2320}{11490}\) = \(\frac{232}{1149}\)

Probability Class 9 Extra Questions Value Based (VBQs)

Question 1.
An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained are given in the following table :

Find the probability of the following events for a driver-selected at random from the city :
(i) being 18 – 29 years of age and having exactly 3 accidents in one year.
(ii) being 30 – 50 years of age and having one or more accidents in a year.
(iii) having no accident in one year.
(iv) Which value would you like to remember from this data?
Solution:
Here, total number of drivers = 2000
(i) The number of drivers in the age group 18 – 29 having exactly 3 accidents = 70
So, P (driver in age group 18 – 29 having exactly 3 accidents in one year) = \(\frac{70}{2000}\) = 0.035
(ii). The number of drivers in the age group 30 – 50 and having one or more than one accidents in one year = 118 + 65 + 20 + 21 = 224
P (driver in age group 30 – 50 having one or more accidents in one year) = \(\frac{224}{2000}\) = 0.112
(iii) The number of drivers having no accident in one year = 395 + 520 + 390 = 1305
So, P (driver having no accident) = \(\frac{1305}{2000}\) = 0.6525
(iv) Most people in India died or injured due to accidents as compared to any other country. So, we should obey the traffic rules as life is very precious.

Question 2.
100 plants were sown in six different colonies A, B, C, D, E, and F. After 31 days, the number of plants survived as follows :

What is the probability of :
(i) more than 80 plants survived in a colony?
(ii) less than 82 plants survived in a colony?
(iii) which values are depicted from the above data?
Solution:
Here, we have total number of colonies = 6
(i) Number of colonies in which more than 80 plants survived = 4 (i.e., B, C, E and F)
∴ P(more than 80 plants survived in a colony) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
(ii) Number of colonies in which less than 82 plants survived = 2 (i.e., A and D)
∴ P (less than 82 plants survived in a colony) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
(iii) In order to keep environment safe, clean and green, we should grow more and more plants.

In this page, we are providing Is Matter Around Us Pure Class 9 Extra Questions and Answers Science Chapter 2 pdf download. NCERT Extra Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 2 Extra Questions and Answers Is Matter Around Us Pure

Extra Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure with Answers Solutions

Is Matter Around Us Pure Class 9 Extra Questions Very Short Answer Type

Class 9 Science Chapter 2 Extra Questions And Answers Question 1.
What is the reason for the difference in properties of solutions, colloids and suspensions?
Answer:
Due to different particle size.

Is Matter Around Us Pure Extra Questions Question 2.
Is rain water or distilled water a pure substance?
Answer:
Yes, because it contains particles (molecules) of only water.

Is Matter Around Us Pure Class 9 Extra Questions Question 3.
Can we separate a mixture of alcohol and water by a separating funnel?
Answer:
No, the two liquids are miscible.

Ch 2 Science Class 9 Extra Questions Question 4.
Give two examples of metalloids.
Answer:
Silicon and germanium.

Class 9 Chemistry Chapter 2 Extra Questions Question 5.
Give an example of a solution in which solid is a solute as well as the solvent.
Answer:
Alloys are solid in solid solutions. For example, brass contains about 30% zinc and 70% copper. Here, zinc is the solute while copper is the solvent.

Class 9 Science Ch 2 Extra Questions Question 6.
What type of liquid mixture will kerosesne oil and water form? How will you separate it?
Answer:
Immiscible. We can separate this mixture by using a separating funnel.

Class 9 Is Matter Around Us Pure Extra Questions Question 7.
Define the term heterogeneous.
Answer:
Heterogeneous means that the substance does not have the same properties or characteristics throughout its bulk.

Chapter 2 Science Class 9 Extra Questions Question 8.
Write the constituent element of potassium hydroxide and sodium chloride.
Answer:
The constituent element of potassium hydroxide is K, H and O and sodium chloride is Na and Cl.

Ncert Class 9 Science Chapter 2 Extra Questions Question 9.
On the basis of composition, how is matter classified?
Answer:

1. Pure substance
2. Mixture.

Extra Questions Of Chapter 2 Science Class 9 Question 10.
What are different categories of pure substances?
Answer:

1. Elements
2. Compounds.

Class 9 Chapter 2 Science Extra Questions Question 11.
What are the different kinds of mixture?
Answer:

1. Homogeneous mixture
2. Heterogeneous mixture.

Is Matter Around Us Pure Class 9 Important Questions With Answers Question 12.
What are the constituents of brass?
Answer:
Brass is an alloy and is a mixture of zinc (30%) and copper (70%).

Class 9 Science Chapter 2 Extra Questions Question 13.
How are elements further classified?
Answer:
Metals, non-metals, metalloids.

Extra Questions Of Is Matter Around Us Pure Question 14.
A solution of water and alcohol contains 30 g of water and 60 g of alcohol. What is the concentration of solution?
Answer:
\(\frac { 30 }{ 30 + 60 }\) x 100 = \(\frac { 30 }{ 90 }\) x 100 = 33.3%.

Class 9th Science Chapter 2 Extra Questions Question 15.
What are aqueous solutions?
Answer:
Solutions in which water is the solvent are called aqueous solutions, e.g., sugar solution, in which sugar is dissolved in water.

Question 16.
What is an unsaturated solution?
Answer:
A solution in which some more solute can be dissolved at any fixed temperature is called an unsaturated solution.

Question 17.
How many gram of water is needed to make 8% mass by mass percentage of sodium carbonate solution if 4 g of sodium carbonate is a variable to make a solution?
Answer:
8% means 8 g in 100 g of solution. So if 4 g Na₂CO₃ is present, it means solution must be 50 g.

Question 18.
What are the conditions required to convert air into liquid air?
Answer:
200 atmospheric pressure and – 200°C.

Question 19.
Define the term inter-conversion of matter.
Answer:
The phenomenon of change of one state of matter into another and back to the original state is called inter-conversion of matter.

Question 20.
Why do fish go in deep water during day light?
Answer:
During day time, the shallow water is warmed and hence it has less dissolved oxygen. Therefore fish tend to go in deep water during day time.

Question 21.
Out of colloid, solution and a suspension which one can be separated by filtration.
Answer:
Suspension.

Question 22.
Out of colloid, solution and a suspension which has the smallest particle?
Answer:
Solution.

Is Matter Around Us Pure Class 9 Extra Questions Short Answer Type 1

Question 1.
Suggest separation technique(s) one would need to employ to separate the following mixtures. [NCERT Exemplar]
(a) Mercury and water
(b) Potassium chloride and ammonium chloride
(c) Common salt, water apd sand
(d) Kerosene oil, water and salt.
Answer:
(a) Separation by using separating funnel
(b) Sublimation
(c) Filtration followed by evaporation
Or
Centrifugation followed by evaporation/distillation
(d) Separation by using separating funnel to separate kerosene oil followed by distillation.

Question 2.
A salt can be recovered from its solution by evaporation. Suggest some other technique for the same? [NCERT Exemplar]
Answer:
Crystallisation.

Question 3.
The ‘sea water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.
Answer:

• Homogeneous – mixture of salts and water only.
• Heterogeneous – contains salts, water, mud, decayed plant, etc.

Question 4.
Fill in the following blanks:
(i) Milk is a ……………… solution but vinegar is a solution.
(ii) Milk is a colloidal solution in which ……………… is the dispersed phase and ……………… is the dispersion medium.
Answer:
(i) colloidal, true
(ii) fat, water.

Question 5.
(a) Classify Brass and Diamond as element and mixture.
(b) How is a chemical change different from a physical change?
Answer:
(a) Brass is homogeneous mixture also called alloy. The constituents are Cu and Zn. Diamond is an element. It is an allotropic form of carbon.

(b) In a chemical change, a new substance is formed as a result of chemical reaction. No new substance is formed in a physical change.

Question 6.
Identify the dispersed phase and dispersion medium in the following examples of colloids:
(a) Fog
(b) Cheese
(c) Coloured gemstone.
Answer:
(a) Fog: Liquid (water drops) acts as dispersed phase and gas (air) as the dispersion medium.
(b) Cheese: Solid (fat) acts as the dispersed phase and water (liquid) as the dispersion medium.
(c) Coloured gemstone: Solids act the dispersed phase as well as the dispersion medium.

Question 7.
Explain, why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do. [NCERT Exemplar]
Answer:
Particle size in a suspension is larger than those in a colloidal solution. Also molecular interaction in a suspension is not strong enough to keep the particles suspended and hence they settle down.

Question 8.
Smoke and fog both are aerosols. In what way are they different? [NCERT Exemplar]
Answer:
Both fog and smoke have gas as the dispersion medium. The only difference is that the dispersed phase in fog is liquid and in smoke it is a solid.

Question 9.
How will you bring about the following separation?
(i) Fine mud particles floating in water.
(ii) Carbon particles present in smoke.
Answer:
(i) By coagulation using alum and then filtering.
(ii) By passing smoke through electric plates maintained at a high potential difference. The colloidal particles of carbon get neutralised and fall down while air escapes out.

Question 10.
Classify the following as Physical or chemical properties. [NCERT Exemplar]
(a) The composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements.
(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
(c) Metallic sodium is soft enough to be cut with a knife.
(d) Most metal oxides form alkalies on interacting with water.
Answer:
Physical properties: (a) and (c)
Chemical properties: (b) and (d)

Question 11.
Suggest a suitable separation technique for the following: [NCERT Exemplar]
(a) Mercury and water
(b) Coloured components from blue ink
(c) Ammonium chloride and potassium chloride
(d) Mixture of alcohol and water.
Answer:
(a) The separation can be done by the use of a separating funnel. Mercury forms the lower layer (heavier) and water the upper layer (lighter).
(b) The separation can be done with the help of chromatography.
(c) Process of sublimation can be used. Ammonium chloride collects as the sublimate while potassium chloride remains in the dish.
(d) Process of fractional distillation can be used. Alcohol (ethyl alcohol) with lower boiling point (78°C or 351 K) gets distilled leaving behind water with higher boiling point (100°C or 373 K) in the distillation flask.

Question 12.
Name the type of colloids in each of the following giving an example of each.

	Dispersed Phase	Dispersing Medium
A	Liquid	Gas
B	Liquid	Liquid
C	Liquid	Solids

Answer:

A	Aerosol	(Fog)
B	Emulsion	(Milk)
C	Gel	(Jelly)

Question 13.
You are given two samples of water labelled as ‘A’ and ‘B’ Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment. [NCERT Exemplar]
Answer:
Sample ‘B’ will not freeze at 0°C because it is not pure water. At 1 atm, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C.

Question 14.
Identify colloids from the following: Copper sulphate solution, milk, smoke, muddy water, butter, sugar solution, face cream, lemonade.
Answer:
Colloids: milk, smoke, muddy water, butter, face cream, lemonade.

Question 15.
What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments? [NCERT Exemplar]
Answer:
Pure gold is very soft as compared to gold alloyed with silver or copper. Thus for providing strength to gold, it is alloyed.

Question 16.
12 mL of dettol is added to a beaker containing 500 mL of water and stirred. State four observations that you make.
Answer:
When dettol is added to a beaker containing of water, the following observations are made.

• An emulsion is formed which is of colloidal nature.
• The colour of emulsion is milky.
• It gives characteristic smell of dettol.
• The solution can pass through a filter paper.

Is Matter Around Us Pure Class 9 Extra Questions Short Answer Type 2

Question 1.
During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution. [NCERT Exemplar]
(a) Are the two solutions of the same concentration?
(b) Compare the mass % of the two solutions.
Answer:
(a) No.
\(Mass\%=\frac{\text { Mass of solute }}{\text { Mass of solute }+\text { Mass of solvent }} \times 100\)

(b) Solution made by Ramesh
Mass % = \(\frac { 10 }{ 10 + 100 }\) = \(\frac { 10 }{ 110 }\) x 100 = 9.09%
solution made by sarika
Mass % = \(\frac { 10 }{ 100 }\) x 100 = 10%
The solution prepared by Sarixa has a higher mass % than that prepared by Ramesh.

Question 2.
State which of the following solutions exhibit Tyndall effect:
Starch solution, Sodium chloride solution, Tincture of iodine, Air.
Answer:
(i) Tyndall effect is shown both by starch solution and air which are heterogeneous mixtures and have the capacity to scatter a beam of light as it passes through them.

(ii) Sodium chloride solution and tincture of iodine (iodine crystals dissolved in ethyl alcohol) are both homogeneous in nature and do not exhibit any Tyndall effect.

Question 3.
While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice. [NCERT Exemplar]
Answer:
Distillation, since acetone is more volatile it will separate out first.

Question 4.
(a) Why is crystallisation technique better than evaporation?
(b) Write any two physical properties of each of metals and non-metals.
(c) Name the technique used to separate butter from curd.
Answer:
(a) Both these techniques are used to separate solid substances from their solutions. But crystallisation is considered better because during evaporation certain solids may decompose or some of them like sugar get charred when the solution is evaporated completely to dryness. As a result of crystallisation, even the shapes of the crystals do not change.

(b) (i) Metals have a shiny surface known as lustre.
(ii) Metals are malleable and ductile.
(iii) Non-metals are mostly poor conductors of electricity.
(iv) Non-metals are generally soft.

(c) Butter can be separated from curd by the process of centrifugation. This is usually done by churning which is very common as well as convenient.

Question 5.
Name the process associated with the following:
(a) Dry ice is kept at room temperature and at one atmospheric pressure.
(b) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
(c) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
(d) A acetone bottle is left open and the bottle becomes empty.
(e) Milk is churned to separate cream from it.
(f) Settling of sand when a mixture of a sand and water is left undisturbed for some time.
Answer:
(a) Sublimation
(b) Diffusion
(c) Dissolution/diffusion
(d) Evaporation, diffusion
(e) Centrifugation
(f) Sedimentation

Question 6.
On dissolving chalk powder in water, a suspension is obtained. Give any four reasons to support the fact that the mixture so obtained is a suspension only.
Answer:
It is supported by the following reasons:

• White particles of chalk powder can be seen with the naked eyes.
• The particles can be separated by ordinary filter paper.
• Upon shaking, a white turbidity reappears in solution.
• Light cannot pass through the suspension which shows that it is of opaque nature.

Question 7.
Give an example for each of the following:
(a) Solid-liquid homogeneous mixture
(b) Gas-gas homogeneous mixture
(c) Liquid-liquid heterogeneous mixture.
Answer:
(a) Mixture of sodium chloride in water.
(b) Air. It is a homogeneous mixture of a number of gases.
(c) Emulsion of oil and water.

Question 8.
What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature?
(b) an aqueous sugar solution is heated to dryness?
(c) a mixture of iron filings and sulphur powder is heated strongly?
Answer:
(a) Solid potassium chloride will separate out.
(b) Initially the water will evaporate and then sugar will get charred.
(c) Iron sulphide will be formed.

Question 9.
(a) Arrange solids, liquids and gases in increasing order of the following properties of matter
(i) rigidity
(ii) diffusion
(iii) compressibility.
(b) Write one example from your daily life which is based on diffusion of gases.
Answer:
(a) (i) Rigidity: Gases < Liquids < Solids
(ii) Diffusion: Solids < Liquids < Gases
(iii) Compressibility: Solids < Liquids < Gases.

(b) Smell of aroma or perfume released in one corner of the room soon spreads in the whole room.

Question 10.
Is air a mixture or a compound? Give three reasons.
Answer:
Air is a mixture and not a compound as discussed below:
(i) The properties of a mixture are in between those of its constituents. The two major components of air are oxygen (20.9% by volume) and nitrogen (78.1% by volume). In oxygen, any fuel bums brightly but in nitrogen it gets extinguished. In contrast, in air the fuel bums slowly.

(ii) The components of a mixture can be separated by simple physical methods. For example, the components of air can be separated by fractional distillation of liquid air.

(iii) The composition of a mixture is variable. The composition of air is also variable. It has more oxygen in the country side than in big cities.

(iv) When air is obtained by mixing its constituent gases, no heat is either evolved or absorbed.

(v) Liquid air does not have a fixed boiling point.

Is Matter Around Us Pure Class 9 Extra Questions Long Answer Type

Question 1.
Classify each of the following, as a physical or a chemical change. Give reasons.
(a) Drying of a shirt in the Sun.
(b) Rising of hot air over a radiator.
(c) Burning of kerosene in a lantern.
(d) Change in the colour of black tea on adding lemon juice to it.
(e) Churning of milk cream to get butter.
Answer:
Physical changes: (a), (b), (e)
Chemical changes: (c), (d)

Question 2.
Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?
Answer:
Part A

Part B

When dilute HCl is added to it, only the iron filings in the mixture react and sulphur remains unreacted
Fe (s) + 2 HCl (aq) → FeCl₂ + H₂ gas
H₂S gas formed has a foul smell and on passing through lead acetate solution, it turns the solution black. Hydrogen gas burns with a pop sound.

Question 3.
Fill in the blanks:
(a) A colloid is a ……………….. mixture and its components can be separated by the technique known as ………………..
(b) Ice, water and water vapour look different and display different ……………….. properties but they are the same.
(c) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for sometime. The upper layer in the separating funnel will be of ……………….. and the lower layer will be that of
(d) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called ………………..
(e) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the ……………….. of light by milk and the phenomenon is called ……………….. This indicates that milk is a ……………….. solution. [NCERT Exemplar]
Answer:
(a) heterogeneous, centrifugation
(b) physical, chemically
(c) water, chloroform (hint-density of water is less than that of chloroform)
(d) fractional distillation scattering, Tyndall effect, colloidal.

Question 4.
Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures. [NCERT Exemplar]
(a) A volatile and a non-volatile component.
(b) Two volatile components with appreciable difference in boiling points.
(c) Two immiscible liquids.
(d) One of the components changes directly from solid to gaseous state.
(e) Two or more coloured constituents soluble in some solvent.
Answer:
(a) Evaporation or distillation
(b) Distillation
(c) By using a separating funnel
(d) Sublimation
(e) Chromatography

Question 5.
Which separation techniques you will apply for the separation of the following mixtures?
(a) Oil from water
(b) Camphor from sand
(c) Sodium chloride from its solution in water
(d) Cream from milk
(e) Metal pieces from engine oil of a car.
Answer:
(a) By the use of a separating funnel.
(b) With the help of sublimation technique
(c) By evaporation crystallisation technique.
(d) By the use of a centrifuge
(e) By the use of filtration technique

Is Matter Around Us Pure Class 9 Extra Questions HOTS

Question 1.
Which of the tubes shown below will be more effective as a condenser in the distillation apparatus? [NCERT Exemplar]

Answer:
The presence of beads in tube (a) would provide a larger surface area for cooling. So, this tube will be more effective as a condenser.

Question 2.
Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, non- sonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
(b) Name a non-metal which exists as a liquid at room temperature.
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
(d) Name a non-metal which is known to form the largest number of compounds.
(e) Name a non-metal other than carbon which shows allotropy.
(f) Name a non-metal which is required for combustion.
Answer:
(a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Sulphur, phosphorus
(f) Oxygen.

Question 3.
The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?
Answer:
‘C’ has made the desired solution.

= \(\frac { 50 }{ 100 }\) x 100
= 50% mass by volume.

Question 4.
On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.
(a) Is this a physical or a chemical change?
(b) Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.
Answer:
(a) Chemical change

(b) Acidic and basic solutions can be prepared by dissolving the products of the above process in water
CaO + H₂O → Ca(OH)₂ (basic solution)
CO₂ + H₂O → H₂CO₃ (acidic solution)

Question 5.
Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g of NaCl + 100 g water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl + 99.99 g of water
(d) 0.10 g of NaCl + 99.90 g of water
Answer:
(c) 0.01 g of NaCl + 99.99 g of water

= \(\frac { 0.01 }{ 0.01 + 99.99 }\) x 100
= \(\frac { 0.01 }{ 100 }\) x 100
= 0.01 g

Question 6.
A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the figure. The filter paper was removed when the water moved near the top of the filter paper. [NCERT Exemplar]

1. What would you expect to see, if the ink contains three different coloured components?
2. Name the technique used by the child.
3. Suggest one more application of this technique.

Answer:

1. Three different bands will be observed.
2. Chromatography
3. To separate the pigments present in chlorophyll.

Question 7.
A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the figure. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?

1. Explain why the milk sample was illuminated. Name the phenomenon involved.
2. Same results were not observed with a salt solution. Explain.
3. Can you suggest two more solutions which would show the solution?

Answer:

1. Milk is a colloid and would show Tyndall effect.
2. Salt solution is a true solution and would not scatter light.
3. Detergent solution, sulphur solution.

Question 8.
Sudha tested the solubility of four salts X, Y, Z and T at different temperatures and collected the following data. (Solubility refers to the amount in grams dissolved in 100 g of water to give a saturated solution.)

Answer the following questions from the table:

1. Which salt has the highest and lowest solubility at 323 K?
2. A student prepared a saturated solution of X at 323 K and then added 25 g water to it. What mass of X must be added to again make the solution saturated?
3. The solubility of which salt is least affected by increase in temperature?
4. What mass of T would be required to make saturated solution in 200 g of water at 290 K?

Answer:
1. At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility.

2. By definition of saturated solution,
100 g of water at 323 K contain salt = 40 g
125 g of water at 323 K contain salt = \(\frac { 40 g }{ 100 g }\) x (125 g) = 50 g
∴ Mass of salt to be added to make the solution again saturated = (50 – 40) = 10 g

3. The data show that the solubility of the salt Y is least affected with increase in temperature.

4. At 290 K, mass of T required to make a saturated solution in 200 g of water = \(\frac { 25 g }{ 100 g }\) x (200 g) = 50 g

Is Matter Around Us Pure Class 9 Extra Questions Value Based (VBQs)

Question 1.
Mallika’s mother was suffering from cold and cough. Mallika prepared tea for her mother. She boiled water in a pan, then she added tea leaves, sugar and milk to it. She filtered the tea in a cup and served it to her mother.
(a) Explain the values shown by Mallika.
(b) Identify solute, solvent, residue and filtrate in this activity.
Answer:
(a) Mallika used the knowledge of chemistry to provide relief to her mother. Actually she prepared an extract of tea leaves which is helpful in curing cold and cough and gives warmth to the body.

(b) Solute: Tea leaves and sugar.
Solvent: water and milk.
Filtrate: homogeneous mixture of water, milk, sugar and extract of tea leaves.

Question 2.
Amit was asked by his teacher to separate a liquid mixture of acetone and ethyl alcohol. He set up a distillation apparatus and tried to distil the mixture. To his surprise, both the liquids got distilled. Teacher told Amit to repeat the experiment by using a fractionating column in the distillation flask. Amit followed the advice of the teacher and he was able to separate the two liquids.

1. Why was Amit not successful in separating the liquid mixture earlier?
2. Why did teacher ask him to use the fractionating column?
3. Which liquid was distilled first?
4. As a student of chemistry, what value based information you have gathered?

Answer:
1. The difference in boiling point temperatures of acetone (56°C) and ethyl alcohol (78°C) is only 22°C.
Therefore, process of simple distillation fails in this case.

2. Fractionating column is quite effective in this case because it obstructs the distillation of ethyl alcohol which is high boiling and at the same time helps in the distillation of acetone which is low boiling.

3. Acetone was distilled first since it has comparatively low boiling point.

4. The process of simple distillation can be used only in case, the liquids present in the mixture differ in their boiling point by 25°C or more.

In this page, we are providing Matter in Our Surroundings Class 9 Extra Questions and Answers Science Chapter 1 pdf download. NCERT Extra Questions for Class 9 Science Chapter 1 Matter in Our Surroundings with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 1 Extra Questions and Answers Matter in Our Surroundings

Extra Questions for Class 9 Science Chapter 1 Matter in Our Surroundings with Answers Solutions

Matter in Our Surroundings Class 9 Extra Questions Very Short Answer Type

Matter In Our Surroundings Class 9 Extra Questions With Answers Question 1.
Define the term ‘matter’.
Answer:
Matter is defined as anything that occupies some space and has definite mass.

Class 9 Science Chapter 1 Extra Questions And Answers Question 2.
What is Law of Conservation of Mass?
Answer:
Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction.

Matter In Our Surroundings Extra Questions Class 9 Question 3.
Define melting point.
Answer:
It is the temperature at which a solid becomes liquid at atmospheric pressure by absorbing heat.

Class 9 Chemistry Chapter 1 Extra Questions Class 9 Question 4.
Out of water and alcohol, which is more volatile?
Answer:
The boiling point of alcohol (78°C or 351K) is lower than that of water (100°C or 373K), therefore, alcohol is more volatile than water.

Matter In Our Surroundings Class 9 Extra Questions With Answers Pdf Question 5.
What is sublimation?
Answer:
Direct conversion of a solid into vapour and vice-versa (i.e., vapour into solid) is called sublimation.

Matter In Our Surroundings Class 9 Extra Questions Question 6.
Is dry ice the same thing as ordinary ice?
Answer:
No, dry ice is solid carbon dioxide while ordinary ice is solid water.

Extra Questions For Class 9 Science Chapter 1 Question 7.
Define latent heat of fusion.
Answer:
It is the heat energy required to convert 1 kg of solid into liquid at its melting point at atmospheric pressure.

Class 9 Science Ch 1 Extra Questions  Question 8.
Define vapourisation.
Answer:
The process of change from liquid state to gaseous (vapour) state is called vapourisation.

Class 9 Chapter 1 Science Extra Questions Question 9.
Give the important properties on the basis of which the three states of matter can be distinguished.
Answer:
The three states of matter can distinguished on the basis of shape, volume, compressibility, packing of molecules, number of free surfaces, etc.

Ncert Class 9 Science Chapter 1 Extra Questions Question 10.
Name the term used for the solid which is directly formed from the gas.
Answer:
Sublimate.

Science Class 9 Chapter 1 Extra Questions Question 11.
Define the term volatile liquid.
Answer:
Those liquids which can change into vapour easily are termed as volatile liquids.

Ch 1 Science Class 9 Extra Questions Question 12.
What is the effect of pressure on boiling point?
Answer:
Boiling point increases with increase in pressure.

Class 9 Science Chapter 1 Extra Questions Question 13.
Name any two substances which sublime.
Answer:
Camphor, napthalene, iodine, ammonium chloride.

Class 9 Science Chapter 1 Short Question Answer Question 14.
Define condensation.
Answer:
The change of a gaseous state to a liquid state on cooling is known as condensation.

Chapter 1 Science Class 9 Extra Questions Question 15.
State the effect of surface area on rate of evaporation.
Answer:
If the surface area is increased, the rate of evaporation increases.

Question 16.
Define evaporation.
Answer:
Evaporation is a physical process in which a liquid changes to its gaseous state, at a temperature lower than its boiling point.

Question 17.
What are the ways in which a gas can be liquefied?
Answer:
Applying pressure and reducing temperature can liquefy gases.

Question 18.
What is plasma?
Answer:
It is a state of matter which consists of super energetic and super excited particles. These particles are in the form of ionised gases.

Question 19.
How do solids, liquids and gases differ in shape and volume?
Answer:
Solids have a definite shape and a fixed volume, liquids have a definite volume but no fixed shape while gases neither have a definite volume nor a definite shape.

Question 20.
Kelvin scale of temperature is regarded as better scale than Celsius. Why?
Answer:
As it has a wide range of measurement and temperature in kelvin scale always has a positive sign, hence regarded as better scale than Celsius.

Matter in Our Surroundings Class 9 Extra Questions Short Answer Type 1

Question 1.
What are characteristics of particles of matter?
Answer:
The particles of matter have following characteristics:

• Particles of matter are very very small.
• Particles of matter have space between them.
• Particles of matter attract each other.
• Particles of matter are constantly moving.

Question 2.
Write four main characteristics of solid state of matter.
Answer:

• Solids have definite mass, volume and shape.
• The particles in solid state are closely packed and empty spaces in them are negligible.
• Solids are rigid.
• Solids can have a number of free surfaces.

Question 3.
Write four main characteristics of liquid state of matter.
Answer:

• Liquids have a definite mass and volume.
• A liquid can take the shape of a container.
• Liquids have only one free surface.
• Liquids show the property of diffusion.

Question 4.
Write four characteristics of gaseous state of matter.
Answer:

• A gas has definite mass but it has neither definite shape nor definite volume.
• Gases can occupy the whole of the space available to them.
• There are larger vacant spaces between the molecules of a gas.
• Gases are highly compressible.

Question 5.
Explain evaporation and its cooling effect in terms of kinetic energy of particles.
Answer:
During evaporation, the molecules which possess higher kinetic energy leave the liquid and go into the space above the liquid as vapour. The remaining molecules possessing lower kinetic energy are left in the liquid state. Consequently, the average kinetic energy decreases which results in the fall of temperature of the liquid.

Question 6.
How is heat transferred when a solid sublimes?
Answer:
Certain solids like iodine, naphthalene, solid CO₂ sublimes on heating. Heat is absorbed by the molecules of these solids rapidly which provides enough kinetic energy to show phase change into gaseous state.

Question 7.
Why do gases diffuse rapidly?
Answer:
Gases diffuse rapidly due to high speed of the particles and large space between them.

Question 8.
For any substance, why does the temperature remain constant during the change of state?
Answer:
On increasing the temperature of a substance, for example a solid, the kinetic energy of the particles increases which is used to overcome the forces of attraction between the particles therefore the temperature remains constant during the change of state.

Question 9.
Explain compressibility in gases with an example.
Answer:
Liquefied petroleum gas (LPG) cylinders are used in our homes for cooking, contains gases in the compressed state. Similarly, compressed natural gas (CNG) is used as a fuel in vehicles. Large volume of gases can be compressed in small cylinders and are transported to distant places.

Question 10.
Why solids cannot be compressed like gases?
Answer:
The particles in solids are so tightly packed that there are no or little interparticle spaces left among them. Therefore solids are not compressible like gases. Gases which have large interparticle spaces are therefore compressible.

Question 11.
Define boiling. Why is boiling considered as bulk phenomenon?
Answer:
Rapid and breaking of bubbles in the bulk of a liquid being heated is called boiling. During boiling particles from the bulk of liquid gain enough energy to get converted to vapour. Therefore it is a bulk phenomenon.

Question 12.
Why do we see water droplets on the outer surface of a glass containing ice-cold water?
Answer:
The water vapour present in air, on coming in contact with the cold glass of water, loses energy and gets converted to liquid state, which we see as water droplets.

Question 13.
Why do we sprinkle water on the roof or open ground in summer?
Answer:
During hot summer evenings, we often sprinkle water on the roof of the house or open ground in front of our house. The water evaporates by absorbing heat from the ground and the surrounding air. By losing heat, the ground becomes cool and we feel comfortable.

Question 14.
Why is ice rubbed on a burnt part of the skin?
Answer:
When a finger or some part of our body gets burnt, we rub the burnt portion with an ice cube. The reason being that due to burning, the temperature of the injured skin increases. When ice is rubbed, the excess heat from the skin is taken away by large latent heat of fusion of water. As a result, the temperature of the injured skin decreases and we feel less pain.

Question 15.
How will you demonstrate that particles of matter are continuously moving?
Answer:
When an incense stick is lit in one corner of a room, we get the smell while sitting at a distance from the stick. This is because the particles of matter are continuously moving. Because of their random motion, the particles of incense mix with the particles of air rapidly and the smell of the incense reaches us even when we are sitting at a distance from the incense stick.

Question 16.
Why do solids expand a bit on heating and contract a bit on cooling?
Answer:
The solid molecules do not have sufficient intermolecular (or interparticle) space thus its expands a bit on heating. The interparticle forces of attraction are very strong which do not let solid particles leave their mean positions. Therefore solid contracts a bit on cooling.

Question 17.
Draw diagram to shown interconversion among states of matter.
Answer:

Question 18.
Why is light not considered matter?
Answer:
Matter occupies space and has mass. Light has neither of the two and that is why it is not considered as matter. It is considers as a form of energy and electromagnetic radiation.

Question 19.
Convert the following temperatures:
(a) – 78.0°C to kelvin
(b) 775 K to °C
(c) 489 K to °C
(d) 24°C to kelvin
Answer:
(a) – 78 + 273 = 195 K
(b) 775 – 273 = 502°C
(c) 489-273 = 216°C
(d) 24 + 273 = 297 K

Question 20.
Mention the difference between gas and vapour.
Answer:
Gas – The gas is a substance which exists in the gaseous state at a temperature equal to or more than the boiling point of its liquid state. For example oxygen, hydrogen, nitrogen, etc.

Vapour – A vapour is a substance which exists in the gaseous state such that its temperature is lower than that of boiling point of its liquid state. For example, water vapour, iodine vapour, etc.

Question 21.
A sample of water under study was found to boil at 102″C at normal temperature and pressure. Is the water pure? Will this water freeze at 0°C? Comment.
Answer:
It’s freezing point will be below 0°C due to the presence of a non-volatile impurity in it.

Question 22.
Water as ice has a cooling effect, whereas water as steam may cause severe burns. Explain these observations. [NCERT Exemplar]
Answer:
In case of ice, the water molecules have low energy while in the case of steam the water molecules have high energy. The high energy of water molecules in steam is transformed as heat and may cause burns. On the other hand, in case of ice, the water molecules take energy from the body and thus gave a cooling effect.

Question 23.
It is a hot summer day, Priyanshi and Ali are wearing cotton and nylon clothes respectively. Who do you think would be more comfortable and why? [NCERT Exemplar]
Answer:
Cotton being a better absorber of water than nylon helps in absorption of sweat followed by evaporation which leads to cooling. So Priyanshi would be more comfortable than Ali.

Question 24.
You want to wear your favourite shirt to a party, but the problem is that it is still wet after a wash. What steps would you take to dry it faster? [NCERT Exemplar]
Answer:
Conditions that can increase the rate of evaporation of water are:

• An increase of surface area by spreading the shirt
• An increase in temperature by putting the shirt under the Sun
• An increase the wind speed by spreading it under the fan.

Matter in Our Surroundings Class 9 Extra Questions Short Answer Type 2

Question 1.
What is evaporation? Why does evaporation cause cooling?
Answer:
The process in which a liquid changes into its vapour state at a temperature below the boiling point is called evaporation. Evaporation is an endothermic process i.e., the liquid absorbs heat during evaporation. This heat may be provided either by the surrounding or by the liquid itself. When the evaporating liquid takes the required heat from other parts of the liquid, the rest of the liquid cools down.

On the other hand, if the liquid takes heat from the surroundings, it causes cooling of the surroundings. For example, on a hot day (sunny day) we perspire. When this sweat evaporates, it absorbs the required heat from our body, and we feel cool.

Question 2.
What factors affect the rate of evaporation?
Answer:
Factors that affect the rate of evaporation are:

• Temperature: Evaporation increases with increase in temperature.
• Humidity: Evaporation decreases with an increase in humidity.
• Wind speed: Evaporation increases with an increase in wind speed.

Question 3.
What is a dry ice and what are its properties?
Answer:
Solid carbon dioxide is known as dry ice. It is stored under high pressure. Solid CO₂ gets converted directly to gaseous state on decrease of pressure to 1 atmosphere without passing through the liquid state (i.e., sublimes). This is the reason that solid carbon dioxide is also known as dry ice.

It is mainly used as a cooling agent because its temperature is very low than ice formed from water. Dry ice is commonly used in theaters and in movies to produce the effect of fog.

Question 4.
Why should we wear cotton clothes in summer?
Answer:
During summer, we perspire more because of the mechanism of our body which keeps us cool. We know that during evaporation, the particles at the surface of the liquid gain energy from the surroundings or body surface and change into vapour. The heat energy equal to latent heat of vapourisation is absorbed from the body leaving the body cool. Cotton, being a good absorber of water helps in absorbing the sweat and exposing it to the atmosphere for easy evaporation.

Question 5.
Give the main postulates of kinetic theory of matter.
Answer:
The main postulates of kinetic theory are:

• All matter is made up of a large number of extremely small particle called molecules.
• The molecules are always in a state of rapid random motion.
• The molecules possess kinetic energy.
• There are attractive forces between the molecules.
• The kinetic energy of molecules increases with increase in temperature.

Question 6.
Identify each of the following changes of state as evaporation, boiling or condensation. Give reason for your answer.
(a) Wet clothes dry when spread on wire.
(b) After a hot shower, your bathroom mirror is covered with water.
(c) Lava flows into the ocean and forms steam.
Answer:
(a) Evaporation, because conversion of liquid water to vapour occur at room temperature.
(b) Condensation, because hot water vapour condense to form liquid water.
(c) Boiling, because heat of lava makes liquid water boil and hence steam is formed.

Question 7.
Why do surgeons often spray some ether on the skin before performing minor surgery?
Answer:
Quite often doctors spray ether on a portion of skin the before performing minor surgery. The reason being that a ether has very low boiling point (308 K). Therefore, it evaporates quite rapidly. The heat energy needed for evaporation is taken from the skin. As a result, the temperature of the skin becomes so low that it almost becomes numb. Due to this numbness, the patient does not feel much pain when a minor cut is made in the skin in order to perform surgery.

Similarly, when a player gets injured during a game, ethyl chloride on the injured portion of the body. Since the boiling point of ethyl chloiide (285.5K or 12.5°C) is very low, it quickly evaporates. The heat energy needed for evaporation is taken from the skin. By losing heat, temperature of the skin becomes so low that it almost becomes numb. Due to this numbness, the player does not feel much pain.

Question 8.
Comment upon the following:
(i) Rigidity
(ii) Compressibility
(iii) Fluidity
Answer:
(i) Rigidity means tendency to maintain shape when some outside force is applied due to strong interparticle force.

(ii) Compressibility means tendency to decrease volume when some outside force is applied. Due to large interparticle distances in gases their volume decreases when some pressure is applied on them therefore, gases possess high compressibility.

(iii) Fluidity means tendency to flow. Due to large interparticle distances and weak forces of attraction gases have highest fluidity.

Question 9.
Comment on the following statements:
(а) Evaporation produces cooling.
(b) Rate of evaporation of an aqueous solution decreases with increase in humidity.
(c) Sponge though compressible is a solid. [NCERT Exemplar]
Answer:
(а) For evaporation to occur, heat energy is needed. This heat energy is taken out from the substance or the surroundings. As a result surrounding becomes cool. Thus, evaporation causes cooling.

(b) By humidity we mean, the amount of water vapours present in the air. With increase in humidity the rate of evaporation decreases. If the humidity of air is already high, it can hold only a little more amount of water vapour to reach that optimum level, therefore the rate of evaporation decreases.

(c) Sponge has large number of minute holes in which air is trapped. When we press it, air expelled and sponge is compressed to a small amount of matter which has a definite shape as well as definite volume.

Matter in Our Surroundings Class 9 Extra Questions Long Answer Type

Question 1.
List any five physical properties of liquids.
Answer:

• Liquids do not have fixed shape or boundaries.
• They have fixed volume.
• They exhibit fluidity i.e., they can flow.
• Less compressible as compared to gases but higher than solids.
• Lower density as compared to solids.
• Compared to solids, liquids have higher kinetic energy but less than gases.
• The intermolecular forces of attraction are weaker than those of solids.
• Show the property of intermixing i.e., can diffuse.

Question 2.
Fill in the blanks:
(а) Evaporation of a liquid at room temperature leads to a ………………… effect.
(b) At room temperature the forces of attraction between the particles of solid substances are ………………… than those which exist in the gaseous state.
(c) The arrangement of particles is less ordered in the state. However, there is no order in ………………… the state.
(d) is the change of gaseous state directly to solid state without going through the ………………… state.
(e) The phenomenon of change of a liquid into the gaseous state at any temperature below its boiling point is called ………………… [NCERT Exemplar]
Answer:
(a) cooling
(b) stronger
(c) liquid, gaseous
(d) Sublimation, liquid
(e) evaporation

Question 3.
Classify the following into osmosis/diffusion [NCERT Exemplar]
(a) Swelling up of a raisin on keeping in water.
(b) Spreading of virus on sneezing.
(c) Earthworm dying on coming in contact with common salt.
(d) Shrinking of grapes kept in thick sugar syrup.
(e) Preserving pickles in salt.
Answer:
(a) Osmosis
(b) Diffusion
(c) Osmosis
(d) Osmosis
(e) Osmosis

Question 4.
You are provided with a mixture of naphthalene and ammonium chloride by your teacher. Suggest an activity to separate them with well labelled diagram. [NCERT Exemplar]
Answer:
Naphthalene is insoluble in water but soluble in ether an organic solvent. It is volatile at room temperature. Ammonium chloride is soluble in water and volatile at higher temperature. It decomposes on heating to dryness.

Question 5.
Why does the temperature of a substance remain constant during its melting point or boiling point?
Answer:
The temperature of a substance remains constant at its melting and boiling points untill all the substance melts or boils because, the heat supplied is continuously used up in changing the state of the substance by overcoming the forces of attraction between the particles. This heat energy absorbed without showing any rise in temperature is given the name latent heat of fusion/latent heat of vapourisation.

Matter in Our Surroundings Class 9 Extra Questions HOTS

Question 1.
Arrange the following substances in increasing order of force of attraction between the particles.
(a) Milk
(b) Salt
(c) Oxygen
Answer:
Oxygen < Milk < Salt.

Question 2.
Explain with an experiment to show gases do not have fixed shape or volume.
Answer:
Experiment: To show gases do not have fixed shape or volume.

Method:

• Take two balloons of different shapes. For example, one round and one heart shape or cylindrical.
• Fill the balloons with air.

Observation: Air takes up the shape of balloon.

Conclusion: This shows air has no definite shape or volume. It takes up the shape of the balloon.

Question 3.
Name the change of state during the following changes:
(a) Drying of wet clothes
(b) Melting of wax
(c) Melting of ice
(d) Formation of clouds
Answer:
(a) Liquid to gaseous state
(b) Solid to liquid state
(c) Solid to liquid state
(d) Liquid to gaseous state

Question 4.
With proper explanation, explain whether the following statements are true or false?
(a) Sublimation occurs only when the solid is heated.
(b) A lighter gas can move downwards and a heavier gas can move upwards.
(c) Interconversion of matter is a constant temperature process.
Answer:
(a) Statement is wrong. Sublimation may occur on its own or by heating, e.g., camphor, naphthalene, iodine, etc., sublime slowly at room temperature.

(b) Statement is true. Diffusion occurs against the law of gravitation. Therefore, lighter gases can also diffuse downwards and the heavier gases can also diffuse upwards. However rate of diffusion of lighter gases is faster than those of heavier gases.

(c) Statement is true. During interconversion of state of matter from solid to liquid or from liquid to gas, it tends to reach its melting point or boiling point. At this point, the temperature remains constant unit it has changed in another state.

Question 5.
What is meant by Bose-Einstein Condense?
Answer:
(a) In 1920, Indian scientist Satyendra Nath Bose did some calculations, based on which Albert Einstein predicted that a new state of matter should exist.

(b) This new state was named as Bose-Einstein Condensate (BEC). In 2001, Cornell, Ketterie and Wieman of USA received Noble Prize for actually making this state in laboratory. BEC is made by cooling gas of very low density to super low temperature.

Question 6.
A student heats a beaker containing ice and water. He measures the temperature of the content of the beaker as a function of time. Which of the following figure would correctly represent the result? Justify your choice. [NCERT Exemplar]

Answer:
Since ice and water are in equilibrium, the temperature would be zero. When we heat the mixture, energy supplied is utilised in melting the ice and the temperature does not change till all the ice melts because of latent heat of fusion. On further heating, the temperature of the water would increase. Therefore the correct option is (d).

Question 7.
Look at the figure and suggest in which of the vessels A, B, C or D the rate of evaporation will be the highest? Explain. [NCERT Exemplar]
Answer:

(c) The rate of evaporation increases with an increase in the surface area of absorption because evapo-ration in a surface phenomenon. Also, with the increase in air speed, the particles of water vapour will move away with the air, which will increase the rate of evaporation.

Matter in Our Surroundings Class 9 Extra Questions Value Based (VBQs)

Question 1.
In a hot summer day, Rajeev wants to watch a movie in a nearest cineplex. His mother wears a nylon saree so Rajeev suggest her to wear cotton saree.
(i) Why does Rajeev suggest her to wear cotton saree and not nylon saree?
(ii) Mention the values exhibited by Rajeev.
Answer:
(i) On a hot summer day, we sweat a lot. Cotton clothes absorb sweat from the body. As the sweat evaporates it results in cooling giving comfort to the body.

(ii) Caring, use of knowledge of science.

Question 2.
Mohan was getting late for school. He tried to sip tea from the cup. His father advised him to use a plate and asked him to sip the tea from the plate. Mohan followed the advice and finished his tea very easily.
(i) Why is siping off tea easier from a plate?
(ii) Mention the values exhibited by Mohan’s father.
Answer:
(i) A plate has a larger surface area than a cup. Evaporation becomes faster in this case. Since cooling is always caused during evaporation the temperature got lowered. Therefore it became easier to sip tea from a plate.

(ii) Caring, use of knowledge of science, helpful.

Here we are providing Statistics Class 9 Extra Questions Maths Chapter 14 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Statistics with Answers Solutions

Extra Questions for Class 9 Maths Chapter 14 Statistics with Solutions Answers

Statistics Class 9 Extra Questions Very Short Answer Type

Statistics Class 9 Extra Questions Question 1.
The points scored by a basketball team in a series of matches are follows:
17, 7, 10, 25, 5, 10, 18, 10 and 24. Find the range.
Solution:
Here, maximum points = 25 and
minimum points = 5
Range = Maximum value – Minimum value
= 25 – 5 = 20

Class 9 Statistics Extra Questions Question 2.
The points scored by a basketball team in a series of matches are as follows:
17, 2, 7, 27, 25, 5, 14, 18, 10. Find the median.
Solution:
Here, points scored in ascending order are 2, 5, 7, 10, 14, 17, 18, 25, 27, we have n = 9 terms

Statistics Class 9 Extra Questions Pdf Question 3.
The scores of an English test (out of 100) of 20 students are given below :
75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. .:: Find the median and mode of the data.
Solution:
The scores of an English test (out of 100) in ascending order are
44, 55, 59, 64, 67, 69, 73, 75, 75, 88, 88, 88, 88, 88, 90, 95, 95, 95, 98, 99
Here, n = 20

= Mean of 10^th and 11^th term Median
= Mean of 88 and 88 = 88
Mode = 88 [∵ 88 occured max. no. of times i.e., 5 times]

Statistics Extra Questions Class 9 Question 4.
Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean.
Solution:
Since mean of 20 observations is 17
∴ Sum of the 20 observations = 17 × 20 = 340
New sum of 20 observations = 340 – 40 + 12 = 312
New mean = \(\frac{312}{20}\) = 15.6

Extra Questions On Statistics Class 9 Question 5.
Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct
mean.
Solution:
Mean of 36 observations = 12
Total of 36 observations = 36 × 12 = 432
Correct sum of 36 observations = 432 – 74 + 47 = 405
Correct mean of 36 observations = \(\frac{405}{36}\) = 11.25

Class 9 Maths Chapter 14 Extra Questions Question 6.
The median of the data 26, 56, 32, 33, 60, 17, 34, 29, 45 is 33. If 26 is replaced by 62, then
find the new median.
Solution:
Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62

Hence, new median is 34.

Statistics Class 9 Extra Questions With Solutions Question 7.
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers.
Solution:
Let x be the mean of 50 numbers.
i. Sum of 50 numbers = 50x
Since each number is subtracted from 53.
According to question, we have

Class 9 Maths Statistics Extra Questions Question 8.
To draw a histogram to represent the following frequency distribution :

Find the adjusted frequency for the class 25-45.
Solution:
Adjusted frequency of a class

Statistics Class 9 Extra Questions Short Answer Type 1

Extra Questions Of Statistics Class 9 Question 1.
For a particular year, following is the distribution of ages (in years) of primary school teachers in a district :

(i) Write the lower limit of first class interval.
(ii) Determine the class limits of the fourth class interval.
(iii) Find the class mark of the class 45 – 50.
(iv) Determine the class size.
Solution:
(i) First class interval is 15 – 20 and its lower limit is 15.
(ii) Fourth class interval is 30 – 35
Lower limit is 30 and upper limit is 35.
(iii) Class mark of the class 45 – 50 = \(\frac{45+50}{2}\) = \(\frac{95}{2}\) = 47.5
(iv) Class size = Upper limit of each class interval – Lower limit of each class interval
∴ Here, class size = 20 – 15 = 5

Statistics Class 9 Important Questions Question 2.
Find the mean of the following distribution :

Solution:

Class 9 Maths Ch 14 Extra Questions Question 3.
In figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table.

Solution:

Questions On Statistics Class 9 Question 4.
Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending
order. The median of the data is 24. Find the value of x.
Solution:
Here, the arranged data is 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43
Total number of observations = 10

But median of data is 24 (given)
⇒ \(\frac{3 x-12}{2}\) = 24
⇒ 3x – 12 = 48
⇒ 3x = 60
⇒ x = 20
∴ The value of x = 20

Statistics Class 9 Extra Questions Short Answer Type 2

Chapter 14 Maths Class 9 Extra Questions Question 1.
Draw a histogram for the given data :

Solution:
Let us represent class-intervals along x-axis and corresponding frequencies along y-axis on
a suitable scale, the required histogram is as under :

Question 2.
Given are the scores (out of 25) of 9 students in a Monday test :
14, 25, 17, 22, 20, 19, 10, 8 and 23
Find the mean score and median score of the data.
Solution:
Ascending order of scores is :
8, 10, 14, 17, 19, 20, 22, 23, 25

Question 3.
Draw a histogram of the weekly pocket expenses of 125 students of a school given below :

Solution:
Here, the class sizes are different, so calculate the adjusted frequency for each class by using the formula.
Minimum class size Frequency density or adjusted frequency for a class

Here, the minimum class size = 10 – 0 = 10

Let us represent weekly pocket money along x-axis and corresponding adjusted frequencies along y-axis on a suitable scale, the required histogram is as given below :

Question 4.
The weight in grams of 35 mangoes picked at random from a consignment are as follows:
131, 113, 82, 75, 204, 81, 84, 118, 104, 110, 80, 107, 111, 141, 136, 123, 90, 78, 90, 115, 110, 98, 106, 99, 107, 84, 76, 186, 82, 100, 109, 128, 115, 107, 115 From the grouped frequency table by dividing the variable range into interval of equal width of 20 grams, such that the mid-value of the first class interval is 70 g. Also, draw a histogram.
Solution:
It is given that the size of each class interval = 20 and the mid-value of the first class interval is 70.
Let the lower limit of the first class interval be a, then its upper limit = a + 20.
\(\frac{a+(a+20)}{2}\) = 70
⇒ a = 70 – 10 = 60
Thus, the first class interval is 60 – 80 and the other class-intervals are 80 – 100, 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200 and 200 – 220.
So, the grouped frequency table is as under :


Let us represent weight (in g) along x-axis and corresponding frequencies along y-axis on a suitable scale, the required histogram is as under :

Statistics Class 9 Extra Questions Long Answer Type

Question 1.
Find the mean salary of 60 workers of a factory from the following table :

Solution:

Hence, mean salary of 60 workers is ₹5083.33.

Question 2.
In a school marks obtained by 80 students are given in the table. Draw a histogram. Also,
make frequency polygon.

Solution:
∴ Lower limit of first class interval is 305 – \(\frac{10}{2}\) = 300
Upper limit of first class interval is 305 + \(\frac{10}{2}\) = 310
Thus, first class interval is 300 – 310

Required histogram and frequency polygon is given on the graph paper.

Question 3.
The following two tables gives the distribution of students of two sections according to the marks obtained by them :

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
The class marks are as under :

Let us take class marks on X-axis and frequencies on Y-axis. To plot frequency polygon of Section-A, we plot the points (5, 3), (15, 9), (25,17), (35,12), (45, 9) and join these points by (15,19). line segments. To plot frequency polygon of Section-B, we plot the points (5, 5), (15, 19), (25, 15), (35, 10), (45, 1) on the same scale and join these points by dotted line segments.

From the above two polygons, clearly the performance of Section A is better.

Statistics Class 9 Extra Questions HOTS

Question 1.
The mean weight of 60 students of a class is 52.75 kg. If mean weight of 25 students of this class is 51 kg, find the mean weight of remaining 35 students of the class.
Solution:
Total weight of 60 students = 60 × 52.75 kg = 3165 kg
Total weight of 25 students = 25 × 51 kg = 1275 kg
∴ Total weight of 35 students = (3165 – 1275) kg = 1890 kg
∴ Mean weight of 35 students = \(\frac{1890}{35}\) = 54 kg

Question 2.
Find the missing frequencies in the following frequency distribution. If it is known that the mean of the distribution is 50.16 and the total number of items is 125.

Solution:
Since total number of items = 125
∴ 17 + f₁ + 32 + f₁ + 19 = 125
f₁ + f₂ = 125 – 17 – 32 – 19
f₁ + f₂ = 57 …..(i)
Now, mean of data = 50.16
We know that

Statistics Class 9 Extra Questions Value Based (VBQs)

Question 1.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of the Indian society is given below :

(i) Represent the information above by a bar graph.
(ii) In the classroom, discuss what conclusions can be arrived at from the graph.
(iii) What steps should be taken to improve the situation ?
Solution:
(i) The required graph is given below :

In the graph, different sections of the society is taken on X-axis and number of girls per thousand boys is taken on the Y-axis.
[Scale : 1 cm = 10 girls.]
(ii) From the graph, the number of girls to the nearest ten per thousand boys are maximum in scheduled tribes whereas they are minimum in urban.
(iii) Pre-natal sex determination should strictly banned in urban.

Question 2.
Shimpi, a class IX student received cash award of 10000 (Ten thousand) in the singing competition. Her father advised her to make a budget plan for spending this amount. She
made the following plan :

Make a bar graph for the above data.
From above answer the following questions :
(i) Which mathematical concepts have been covered in this ?
(ii) How will you rate her budget plan ? In your opinion which head has been given
(a) more than deserved and (b) less than it deserved ?
(iii) Which values are depicted in her plan?
Solution:
The bar graph of given data is given below :

In the graph, head is taken on X-axis and amount is taken on Y-axis.
(i) Representation of data using bar graph.
(ii) Very good
(a) Picnic for family
(b) Tuition fee for needy child
(iii) Help the needy people and respect the elders.