Category: CBSE Class 12

NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases

Multiple Choice Questions

Question 1.
The term ‘Health’ is defined in many ways. The most accurate definition of the health would be
(a) health is the state of body and mind in a balanced condition
(b) health is the reflection of a smiling face
(c) health is a state of complete physical, mental and social well-being
(d) health is the symbol of economic prosperity.
Answer:
(c) : Health could be defined as a state of complete physical, mental and social well¬being. When people are healthy, they are more efficient at work. This increases productivity and brings economic prosperity. Health also increases longevity of people and reduces infant and maternal mortality.

Question 2.
The organisms which cause diseases in plants and animals are called
(a) pathogens
(b) vectors
(c) insects
(d) worms.
Answer:
(a) : A wide range of organisms belonging to bacteria, viruses, fungi, protozoans, helminths, etc. could cause diseases in pants and animals including humans. Such disease causing organisms are called pathogens. Most parasites are therefore pathogens as they cause harm to the host by Jiving in (or on) them. The pathogens can enter our body by various means, multiply and interfere with normal vital activities, resulting in morphological and functional damage.

Question 3.
The chemical test that is used for diagnosis of typhoid is
(a) ELISA-Test
(b) ESR – Test
(c) PCR – Test
(d) Widal-Test.
Answer:
(d) :Salmonella typhi is a pathogenic bacterium which causes typhoid fever in human beings. These pathogens generally enter the small intestine through food and water contaminated with them and migrate to other organs through blood. Sustained high fever (39 to 40°C), weakness, stomach pain, constipation, headache and loss of appetite are some of the common symptoms of the disease. Intestinal perforation and death many occur in severe cases. Typhoid fever could be confirmed by Widal test. Widal test is an agglutination test for the presence of antibodies against the Salmonella organisms that cause typhoid fever.

Question 4.
Diseases are broadly grouped into infectious and non-infectious diseases. In the list given below, identify the infectious diseases.
(1) Cancer
(2) Influenza
(3) Allergy
(4) Small pox
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iv)
Answer:
(d) : Diseases can be broadly grouped into infectious and non-infectious. Diseases which are easily transmitted from one person to another, are called infectious diseases. Influenza and small pox are infectious diseases which are spreaded by direct contact, inhalation and droplet infections. Cancer and allergy are non-communicable diseases, as these diseases remain confined to the person who suffer from them. They are not transmitted from infected person to a healthy person.

Question 5.
The sporozoites that cause infection, when a female Anopheles mosquito bites a person, are formed in
(a) liver of the person
(b) RBCs of mosquito
(c) salivary glands of mosquito
(d) intestine of mosquito.
Answer:
(d) : When a female Anopheles mosquito bites an infected person, gametocyte of Plasmodium enter the mosquito’s gut along with blood meal and fuse to form zygote which elongates and becomes motile, called ookinets. Ookinetes bore through the wall of gut and change to oocysts. Inside oocyst, sporozoites are formed. Mature infective stage(sporozoite) are released in the body cavity of mosquito and migrate to salivary gland of mosquito. When these mosquitoes bite a human, the sporozoites are introduced into his/her body, thereby initiating the disease in the host’s body.

Question 6.
The disease chikungunya is transmitted by
(a) houseflies
(b) Aedes mosquitoes
(c) cockroach
(d) female Anopheles.
Answer:
(b) : Chikungunya is caused by chikungunya virus. The virus was first isolated from human patients and Aedes aegypti mosquitoes from Tanzania in 1952. The disease is transmitted by the bite of Aedes aegypti mosquito.

Question 7.
Many diseases can be diagnosed by observing the symptoms in the patient. Which group of symptoms are indicative of pneumonia?
(a) Difficulty in respiration, fever, chills, cough, headache.
(b) Constipation, abdominal pain, cramps, blood clots.
(c) Nasal congestion and discharge, cough, sore throat, headache.
(d) High fever, weakness, stomach pain, loss of appetite and constipation.
Answer:
(a) : Bacteria like Streptococcus pneumoniae and Haemophilus influenzae are responsible for the disease pneumonia in humans which infects the alveoli (air filled sacs) of the lungs. As a result of the infection, the alveoli get filled with fluid leading to severe problems in respiration. The symptoms of pneumonia include fever, chills, cough and headache.

Question 8.
The genes causing cancer are
(a) structural genes
(b) expressor genes
(c) oncogenes
(d) regulatory genes.
Answer:
(c) : Cancer is one of the most serious medical problems caused by the abnormal cell growth and proliferation due to the loss of regulation. Cancer-causing genes are oncogenes. Oncogene results from mutation of a normal gene. It is capable of both initiation and continuation of malignant transformation of normal cells.

Question 9.
In malignant tumours, the cells proliferate, grow rapidly and move to other parts of the body to form new tumours. This stage of disease is called
(a) metagenesis
(b) metastasis
(c) teratogenesis
(d) mitosis.
Answer:
(b) : There are two major types of tumours with respect to overall form or growth pattern. If the tumour cells remain in place to form a compact mass, the tumour is benign. In contrast, cells from malignant or cancerous tumours can actively spread, throughout the body in a process known as metastasis, often by floating in the blood and establishing secondary tumours.

Question 10.
When an apparently healthy person is diagnosed as unhealthy by a psychiatrist, the reason could be that
(a) the patient was not efficient at his work
(b) the patient was not economically pros-perous
(c) the patient shows behavioural and social maladjustment
(d) he does not take interest in sports.
Answer:
(c) : If a patient shows behavioural and social maladjustment then, he cannot be considered as healthy. Health is the condition in which the organism (and all of its parts) performs its vital functions normally. It is a state of physical, social and mental well-being and not merely the absence of disease.

Question 11.
Which of the following are the reason(s) for rheumatoid arthritis? Choose the correct option.
(i) The ability to differentiate pathogens or foreign molecules from self cells increases.
(ii) Body attacks self cells.
(iii) More antibodies are produced in the body.
(iv) The ability to differentiate pathogens or foreign molecules from self cells is lost.
(a) (i)and(ii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(iv) (d) (i) and (iii)
Answer:
(b) : Rheumatoid arthritis is a disorder of the body’s defence mechanisms in which an immune response is elicited against its own tissues, which are thereby damaged or destroyed i.e., autoimmune disease.
Rheumatoid arthritis is the second most common form of arthritis (after osteoarthritis). It typically involves the joints of the fingers,
wrists, feet, and ankles, with later involvement of the hips, knees, shoulders, and neck. It is a disease of the synovial lining of joints; the joints are initially painful, swollen, and stiff and are usually affected symmetrically. As the disease progresses the ligaments supporting the joints are damaged and there is erosion of the bone, leading to deformity of the joints. Tendon sheath can be affected, leading to tendon rupture.

Question 12.
AIDS is caused by HIV. Among the following, which one is not a mode of transmission of HIV?
(a) Transfusion of contaminated blood.
(b) Sharing the infected needles.
(c) Shaking hands with infected persons.
(d) Sexual contact with infected persons.
Answer:
(c) : AIDS is a result of an infection by the human immunodeficiency virus (HIV) :
HIV can be transmitted by the following ways.

• Transfusion of infected blood.
• Use of contaminated needles and syringes to inject drugs or vaccines.
• Use of contaminated razors.
• Use of contaminated needles for boring pinnae.
• Sexual intercourse with an infected partner without a condom.
• From infected mother to child through placenta.
• Artificial insemination.
• Organ transplantation.

Question 13.
‘Smack’is a drug obtained from the
(a) latex of Papaver somniferum
(b) leaves of Cannabis sativa
(c) flowers of Datura
(d) fruits of Erythroxylum coca.
Answer:
(a) : Heroin commonly called smack is chemically diacetylmorphine which is a white, odourless, bitter crystalline compound. This is obtained by acetylation of morphine, which is extracted from the latex of poppy plant (Papaver somniferum).

Question 14.
The substance produced by a cell in viral infection that can protect other cells from further infection is
(a) serotonin
(b) colostrum
(c) interferon
(d) histamine.
Answer:
(c) : Interferons are glycoproteins released by living cells in response to viral attack, and make the surrounding cells resistant to viral infection by inhibiting multiplication of viral particles.
Interferons are divided into three groups based on the cell of origin, namely leucocyte (alpha interferon), fibroblast (beta interferon) and lymphocyte (gamma interferon). They are cytokine (chemical mediators) barriers.

Question 15.
Transplantation of tissues/organs to save certain patients often fails due to rejection of such tissues/organs by the patient. Which type of immune response is responsible for such rejections?
(a) Auto immune response
(b) Humoral immune response
(c) Physiological immune response
(d) Cell-mediated immune response
Answer:
(d) : Cell mediated immune response is responsible for graft rejection. Tissue matching and blood group matching are essential before undertaking any graft/transplant and even after this the patient has to take immuno¬suppressants throughout his/her life as body is able to differentiate ‘self’ from ‘non-self’.

Question 16.
Antibodies present in colostrum which protect the new born from certain diseases is of
(a) IgGtype
(b) IgA type
(c) IgDtype
(d) IgEtype.
Answer:
(b) : IgA antibody is the major antibody present in the colostrum. It protects the infant from inhaled and ingested pathogens. It is the second most abundant class of immunoglobulins, constituting about 13% of serum immunoglobulins.

Question 17.
Tobacco consumption is known to stimulate secretion of adrenaline and nor-adrenaline. The component causing this could be
(a) nicotine
(b) tannic acid
(c) curaimin
(d) catechin.
Answer:
(a) : Tobacco contains a large number of chemical substances including nicotine, an alkaloid. Nicotine stimulates adrenal gland to release adrenaline and nor-adrenaline into blood circulation, both of which raise blood pressure and increase heart rate.

Question 18.
Anti venom against snake poison contains
(a) antigens
(b) antigen-antibody complexes
(c) antibodies
(d) enzymes.
Answer:
(c) : In case of snakebite, the injection which is given to a patient, contains preformed antibodies against the snakevenom. This type of immunisation is called passive immunisation.

Question 19.
Which of the following is not a lymphoid tissue?
(a) Spleen
(b) Tonsils
(c) Pancreas
(d) Thymus
Answer:
(c) : Lymphoid organs are the organs where origin and/or maturation
and proliferation of lymphocytes occur. The primary lymphoid organs are bone marrow and thymus, where immature lymphocytes differentiate into antigen – sensitive lymphocytes. After maturation the lymphocytes migrate to secondary lymphoid organs like spleen, lymph nodes, tonsils, Peyer’s patches of small intestine and mucosa associated lymphoid tissue (MALT). The secondary lymphoid organs provide the sites for interaction of lymphocytes with the antigen, which then proliferate to become effector cells. Pancreas is not a lymphoid tissue.

Question 20.
Which of the following glands is large sized at birth but reduces in size with ageing?
(a) Pineal
(b) Pituitary
(c) Thymus
(d) Thyroid _____
Answer:
(c) : The thymus is a lobed organ located near the heart and beneath the breastbone. The thymus is quite large at the time of birth but keeps reducing in size with age and by the time puberty is attained, it reduces to a very small size.

Question 21.
Haemozoin is a
(a) precursor of haemoglobin
(b) toxin released from Streptococcus infected cells.
(c) toxin released from Plasmodium infected cells.
(d) toxin released from Haemophilus infected ‘cells.
Answer:
(c) : Plasmodium enters the human body as sporozoites (infectious form) through the bite of infected female Anopheles mosquito. The parasites initially multiply within the liver cells and then attack the red blood cell (RBCs) resulting in their rupture. The rupture of RBCs is associated with release of a toxic substance, haemozoin, which is responsible for the chill and high fever recurring every three to four days.

Question 22.
One of the following is not the causal organism for ringworm.
(a) Microsporum
(b) Trichophyton
(c) Epidermophyton
(d) Macrosporum
Answer:
(d) : Many fungi belonging to the genera Microsporum, Trichophyton and Epidermophyton are responsible for ringworms which is one of the most common infectious diseases in man. Appearance of dry, scaly lesions on various parts of the body such as skin, nails and scalp are the main symptoms of the disease.

Question 23.
A person with sickle cell anaemia is
(a) more prone to malaria
(b) more prone to typhoid
(c) less prone to malaria
(d) less prone to typhoid.
Answer:
(c) : Sickle cell trait protects against malaria. Malarial parasite is unable to penetrate the erythrocyte membrane of sickle shaped erythrocytes, reducing the number of parasites that actually infect the host thus conferring some protection against the disease.

Very Short Answer Type Questions

Question 1.
Certain pathogens are tissue/organ specific. Justify the statement with suitable examples.
Answer:
Some pathogens attack specific organs or tissues, such as :

1. Vibrio cholerae – It is a bacterium which causes cholera. It attacks digestive tracts and intestine and spreads through contaminated food and drinks.
2. Streptococcus pneumoniae – It is a bacterium which causes pneumonia. It is a serious disease of lungs characterised by
   accumulation of mucus/fluid in alveoli and bronchioles hence, breathing becomes difficult.

Question 2.
The immune system of a person is suppressed. In the ELISA test, he was found positive to a pathogen.
(a) Name the disease the patient is suffering from.
(b) What is the causative organism?
(c) Which cells of body are affected by the pathogen?
Answer:
(a) The patient must be suffering from Acquired Immune Deficiency Syndrome (AIDS). .
(b) The causative agent is Human Immuno-deficiency Virus (HIV).
(c) Virus affects helper T cells of the body.

Question 3.
Where are B-cells and T-cells formed? How do they differ from each other?
Answer:
B-cells and T-cells are lymphocytes which are responsible for immune response in the body. Both B-cells and T-cells are formed in bone marrow.
Differences between B-cells and T-cells are

B-cells	T-cells
(1) Bone marrow is the site of both formation and maturation.	They are produced in bone marrow but mature in thymus.
(2) B-cells are responsible for humoral or antibody mediated immunity.	T-cells are responsible for cell mediated immunity.

Question 4.
Given below are the pairs of pathogens and the diseases caused by them. Which out of these is not a matching pair and why?
(a) Virus common cold
(b) Salmonella typhoid
(c) Microsporum filariasis
(d) Plasmodium malaria
Answer:
(c) : Microsporum is a fungus and causes ringworm. It attacks the hair and skin.Filariasis is caused by Wuchereria bancrofti, a helmirith, which results in swelling of limbs, scrotal sacs and breasts.

Question 5.
What would happen to immune system, if thymus gland is removed from the body of a person?
Answer:
T-lymphocytes get differentiated in thymus gland and are responsible for cell mediated immunity (CMI). If thymus gland is removed, then CMI will weaken and person will become more susceptible to infectious diseases.

Question 6.
Many microbial pathogens enter the gut of humans along with food. What are the preventive barriers to protect the body from such pathogens? What type of immunity do you observe in this case?
Answer:
Physiological barriers of innate immunity protect the body against the pathogens which enter the gut along with food. These barriers are as follows:

1. Saliva contains lysozyme which kills the microorganisms that are not the normal inhabitants of the buccal cavity and come with food and drinks.
2. Bile, a bitter alkaline secretion of the liver, checks the growth of foreign bacteria on semidigested food, the chyme, in the intestine.
3. Acidity of gastric juice kills most of the microorganisms entering the body through digestive tract.

Question 7.
Why is mother’s milk considered the most appropriate food for a new born infant?
Answer:
Mother’s milk (colostrum) contains all the necessary nutrients needed by the newborn and has abundant antibodies (IgA) which provide passive immunity and protect the infant for few months after birth.

Question 8.
What are interferons? How do interferons check infection of new cells?
Answer:
Interferons are cytokine barriers of innate immunity. These are proteins secreted by virus infected cells which protect non- infected cells from further viral infections.

Question 9.
In the figure, structure of an antibody molecule is shown. Name the parts A, B and C

Answer:
A – Variable region of heayy chain
B – Constant region of light chain .
C – Disulphide bond

Question 10.
If a regular dose of drug or alcohol is not provided to an addicted person, he shows some withdrawal symptoms. List any four such withdrawal symptoms.
Answer:
Withdrawal symptoms shown by an addicted person when not provided with drug or alcohol includes anxiety, nausea, muscle twitch, sweating, nervousness and epilepsy etc.

Question 11.
Why is it that during changing weather, one is advised to avoid closed, crowded and airconditioned places like cinema halls etc.?
Answer:
It is advised to avoid overcrowded places during changing weather because chances of getting infections are very high during this time. This is because changing seasons are the time when infectious agents are more prevalent as moist condition favours pathogen growth. Moreover, people are more vulnerable as their body system is busy in adapting to the changing environmental conditions.

Question 12.
The harmful allele of sickle cell anaemia has not been eliminated from human population. Such afflicted people derive some other benefit. Discuss.
Answer:
Sickle cell anaemia is caused due to a recessive allele in haemoglobin. It is lethal
in homozygous (HbsHbs) condition and the person dies in early age but in heterozygous condition (HbAHbs) person is carrier of disease. Despite having harmful effect, the allele for sickle-cell anaemia continues to persist in human population because it has survival value in malaria infested areas like tropical Africa. Heterozygous individuals show resistance to malarial infection, because malarial parasite is unable to penetrate the membrane of sickle shaped erythrocytes.

Question 13.
Lymph nodes are secondary lymphoid organs. Explain the role of lymph nodes in our immune response.
Answer:
The lymph nodes are small solid structures located at different points along the lymphatic system. Lymph nodes serve to trap the microorganisms or other antigens, which happen to get into the lymph and tissue fluid. Antigens trapped in the lymph nodes are responsible for the activation of lymphocytes present there and cause the immune response.

Question 14.
Why is an antibody molecule represented as H2L2?
Answer:
Antibody molecule is made up of four polypeptide chains-2 Heavy (H) and 2 Light (L), therefore it is represented as H:L2.

Question 15.
What does the term ‘memory’ of the immune system mean?
Answer:
Memory response is a feature of acquired immunity. It develops during first encounter between specific foreign agent and body’s immune system. Second encounter with the same pathogen generates quicker and heightened immune response due to activated memory cells.

Question 16.
lf a patient is advised anti retroviral therapy, which infection is he suffering from? Name the causative organism.
Answer:
The patient must be suffering from Required Immune Deficiency Syndrome (AIDS) caused by Human Immunodeficiency Virus (HIV), which is a member of retrovirus group

short Answer Type Questions

Question 1.
Differentiate between active immunity and passive immunity.
Answer:
Differences between active immunity and passive immunity are as follows:

Active immunity	Passive immunity
(1) It develops in response to vaccine or infection.	It develops when readymade antibodies are injected from outside.
(2) It is long lasting.	It lasts for short period only.
(3) Antibodies are produced within the body.	Antibodies are injected from outside.
(4) It has no side effects.	It may cause reaction.
(5) Immunity is not immediate. A time lapse occurs for its development.	Immunity develops immediately

Question 2.
Differentiate between benign tumour and malignant tumour.
Answer:
Difference between benign tumour and malignant tumour are as follows:

Benign tumour	Malignant
(1) It is non-cancerous tumour	It is cancerous tumour.
(2) It does not show metastasis and remains confined to the affected organ.	It shows metastasis and spreads to other organs of the body.
(3) Rate of growth is slow. It shows definite growth.	Rate of growth is rapid. It shows indefinite growth.

Question 3.
Do you consider passive smoking is more dangerous than active smoking? Why?
Answer:
Passive smoking is inhalation of air containing tobacco smoke exhaled by an active smoker. Passive smoker is at higher risk as compared to an active smoker, because active smoker might inhale 10% of the smoke, but passive smoker inhales 90% of the smoke.

Question 4.
“Prevention is better than cure”. Comment.
Answer:
Prevention is always better than cure.This is because there are some diseases which have detrimental effects on our body and lives. Certain diseases like cancer and AIDS etc., are fatal. They can be avoided if our health systems focus on prevention of such diseases rather than concentrating on treatment methods. Many people can be saved from contracting diseases. Similar conditions apply for communicable, infectious diseases. If preventive measures are undertaken, their spread, can be checked and controlled easily.

Question 5.
Explain any three preventive measures to control microbial infections.
Answer:
Maintenance of personal and public hygiene is important to control microbial infections. Some of the measures that can be undertaken are as follows:

1. Eradication of vectors and destroying their breeding sites
2. Use of mosquito nets and repellents.
3. Proper disposal of waste.
4. Periodic cleaning of water reservoirs.
5. Checking water stagnation and garbage accumulation.
6. Implementation of vaccination and immunisation programmes for diseases.

Question 6.
In the given flow diagram, the replication of retrovirus in a host is shown. Observe and answer the following questions.
a. Fill in (1) and (2).
b. Why is the virus called retrovirus?
c. Can the infected cell survive while viruses are being replicated and released?

Answer:
(a)
(1) – Viral DN A is produced by reverse transcriptase.
(2) – New viral RNA is produced by the infected cell.
(b) HIV is called retrovirus because viral DNA is produced from viral RNA by reverse transcription.
(c) Infected cell survives but T-lymphocytes decrease in number due to replication and release of virus.

Question 7.
“Maintenance of personal and public hygiene is necessary for prevention and control of many infectious diseases”. Justify the statement giving suitable examples.
Answer:
Maintenance of personal and public hygiene is very important for prevention and control of many infectious diseases because diseases spread through contaminated food, water, air and through vectors. Measures for personal hygiene include keeping the body clean, consumption of clean drinking water, food, etc.
Public hygiene includes proper disposal of waste and excreta, periodic cleaning and disinfection of water reservoirs, pools, and tanks and observing standard practices of hygiene in public catering.
In case of air-borne diseases such as pneumonia and common cold, close contact with the infected persons or their belongings should be avoided. Vector borne diseases can be controlled by eliminating the vectors and their breeding places. This can be achieved by avoiding stagnation of water in and around residential areas, regular cleaning of household coolers, use of mosquito nets, etc.

Question 8.
The following table shows certain diseases, their causative organisms and symptoms. Fill the gaps.”

Name of the Disease	Causative organism	Symptoms
(1) Ascariasis	Ascaris	—
(2) —	Trichophyton	Appearance of dry, scaly lesions on various parts of the body
(3) Typhoid		High fever, weakness, headache, stomach pain, constipation
(4) Pneumonia	Streptococcus pneumoniae	—
(5)	Rhino viruses	Nasal congestion and discharge, sorethroat, cough, headache
(6) Filariasis	—	Inflammation in lower limbs

Answer:
(i) Abdominal pain, indigestion, nausea, vomiting etc.
(ii) Ringworm
(iii) Salmonella typhi
(iv) Difficulty in breathing, fever, cough, headache, chills, lips and finger nails may turn grey to bluish in colour
(v) Common cold
(vi) Wuchereria bancrofti

Question 9.
The outline structure of a drug is given below.
a. Which group of drugs does this represent?
b. What are the modes of consumption of these drugs?
c. Name the organ of the body which is affected by consumption of these drugs.

Answer:
(a) Cannabinoids
(b) Cannabinoids are generally taken by inhalation and oral ingestion.
(c) They affect cardiovascular system of the body.

Question 10.
Give the full form of CT and MRI. How are they different from each other?
Answer:
CT (Computed Tomography) is recon-struction of three dimensional image made by X rays directly on a computer instead of a photographic film. It is radiological invasive technique. It is used to diagnose diseases of brain, spinal cord, chest and abdomen and detection of tumours. MRI (Magnetic Resonance Imaging) is a non invasive, non radiological technique in which the image is reconstructed on computer by detecting MRI signals generated by nuclei of hydrogen atoms in a magnetic field using non-ionising radiations. It is used to obtain multiplanar imaging of soft living tissues like in case of tumours, muscular disorders, cardiovascular disorders, haemorrhage with superior resolution.

Question 11.
Many secondary metabolites of plants have medicinal properties. It is their misuse that creates problems. Justify the statement with an example.
Answer:
Drugs like barbiturates, amphetamines, benzodiazepines, lysergic acid diethylamides (LSD) and other similar drugs, that are normally used as medicines to help patients coping with mental illnesses like depression and insomnia are secondary metabolites of plants. Misuse of plant metabolites can impair one’s .physical, physiological or functional behaviour creating problem for the society. For example, Cocaine is natural alkaloid obtained from leaves of plant Erythroxylum coca, native to South America. Cocaine has vasoconstrictor properties and therefore, is a good local anaesthetic. It is chewed, eaten or sniffed in its powdered form or taken as a drink. It is a powerful CNS stimulant (interferes with the transport of the neurotransmitter dopamine) and induces a sense of well being and pleasure and delays fatigue. Its overdose causes hallucinations, headache, insomnia, convulsions, etc.

Question 12.
Why cannabinoids are banned in sports and games?
Answer:
Cannabinoids may enhance the performance of some athletes. Smoking cannabis may decrease anxiety, fear, depression, tension, etc., and improve self confidence, relaxation, well-being and sleep, etc. These effects may allow the athletes to perform better, especially under pressure and may reduce stress, experienced before and during competition. The increase in ‘risk taking’ associated with cannabinoids may possibly improve training and performance giving athlete a competitive edge. Moreover, cannabinoids increase bronchodilation (which may improve oxygenation of tissue) and have analgesic (pain relieving) properties which could permit athletes to work during injury. Moreover, long term use of cannabinoids causes various hazards to the health of abuser. Therefore, use of cannabinoids by sportsperson is not justified, as it will lead to unfair competition. Hence, their use is banned in sports and games.

Question 13.
What is secondary metabolism?
Answer:
Secondary metabolism consists of metabolic pathways and products of metabolism that seem to have no direct function in growth and development of plants and are not absolutely required for their survival. Secondary metabolism is responsible for many bioactive compounds used medicinally.

Question 14.
Drugs and alcohol give short-term ‘high’ and long-term’damages’. Discuss.
Answer:
Curiosity, need for adventure and excitement, peer pressure, etc., constitute some common causes which motivate youngsters towards drug and alcohol abuse. Short term effects of drugs/alcohols are generally pleasing and relaxing which habituate the abuser for using these again and again. But long term effects of drugs or alcohol are devastating and detrimental. For example, alcohol when taken occasionally in small quantities, acts as sedative, analgesic and anaesthetic. It stimulates the secretion of gastric juices. But, if a person becomes addicted to it, he consumes high concentration alcohol frequently. Heavy drinking causes depression. Suicide attempt is much common in alcoholics than in the rest of society.

Sexual relationship is usually deteriorated because of erectile dysfunction or rejection by the partner.
Caffeine is a mild stimulant and taken as beverages- tea, coffee, cocoa and cola drinks. Caffeine is CNS stimulant and thus increases alertness and thought. It also acts as a cardiac and respiratory stimulant. It is a mild diuretic and also increases basal metabolic rate. Excessive use may cause anxiety, irritability, diarrhoea, irregular heart beat (arrhythmia) and decreased concentration. It also causes indigestion and disturbs pancreatic and renal functions. If taken empty stomach, it can cause stomach ulcers.

Question 15.
Diseases like dysentery, cholera, typhoid etc., are more common in over crowded human settlements. Why?
Answer:
Overcrowded areas are generally unhygienic. This is because public hygeine is not maintained properly. Water bodies in these areas are generally contaminated with disease causing microbes. Water stagnation and garbage accumulation is often not checked.
Diseases like dysentery, cholera, typhoid etc., are water borne infectious diseases and may easily spread in overcrowded human dwellings.

Question 16.
From which plant cannabinoids are obtained? Name any two cannabinoids. Which part of the body is effected by consuming these substances?
Answer:
Cannabinoids are obtained from leaves, resins and flowers of hemp plants e.g., Cannabis sativa. Examples of cannabinoids are bhang, ganga, charas etc. They interact with receptors present in the brain and affect cardiovascular system of the body.

Question 17.
In the metropolitan cities of India, many children are suffering from allergy/asthma. What are the main causes of this problem. Give some symptoms of allergic reactions.
Answer:
Children living in metropolitan cities are more prone to allergies/asthma, mainly due to life style and increased level of pollution. Suspended particulate matter released by vehicles add more to the increasing pollution levels. Moreover children living in metropolitan cities are overprotected in their early childhood days and are more sensitive to allergens.

Allergy is hypersensitive response of the body against foreign particles (allergens). Common symptoms are sneezing, watery eyes, running nose, difficulty in breathing, irritation of throat, trachea and skin etc. Asthma related problems worsen due to changing weather conditions and increased pollution levels.

Question 18.
What is the basic principle of vaccination?How do vaccines prevent microbial infections? Name the organism from which hepatitis B vaccine is produced.
Answer:
Vaccine is a preparation of antigenic proteins or inactivated/live but weakened germs of a disease which on inoculation into a healthy person provides temporary or permanent, active immunity by inducing antibody formation.

Vaccination is based on the property of ‘memory’ of the immune system. When a vaccine is injected, it generates primary immune response of low intensity and also produces B cells and T cells. When vaccinated person is again attacked by same pathogen, the existing memory T and B cells recognise the antigen quickly and attack the invaders with a massive production of lymphocytes and antibodies. Hepatitis B vaccine is second generation vaccine that has been produced from yeast by recombinant DNA technology.

Question 19.
What is cancer? How is a cancer cell different from the normal cell? How do normal cells attain cancerous nature?
Answer:
Cancer is an abnormal and uncontrolled division of cells, that invade and destroy surrounding tissues. Cancer cells differ from normal cells in the following ways:

1. Cancer cells multiply in an uncontrolled manner.
2. They do not exhibit property of contact inhibition.
3. They show metastasis.
4. They have indefinite life span.

Cancer causing agents known as carcinogens include physical, chemical and biological agents which activate the proto-oncogenes or cellular oncogenes to oncogenes and cause deviation in growth pattern leading to uncontrolled growth of cells. They alter the behaviour of normal cells and make them cancerous.

Question 20.
A person shows strong unusual hypersensitive reactions when exposed to certain substances present in the air. Identify the condition. Name the cells responsible for such reactions. What precaution should be taken to avoid such reactions.
Answer:
Hypersensitive reaction to foreign substances is known as allergy. Substances which cause allergic reactions are called allergens. The common allergens are dust, pollen, feathers, paint. In allergic chemicals called histamine and serotonin are released from mast cells. Allergy can be avoided by reducing an exposure to allergen and by taking drugs like antihistamines, adrenaline and steroids.

Question 21.
For an organ transplant, it is an advantage to have an identical twin. Why?
Answer:
For an organ transplantation, tissue matching(histocompatibility)andblood group compatibility of donor and recipient are very important. If they do not match, organ may be rejected. This is because, immune system recognises the protein in the transplanted tissue or organ as foreign and initiates cellular immunity. Chances of matching of tissue as well as blood group are very high if donor and recipient are identical twins, because of genetic similarity. Transplantation between identical twins is known as isograft.

Question 22.
What are lifestyle diseases? How are they caused? Name any two such diseases.
Answer:
Diseases which occur due to improper changes in the life style are known as lifestyle diseases. These diseases may be caused due to over-eating or crash dieting, lack of exercise, sedentary lifestyle, smoking, alcoholism, drug addiction etc. Hypertension and obesity are common lifestyle diseases.

Question 23.
If there are two pathogenic viruses, one with DNA and other with RNA, which would mutate faster? And why?
Answer:
Pathogenic virus with RNA as genetic material will mutate faster than the one with DNA as genetic material, because

1. RNA is chemically and structurally unstable. Uracil present in RNA is less stable than thymine in DNA.
2. RNA is highly reactive, labile and easily degradable.

Long Answer Type Questions

Question 1.
Represent schematically the life cycle of a malarial parasite.
Answer:

Question 2.
Compare the life style of people living in the urban areas with those of rural areas and briefly describe how the lifestyle affects their health.
Answer:
Lifestyle affects human health in various ways which are as follows:
(1) Food habits – People living in urban areas usually prefer fast food like pasta, pizza, burger, noodles etc., as compared to fresh, green leafy vegetables, fruits which are more consumed by villagers, because of this they suffer from obesity, hypertension and some cardiovascular problems.

(2) Residential areas — Due to lack of space in urban areas, residential areas are generally overcrowded. Sometimes there is no facility for proper disposal of garbage, no proper ventilation etc., therefore people become prone to food, water, and air borne diseases, like cholera, tuberculosis, typhoid, pneumonia etc. On the other hand, villages have clean air, human dwellings are not crowded and the environment is free of pollution. But due to lack of personal and public hygiene and malnutrition, villagers frequently suffer from various diseases.
Moreover, medical facilities in rural areas are not up to the mark. People of urban areas are exposed to good education, proper medical facilities etc.

(3) Exercise and sleep — Because of busy lifestyle, people in urban areas do not get enough rest and have no time for exercise. Whereas people in rural areas do a lot of physical work and have enough time to rest as well.

(4) Due to overprotected environment provided during early life, people in urban areas sometimes have low immunity and are susceptible to certain diseases and are affected by allergens. However, people in rural areas are exposed to harsh environmental conditions therefore develop better immunity.

Question 3.
Why do some adolescents start taking drugs? How can this be avoided?
Answer:
Reasons for drug and alcohol abuse among adolescents are :
(1) Curiosity : Adolescents want to have personal experience of smoking cigarette and taking alcohol and drugs.

(2) Adventure and excitement : A child may go in for use of drug, smoking and alcoholic drink for the sake of adventure and excitement.

(3) Family set up : In certain families, use of alcohol, tobacco, sleeping pills and pain killers are common. It induces the youngsters to taste the
same.

(4) Group or peer pressure : Friends and peer groups often motivate some adolescents to take drugs, alcohol as a defiance of authority and feeling of independence.

(5) Progressiveness : There is a false perception that taking of drugs, alcohol or smoking is a sign of progressiveness in society.

Various measures which help to avoid drug abuse are:

(1) Discipline : Good nurturance with consistent discipline but without suffocating strictness reduces the risk of addictions.

(2) Communication : The child must be able to communicate with the parents seeking clarification of all doubts and discussing problems that arise in studies or develop in the class, with friends, siblings and others.

(3) Independent working : Give responsibility to the child for small tasks and allow him/ her to perform independently. Of course, provide guidance where required.

(4) Education and counselling : Stresses, failures, disappointments and problems are part of life. A child has to be trained, educated and counselled to face them as and when they come.

(5) Looking for danger signs : Teachers and parents should always be careful to look for and identify danger signs that can indicate tendency to go in for addiction.

Question 4.
In your locality, if a person is addicted to alcohol, what kind of behavioural changes do you observe in that person? Suggest measures to overcome the problem.
Answer:
Behavioural changes which can be observed in an alcoholic are as follows:
(1) Reckless behaviour, vandalism and violence.
(2) Drop in academic performance and unexplained absence from school/ college.
(3) Lack of interest in personal hygiene, isolation, depression, fatigue, aggressive-ness.
(4) Change in eating and sleeping habits. Measures to overcome the problem of alcohol addiction are as follows:

• Seeking professional advice : Highly qualified psychologists, psychiatrists and de-addiction and rehabilitation programmes can help individuals who are suffering from alcohol abuse.
• Avoid undue peer pressure : Every person has his/her own choice and personality, which should be kept in mind. So he/ she should not be pressed unduly to do beyond his/her capacities, in work condition and other in social get together or activities.
• Education and counselling : Helps to overcome the problems, like stresses, disappointments and failure in life. One should utilise one’s energy in some beneficial activities like sports, music, reading, yoga and other extra curricular activities.
• Seeking help from parents and peers : In case of minors, whenever, there is any problem, one should seek help and guidance from parents and peers. Help should be taken from close and trusted friends. This would help young to share their feelings of anxiety and wrong doings.

Question 5.
What are the methods of cancer detection? Describe the common approaches for treatment of cancer. Answer:
Detection and diagnosis of cancer depends upon histological features of malignant structure. Few methods of cancer detection are as follows:
(1) Bone marrow biopsy and abnormal count of WBCs in leukemia.
(2) Biopsy of tissue, direct or through endoscopy. Pap test (cytological staining) is used for detecting cancer of cervix and other parts of genital tract.
(3) Techniques such as radiography (use of X-rays), CT Scan (computed tomography), MRI Scan (magnetic resonance imaging)
are very useful to detect cancers of the internal organs. Mammography is radiographic examination of breasts for possible cancer.
(4) Monoclonal antibodies coupled to appropriate radioisotopes can detect cancer specific antigens and hence cancer.
Treatment of cancer includes following :
(1) Chemotherapy : In chemotherapy a variety of anti-cancer drugs are used that produce more injury to cancer cells than to normal cells. These drugs interfere with the cell division and growth and affect both normal and cancerous cells. Vincristine and vinblastine from Catharanthus roseus (Vinca rosea) are effective in leukaemia control. Taxol is another anti-cancer drug obtained from Taxus baccata. Tetrathiomolybdate is the new anti-cancer drug. It arrests the tumour growth by starving cancer cells of copper.

(2) Radiotherapy : It is used in addition to chemotherapy. The basic priniciple there is to bombard cancer cells with rays that damage or destroy the ability of cancer cell to grow and divide by damaging the DNA within the tumour cells, but produce minimum damage to the surrounding normal tissue.

(3) Surgery : It is removal of the cancerous cells surgically and has only limited usefulness. In breast tumour and uterine tumour, it is most effective but other treatments are also given to kill any cells that may have been left.

(4) Immunotherapy : Immunotherapy is a form of treatment that enhances the body’s ability to recognise cancer cells and destroy them. It can be given intravenously or by subcutaneous injection.

(5) Blood and marrow transplant : High dose chemotherapy or radiation therapy can destroy bone marrow’s ability to make blood cells. A blood or marrow transplant can be used to replace marrow stem cells which produce blood cells.

Question 6.
Drugs like LSD, barbiturates, amphetamines, etc., are used as medicines to help patients with mental illness. However, excessive doses and abusive usage are harmful. Enumerate the major adverse effects of such drugs in humans.
Answer:
LSD is a psychedelic drug or hallucinogen which induces behavioural abnormalities, they cause optical or auditory hallucinations, horrible dreams, emotional outburst and severe damage to central nervous system.

Barbiturates are sedatives which depress CNS activity, give feeling of calmness, relaxation and drowsiness but high doses induces deep sleep.

Amphetamines are synthetic drugs and are CNS stimulants. They cause alertness, self-confidence, talkativeness, wakefulness and increased work capacity. High doses cause euphoria, marked excitement and sleeplessness which may lead to mental confusion. Their use may produce after effects like nausea and vomiting. Amphetamine is one of the drugs included in the ‘dope test’ for athletes.

Question 7.
What is Pulse Polio Programme of Government of India? What is OPV? Why is it that India is yet to eradicate Polio?
Answer:
Pulse polio programme is a programme that involves vaccination of children in the age group of 0-5 years with polio vaccine. It began in 1995 with the aim to make India a polio-free country. Oral Polio Vaccine (OPV) is a live-attenuated vaccine, produced by the passage of the virus through non-human cells at a sub-physiological temperature, which produces spontaneous mutations in the viral genome. OPV also proved to be superior in administration, eliminating the need for sterile syringes and making the vaccine more suitable for mass vaccination campaigns. OPV also provides long lasting immunity than the Salk vaccine. One dose of OPV produces immunity to all three poliovirus serotypes in approximately 50% of recipients. WHO declared India and the entire South-East Asia region polio free in March, 2014. Last case of polio in India was reported in January, 2011.

Question 8.
What are recombinant DNA vaccines? Give two examples of such vaccines. Discuss their advantages.
Answer:
Recombinant vaccines are vaccines having a gene encoding the protein of pathogen that causes immunogenic reactions in the host so as to produce antibody against the disease. These vaccines are produced from yeast through recombinant DNA technology. Examples are Hepatitis B vaccine and herpes virus vaccine.

Advantages – By rDNA technology the vaccines can be produced on larger scale so providing greater availabilitv for immunisation. These vaccines are highly specific, pure and elicit strong immune response.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases help you. If you have any query regarding.NCERT Exemplar Solutions for Class 12 Biology chapter 8 Human Health and Diseases, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution

Multiple Choice Questions

Question 1.
Which of the following is used as an atmospheric pollution indicator?
(a) Lepidoptera
(b) Lichens
(c) Lycopersicon
(d) Lycopodium
Answer:
(b): Lichens are the association between specific ascomycetes and certain genera of either green algae or cyanobacteria. Lichens can be used as industrial pollution indicators. They never grow in area that are polluted.

Question 2.
The theory of spontaneous generation stated that
(a) life arose from living forms only
(b) life can arise from both living and non-living
(c) life can arise from non-living things only
(d) life arises spontaneously, neither from living nor from the non-living.
Answer:
(c) : The theory of spontaneous generation states that life originated from non-living things in a spontaneous manner. This concept was held by early Greek philosophers like Thales, Anaximander, Xanophanes,Empedocles, Plato, Aristotle, etc. In ancient Egypt, it was believed that the mud of the Nile could give rise to frogs, toads, snakes, mice and even crocodiles when warmed by the sun. Van Helmont (1577-1644) held that human sweat and wheat grains could give rise to organisms.

Question 3.
Animal husbandry and plant breeding programmes are the examples of
(a) reverse evolution
(b) artificial selection
(c) mutation
(d) natural selection.
Answer:
(b) : Animal husbandry and plant breeding programmes are the examples of artificial selection. Animals or plants with desirable characteristics are interbred with the aim of altering the genotype and producing a new strain of the organism for a specific purpose. Traditional breeding techniques have been supplemented by more recent methods of genetic engineering, genetic testing and embryo manipulation. Sequencing of genomes of commercially important animals and plant species has enabled the inheritance of desired genes to be monitored directly by molecular methods.

Question 4.
Palaentological evidences for evolution refer to the
(a) development of embryo
(b) homologous organs
(c) fossils
(d) analogous organs.
Answer:
(c) : Fossils are the remains or traces of any organism that lived in the geological past. In general only the hard parts of organisms become fossilised (e.g., bones, teeth, shells and wood) but under certain circumstances the entire organism is preserved. Fossil records such as number and nature of fossils in early rocks, distribution of fossils in successive strata, etc., have helped the scientists conclude that evolution has taken place from simple to complex forms in a gradual manner

Question 5.
The bones of forelimbs of whale, bat, cheetah and man are similar in structure, because
(a) one organism has given rise to another
(b) they share a common ancestor
(c) they perform the same function
(d) they have biochemical similarities.
Answer:
(b) : The bones of forelimbs of whale, bat, cheetah and man are similar in structure, because they are homologous organs. The organs which have the same fundamental structure but are different in functions are called homologous organs. These organs follow the same basic plan of organisation during their development. But in the adult condition, these organs are modified to perform different functions as an adaptation to different environments. The homologous structures are a result of divergent evolution.

Question 6.
Analogous organs arise due to
(a) divergent evolution
(b) artificial selection
(c) genetic drift
(d) convergent evolution.
Answer:
(d) : The organs which have similar functions but are different in their structural details and origin are called analogous organs. The analogous structures are the result of convergent evolution. The wings of an insect are analogous to wings of a bird. It is due to the fact that the basic structure of the wings of insects is different from the wings of bird. However, their function is similar.

Question 7.
(p+q)2 = p² + 2pq + q² = 1 represents an equation used in
(a) population genetics
(b) mendelian genetics
(c) biometrics
(d) molecular genetics.
Answer:
(a) : Hardy-Weinberg principle describes a theoretical situation in which a population is undergoing no evolutionary changes. According to Hardy-Weinberg principle, the balance in the relative numbers of alleles that is maintained within a large population over a period of time assuming that:

1. there is no genetic recombination
2. there is no natural selection
3. there is no gene migration
4. there is no mutation. In such a stable population, for a gene with two alleles A (dominant) and a (recessive), if the frequency of A is p and a is q, then the frequencies of the three possible genotypes (AA, Aa and aa) can be expressed by the Hardy-Weinberg equation; p² + 2pq + q² = 1.

Question 8.
Appearance of antibiotic-resistant bacteria is an example of
(a) adaptive radiation
(b) transduction
(c) pre-existing variation in the population
(d) divergent evolution.
Answer:
(c) : When a bacterial population encounters a particular antibiotic, those sensitive to it die. But some bacteria having favourable mutations become resistant to the antibiotic. Such resistant bacteria survive and multiply quickly. Soon the resistance providing genes become widespread and entire bacterial population becomes resistant.

Question 9.
Evolution of life shows that life forms had a trend of moving from
(a) land to water
(b) dryland to wet land
(c) fresh water to sea water
(d) water to land.
Answer:
(d) : Life’ originated in the ocean persumably about 4200 million years ago in precambrian era. The effect of electrical discharges and UV radiations on the mixture of methane, water, ammonia and hydrogen present in the sea gave rise to organic molecules like coacervates and microspheres. These protocells collected more substances forming cytoplasm and became first living beings. They were prokaryotic and chemoheterotrophic. When supply of existing organic molecules was exhausted, some heterotrophs evolved into autotrophs capable of synthesising food by chemosynthesis or photosynthesis. With + the advent of photoautotrophs, oxygen was added in sufficient amount in the primitive atmosphere. Eukaryotes originated in the ancient ocean, presumably about 1.5 billion years ago. These later on gave rise to complex forms. With the advent of oxygen, ozone layer was formed in the atmosphere which masked the land against the harmful UV radiations of sun and made it inhabitable. In the mean time, some complex plants could grow on land near water i.e., on boundary line and animal started visiting land, for food etc. Complex plants and animals developed jacketed sex organs and shelled eggs, respectively which reduced their dependence on water for completing their life cycles and they moved to land permanently. In this way, life forms moved from water to land.

Question 10.
Viviparity is considered to be more evolved because
(a) the young ones are left on their own
(b) the young ones are protected by a thick shell
(c) the young ones are protected inside the mother’s body and are looked after they are born leading to more chances of survival
(d) the embryo takes a long time to develop.
Answer:
(c)

Question 11.
Fossils are generally found in
(a) sedimentary rocks
(b) igneous rocks
(c) metamorphic rocks
(d) any type of rock.
Answer:
(a) : Sedimentary rocks are formed by gradual settling down or sedimentation of fragments or earth’s material in regions such as lake or sea. The animals or plants are preserved and fossilised when they are buried in the lava of volcano, in the ice, in an oil rich soil, in swamps, in desiccated deserts, in rocks, or under water, etc. Of all the media mentioned above the most common is water. Dead remains of aquatic animals and plants settle down at the bottom. Remains of terrestrial organisms are also brought to seas and big lakes by rivers and streams. Mud and sand settle down continuously at the bottom.

Sedimentation (deposition of layers) of mud and sand occurs. Fine mineral particles may penetrate the dead bodies. Decay and disintegration of organic remains take place to leave only the harder parts, impression, moulds, casts, etc. The sedimented mud and sand harden with time to form rocks.

Question 12.
FortheMN-blood group system, thefrequencies of M and N alleles are 0.7 and 0.3, respectively. The expected frequency of MN-blood group bearing organisms is likely to be
(a) 42%
(b) 49%
(c) 9%
(d) 58%.
Answer:
(a) : According to Hardy-Weinberg principle, the expected frequency of MN- blood group bearing organism is likely to be
p² + 2pq + q² = l
49 + .09 + 2x = 1
.58 + 2x = 1
2x = 1 – 59
= 42%

Question 13.
Which type of selection is industrial melanism, observed in moth, Biston betularia?
(a) Stabilising
(b) Directional
(c) Disruptive
(d) Artificial
Answer:
(b) : In directional selection, the population changes towards one particular direction. It means this type of selection
favours small or large-sized individuals and more individuals of that type will be present in next generation. The mean size, of the population changes. Example are evolution of DDT resistant mosquitoes, industrial melanism in peppered moth and evolution of giraffe.

Question 14.
The most accepted line of descent in human evolution is –
(a) Australopithecus—>Ramapithecus – ? Homo sapiens —»Homo habilis
(b) Homo erectus —> Homo habilis —> Homo sapiens
(c) Ramapithecus —» Homo habilis —> Homo erectus —> Homo sapiens
(d) Australopithecus —> Ramapithecus —> Homo erectus —> Homo habilis —> Homo sapiens.
Answer:
(c)

Question 15.
Which of the following is an example for link species?
(a) Lobe fish
(b) Dodo bird
(c) Sea weed
(d) Chimpanzee
Answer:
(a)

Question 16.
Match the scientists listed under Column ‘A’ with ideas listed under Column ‘B’.

Answer:
(a) A-(i); B-(iv); C-(ii); D-(iii)
(b) A-(iv); B-(i); C-(ii); D-(iii)
(c) A-(ii); B-(iv); C-(iii); D-(i)
(d) A-(iv); B-(iii); C-(ii); D-(i) miim (b)

Question 17.
In 1953, S. L. Miller created primitive earth conditions in the laboratory and gave experimental evidence for origin of first form of life from pre-existing non-living organic molecules. The primitive earth conditions created include
(a) low temperature, volcanic storms, atmosphere rich in oxygen
(b) low temperature, volcanic storms, reducing atmosphere
(c) high temperature, volcanic storms, non-reducing atmosphere
(d) high temperature, volcanic storms, reducing atmosphere containing CH4, NH3, etc.
Answer:
(d)

Question 18.
Variations during mutations of meiotic recombinations are
(a) random and directionless
(b) random and directional
(c) random and small
(d) random, small and directional.
Answer:
(a): Hugo de Vries (1901) put forward a theory of evolution, called mutation theory. The theory states that evolution is a jerky process where new varieties and species are formed by mutations (discontinuous variations) that function as raw material of evolution. Mutations appear all of a sudden. They become operational immediately.

Short Answer Type Questions

Question 1.
What were the characteristics of life forms that had been fossilised?
Answer:

1. Presence of hard parts in the body like bones, teeth, shell etc.
2. Buried in a medium where no oxidation of substances occur.

Question 2.
Did aquatic life forms get fossilise? If, yes where do we come across such fossils?
Answer:
Yes, aquatic life forms have better chances of getting fossilised, as their dead bodies are buried at the bottom of water in sedimentary rocks. They get exposed when the rocks come to the surface because of upheavals in the earth crust.

Question 3.
What are we referring to when we say ‘simple organisms’or’complex organisms’?
Answer:
Simple organisms are those organisms which show cellular level of body organisation,
i. e., their body is equivalent to a single cell. These organisms do not show ageing and natural death. As soon as the cell grows to maximum size, it divides to form daughter cells.
Complex organisms have higher level of body organisation as compared to unicellular organisms like tissue level, organ level and organ system level.

Question 4.
How do we compute the age of a living tree?
Answer:
We can compute the age of a living tree by counting the number of annual growth rings of tree. This technique is known as dendrochronology. It depends on the fact that trees in the same locality show a characteristic pattern of growth rings resulting from climatic conditions. Thus, it is possible to assign a definite date for each growth ring in living trees, and to use the ring patterns to date fossil trees or specimens of wood (e.g., used for building or objects on archaeological sites) with life spans that overlap those of living trees.

Question 5.
Give an example for convergent evolution and identify the features towards which they are converging.
Answer:
Tendrils in pea and cucurbits show convergent evolution because tendrils in both provide support to the plant (same function) but pea tendrils develop from leaves, whereas cucurbit tendrils are stem tendrils (different origin).

Question 6.
How do we compute the age of a fossil?
Answer:
Age of fossil can be computed by one of the following methods:

1. Radioactive carbon dating method
2. Uranium-lead method
3. Potassium-argon method
4. Electron spin resonance method.

Question 7.
What is the most important pre-condition for adaptive radiation?
Answer:
Most important pre-condition for adaptive radiation is that it should occur in s^me geographical area and is shown by different members of same ancestor. It is radiated to different habitats and develops different habits.

Question 8.
How do we compute the age of a rock?
Answer:
Age of rock can be determined by quantity of uranium and lead in the rocks and potassium-argon transformations.

Question 9.
When we talk of functional macromolecules (e.g. proteins as enzymes, hormones, receptors, antibodies etc), towards what are they evolving?
Answer:
Functional macromolecules are evolving towards creation of a complex organism. There are various evidences that are common to simple and complex forms of life which indicate common ancestry. E.g., histone protein tends to be well preserved among all eukaryotes with only one or two amino acids different.

Question 10.
In a certain population, the frequency of three genotypes is as follows:
Genotypes: BB Bb bb
Frequency: 22% 62% 16%
What is the likely frequency of B and b alleles?
Answer:
Frequency of allele B = 100% of BB + 50% of Bb
i. e., 22% + 31% = 53%
Frequency of allele b = 100 % of bb + 50% of Bb
i.e., 16% + 31% = 47%

Question 11.
Among the five factors that are known to affect Hardy-Weinberg equilibrium, three factors are gene flow, genetic drift and genetic recombination. What are the other two factors?
Answer:
The other two factors are mutation and natural selection.

Question 12.
What is founder effect?
Answer:
Founder effect is the phenomenon occurring when a population is founded by a small sample of the entire species, perhaps just a handful of individuals. Chance dictates that these founder members will be genetically unrepresentative of the species as a whole, and that the genetic make up of the new population will differ markedly from the main species population.

Question 13.
Who among the Dryopithecus and Ramapithecus was more man like?
Answer:
Ramapithecus was more man like in comparison to Dryopithecus, which was more ape-like.

Question 14.
By what Latin name the first hominid was known?
Answer:
As per the latest discoveries, Sahelanthropous tchadensis is considered the earliest hominid.

Question 15.
Among Ramapithecus, Australopithecus and Homo habilis – who probably did not eat meat?
Answer:
Ramapithecus probably did not eat meat. It was more man-like and lived on the tree tops but also walked on the ground. Its jaws and teeth were like those of humans. Its small canines and large molars suggest that Ramapithecus ate hard nuts and seeds like modern man.

Short Answer Type Questions

Question 1.
Louis Pasteur’s experiments, if you recall, proved that life can arise from only pre-existing life. Can we correct this as life evolves from pre¬existing life or otherwise we will never answer the question as to how the first forms of life arose? Comment.
Answer:
If we trace back the origin of first living organism, this fact will be revealed that life originated on early earth through physio- chemical processes in which atoms combined to form molecules which in turn reacted to produce inorganic and organic compounds.

These compounds then interacted to produce macromolecules which organised to form first living cells. Thus, first life form originated through abiogenesis. Once life originated, abiogenesis could not follow and new life forms then further evolved from pre-existing life (i.e., biogenesis).

Louis Pasteur in his experiments, disproved origin of life through abiogenesis and proved that life can arise from pre-existing life only. This statement holds true to its meaning but contradicts the mechanism of origin of first life form. Hence, it could be corrected as “life evolves from pre-existing life”.

Question 2.
The scientists believe that evolution is gradual. But extinction, part of evolutionary story, are ‘sudden’ and ‘abrupt’ and also group-specific. Comment whether a natural disaster can be the cause for extinction of species.
Answer:
Yes, natural disaster can lead to mass extinction of species from time to time, which is always ‘sudden’, ‘abrupt’ and ‘group specific’. For example extinction of dinosaurs is a good example of mass extinction triggered by a massive meteorite/asteroid impact. Asteroid impact was confirmed by the discovery of high concentrations of metal iridium in a thin layer of earth in almost all parts of the world as iridium is rare on earth and abundantly present in meteorites.

This natural disaster lead to the extinction of many contemporary species present on earth which included, non-avian dinosaurs, benthic foraminiferans, certain marine invertebrate groups etc.

Question 3.
Why is nascent oxygen supposed to be toxic to aerobic life forms?
Answer:
Nascent oxygen in highly reactive and can oxidise various biomolecules like DNA, RNA and proteins etc. If present in the cells of aerobic life forms, it can cause mutations and undesirable changes in their metabolic pathways. Therefore it has been considered as major toxic pollutant.

Question 4.
While creation and presence of variation is directionless, natural selection is directional as it is in the context of adaptation. Comment.
Answer:
The creation and presence of variations is directionless in regard that they occur randomly and spontaneously. Variations in organisms develop due to various reasons like genetic recombination, genetic drift, gene migration, mutations etc. All these variations are non-directional. As a result of variations, population becomes heterogeneous. Natural selection favours more adapted members oven less adapted ones and allows them to multiply at a faster rate to increase their population. This provides direction in a specific way towards evolution.

Question 5.
The evolutionary story of moths in England during industrialisation reveals, that’evolution is apparently reversible’ Clarify this statement.
Answer:
Peppered moth (Biston betularia) exists in two forms, white and melanic. Before industrialisation the trees were covered with whitish lichens. White moths could easily escape from their predators as they remained unnoticed in the surroundings. After industrialisation the bark of the trees got covered with black smoke and white moth were selectively picked up by the predators whereas the black ones escaped. Therefore, the population of white moths decreased. This showed that evolution is reversible and in a mixed population, those that can better adapt to the prevalant environmental conditions, survive and increase their population size but no variant is completely wiped out.

Question 6.
Comment on the statement that “evolution and natural selection are end results or consequences of some other processes but themselves are not processes”.
Answer:
All organisms possess enormous fertility. They multiply in geometric ratio, but the limited food supply, space and ecological niche begins intraspecific or interspecific struggle amongst them. For survival during struggle, organisms develop certain variations due to genetic recombinations, mutations etc., and become heterogenous. The organisms which have favourable variations would survive (survival of the fittest). Those who keep changing themselves according to the changing environment are selected by nature (natural selection). Selected organisms transfer their useful variations to next generation. Accumulation of continuous ’”variations generation after generation leads to evolution of new species. This clearly shows that evolution and natural selection are consequences of some other processes.

Question 7.
State and explain any three factors affecting allele frequency in populations.
Answer:
(1) Gene flow – It is also known as gene migration. It refers to the movement of alleles from one population to another as a result of interbreeding between members of the two population. The genes of two populations intermingle through breeding and result causes variations in the offspring.

(2) Genetic recombination – During sexual reproduction, fusion of male and female gametes takes place. These gametes are formed by reduction division or meiosis. Crossing over during meiosis leads to genetic variation as exchange of genetic material between non-sister chromatids of homologous chromosomes takes place. Hence, new association of alleles are formed in the gametes. Random fusion of gametes leads to genetic variations in the offspring.

(3) Mutations – Mutation is a sudden random change in genetic material of a cell. Somatic mutations affect the non- reproductive cells and are therefore restricted to a single organism but germline mutations which occur in reproductive cells (gametes) are transmitted to the offspring of organism. Mutations are caused by various chemicals or forms of radiations.

Question 8.
Gene flow occurs through generations. Gene flow can occur across language barriers in humans. If we have a technique of measuring specific allele frequencies in different population of the world, can we not predict human migratory patterns in pre-history and history? Do you agree or disagree? Provide explanation to your answer.
Answer:
Gene flow or gene migration refers to movement of alleles from one population to another as a result of interbreeding between members of two populations. Gene flow occurs through geographical barriers over generations. Changing gene frequencies would indicate that evolution is in progress. By studying specific allele frequencies of specific genes human migratory patterns and evolutionary trends in history and pre-history can be traced as gene migration (affected by human migration) contributes to changing gene frequency in an evolving population. Actually scientists are doing so, under Human Genome Project.

Question 9.
How do you express the meaning of words like race, breed, cultivars or variety?
Answer:
Race is a category used in the classification of organisms that consist of a group t>f individuals within a species that are geographically, ecologically, physiologically distinct from other members of the species e.g., negroid, mongoloid, etc. The term is frequently used in the same sense as subspecies. Breed is a group of animals similar in most characters and are related by descent, e.g., breeds of cow – Jersey, Holstein etc. Variety is a category used in the classification of plants and animals below the species level. A variety consists of a group of individuals that differ distinctly from but can interbreed with other varieties of the same species e.g., cabbage, cauliflower. Cultivar is a plant that has been developed and maintained by cultivation as a result of agricultural or horticultural practices. The term is derived from cultivated variety e.g., Pusa sem of flat beans.

Question 10.
When we say “survival of the fittest”, does it mean that
a. those which are fit only survive, or
b. those that survive are called fit? Comment.
Answer:
(a) In the struggle for existence, the individuals which have more favourable variations will have a competitive advantage over others which have less favourable or unfavourable variations. They are considered fit and thus survive and reproduce.

Question 11.
Enumerate three most characteristic criteria for designating a Mendelian population.
Answer:
Characteristic criteria for designating a Mendelian population are as follows :

1. Population must be large enough and there should be free flow of genetic materials among individuals through sexual reproduction.
2. Random mating of organisms takes place.
3. There should be either nil or negligible migration.

Question 12.
“Migration may enhance or blur the effects of selection” Comment.
Answer:
Migration involves entry or exit of some members of a species into or from an area. It may cause improvement of race if there is entry of individuals with more desirable genes, or may cause deterioration of race if there is entry of individuals with undesirable traits or exit of individuals with desirable traits. Hence, migration may enhance or blur the effects of selection.

Long Answer Type Questions

Question 1.
Name the law that states that the sum of allelic frequencies in a population remains constant. What are the five factors that influence these values?
Answer:
Hardy-Weinberg Principle states that sum of allele frequencies in a population is stable and remains constant from generation to generation i.e., the gene pool (total genes and their alleles in a population) remain constant. The five factors, called evolutionary agents, which influence the allele frequencies in a population are :

1. Gene flow – When migration of a section of population to another place occurs, gene frequencies change in the original as well as new population. New alleles are added to the new population and these are lost from the old population. Hence, gene flow changes the allele frequency in a population.
2. Genetic drift – It refers to the elimination of genes of certain traits when a section of population migrates or dies of natural calamity. It affects the gene frequency of a population.
3. Genetic recombinations – During sexual reproduction, fusion of male and female gametes takes place. The gametes are formed by reduction division or meiosis. Crossing over during meiosis leads to genetic variation as exchange of genetic material between non-sister chromatids of homologous chromosomes takes place and new association of alleles are formed in the gametes. The offsprings formed by fusion of gametes show new combination of traits.
4. Mutations – The sudden heritable change in genetic material which is directionless is called mutation. It alters the genetic make up of an individual.
5. Natural selection – It is the phenomenon by which some members of a population having desirable or favourable traits are favoured and survive to produce new offspring.

Question 2.
Explain divergent evolution in detail. What is the driving force behind it?
Answer:
Similarity in basic structural plan and developmental origin and difference in appearance and functioning of those structures due to adaptations of different needs is called divergent evolution. The driving force behind it is adaptation to new habitat and the prevailing environmental conditions there. As the original population increases in size, it spreads out from its centre of origin to exploit new habitats and food resources. In time this results in a number of populations each adapted to this particular habitat. Eventually these populations differ from each other sufficiently to become new species.
A good example of this process is the evolu tion of the Australian marsupials into species adapted as carnivores, herbivores, burrowers, fliers, etc.

Question 3.
You have studied the story of Pepper moths in England. Had the industries been removed, what impact could it have on the moth population? Discuss.
Answer:
Peppered moth, Biston betularia lives in all parts of England. In the population of peppered moth, two variants exist – dark or melanic and light coloured. Before industrialisation, light coloured moths were prevelant because they blended with the lichen covered bark of trees and remained unnoticed by predatory birds. The melanic moths were easily detected and preyed upon by predatory birds. Hence, the population of light coloured moth was much more as compared to dark  or melanic moth. With industrialisation, the pale tree trunks became more and more blackened due to industrial smoke and soot. As a result the light moths stood out in contrast to their background, increasing the possibility of being easily detected and eaten by their predators, such as birds, in much greater number than the dark melanic variety. Decrease in the number of light moths and increase in the number of dark moths was the ultimate result.

Therefore, evolution favoured the melanic moths to reproduce more successfully for their adaptation in the polluted areas of England. If the industries are removed the area will again become free of pollution leading to growth of lichens. As stated earlier, light coloured moth would easily camouflage with a light background and dark variants would be spotted easily by predators and eaten more frequently. Hence, the population of light coloured moth will again increase.

Question 4.
What are the key concepts in the evolution theory of Darwin?
Answer:
Key concepts in the evolution theory of Darwin are as follows:
(1) Overproduction or enormous fertility: Living beings have an innate ability of producing their own progeny for the continuity of race.

(2) Struggle for existence : According to Darwin, individuals multiply in geometric ratio whereas space and food remain almost constant.

(3) Variations and heredity : The everlasting competition among the organisms compells them to change according to the conditions so that they can utilise the natural resources and can survive successfully. According to Darwin, the variations are gradual (continuous) and those which are helpful in the adaptations of an organism towards its surroundings would be passed on to the next generation, while the others disappear.

(4) Survival of the fittest or Natural selection: During the struggle for existence only those individuals could survive which exhibit such variations that are more beneficial in facing the hardships and rigours of environment or which change to adapt themselves to the changing conditions. This is known as natural selection.

(5) Origin of species : Natural selection results in modification of traits within a lineage which over a period of time can bring about evolution.

Question 5.
Two organisms occupying a particular geographical area (say desert) show similar adaptive strategies. Taking examples, describe the phenomenon.
Answer:
The phenomenon indicated in the question is convergent evolution wherein organisms which are not closely related, evolve similar traits independently as a result of adaptation to similar environment. There is specific type of environment existing in every geographical area. The environmental factors influence all the organisms living in that area. This can be explained by following examples :
(1) Pectoral fins of sharks and flippers of dolphins are analogous organs. Pectoral fins of sharks are not pentadactyle. The flippers of dolphins are pentadactyle. Thus, basic structure of pectoral fins of sharks and flippers of dolphins are different but both are useful in swimming.

(2) Stings of honey bee and scorpion are analogous structures. The sting of honey bee is a modification of its ovipositor (structure that helps in egg laying) while that of scorpion is modified last abdominal segment. Stings of both arthropods perform similar function.

(3) Leaves are plant organs specialised for photosynthesis. However, there are plants where the leaves are modified or reduced in response to hot and dry environment. The function of leaves is then taken over by other organs like stipules (e.gLathyrus aphaca), petiole (e.g., Acacia auriculiformis) and stem branches (e.g, Ruscus, Asparagus). They are analogous amongst them as well as to the leaves.

Question 6.
We are told that evolution is a continuing phenomenon for all living things. Are humans also evolving? Justify your answer.
Answer:
New researches suggest that human beings are also evolving like all other organisms. By studying specific evolutionary trends in various human characters it can be provfed that human beings are evolving :
(1) Lactose tolerance : Historically the gene that regulated human’s ability to digest lactose was shut down as infants are weaned off of their mother’s breast milk. However, adult human in regions of Africa and Northern Europe developed the ability to tolerate lactose in their diets as recent as 5,000 or 6,000 years ago due to mutations.

(2) Wisdom teeth : Our ancestors had much bigger jaws than we do. Today our jaws are much smaller and wisdom teeth may not erupt in some. Estimates say that they will disappear in the coming population.

Question 7.
Had Darwin been aware of Mendel’s work, would he been able to explain the origin of variations. Discuss.
Answer:
Mendel proposed certain laws of heredity which form the basis of inheritance of characters. He is commonly called ‘father of genetics’ because of his contribution to genetics. Darwin’s theories explain the survival of the fittest, but does not explain the mode of arrival of fittest. He did not explain about any particular substance for inheritance of continuous useful variations from one generation to the other. Whereas Mendel stated that inheritance of characters is particulate and every character is of discrete units called factors (now called alleles). If Darwin had been aware of Mendel’s work, he could have been able to explain the origin of variations and their inheritance in the light of changes in the factors of Mendel.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 7 Evolution, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 5 Principles of Inheritance and Variation

Multiple Choice Questions

Question 1.
In a DNA strand, the nucleotides are linked together by
(a) glycosidic bonds
(b) phosphodiester bonds
(c) peptide bonds
(d) hydrogen bonds.
Answer:
(b) : A nucleotide has three components – nitrogenous base, a pentose sugar (ribose in case of RNA and deoxyribose for DNA) and a phosphate group. There are two types of nitrogenous bases – purines (adenine and guanine) and pyrimidines (cytosine, thymine and uracil). A nitrogenous base is linked to the pentose sugar through a N-glycosidic linkage to form a nucleoside. When a phosphate group is linked to 5′ -OH of a nucleoside through phosphoester linkage, a correspondingnucleotide is formed. The nucleotides are linked through 3′ – 5′ phosphodiester linkage to form a dinucleotide.

Question 2.
A nucleoside differs from a nucleotide. It lacks the
(a) base
(b) sugar
(c) phosphate group
(d) hydroxyl group.
Answer:
(c) : A purine or pyrimidine base joined with a pentose sugar, either ribose or deoxyribose, is a nucleoside. A nucleotide is a nucleoside with one or more phosphate groups attached to the sugar.

Question 3.
Both deoxyribose and ribose belong to a class of sugars called
(a) trioses
(b) hexoses
(c) pentoses  
(d) polysaccharides.
Answer:
(c) : Pentose sugar is a sugar molecule with five carbon ring structure. Both ribose and deoxyribose are pentoses, the latter is formed by deoxygenation of the former at one carbon atom.

Question 4.
The fact that a purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix
(a) the antiparallel nature
(b) the semi-conservative nature
(c) uniform width throughout DNA
(d) uniform length in all DNA.
Answer:
(c) : In DNA chain, the base pairs on the two strands are complementary and a specific purine pairs with a specific pyrimidine. This makes the two chains uniformly (2nm) thick. A larger sized purine lies opposite to the smaller-sized pyrimidine i.e., A pairs with T and C pairs with G giving uniform width throughout DNA.

Question 5.
The net electric charge on DNA and histones is
(a) both positive
(b) both negative
(c) negative and positive, respectively
(d) Zero
Answer:
(c) : DNA is much more organised in eukaryotic chromatin and is associated with a variety of proteins, most prominent of which are histones. Histones are rich in the basic amino acid residues lysines and arginines. Both these amino acid residues carry positive charges in their side chains. Thus, histones are positively charged. Histones are organised to form a unit of eight molecules called as histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.

Question 6.
The promoter site and the terminator site for transcription are located at
(a) 3′ (downstream) end and 5′ (upstream) end, respectively of the transcription unit
(b) 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit
(c) the 5′(upstream) end
(d) the 3′ (downstream) end.
Answer:
(b)

Question 7.
Which of the following statements is the most appropriate for sickle cell anaemia?
(a) It cannot be treated with iron supple­ments.
(b) It is a molecular disease.
(c) It confers resistance to acquiring malaria.
(d) All of the above
Answer:
(d) : Sickle cell anaemia is an autosomal hereditary disorder in which the erythrocytes become sickle shaped under oxygen deficiency, such as during strenuous exercise and at high altitudes. The disorder or disease is caused by the formation of an abnormal haemoglobin molecule called haemoglobin-S. Sickle cell trait protects against malaria. Several studies have suggested that, sickle haemoglobin might get in the way of Plasmodium parasite infecting RBCs, reducing the number of parasites that actually infect the host cell and thus confer some protection against the disease.

Question 8.
One of the following is true with respect to AUG:
(a) It codes for methionine only.
(b) It is also an initiation codon.
(c) It codes for methionine in both prokaryotes and eukaryotes.
(d) All of the above
Answer:
(d) AUG has dual functions : it acts as initiation codon and also codes for methionine (both in prokaryotes and eukaryotes, although formylated in prokaryotes).

Question 9.
The first genetic material could be
(a) protein
(b) carbohydrates
(c) DNA
(d) RNA.
Answer:
(d) : The first genetic material could be RNA. Evidences suggest that, metabolism, splicing and translation all have evolved around RNA. The first biocatalysts were also RNAs. Even now, some enzymes are made of RNAs, e.g., ribozyme. Gradually during evolution, genetic storage function of RNA was taken over by DNA as RNA being single stranded was more reactive, hence unstable. DNA being double stranded with complementary strands is not only more stable but also resists changes as it has evdlved a process of repair. For biocatalysis, RNA was replaced by protein enzymes. The latter were more stable, efficient and occur in many forms.

Question 10.
With regard to mature mRNA in eukaryotes
(a) exons and introns do not appear in the mature RNA
(b) exons appear but introns do not appear in the mature RNA
(c) introns appear but exons do not appear in the mature RNA
(d) both exons and introns appear in the mature RNA.
Answer:
(b) : Eukaryotic transcripts possess extra segments called introns or intervening sequences or noncoding sequences. They do not appear in mature or processed RNA because post transcriptional processing of the transcript includes splicing i.e., the’ process of removal of introns and fusion of exons to form functional RNAs.

Question 11.
The human chromosome with the highest and least number of genes in them are respectively
(a) chromosome 21 and Y
(b) chromosome 1 and X
(c) chromosome 1 and Y
(d) chromosome X and Y.
Answer:
(c) : Chromosome 1 has 2968 genes while Y-chromosome has 231 genes which respectively are the maximum and minimum number of genes present in a chromosome.

Question 12.
Who amongst the following scientists had no contribution in the development of the double helix model for the structure of DNA?
(a) Rosalind Franklin
(b) Maurice Wilkins
(c) Erwin Chargaff
(d) Meselson and Stahl
Answer:
(d)

Question 13.
DNA isa polymer of nucleotides whicharelinked to each other by 3′-5’phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?
(a) Replace purine with pyrimidines.
(b) Remove/Replace 3′ OH group in deoxyribose.
(c) Remove/Replace 2′ OH group with some other group in deoxyribose.
(d) Both ‘b’ and ‘c’.
Answer:
(b) : If 3′ OH group is removed/replaced in deoxyribose, there will be no formation of phosphodiester bonds which is formed between 3′ hydroxyl (OH) group of one nucleotide and 5′ phosphate of the other, and hence polymerisation of nucleotides will be prevented.

Question 14.
Discontinuous synthesis of DNA occurs in one strand, because
(a) DNA molecule being synthesised is very long
(b) DNA dependent DNA polymearse catalyses polymerisation only in one direction (5′->3’)
(c) it is a more efficient process
(d) DNA ligase has to have a role.
Answer:
(b) :The DNA-dependent DNA poly­merases catalyse polymerisation only in one direction, that is 5′ —> 3′. Consequently, on one strand (the template with polarity 3′ —> 5′), the replication is continuous in 5′ —» 3′ direction, while on the other (the template with polarity 5′ —» 3′), it is discontinuous. A RNA primer is added the 3′ end of the opened region of 5′ —» 3′ strand and on this primer, a short stretch of DNA is synthesised in 5′ —> 3′ direction. Then again, when further uncoiling occurs then a new primer is added at 3′ end which is elongated, and the process continues. These discontinuously synthesised fragments are later joined by the enzyme DNA ligase.

Question 15.
Which of the following steps in transcription is catalysed by RNA polymerase?
(a) Initiation
(b) Elongation
(c) Termination
(d) All of the above
Answer:
(d) : There is single DNA-dependent RN Apoly merase that catalyses transcription of all types of RNA in bacteria. RNA polymerase binds to promoter and initiates transcription (initiation). It uses nucleoside triphosphates as substrate and polymerises them in a template dependent fashion following the rule of complementarity. It somehow also facilitates opening of the helix and continues elongation. Only a short stretch of the prepared RNA remains bound to the enzyme.Once the polymerase reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription. The RNA polymerase alone is only capable to catalyse the process of elongation. It associates transiently with initiation factor(o) and termination factor(р)  to initiate and terminate the transcription respectively. Association with these factors alters the specificity of the RNA polymerase to either initiate or terminate. In eukaryotes, several initiation factors and termination factors along with three types of RNA polymerase each polymerising a specific RNA perform these functions.

Question 16.
Control of gene expression takes place at the level of
(a) DNA replication
(b) transcription
(c) translation
(d) none of the above
Answer:
(b & c) : Regulation of gene expression refers to a very broad term that may occur at various levels. In eukaryotes, the gene expression can be regulated at transcriptional level, post transcriptional processing level, during transport of mRNA from nucleus to the cytoplasm, and at translational level whereas in prokaryotes, control of the rate of transcriptional initiation is the predominant site for control of gene expression.

Question 17.
Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. Which of the following statements is correct about regulatory proteins?
(a) They only increase expression.
(b) They only decrease expression.
(c) They interact with RNA polymerase but do not affect the expression.
(d) They can act both as activators and as repressors.
Answer:
(d) : In a transcription unit, the activity of RNA polymerase at a given promoter is regulated by interaction with accessory proteins, which affect its ability to recognise start sites. These regulatory proteins can act both positively (activators) and negatively (repressors).

Question 18.
Which was the last human chromosome to be completely sequenced:
(a) Chromosome 1
(b) Chromosome 11
(c) Chromosome 21
(d) Chromosome X
Answer:
(a) : The sequencing of chromosome 1 was completed in May 2006. This was the last of the 24 human chromosomes (22 autosomes and X and Y) to be sequenced.

Question 19.
Which of the following are the functions of RNA?
(a) It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.
(b) It carries amino acids to ribosomes.
(c) It is a constituent component of ribosomes.
(d) All of the above.
Answer:
(d) : mRNA carries genetic information from DNA to ribosomes for synthesising polypeptide chains. fRNA carries amino acids to ribosomes attached to mRNA for translation. rRNA, is a vital component of ribosomes.

Question 20.
While analysing the DNA of an organism, a total number of 5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff’s rule it can be concluded that
(a) it is a double stranded circular DNA
(b) it is a single stranded DNA
(c) it is a double stranded linear DNA
(d) no conclusion can be drawn.
Answer:
(b) : It cannot be a double stranded DNA because as per Chargaff’s rule for a dsDNA, the ratios between adenine and thymine, and guanine and cytosine are constant and equal. Hence, it is a ssDNA.

Question 21.
In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called
(a) A-DNA
(b) B-DNA
(c) c DNA
(d) rDNA.
Answer:
(c) : DNA synthesis on RNA template occurs in retroviruses or RNA viruses. They have RNA as their genetic material. When they infect a host cell, they first synthesise DNA on RNA template which is known as cDNA or complementary DNA. This cDNA then continues the process of infection.Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription. Which of the following statements is correct about regulatory.

Question 22.
If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of ^,5N/¹⁵N: ¹⁵N/¹⁴N: ¹⁴N/¹⁴N containing DNA in the fourth generation would be
(a) 1:1:0
(b) 1:4:0
(c) 0:1:3
(d) 0:1:7
Answer:
(d) : In Meselson and Stahl’s experiment, parent DNA isolated from E.coli is grown in heavy ¹⁵N medium. It was then put in light ^,4N medium.Then, when replication occurs the new strand synthesised would have ^l4N. The process can be represented as follows:

Question 23.
If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is 5′ – AT G A A T G – 3′, the sequence of bases in its RNA transcript would be
(a) 5′-AUGAAUG-3′
(b) 5′-UACUUAC-3′
(c) 5′-C All UC AU-3′
(d) 5′ – G U A A G U A – 3′.
Answer:
(a) : The sequence of bases in RNA transcript is similar (not complementary) to the coding DNA strand except that in RNA, uracil is present in place of thymine. So, the correct sequence of bases is:
5′ – AUGAAUG – 3′

Question 24.
The RNA polymerase holoenzyme transcribes
(a) the promoter, the structural gene and the terminator region
(b) the promoter and the terminator region
(c) the structural gene and the terminator region
(d) the structural gene only.
Answer:
(d) : Transcription involves three separate processes : initiation, elongation and termination. Initiation begins when the RNA polymerase binds to the promoter which serves only as a target site for binding of the RNA polymerase and is not transcribed. Each gene contains a specific promoter region for guiding the beginning of transcription. This is followed by region of gene (structural gene) that is transcribed and ends with a terminator that stops transcription and is not transcribed.

Question 25.
If the base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of fRNA pairing with it must be
(a) 5′- UAC – 3′
(b) 5′-CAU-3′
(c) 5’-AUG-3′
(d) 5′- GUA – 3′.
Answer:
(b) : The first base of anticodon in 5′ – 3′ direction binds with the third base in codon (reading in 5′ – 3′ direction). Thus, if the base sequence in codon of mRNA is 5′- AUG – 3′, the complementary anticodon will be 3′-UAC – 3′ or 5′- CAl – 3′.

Question 26.
The amino acid attaches to the fRNA at its
(a) 5′-end
(b) 3′-end
(c) Anticodon site
(d) DHU loop.
Answer:
(b) : fRNA has an anticodon loop that has bases complementary to the code, it also has an amino acid acceptor site at which it binds to amino acids. This site lies at the 3′ end opposite the anticodon. fRNAs are specific for each ahrino acid.

Question 27.
To initiate translation, the mRNA first binds to
(a) the smaller ribosomal sub-unit,
(b) the larger ribosomal sub-unit
(c) the whole ribosome
(d) no such specificity exists.
Answer:
(a) : The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. It has two subunits, a large subunit and a small subunit. To start translation, mRNA should bind to small ribosomal unit.

Question 28.
In E. coli, the lac operon gets switched on when
(a) lactose is present and it binds to the repressor
(b) repressor binds to operator
(c) RNA polymerase binds to the operator
(d) lactose is present and it binds to RNA polymerase.
Answer:
(a) : The repressor of the operon is synthesised (all-the-time-constitutively) from the / gene. The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon. In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase access to the promoter and transcription proceeds. Thus, the lac operon gets switched on.

Very Short Answer Type Questions

Question 1.
What is the function of histones in DNA packaging?
Answer:
Maim Histones are positively charged basic proteins which help in condensation of DNA. Negatively charged DNA gets wrapped around positively charged histone octamer to form nucleosome.

Question 2.
Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?
Answer:
Heterochromatin is thicker, densely packed, dark stained part of chromatin. Euchromatin is thinner, loosely packed lightly stained part of chromatin. Transcriptionally active part is euchromatin.

Question 3.
The enzyme DNA polymerase in coli is a DNA dependent polymerase and also has the ability to proofread the DNA strand being synthesised. Explain. Discuss the dual polymerase.
Answer:
DNA polymerase serves as a dual polymerase which serves two functions. It primary function is to add nucleotides according to the template strand in 5′ —> 3′ direction. It simultaneously proof reads the newly formed double strand, as it passes through the polymerase molecule. If the wrong base is inserted, the bond is unstable and spontaneous melting of the newly formed stretch occurs. The new strand gets exposed to the 3′ exonuclease site of the enzyme which removes the mismatched base and some additional nucleotides from the 3’ end. Then, the polymerase activity is continued again. This reduces the chances of error in DNA replication from about one in a million to about one in a hundred million base pairs.

Question 4.
What is the cause of discontinuous synthesis of DNA on one of the parental strands of DNA? What happens to these short stretches of synthesised DNA?
Answer:
New DNA strand is always formed in 5′-3′ direction during DNA replication, over DNA template with 3′-5′ direction. When DNA opens during replication, 3′-5′ strand forms a continuous strand called leading strand. The other template strand with 5′-3′ orientation produces a new short strand over the fork every time when it opens. These short segments are called Okazaki fragments. These segments join together with DNA ligases and form lagging strand.

Question 5.
Given below is the sequence of coding strand of DNA in a transcription unit:
3’AATGCAGCTATTAGG-5′ Write the sequence of
(a) its complementary strand
(b) the mRNA
Answer:
(a) Sequence of complementary strand will be : 5′ -TTACGTCGATAATCC3′
(b) Sequence of mRNA will be : 3′ – A A U G C A G C U A U U A G G 5′

Question 6.
What is DNA polymorphism? Why is it important to study it?
Answer:
Variation at genetic level arisen due to mutations, is called polymorphism. Such variations are unique at particular site of DNA. The polymorphism (variations) in DNA sequences is the basis of genetic mapping and DNA fingerprinting.

Question 7.
Based on your understanding of genetic code, explain the formation of any abnormal haemoglobin molecule. What are the known consequences of such a change?
Answer:
Haemoglobin is made of four polypeptide chains, two a-chains-which are 141 amino acids long and two (3-chains which are 146 amino acids long. In case of sickle cell anaemia, a fault occurs at the sixth amino acid in the (5-chain. The, amino acid should be glutamic acid. In HbS however, it is replaced by valine.

Question 8.
Sometimes cattle or even human beings give birth to their young ones that are having extremely different sets of organs like limbs/ position of eye(s) etc. Comment.
Answer:
Development of organs in an organism is regulated by expression of different sets of genes in a definite sequence in an orderly manner. Any disturbance in co-ordination ‘         and expression of genes will lead to abnormal  i. organ formation.

Question 9.
In a nucleus, the number of ribonucleoside triphosphates is 10 times the number of deoxy ribonucleoside triphosphates, but only deoxy ribonucleotides are added during the DNA replication. Suggest a mechanism.
Answer:
Recent studies show that there is a region in the active site cleft of the polymerase enzyme that monitors the 2′ and 3′ substituents of the incoming nucleotide and identifies the sugar component. It ensures that only deoxyribonucleotides are picked up during  replication from the nuclear pool.

Question 10.
Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them.
Answer:

1. DNA helicase which stimulates the separation of the two strands.
2. Topoisomerase which changes the degree of supercoiling in DNA by cutting one or both strands, which forms
3. Primase which forms RNA primer strands on single-stranded DNA templates.

Question 11.
Name any three viruses which have RNA as the genetic material.
Answer:

1. Tobacco Mosaic Virus
2. Human Immunodeficiency Virus
3. Influenza Virus

Short Answer Type Questions

Question 1.
Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as the genetic material.
Answer:
In 1928, Frederick Griffith performed the transformation experiment using Streptococcus pneumoniae. When he injected heat killed, virulent S strain along with non-virulent, live R strain in mice, then the mice died. It showed that something from dead S strain transformed the non-virulent R strain into virulent one. This phenomenon was called transformation by him. Transformation is the phenomenon by which the transforming principle (as named by Griffith), isolated from one type of cell, when introduced into another type, is able to bestow some of the properties of the former to the latter. This discovery started the quest for identification of this transforming principle. Avery, MacLeod and McCarty later purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells. They discovered that DNA alone (neither proteins nor RNAs) from S bacteria caused R bacteria to become transformed. They discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation.

Question 2.
Who revealed biochemical nature of the transforming principle? How was it done?
Answer:
Refer answer 1.

Question 3.
Discuss the significance of heavy isotope of nitrogen in the Meseison and Stahl’s experiment.
Answer:
¹⁵N is not a radioactive isotope. It is a heavy isotope of N and can be separated from ¹⁴N by density gradient centrifugation. This helped Meseison and Stahl to prove that DNA replicates semiconservatively. Meseison and Stahl extracted bacterial DNA and centrifuged it in caesium chloride solution. Depending on the mass of the molecule, the DNA would settle out at a particular point in the tube. They first grew Escherichia coli bacteria in a medium containing heavy isotope ¹⁵N for several generations.

The ¹⁵N bacteria were then transferred to a growth medium containing the normal, lighter isotope of nitrogen, ¹⁴N, where they reproduced by cell division. Extracts of DNA from the first generation offspring were shown to have a lower density, since half the DNA was made up of the original strand containing ¹⁵N and the other half was made up of the new strand containing ¹⁴N. At succeeding generation times, the DNA extracts were found to separate at lower densities indicating that have a lower proportion of ^,5N as more ¹⁴N had incorporated into the bacterial DNA. This was conclusive evidence for the semi­conservative method of DNA replication.

Question 4.
Define a cistron. Giving examples, differentiate between monocistronic and polycistronic ‘transcription unit.
Answer:
Cistron is a length of DNA that contains the information for coding a specific polypeptide chain or codes for a functional RNA molecule (i.e., mRNA, transfer RNA or ribosomal RNA).

Monocistronic transcription unit is a type of messenger RNA that can encode only one polypeptide per RNA molecule. In eukaryotic cells, virtually all messenger RNAs are monocistronic.

Polycistronic transcription unit is a type of messenger RNA that can encode more than one pqlypeptide separately within the same RNA molecule. Bacterial messenger RNA is generally polycistronic.

Question 5.
Give any six features of the human genome.
Answer:
Features of the human genome are as follows:

• Human genome has 3.1647 billion nucleotide base pairs.
• The average gene size is 3000 base pairs. The largest gene is that of Duchenne muscular dystrophy on X-chromosome. It has 2.4 million (2400 kilo) base pairs. (3-globin and insulin genes are less than 10 kilobases.
• The human genome consists of about 30,000 genes. Previously, it was estimated to contain 80,000 to 100,000 genes.
• Chromosome 1 has 2968 genes while Y-chromosome has 231 genes. They are the maximum and minimum genes for the human chromosomes.
• The function of over 50% of discovered genes is unknown.
• Less than 2% of the genome represents structural genes that code for proteins.

Question 6.
During DNA replication, why is it that the entire molecule does not open in one go? Explain replication fork. What are the two functions that the monomers (dNTPs) play?
Answer:
Whole of DNA does not open in one stretch due to very high energy requirement. The point of separation proceeds slowly towards both the directions in DNA strands. In each direction, it gives the appearance of Y-shaped structure called replication fork. Deoxyribonucleoside triphosphates (dNTPs) serve dual purposes. In addition to acting as substrates, they provide energy for polymerisation reaction (the two terminal phosphates in a deoxynucleoside triphosphates are high-energy phosphates, same as in case of ATP).

Question 7.
Retroviruses do not follow central dogma. Comment.
Answer:
Retroviruses follow reverse of central dogma. Temin and Baltimore reported that double stranded RNA of Rous Sarcoma
Virus (RSV) operates a central dogma reverse (inverse flow of information). RNA of these viruses first synthesises DN A through reverse transcription or feminism. DNA then transfers information to RNA which takes part in translation of coded information to form polypeptide. The mechanism is characteristic of retroviruses, e.g., HIV.

Question 8.
In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive bases increases from 0.34 nm to 0.44 nm. Calculate the length of DNA double helix (which has 2 x 10⁹ bp) in the presence of saturating amount of this compound.
Answer:
In the given question, the distance between two consecutive base pairs is 0.44 nm (0.44 x 10⁹m).
Total number of base pair is 2 x 10⁹ bp.
Length of DNA double helix is calculated by multiplying the total number of bp with distance between two consecutive bp i.e., 2 x 10⁹ bp x 0.44 x 10^-9 m/bp.Therefore, length of DNA double helix is 0.88 m.

Question 9.
What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?
Answer:
Acidic amino acids are negatively charged. DNA is also negatively charged, Therefore, acidic amino acids will not be able to hold DNA over them. This will lead to failure of packaging of DNA material resulting no chromatin formation.

Question 10.
Recall the experiments done by Frederick Griffith, Avery, MacLeod and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the R-strain into virulent
Answer:
RNA is labile and prone to degradation Recall the experiments done by Frederick Griffith, Avery, MacLeod and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Pneumococcus have transformed the R-strain into virulent form.

Question 11.
You are repeating the Hershey-Chase experiment and are provided with two isotopes ³²P and ¹⁵N (in place of ³⁵S in the original experiment). How do you expect your results to be different?
Answer:
³²P is a radioactive isotope but ¹⁵N is not. ¹⁵N is heavier isotope of nitrogen. Even if it can be detected, ¹⁵N will not be able to differentiate between protein and genetic material because it will get incorporated both in DNA as well as proteins. Therefore, the experiment will not give any conclusive result.

Question 12.
There is only one possible sequence of amino acids when deduced from a given nucleotide sequence. But multiple nucleotide sequences can be deduced from a single amino acid sequence. Explain this phenomena.
Answer:
There are 64 triplet codons and only 20 amino acids. Some amino acids are coded by more than one codons. Except tryptophan (UGG) and methionine (AUG) which have single codon each, all other amino acids involved in protein synthesis have more than one codon. In degenerate codons, mostly the first two nitrogen bases are similar while the third one is different. Thus as a codon codes only for a specific amino acid we can deduce a single amino acid sequence from a nucleotide sequence. But, as an amino acid can be coded by multiple codons thus many nucleotide sequences can be coded from a given amino acid sequence.

Question 13.
A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do you think the statement is correct? Defend your answer.
Answer:
Sometimes single base mutation may not cause formation of new amino acid. It occurs usually when change’ occurs in third base of triplet codon, e.g. G G A can be replaced by GGU, GGC, and GGG without affecting the incorporation of amino acid glycine. Such mutations are called silent mutations and do not cause any change in the phenotype of the organism.

Question 14.
A low level of expression of lac operon occurs at all the time. Can you explain the logic behind this phenomenon?
Answer:
The lac operon synthesises the permease enzyme which is responsible for transporting lactose into the cell. If permease will not be present, then the lactose cannot enter the cell and cannot act as inducer for lac operon. Hence, a low level of expression of lac operon is always maintained in the cell, so that whenever lactose is present, it can enter the cell and induce lac operon.

Question 15.
How has the sequencing of human genome opened new windows for treatment of various genetic disorders. Discuss amongst your classmates.
Answer:
Human genome sequencing has been very useful in treatment of genetic disorders, such as:

• It has been found that more than 1200 genes are responsible for common human cardiovascular diseases, endocrine diseases (like diabetes), neurological disorders (like Alzheimer’s disease), cancer and many more. Thus, the medicines that specifically target these genes can be made.
• Efforts are in progress to determine genes that will change cancerous cells into normal.
• All the genes or transcripts in a particular t tissue, organ or tumor can be analysed toknow the cause or effect produced in it.

Question 16.
The total number of genes in humans is far less (< 25,000) than the previous estimate (up to 1,40,000 gene). Comment.
Answer:
At the start of genome sequencing project, scientists estimated that human genome has about 1,40,000 genes. The number gradually fell down to 30,000 and finally has been found to be between 20,000 to 25,000. The estimates were so high because scientists have found E.coli (single celled) to have about 4300 genes and Caenorhabditis elegans (about 1000 cell) to have about 19000 genes. Thus, asper the assumption that the more complex an organism the more is number of genes, the human gene number has been estimated to be more than one lakh. But, number of genes has no relationship with complexity of organism. Besides, it also has been found that genome size and gene number is unrelated because only 2% of the genome actually consists of structural genes, rest 98% is non-coding DNA.

Question 17.
Now, sequencing of total genome is getting less expensive day by the day. Soon it may be affordable for a common man to get his genome sequenced. What in your opinion could be the advantage and disadvantage of this development?
Answer:
Advantages of genome sequencing are as follows:

1. It will provide information about genetic abnormalities.
2. There will be possibility of transfer of certain desired genes to the progeny.
3. Proneness to certain diseases will make a person avoid certain factors causing those diseases.
4. Information about timing of normal appearance of degenerative diseases and how to postpone them can be gathered.
5. Metobolic defects can be taken care of. Disadvantages of genome sequencing are as follows:
6. Knowing about disease much in advance may lead the person under depression.
7. People will become keen to improve their genetic constitution.

Question 18.
Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage?
Answer:
Bacteriophages do not have repetitive sequences like VNTR in their genome. Bacterial genome is small in size and with all coding sequences. Therefore DNA finger­printing cannot be done in bacteriophage.

Question 19.
During in vitro synthesis of DNA, a researcher used 2′, 3′ – dideoxycytidine triphosphate as raw nucleotide in place of 2′ -deoxycytidine. What would be the consequence?
Answer:
2′-3′ dideoxycytidine triphosphate cannot form an ester bond necessary for chainformation of DNA, where as 2′ deoxycytidine phosphate can develop such bond. Chain formation of DNA or polymerisation of DNA will stop due to presence of 2′-3′- dideoxycytidine.

Question 20.
What background information did Watson and Crick have made available for developing a model of DNA? What was their contribution?
Answer:
The information already available with them for developing a model of DNA was –

1. Erwin Chargaff’s generalisations about DNA structure e., the purine and pyrimidines are always in equal amounts A + G = T + C, and the amount of adenine is always equal to that of thymine and the amount of guanine is always equal to that of cytosine A = T and G = C.
2. X Ray diffraction pictures of crystalline DNA produced by Maurice Wilkins and Rosalind Franklin.

Contributions of Watson and Crick to this background data are:

1. DNAis a double helical structure with two chains running in antiparallel direction.
2. Complementary base pairing rule
3. Semi conservative replication
4. Mutations occur due to tautomeric changes in nitrogen bases.

Question 21.
What are the functions of (i) methylated guanosine cap, (ii) poly-A “tail” in a mature RNA?
Answer:
Methylated guanosine cap helps in attachment of mRNA to smaller subunit of ribosome during initiation of translation process.
Poly-A-tail provides longevity to mRNA’s life. Tail length and longevity of mRNA are positively correlated.

Question 22.
Do you think that the alternative splicing of exons may enable a structural gene to code for several isoproteins from one and the same gene? if yes, how? if not, why so?
Answer:
Alternative splicing is a controlled process which causes different mRNAs to be produced from a single gene by including some exons in one mRNA under some conditions, while including other exons under other conditions. This greatly increases the number of proteins that can be encoded by < 2$,000 functional genes of the genome. This alternative splicing of exons is sex-specific, tissue-specific and even developmental stage- specific. By alternative splicing of exons, a single gene may encode for several isoproteins and/or proteins of similar class.

Question 23.
Comment on the utility of variability in number of tandem repeats during DNA fingerprinting.
Answer:
Number and type of tandem repeats are specific for every individual. An individual gets them from two parents, but are unique for the individual. Therefore, they can be used to identify the individuals, and their relatives by comparing the repeats using DNA finger printing process.

Long Answer Type Questions

Question 1.
Give an account of Hershey and Chase experiment. What did it conclusively prove? If both DNA and proteins contained phosphorus and sulphur do you think the result would have been the same?
Answer:
Hershey and Chase experiment is based on the fact that phosphorus is present in DNA but not in the protein, and similarly sulphur is present in proteins but not in DNA. They incorporated radioactive isotope of phosphorus (³²P) into phage DNA and that of sulphur (³⁵S) into proteins of a separate phage culture. These phage types were used independently to infect the bacterium Escherichia coli. After some time, this mixture was agitated in a blender to separate the empty phage capsids from the surface of bacterial cells and the two were separated by centrifugation. Hershey and Chase showed that when ³²P was used, all radioactivity was associated with bacterial cells and if followed, appeared in the progeny phage.
However, when ³⁵S was used, all radioactive material was limited to phage ‘ghosts’ (empty viral protein coats). These results indicated that the DNA of the bacteriophage and not the protein enters the host, where viral replication takes place. Therefore, DNA is the genetic material of T₂ bacteriophage. It directs protein coat synthesis and allows replication to occur. If both DNA and proteins contained phosphorus and sulphur, they could not differentiate or separate protein and DNA and could not have concluded anything.

Question 2.
During course of evolution why DNA was chosen over RNA as genetic material? Give reasons by first discussing the desired criteria in a molecule that can act as genetic material and in the light of biochemical differences between DNA and RNA.
Answer:
A molecule that can act as a genetic material must fulfill the following criteria:

• It should be able to generate its replica (replication).
• It should chemically and structurally be stable.
• It should provide the scope for slow changes (mutation) that are required for evolution.
• It should be able to express itself in the form of ‘Mendelian Characters’.

Though RNA is known to be the genetic material in some viruses and early cells, it is not a very suitable genetic material because 2′ – OH group present in every nucleotide of RNA is a reactive group. It means RNA is highly reactive, labile and easily degradable. RNA functions as an enzyme and is, therefore, reactive and unstable.Uracil present in RNA is less stable as compared to thymine (methyl uracil) of DNA.

Being unstable, RNA mutates at a much faster rate, that is why RNA viruses have shorter life span and mutate and evolve very fast. Such ‘rapid changes are harmful to higher forms of life.

DNA is the genetic material of most of the organisms because

• DNA is chemically less reactive and structurally more stable as its nucleotides are not exposed except when they have to express their effect or to be replicated.
• They are comparatively more stable than RNA. Heat which killed bacteria in Griffith’s experiment did not destroy their DNAs.
• Presence of thymine in DNA instead of uracil, provides stability to DNA.
• Hydrogen bonding between purines and pyrimidines and their stacking make DNA more stable for storage of genetic information.
• DNA is capable of undergoing slow mutations required of genetic material.
• It has a power of repairing.
  Since DNA is more stable while RNA is more reactive, both the types of nucleic acids have been retained in genetic expression. DNA which is stable enough not to change with different stages of life cycle, age or with change in metabolism of the organism, is retained as better genetic material for the storage of genetic information. It expresses genetic information by protein synthesis through RNA which is more reactive, exposed for quicker action of protein synthesising machinery and thus is better for the transmission of genetic information.

Question 3.
Give an account of post transcriptional modifications of a eukaryotic mRNA.
Answer:
Transcription in eukaryotes occurs within the nucleus. The primary mRNA transcript is longer and localised in the nucleus, where it is also called heterogenous nuclear RNA (/mRNA) or pre-mRNA.
The mRNA is processed from the primary RNA transcript in a process called maturation. Initially, at the 5′ end, a cap (consisting of 7-methyl guanosine or 7 mG) and a tail of poly A at the 3′ end are added. The cap is a chemically modified molecule of guanosine triphosphate (GTP). The primary mRNAs are made up of two types of segments; non¬coding introns and the coding exons. The introns are removed by a process called RNA splicing. Of a pair of small nuclear ribonucleoprotein (snRNPs), one binds to 5′ splice site and the other to 3′ splice site. A spliceosome forms because of interaction between snRNPs and other proteins. This spliceosome uses energy of ATP to cut the RNA and to release the introns. The enzyme ligase joins two adjacent exons to produce mature mRNA.

Question 4.
Discuss the process of translation in detail.
Answer:
The process of decoding of the message from mRNA to protein with the help of fRNA, ribosome and enzyme is called translation (protein synthesis). Protein synthesis occurs over ribosomes. The ribosomes are formed of two subunits. The rosette group formed by ribosomes is called polyribosome. In a polyribosome, ribosomes are held together by strand of /mRNA.
The steps involved in polypeptide synthesis are:
(1) Activation of amino acids : In the presence of ATP, an amino acid combines with its specific aminoacyl-fRNA synthetase.

(2) Charging of fRNA : The complex formed in the above step reacts with tRNA specific for the amino acid to form aminoacyl- tRNA complex.

(3) Initiation : The small subunit of ribosome (with the mRNA) attaches to the large subunit in such a way that the initiation codon (AUG) comes on the P-site of ribosome.

(4) Elongation : Amino acids carried by the tRNA are added one by one at A site of ribosome in the sequence of the codons and become joined together to form

(5) Termination : When the termination codon present on the /mRNA is reached, the polypeptide synthesis stops and the polypeptide is released.

Question 5.
Define an operon. Giving an example, explain an inducible operon.
Answer:
Operon is a functionally integrated genetic unit for the control of gene expression in bacteria, as proposed in the Jacob-Monod hypothesis. Typically, it comprises a closely linked group of structural genes, coding for protein, and adjacent loci controlling their expression-an operator site and a promoter site. Inducible operons are those which normally remain switched off but can be induced to function under certain conditions. Lac operon is an inducible operon. The ; gene of the lac operon codes for a repressor molecule that inhibits synthesis of the structural genes. Lac operon gets switched ‘on’ in the presence of lactose. The repressor molecule coded by i gene is inactivated by interaction with the inducer (lactose). This allows RN Apolymerase access to the promoter, and transcription proceeds. The operon gets switched ‘off’ in the absence of lactose. The repressor molecule binds again with the operator region of the operon and prevents RNA polymerase from transcribing the operon.

Question 6.
There is a paternity dispute for a child’. Which technique can solve the problem. Discuss the principle involved.
Answer:
The paternity dispute of a child can be solved by DNA fingerprinting.
The procedure in DNA-fingerprinting includes the following:

1. Extraction – DNA is extracted from the cells in a high-speed, refrigerated centrifuge.
2. Amplification – Many copies of the extracted DNA are made by polymerase chain reaction.
3. Restriction digestion – DNA is cut into fragments with restriction enzymes into precise reproducible sequences.
4. Separation of DNA sequences/restriction fragments – The cut DNA fragments are introduced and passed through electrophoresis containing agarose polymer gel, the separated fragments can be visualised by staining them with a dye that shows fluorescence under ultraviolet radiation.
5. Southern blotting – The separated DNA sequences are transferred on a nitrocellulose or nylon membrane.
6. Hybridisation – The nylon membrane is immersed in a bath and radioactive probes are added, these probes target a specific nucleotide sequences that is complementary to them.
7. Autoradiography – The nylon membrane is pressed on an X-ray film and dark bands develop at the probe sites.

After hybridisation with the radio labelled Variable Number of Tandem Repeats (VNTR) probe and autoradiography, bands of various sizes are formed. The bands form a characteristic pattern which varies from individual to individual. From the patterns developed by the samples A and B it can be confirmed to whether they belong to one individual or two different individuals. If the banding patterns are similar they belong to the same individual but if banding patterns As. are dissimilar then A and B are from different individuals.

Question 7.
Give an account of the methods used in sequencing the human genome.
Answer:
The methods involved two major approaches :
(1) One approach called as Expressed Sequence Tags (ESTs), focussed on identifying all the genes that are expressed as RNA.
(2) Second approach called as sequence annotation, was to simply sequence the whole set of genome, that included all the coding and non-coding sequences and later assigning functions to different regions in the sequences. The steps involved are as follows :

• The total DNA from the cell is isolated and converted into random fragments of relatively smaller sizes.
• These fragments are then cloned in suitable hosts using specialised vectors. The commonly used hosts are bacteria and yeast and the vectors are bacterial artificial chromosomes (BAC) and yeast artificial chromosomes (YAC).
• The fragments are then sequenced using automated DNA sequencers, which work on the principle developed by Frederick Sanger.
• The sequences were then arranged on the basis of certain overlapping regions present in them. This requires the generation of overlapping fragments for sequencing.
• Specialised computer based programmes were developed for alignment of the sequences.
• These sequences were annotated and assigned to the respective chromosomes.
• The next task was to assign the genetic and physical maps on the genome, this was generated using the information on polymorphism of restriction endonuclease recognition sites and certain repetitive DNA, sequences, called microsatellites.

Question 8.
List the various markers that are used in DNA fingerprinting.
Answer:
Various markers used in DNA fingerprinting are as follows:

1. Restriction fragment length polymor­phism (RFLPs)
2. Variable number of tendem repeats (VNTRs)
3. Short tendem repeats
4. Microsatellite DNA
5. Single nucleotide polymorphism (SNPs)

Question 9.
Replication was allowed to take place in the presence of radioactive deoxynucleotide precursors in coli that was a mutant for DNA ligase. Newly synthesised radioactive DNA was purified and strands were separated by denaturation. These were centrifuged using density gradient centrifugation. Which of the following would be a correct result?


Answer:
(a) Two peaks will be obtained, one of high molecular weight strands that are long, continuous strands which were synthesised on leading strand. The other peak is obtained of low molecular weight strands which is due to short Okazaki fragments synthesised on lagging strand because mutated DNA ligase will not join the Okazaki fragments.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance help you. If you have any query regarding .NCERT Exemplar Solutions for Class 12 Biology chapter 6 Molecular Basis of Inheritance, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction

These Solutions are part of NCERT Exemplar Solutions for Class 12 Biology. Here we have given NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction

Multiple Choice Questions

Question 1.
Choose the incorrect statement from the following. 
(a) In birds and mammals internal fertilisation takes place.
(b) Colostrum contains antibodies and nutrients.
(c) Polyspermy in mammals is prevented by the chemical changes in the egg surface.
(d) In the human female implantation occurs almost seven days after fertilisation.
Answer:
(c) : Polyspermy is the entry of several sperms into the egg during fertilisation. It occurs in animals with yolky eggs (e.g., birds). In humans, although only one sperm nucleus actually fuses with the egg nucleus. Due to acrosomal reaction, plasma membrane of the sperm fuses with the plasma membrane of the secondary oocyte, so that the sperm contents enter the oocyte. Binding of the sperm to the secondary oocyte induces depolarisation of the oocyte plasma membrane. Depolarisation  prevents polyspermy.

Question 2.
Identify the correct statement from the following.
(a) High levels of estrogen triggers the ovulatory surge.
(b) Oogonial cells start to proliferate and give rise to functional ova in regular cycles from puberty onwards.
(c) Sperms released from seminiferous tubules are highly motile.
(d) Progesterone level is high during the post ovulatory phase of menstrual cycle.
Answer:
(d)

Question 3.
Spot the odd one out from the following structures with reference to the male reproductive system.
(a) Rete testis
(b) Epididymis
(c) Vasa efferentia
(d) Isthmus
Answer:
(d) : Females have two oviducts (or Fallopian tubes). Each oviduct has three regions, infundibulum, ampulla and isthmus. Isthmus is the narrow part closest to uterus, while infundibulum is the first part of oviduct that is encountered by released ova.

Question 4.
Seminal plasma, the fluid part of semen, is contributed by
(1) seminal vesicle
(2) prostate
(3) urethra
(4) bulbourethral gland
(a) (i) and (ii)
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) : Seminal plasma is the fluid part of semen and is contributed by seminal vesicles, prostate gland and bulbourethral glands. Seminal vesicles contribute fructose, citric acid and other nutrients as well as fibrinogen and prostaglandins. Secretions from prostate gland contain calcium ions, phosphate ion etc. and are alkaline in nature. Bulbourethral glands secrete alkaline mucus which is important for the lubrication of penis.

Question 5.
Spermiation is the process of the release of sperms from
(a) seminiferous tubules
(b) vas deferens
(c) epididymis
(d) prostate gland.
Answer:
(a) : Spermiation is the release of sperm from seminiferous tubules. From here, they will be transported to vasa efferentia and then to epididymis where maturation of sperm occurs.

Question 6.
Mature Graafian follicle is generally present in the ovary of a healthy human female around
(a) 5-8 day of menstrual cycle
(b) 11-17 day of menstrual cycle
(c) 18-23 day of menstrual cycle
(d) 24-28 day of menstrual cycle.
Answer:
(b) : Starting from puberty, monthly changes in ovaries and uterus also starts. The layer of cells surrounding primary oocyte are called granulosa cells and together, they are  called primary follicle. The growth of the follicles is under the influence of pituitary hormones FSH and LH. Further growth is also stimulated by estrogen. The mature Graafian follicle is present in the ovary around 11-17 day of menstrual cycle and on rupture of which, ovum is released (because of LH surge, 1-2 days before ovulation).

Question 7.
Acrosomal reaction of the sperm occurs due to
(a) its contact with zona pellucida of the ova
(b) reactions within the uterine environment of the female
(c) reactions within the epididymal environ­ment of the male
(d) androgens produced in the uterus.
Answer:
(a) : When head of sperm binds to zona pellucida (ZP) of ovum the acrosome release its contents by exocytosis. The content inside acrosome includes various hydrolytic enzymes like hyaluronidase, corona penetrating enzymes etc. This is called acrosomal reaction. It helps the sperm to reach the plasma membrane of ovum, by dissolving corona radiata and zona pellucida.

Question 8.
Which one of the following is not a male accessory gland?
(a) Seminal vesicle
(b) Ampulla
(c) Prostate
(d) Bulbourethral gland
Answer:
(b) : Ampulla is the part of oviduct between the infundibulum and isthmus. It  is at tfre ampullary-isthmus junction, where fertilisation take place.

Question 9.
The immature male germ cell undergoes division to produce sperms by the process of spermatogenesis. Choose the correct one with reference to above.
(a) Spermatogonia have 46 chromosomes and always undergo meiotic cell division.
(b) Primary spermatocytes divide by mitotic cell division.
(c) Secondary spermatocytes have 23 chromo­somes and undergo second meiotic division.
(d) Spermatozoa are transformed into spermatids.
Answer:
(c) : Spermatogonia are diploid cells on the inside wall of seminiferous tubules that multiply by mitotic divisions. Some of the spermatogonia called primary spermatocyte undergo meiosis-I to give rise to secondary spermatocytes (haploid). Each secondary spermatocyte undergoes meiosis-II to give rise to two haploid spermatids which are transformed to spermatozoa by spermiogenesis.

Question 10.
Match between the following representing parts of the. sperm and their functions and choose the correct option.

(a) A-{ii), B-(iv), C-(i), D-(iii)
(b) A-(iv), B-(iii), C-(i), D-(ii)
(c) A-(iv), B-(i), C-(ii), D-(iii)
(d) A-(ii), B-(i), C-(iii), D-(iv)
Answer:
(b)

Question 11.
Which among the following has 23 chromo­somes?
(a) Spermatogonia
(b) Zygote
(c) Secondary oocyte .
(d) Oogonia
Answer:
(c) : Spermatogonia are the cells on the inside wall of seminiferous tubules and consist of 46 chromosomes. Oogonia are also diploid cells and formed in the foetal ovary. Zygote is also diploid and fertilised ovum formed by fusion of male and female gametes. Secondary oocyte has 23 chromosomes and is formed by meiosis-I of primary oocyte.

Question 12.
Match the following and choose the correct option.

(a) A-(ii), B-(i), C-(iii), D-(iv)
(b) A-(iii), B-(iv), C-(ii), D-(i)
(c) A-(iii), B-(i), C-(ii), D-(iv)
(d) A-(ii), B-(iv), C-(iii), D-(i)
Answer:
(b)

Question 13.
Which of the following hormones is not secreted by human placenta?
(a) hCG
(b) Estrogens
(c) Progesterone
(d) LH
Answer:
(d) : LH is the hormone that is not secreted by human placenta. It is secreted by anteribr pituitary. Initially during pregnancy, hCG performs the function of LH of maintaining corpus luteum for the secretion of progesterone. Later (around 16^th week) in pregnancy, when placenta itself starts secreting estrogen and progesterone, corpus luteum regresses.

Question 14.
The vas deferens receives duct from the seminal vesicle and opens into urethra as
(a) epididymis
(b) ejaculatory duct
(c) efferent ductule
(d) Ureter
Answer:
(b)

Question 15.
Urethral meatus refers to the
(a) urinogenital duct
(b) opening of vas deferens into urethra
(c) external opening of the urinogenital duct
(d) muscles surrounding the urinogenial duct.
Answer:
(c) : Urethral meatus refers to the external opening of urinogenital duct, through which in males, urine and semen both exits the body.

Question 16.
Morula is a developmental stage
(a) between the zygote and blastocyst
(b) between the blastocyst and gastrula
(c) after the implantation
(d)between implantation and parturition.
Answer:
(a) : After fertilisation, zygote is formed. Zygote undergoes, mitotic divisions, called cleavage, to form 2, 4, 8 and 16 daughter cells called blastomeres. These divisions start when zygote is moving towards uterus. It is called morula at 8-16 cell stage and it continues to divide to form blastocyst. Blastocyst has two type of cells : trophoblast cells and inner cell mass. Around 7 day after fertilisation, implantation through trophoblast cells occurs.

Question 17.
The membranous cover of the ovum at ovulation is
(a) corona radiata
(b) zona radiata
(c) zona pellucida
(d) Chorion
Answer:
(a) : The outermost membranous cover of the ovum at ovulation is corona radiata. It is formed by follicular cells. Inner to corona radiata is zona pel lucid a, which is made up of three different glycoproteins secreted by the ovum itself.

Question 18.
Identify the odd one from the following.
(a) Labia minora
(b) Fimbriae
(c) Infundibulum
(d) Isthmus
Answer:
(a) : Labia minora are paired folds of tissues under labia majora which in turn surrounds the vaginal opening. Fimbriae, infundibulum and isthmus, along with ampulla are parts of oviduct (or Fallopian tube).

Very Short Answer Type Questions

Question 1.
Given below are the events in human reproduction. Write them in correct sequential order. Insemination, gametogenesis, fertilisation, parturition, gestation, implantation
Answer:
Gametogenesis, insemination, fertilisa­tion, implantation, gestation, parturition.

Question 2.
The path of sperm transport is given below. Provide the missing steps in blank boxes.

Answer:
Seminiferous tubules —>Rete testis —> Vasa efferentia —> Epididymis —> Vas deferens -> Urethra

Question 3.
What is the role of cervix in the human female reproductive system?
Answer:
Role of cervix in human female reproductive system:

1. Cervical canal e., the cavity of cervix and vagina together form birth canal to facilitate parturition.
2. It regulates the passage of sperms into the uterus.

Question 4.
Why are menstrual cycles absent during pregnancy?
Answer:
Levels of progesterone and estrogen are high during pregnancy. Their high levels suppress the release of gonadotropin (FSH) responsible for transformation of primary follicles into Graafian follicles, and for ovulation. And, no ovulation means, no menstrual cycle.

Question 5.
Female reproductive organs and associated functions are given below in column A and B. Fill the blank boxes.

Answer:
a – Fertilisation, b – Uterus

Question 6.
From where the parturition signals arise- mother or foetus? Mention the main hormone involved in parturition.
Answer:
Parturition is induced by complex neuroendocrine mechanism. Signals for parturition originate from fully developed foetus and placenta which induce mild uterine contractions called foetal ejection reflex. Oxytocin from maternal pituitary induces strong uterine contractions of myometrium, which leads to expulsion of baby (parturition) through birth canal.

Question 7.
What is the significance of epididymis in male fertility?
Answer:
Epididymis helps in storage, nutrition and physiological maturation of sperms. It also aids motility to sperms.

Question 8.
Give the names and functions of the hormones involved in the process of spermatogenesis. Write the names of the endocrine glands from where they are released.
Answer:
The name of hormones, its endocrine glands and functions of the hormones involved in the process of spermatogenesis are given in the following table:

Hormone	Endocrine gland	Function
GnRH (Gonado­ tropin releasing hormone)	Released from hypothalamus	Stimulates the anterior pituitary to release FSH and LH.
Follicle stimulating hormone (FSH)	Released from anterior pituitary	Stimulates Sertoli cells to secrete certain factors which help in spermatogenesis.
Luteinising hormone (LH)	Released from anterior pituitary	Stimulates Leydig’s cells of testes to secrete androgens (testosterone) which regulate

Question 9.
The mother germ cells are transformed into a mature follicle through series of steps. Provide the missing steps in the blank boxes.

Answer:

Question 10.
During reproduction, the chromosome number (2n) reduces to half (n) in the gametes and again the original number (2n) is restored in the offspring. What are the processes through which these events take place?
Answer:
Chromosome number is reduced to half (n) during the process of gametogenesis, and it is again restored to (2n) as a result of fertilisation.

Question 11.
What is the difference between a primary oocyte and a secondary oocyte?
Answer:
Primary oocyte is a diploid structure surrounded by single layer of follicular  through mitosis and differentiation. Secondary oocyte is a hapoid structure and is surrounded by a few layer of granular cells and theca. It is formed from primary oocyte after it undergoes first meiotic division.

Question 12.
What is the significance of ampullary-isthmic junction in the female reproductive tract?
Answer:
Ampullary-isthmic junction of oviduct is a site for fertilisation.

Question 13.
How does zona pellucida of ovum help in preventing polyspermy?
Answer:
During fertilisation, a sperm comes in contact with zona pellucida layer of ovum and induces changes in the membrane known as cortical reaction. This blocks the entry of additional sperms (polyspermy) and maintains monospermy.

Question 14.
Mention the importance of LH surge during menstrual cycle.
Answer:
Rapid secretion of LH leading to its maximum level during mid cycle is known as LH surge. It induces rupture of Graafian follicle and release of ovum (ovulation).

Question 15.
Which type of cell division forms spermatids from the secondary spermatocytes?
Answer:
Secondary spermatocytes undergo meiosis II to form spermatids. Secondary spermatocytes and spermatid both are haploid and this meiosis II is known as equational division.

Short Answer Type Questions

Question 1.
A human female experiences two major changes, menarche and menopause during her life. Mention the significance of both the events.
Answer:
The first menstrual cycle in females is known as menarche. It indicates the attainment of sexual maturity and puberty stage in females. It signifies the maturation and readiness of the female reproductive system for child bearing. It occurs usually at the age of 10-14 years.Menopause is the period of end of cyclic change in females. It indicates the end of reproductive cycle or fertile period in females. It occurs between 45-55 years of age.

Question 2.
(a) How many spermatozoa are formed from one secondary spermatocyte?
(b) Where does the first cleavage division of zygote take place?
Answer:
(a) A secondary spermatocyte gives rise to two spermatids, which get  transformed into two spermatozoa.
(b) First cleavage division of zygote takes place in Fallopian tubes of females.

Question 3.
Corpus luteum in pregnancy has a long life. However, if fertilisation does not take place, it remains active only for 10-12 days. Explain.
Answer:
Zygote formed after fertilisation gets implanted in the inner lining of uterus, called endometrium. Neural signals are conducted to hypothalamus to sustain LH secretion. This helps in maintaining corpus luteum and continue the secretion of progesterone during gestation period. But when fertilisation does not take place, neural signals are not sent to hypothalamus. As a result corpus luteum starts degenerating and can stay alive only for 10-12 days.

Question 4.
What is foetal ejection reflex? Explain how it leads to parturition?
Answer:
Foetal ejection reflex are the uterine contractions induced by fully developed foetus and placenta which signals for parturi­tion.
This stimulates the posterior pituitary of the mother to release oxytocin. Oxytocin causes stronger uterine contractions of smooth muscles of myometrium called labour pains, which further stimulate more secretion of oxytocin. The stimulatory reflex  continues resulting in stronger and stronger contractions and leading to expulsion of the baby (parturition) through birth canal.

Question 5.
Except endocrine function, what are the other functions of placenta?
Answer:
Functions of placenta other than endocrine functions are:

1. Nutritive organ – Food materials pass from the mother’s blood into the foetal blood through the placenta.
2. Digestive organ – The trophoblast of the placenta digests (breaks down) proteins before passing them into the foetal blood.
3. Respiratory organ – Oxygen diffuses from the maternal blood into the foetal blood through the placenta. Carbon dioxide diffuses from the foetal blood into the maternal blood also through the placenta for elimination by the mother’s lungs. Foetal haemoglobin has a greater affinity for oxygen than adult haemoglobin.
4. Excretory organ – Nitrogenous wastes, such as urea, pass from the foetal blood into the maternal blood via placenta for elimination by mother’s kidneys.
5. Storage organ – The placenta stores glycogen for the foetus before liver is formed.
6. Barrier – Placenta serves as an efficient barrier and allows those materials to pass into the foetal blood that are necessary.

Question 6.
Why doctors recommend breast feeding during initial period of infant growth?
Answer:
Human milk consists of water, minerals, fats, proteins and sugar necessary for development of the child. The milk produced’ during initial days of lactation is called colostrum. It is rich in proteins (lactalbumin and lactoprotein) and various other nutrients. It also contains certain antibodies (IgA), which provide passive immunity to the baby.

Question 7.
What are the events that take place in the ovary and uterus during follicular phase of the menstrual cycle.
Answer:
(1) Changes inside the ovary during follicular phase :-

• Primary follicle gets transformed into Graafian follicle.
• Increase in level of estrogen

(2) Changes in uterus during follicular phase :-

• Proliferation of endometrium.
• Endometrium becomes highly vascular and glandular.

Question 8.
Given is a flow chart showing ovarian changes during menstrual cycle. Fill in the spaces giving the name of the hormones responsible for events shown.

Answer:
a – FSH and LH
b – LH
c – Progesterone

Question 9.
Give a schematic labelled diagram to represent oogenesis (without descriptions).
Answer:

Question 10.
What are the changes in the oogonia during the transition of a primary follicle to Graafian follicle?
Answer:
Oogonia divide by mitosis forming primary oocyte which then gets surrounded by a layer of granulosa cells to form primary follicle. The primary follicles are surrounded by more layers of granulosa cells called secondary follicles.The secondary follicle soon changes into a tertiary follicle which is characterised by a fluid filled cavity called follicular antrum. The tertiary follicle is further converted into mature follicle or Graafian follicle. The formation of Graafian follicle through various stages is called folliculogenesis.

Long Answer Type Questions

Question 1.
What role does pituitary gonadotropins play during follicular and ovulatory phases of menstrual cycle? Explain the shifts in steroidal secretions.
Answer:
The menstrual cycle consists of four phases: menstrual phase, follicular phase (proliferative phase), ovulatory phase and secretory phase. Hormonal control during follicular phase:

• FSH from pituitary transforms the primary follicle into Graafian follicle.
• FSH also stimulates Graafian follicular cells to secrete estrogen.
• Estrogen causes proliferation of endo­metrium of the uterine wall.

Hormonal control during ovulatory phase :

• Pituitary glands are stimulated by estrogen to release LH.
• LH surge (maximum level of LH) causes ovulation from the ovary on 14^th day of menstruation cycle. LH also induces transformation of Graafian follicle into corpus luteum, inside the ovary. LH stimulates corpus luteum to secrete progesterone to help implantation,placentation, and maintenance of pregnancy.
• In menstrual phase, there is reduction of progesterone and estrogen. Gonado­tropin releasing hormone (GnRH) stimulates the release of FSH and LH. FSH stimulates the ovarian follicles to produce estrogens during proliferative phase. LH stimulates the ovulation in ovulatory phase.
• LH develops corpus luteum which causes increased production of progesterone in secretory phase.

Question 2.
Meiotic division during oogenesis is different from that in spermatogenesis. Explain how and why?
Answer:
Primary spermatocyte divides by meiosis I to form two secondary spermato­cytes. Primary oocyte undergoes meiosis I to form one secondary oocyte and one polar body.
Secondary spermatocyte divides by meiosis II to produce two spermatids. Secondary oocyte divides by meiosis II to form one ovum and one polar body.
A spermatocyte forms four spermatozoa. An oocyte forms only one egg or ovum. Unequal cell divisions during oogenesis, makes the ovum much larger than the other three polar bodies. Ovum has more cytoplasm and more organelles, it has a better chances of surviving. The male makes million of tiny sperms while, the female makes one egg per month because the sperms have to search for the eggs and only a few succeed in this task.

Question 3.
The zygote passes through several developmental stages till implantation. Describe each stage briefly with suitable diagrams.
Answer:

1. The zygote moves from isthmus to uterus and undergo mitotic divisions called cleavage divisions.
2. It forms 2, 4, 6, 8,16 daughter cells, called  blastomeres.
3. The embryo with 8-16 blastomeres is called morula.
4. The morula continues to get transformed into blastocyst.
5. The blastomeres in the blastocyst are arranged into an outer layer called trophoblast, and an inner group of cells attached to trophoblast called inner cell mass.
6. The trophoblast layer gets attached to the endometrium and inner cell mass gets differentiated as the embryo.
7. After attachment uterine cells divide rapidly and cover the blastocyst.
8. As a result the blastocyst becomes embedded in the endometrium of the uterus.
9. This is known as implantation, and it leads to pregnancy.
   

Question 4.
Draw a neat diagram of thefemale reproductive system and label the parts associated with the following (a) production of gamete, (b) site of fertilisation(c) site of implantation and, (d) birth canal.
Answer:

Question 5.
With a suitable diagram, describe the organisation of mammary gland.
Answer:
Mammary glands are paired structures consisting of glandular tissue, the fibrous tissue and variable amount of fat. They are compound tubulo-alveolar modified sweat glands. The glandular tissue of breast consists of about 15-20 lobes of milk glands. Each lobe is formed of many lobules containing cluster of cells called alveoli. The cells of alveoli secrete milk, which is stored in the lumen (cavities) of alveoli. The alveoli open into mammary  tubules, the tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla, which is connected to lactiferous duct, through which milk is sucked out. Mammary glands are functional in females and vestigial in males. Breasts in females are small sized upto puberty. The size increases after puberty under stimulation of estrogen. It further increases during pregnancy and after child birth under the stimulation of prolactin and progesterone.

We hope the NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction help you. If you have any query regarding NCERT Exemplar Solutions for Class 12 Biology chapter 3 Human Reproduction, drop a comment below and we will get back to you at the earliest.

Here we are providing 1 Mark Questions for Economics Class 12 Chapter 3 Liberalisation, Privatisation and Globalisation: An Appraisal are the best resource for students which helps in class 12 board exams.

One Mark Questions for Class 12 Economics Chapter 3 Liberalisation, Privatisation and Globalisation: An Appraisal

Question 1.
When were economic reforms introduced in India?
Answer:
Economic reforms were introduced in India in 1991. Economic reforms refer to all those measures that aim at rendering the economy more efficient, competitive and developed.

Question 2.
List any two reasons which led to economic reforms in India.
Answer:
The reasons which led to economic reforms in India include:
(i) Unfavourable Balance of Payment
(ii) Inflation
(iii) Falling foreign exchange reserves

Question 3.
What are the three broad components of New Economic Policy, 1991?
Answer:
The three broad components of New Economic Policy are:
(i) Liberalisation
(ii) Privatisation
(iii) Globalisation

Question 4.
Define liberalisation.
Answer:
Liberalisation means liberating the trade and industry of an economy from unnecessary restrictions and making the industries more competitive.

Question 5.
State any two reforms introduced under liberalisation.
Answer:
The reforms introduced under liberalisation include:
(i) Deregulation of industrial sector
(ii) Trade and investment policy reforms
(iii) Tax reforms

Question 6.
What is fiscal policy?
Answer:
It refers to the revenue and expenditure policy of the government to achieve balanced development in the economy.

Question 7.
Define direct tax. Give two examples.
Answer:
Direct taxes are those taxes levied immediately on the property and income of persons, and are paid directly by the consumers to the state. For example, income tax, property tax.

Question 8.
Define indirect tax. Give two examples.
Answer:
Indirect tax is a tax collected by an intermediary (seller) from the person who bears tne ultimate economic burden of the tax (buyer). For example, excise duty, sales tax.

Question 9.
What was the consequence of devaluation of rupee?
Answer:
Devaluation of rupee led to huge inflow of foreign exchange in India.

Question 10.
List the aims of trade policy reforms.
Answer:
The aims of trade policy reforms were:
(i) Removal of quantitative restrictions
(ii) Reduction in tariff rates
(iii) Removal of import licensing

Question 11.
For what categories of products was industrial licensing not abolished?
Answer:
Industrial licensing was not abolished for product categories such as alcohol, cigarettes, hazardous chemicals, industrial explosives, electronics, aerospace and drugs and pharmaceuticals.

Question 12.
Define privatisation.
Answer:
Privatisation means the induction of private management and control in the public sector enterprises.

Question 13.
What is disinvestment?
Answer:
Disinvestment involves selling a part of the Public Sector Undertaking’s equity to the public to promote privatisation.

Question 14.
State the purpose for undertaking disinvestment.
Answer:
Disinvestment was undertaken:
(i) to maintain fiscal discipline; and
(ii) to facilitate modernisation.

Question 15.
Define globalisation.
Answer:
Globalisation means unification or integration of the domestic economy with the world economy.

Question 16.
What is outsourcing?
Answer:
It is the practice of hiring external sources, mostly from other countries, for regular services.

Question 17.
List a few services which are being outsourced by companies in developed countries to India.
Answer:
A few services which are being outsourced by companies in developed countries to India are:
(i) Record keeping
(ii) Accountancy
(iii) Banking services
(iv) Music recording
(v) Film editing
(vi) Clinical advice

Question 18.
How are WTO and GATT related?
Answer:
GATT was established in 1948. WTO was founded in 1995 as the successor organisation to GATT.

Question 19.
Where is the headquarters of WTO?
Answer:
The headquarters of WTO is in Geneva.

Question 20.
What has been the impact of economic reforms on GDP?
Answer:
The overall GDP growth has increased as a result of economic reforms.

Question 21.
List the areas which were ignored during the reform period.
Answer:
The sectors which were ignored during the reform period are:.
(i) Agriculture
(ii) Industry
(iii) Employment
(iv) Infrastructure
(v) Fiscal management

Question 22.
Name the sector that benefited the most with the introduction of economic reforms in India.
Answer:
Service (tertiary) sector benefitted the most with the introduction of economic reforms in India

Question 23.
Define GST.
Answer:
GST (Goods and Services Tax) is an indirect tax for the whole nation, which will make India one unified common market.

Question 24.
Why is GST implemented?
Answer:
(i) GST will create a simpler tax system.
(ii) It increases overall transparency and compliance.

Question 25.
When was GST implemented in India?
Answer:
1st July 2017

Question 26.
Who is the head of the GST Council?
Answer:
Finance Minister

Question 27.
Which constitutional amendment is done to pass the GST bill?
Answer:
101 st

Question 28.
What type of goods are not covered under the GST bill?
Answer:
(i) Cooking gas
(ii) Liquor
(iii) Petrol

Question 29.
List the main categories of GST.
Answer:
(i) CGST
(ii) SGST
(iii) I GST

Question 30.
What is demonetisation?
Answer:
Demonetisation is the act of stripping a currency unit of its status as legal tender.

Question 31.
When did demonetisation take place in India?
Answer:
8th November, 2016

Question 32.
What was main motive behind demonetisation?
Answer:
To curb black money, terror funding and to stop the use of fake currency available in the market

Question 33.
When did demonetisation take place in India for the first time in history?
Answer:
In 1946

Question 34.
Which currency notes were affected due to demonetisation in November 2016?
Answer:
₹ 500 & ₹ 1,000 notes
(Liberalisation, Privatisation and Globalisation: An Appraisal)

Question 35.
Which currency notes were newly implemented after demonetisation in November 2016?
Answer:
X 200 & X 2,000 notes

Question 36.
What was the last date of tendering old currency?
Answer:
30th December, 2016

Question 37.
State one positive effect of demonetisation in India?
Answer:
Over fake currency

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 3 Electrochemistry. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 3 Important Extra Questions Electrochemistry

Electrochemistry Important Extra Questions Very Short Answer Type

Question 1.
Write the name of the cell which Is generally used in hearing aids. Write
the reactions taking place at the anode and the cathode of this cell. (CBSE AI 2017)
Answer:
Mercury cell

At Cathode: HgO + H₂O + 2e^– → Hg (I) + 2OH^–

Question 2.
Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell. (CBSE Al 2017)
Answer:
Dry cell
Anode: Zn (s) → Zn²⁺ + 2e^–
Cathode: MnO₂ + NH₄⁺ + e^– → MnO(OH) + NH₃

Question 3.
Write the name of the cell which is generally used in inverters. Write the reactions taking place at the anode and the cathode of this cell. (CBSE Al 2017)
Answer:
Lead storage battery
Anode: Pb (s) + SO₄^2-(aq) → PbSO₄(s) + 2e^–
Cathode: PbO₂ + SO₄^2-(aq) + 4H⁺ + 2e^– → PbSO₄(s) + 2H₂O(l)

Question 4.
From the given cells:
Lead storage cell, Mercury cell, Fuel cell and Dry cell Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life? (CBSE Delhi 2016)
Answer:
(i) Mercury cell
(ii) Fuel cell
(iii) Lead storage cell
(iv) Dry cell

Question 5.
What happens if external potential applied becomes greater than E°el, of electrochemical cell? (CBSE Al 2016)
Answer:
If E°_cell (external) is greater than E°_cell, the cell starts acting as an electrolytic cell. In this case, electrical energy is used to carry out non-spontaneous chemical reaction.

Question 6.
Using the E° values of A and B, predict which one is better for coating the surface of iron [E°_(Fe²⁺/Fe) = – 0.44V] to prevent corrosion and why?
Given: E°_(A²⁺|A) = -2.37 V and E°_(B²⁺|B) = – 0.14 V (CBSE Al 2016)
Answer:
‘A’ will prevent iron from rusting. So, we can coat the iron surface with metal A because it has more negative value.

How to calculate emf of a cell Calculate the emf of the cell in which the following reaction takes place.

Question 7.
Given that the standard electrode potentials (E°) of metals are:
K⁺/K = – 2.93V, Ag⁺/Ag = 0.80V, Cu²⁺/Cu = 0.34V, Mg²⁺/Mg = – 2.37V, Cr³⁺/Cr = – 0.74V and Fe²⁺/Fe = – 0.44V. (CBSE Al 2010)
Arrange the metals in the increasing order of their reducing power.
Answer:
Ag⁺/Ag < Cu²⁺/Cu < Fe²⁺/Fe < Cr³⁺/Cr < Mg²⁺/Mg < K⁺/K.

Question 8.
What is change in free energy for
(a) galvanic cell and
(b) electrolytic cell?
Answer:
(a) For a galvanic cell, free energy decreases, i.e. ΔG < 0.
(b) For electrolytic cell, free energy increases, i.e. ΔG > 0.

Question 9.
What is role of ZnCl₂ in a dry cell?
Answer:
ZnCl₂ combines with NH₃ produced to form the complex [Zn(NH₃)₂Cl₂], otherwise the pressure developed due to NH₃ would crack the seal of the cell.

Question 10.
When the silver electrode having reduction potential 0.80 V is attached to NHE, will it act as anode or cathode?
Answer:
It will act as cathode.

Question 11.
What is the effect of carbon dioxide in water on corrosion?
Answer:
The presence of carbon dioxide in water increases rusting of iron. Water containing CO₂ acts as an electrolyte and increases the flow of electrons from one place to another.

Question 12.
Why is it not possible to measure the voltage of an isolated half reaction?
Answer:
It is not possible to measure the voltage of an isolated half reaction because neither the oxidation nor the reduction can occur by itself. Therefore, we can only calculate the relative electrode potential by connecting it to some standard electrode.

Question 13.
Why does a dry cell become dead after a long time, even if it has not been used?
Answer:
A dry cell becomes dead after a long time because the acidic NH₄Cl corrodes the zinc container.

Question 14.
Why does the cell potential of mercury cell remain constant throughout its life? (CBSE AI 2015)
Answer:
This is because the overall cell reaction does not involve any ion in the solution whose concentration changes during its life time.

Question 15.
How can you increase the reduction potential of an electrode?
Answer:
By increasing the concentration of the ions

Question 16.
The E° values of MnO₄^–, Ce⁴⁺ and Cl₂ are 1.507, 1.61 and 1.358 V respectively. Arrange these in order of increasing strength as oxidising agent.
Answer:
Cl₂ < MnO₄^– < Ce⁴⁺.

Question 17.
E° values for Fe³⁺/Fe²⁺ and Ag⁺/Ag are respectively 0.771 V and 0.800 V. Is the reaction:
Fe³⁺ + Ag → Fe²⁺ + Ag⁺ spontaneous or not?
Answer:
E° for the reaction is 0.771 – 0.800 = – 0.029 V.
Therefore, the reaction is not spontaneous.

Question 18.
What is the use of platinum foil in the hydrogen electrode?
Answer:
It is used for inflow and outflow of electrons.

Question 19.
What is meant by limiting molar conductivity? (CBSE 2010)
Answer:
The molar conductivity at infinite dilution or when concentration approaches zero is called limiting molar conductivity.

Question 20.
Express the relation among the conductivity of solution in the cell, the cell constant and the resistance of solution in the cell. (CBSE Delhi 2011)
Answer:
The conductivity (K), cell constant (G°) and resistance (R) of the solution are related as:
K = G° × 1/R

Question 21.
Explain the relation between conductivity and molar conductivity of a solution held in a ceil. (CBSE Delhi 2011)
Answer:
Conductivity (κ) and molar conductivity are related as:
∧_m = \(\frac{\kappa \times 1000}{M}\)
where M is the molarity of the solution.

Question 22.
Is it safe to stir 1 M AgNO₂ solution with a copper spoon?
Given E° (Ag⁺/Ag) = 0.80 V, E° (Cu²⁺/Cu) = 0.34 V. Explain.
Answer:
No, copper spoon will dissolve as Cu²⁺ ions because copper has more tendency to get oxidised than silver.

Question 23.
The e.m.f. of the cell:
Zn | Zn²⁺ (1M) || H⁺ (1M) | H₂ (1 atm),
Pt is 0.76 V. What is the electrode potential of Zn²⁺/Zn electrode?
Answer:
E°_cell = E°(H⁺/H₂) – E°(Zn²⁺/Zn)
0. 76 = 0 – E°(Zn²⁺/Zn)
∴ E°(Zn²⁺/Zn) = – 0.76 V.

Question 24.
Write relationship between
(i) standard free energy change and e.m.f. of a cell.
(ii) standard free energy change and equilibrium constant.
Answer:
(i) ∆G° = – nFE°_cell
(ii) ∆G° = – RT In K_c

Question 25.
Give the units of specific conductance and molar conductance.
Answer:
Specific conductance: ohm^-1 cm^-1
Molar conductance: ohm^-1 cm² mol^-1

Question 26.
Give one example each of primary cell and secondary cell.
Answer:
Primary cell: Dry cell
Secondary cell: Lead storage battery

Question 27.
How are cell constant and specific conductance related to one another?
Answer:
Specific conductance = Cell constant × Conductance.

Question 28.
Why is the equilibrium constant K related to only E°_cell and not E_cell?
Answer:
This is because at equilibrium E_cell = 0.

Question 29.
What is the effect of decreasing concentration on the molar conductivity of weak electrolytes?
Answer:
With the decrease in concentration of weak electrolytes, the molar conductivity increases.

Question 30.
Why is it not possible to determine the potential of a single electrode?
Answer:
Oxidation and reduction cannot occur alone. Therefore, it is not possible to measure single electrode. Moreover, it is a relative tendency and can be measured with respect to a reference electrode only.

Question 31.
Suggest a metal which can be used for cathodic protection of iron?
Answer:
Zinc.

Question 32.
Write the overall cell reaction for lead storage battery.
Answer:
The overall cell reaction is:
Pb (s) + PbO₂ (s) + 4H⁺ (aq) + 2SO₄^2- (aq) → 2PbSO₄ (s) + 2H₂O.

Electrochemistry Important Extra Questions Short Answer Type

Question 1.
The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1? (CBSE AI 2008; CBSE Delhi 2007, 2008, 2012)
Ans.
Conductivity, K = 0.146 × 10^-3 S cm^-1
Resistance, R = 1500 ohm

= Conductivity (κ) × Resistance (R)
∴ Cell constant = 0.146 × 10^-3 ohm^-1 cm^-1 × 1500 ohm = 0.219 cm^-1.

Question 2.
The conductivity of 0.20 M KCl solution at 298 K is 0.025 S cm^-1. Calculate its molar conductivity.
Answer:
Molar conductivity
∧_m = \(\frac{\kappa \times 1000}{\mathrm{C}}\)
k = 0.025 cm^-1, C = 0.20 M
∴ ∧_m = \(\frac{0.025 \times 1000}{0.20}\) = 125.0 S cm² mol^-1

Question 3.
Resistance of a conductivity cell filled wIth 0.1 M KCl solution is 100. If the resistance of the same cell when filled with 0.02 M KCl solution is 520Ω, calculate the conductivity and molar conductivity of 0.02 M KCl solution. (The conductivity of 0.1 M KCl solution is 1.29 S m^-1.) (CBSE AI 2006, CBSE Delhi 2014)
Answer:
Step 1. Let us first calculate the cell constant.
Cell constant, G* = Conductivity (κ) × Resistance (R)
Resistance of 0.1 M KCl solution = 100 Ω
Conductivity of 0.1 M KCl solution = 1.29 S m^-1
∴ Cell constant = 1.29 (S m^-1) × 100Ω = 129 m^-1
or = 1.29 cm^-1

Step II. Calculation of conductivity of 0.02 M KCl soLution.
Resistance of solution = 520 Ω
Cell constant (G*) = 1.29 cm^-1

Step III. Calculation of molar conductivity.
∧_m = \(\frac{1000 \times \kappa}{C}\)
C = 0.02 M, κ = 0.248 × 10^-2 S cm^-1
∴ ∧_m = \(\frac{1000 \times 0.248 \times 10^{-2}}{0.02}\) – 124 S cm² mol^-1

Question 4.
The electrical resistance of a column of 0.05 M NaOH⁺ solution of diameter 1 cm and length 50 cm is 5.55 × 10³ ohm. Calculate its
(i) resistivity
(ii) conductivity, and
(iii) molar conductivity. (CBSE AI 2012)
Answer:
Cell constant, G* = \(\frac{1}{a}\)
l = 50 cm, diameter = 1 cm
∴ radius = 0.5 cm
Area of cross-section, a = πr²
= 3.14 × (0.5)² = 0.785 cm²
∴ G* = \(\frac{50}{0.785}\) = 63.694 cm^-1

(i) Resistivity,

(ii) Conductivity,
κ = \(\frac{1}{\rho}=\frac{1}{87.135}\) = 1.48 × 10^-2 cm^-1

(iii) Molar conductivity,
∧_m = \(\frac{\kappa \times 1000}{\mathrm{C}}\)
c = 0.05 M
∴ ∧_m = \(\frac{1.148 \times 10^{-2} \times 1000}{0.05}\) = 229.6 s cm² mol^-1

Question 5.
The molar conductivities at infinite dilution for sodium acetate, hydrochloric acid and sodium chloride are 91.0, 425.9 and 126.4 S cm² mol^-1 respectively at 298 K. Calculate the molar conductivity of acetic acid at infinite dilution. (CBSE Delhi 2010)
Answer:
Molar conductivity at infinite dilution for acetic acid can be calculated as:
∧° (CH₃COOH ) = λ°_H⁺ + ∧°CH₃COO^–
= λCH₃COO^– + λ°_Na⁺ + λ°_H⁺ + λ°_cl^– – λ°_Na⁺ – λ°_cl^–
∧°(CH₃COOH⁺) = ∧°(CH₃COONa) + ∧° (H⁺Cl) – ∧° (NaCl)
∧°(CH₃COONa) = 91.0 S cm² mol^-1
∧°(H⁺Cl) = 425.9 S cm² mol^-1
∧°(NaCl) = 126.4 S cm² mol^-1
∴ ∧°(CH₃COOH⁺) = 91.0 + 425.9 – 126.4 = 390.5 S cm² mol^-1.

Question 6.
(a) Write the reaction that occurs at anode on electrolysis of concentrated H₂SO₄ using platinum electrodes.
(b) What is the effect of temperature on ionic conductance? (CBSE Al 2019)
Answer:
(a) At anode:
2SO₄^2-(aq) → S₂O₈^2-(aq) + 2e^–
(b) Ionic conductance will increase with increase in temperature.

Question 7.
Write anode and cathode reactions that occur in dry cell. How does a dry cell differ from a mercury cell? (CBSE Al 2019)
Answer:
Anode: Zn → Zn²⁺ + 2e^–
Cathode:
NH₄⁺(aq) + MnO₂(s) + e^– → MnO(OH) + NH₃
The potential of a dry cell decreases slowly but continuously, while the potential of mercury cell remains constant throughout its life.

Question 8.
Calculate the limiting molar conductivity of CaS04 if limiting molar conductivities of calcium and sulphate ions are 119.0 and 106.0 S cm² mol^-1 respectively. (CBSE Sample Paper 2012)
Answer:
∧°m (CaSO₄) = λm (Ca²⁺) + λ°m (SO₄^2-)
= 119.0 + 106.0
= 225.0 S cm^2- mol^-1.

Question 9.
Arrange the following solutions in the decreasing order of specific conductance.
(i) 0.01M NaCl (ii) 0.05M NaCl (iii) 0.1M NaCl (iv) 0.5M NaCl
Ans.
(iv) 0.5M NaCl > (iii) 0.1M NaCl > (ii) 0.05M NaCl > (i) 0.01M NaCl

Question 10.
Write the Nernst equation and calculate the e.m.f. of the following cell at 298 K:
Cu(s) | Cu²⁺ (0.130 M) | | Ag⁺ (1.0 × 10^-4 M)) | Ag (s)
Given: E^⊖_(Cu²⁺/Cu) = + 0.34V and E^⊖_(Ag²⁺/Ag) = + 0.80V (CBSE Al 2004)
Answer:
The electrode reactions and cell reaction are:

Question 11.
Calculate E^⊖_cell, for the following reaction at 298K:
2Cr(s) + 3Fe²⁺ (0.01 M) → 2Cr³⁺ (0.01M) + 3Fe(s)
(Given: E_cell = 0.261v)                (CBSE AI 2016)
Answer:

Question 12.
For the reaction:
2AgCl (s) + H₂ (g) (1atm) → 2Ag(s) + 2H⁺ (0.1M) + 2Cl^– (0.1M)
ΔG° = – 43600 J at 25°C
Calculate the e.m.f of the cell. (CBSE AI 2018)
Answer:

Question 13.
Write the name of two fuels other than hydrogen used in fuel cell. Write two advantages of fuel cell over an ordinary cell. (CBSE Al 2019)
Answer:
Methane, oxygen
Advantages of fuel cells:
(i) Fuel cells have greater efficiency than ordinary cells.
(ii) These do not produce any harmful by-product and therefore, do not cause any pollution.

Question 14.
Define conductivity and molar conductivity for the solution of an electrolyte. Why does the conductivity of solution decrease with dilution? (CBSE 2019C)
Answer:
Conductivity of a solution at any given concentration is the conductance of one unit volume of solution kept between two platinum electrodes with unit area of cross section and at a distance of unit length.

Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. Therefore,
∧_m = \(\frac{\mathrm{κA}}{\mathrm{l}}\)
Since l = 1 and A = V (Volume containing 1 mole of electrolyte).
∧_m = κV
Conductivity always decreases with decrease in concentration both for weak and strong electrolytes. It is because the number of ions per unit volume that carry the current in solution decreases on dilution.

Question 15.
Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity? (CBSE AI 2012)
Answer:
The conductivity (κ), cell constant (G*) and resistance (R) of solution are related as:

Molar conductivity (∧_m) is related to conductivity (κ) as:
∧_m = \(\frac{\kappa \times 1000}{M}\)

Question 16.
The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm² mol^-1. Calculate the conductivity of this solution. (CBSE AI 2012)
Answer:
∧_m = 138.9 S cm² mol^-1, M = 1.5 M
138.9 = \(\frac{\kappa \times 1000}{1.5}\)
∴ κ = \(\frac{138.9 \times 1.5}{1000}\) = 0.208 ohm^-1 cm^-1.

Electrochemistry Important Extra Questions Long Answer Type

Question 1.
(a) Following reaction takes place in the cell:
Zn (s) + Ag₂O (s) + H₂O (l) → Zn²⁺ (aq) + 2Ag (s) + 20H^– (aq)
Calculate Δ_rG° of the reaction.
[Given: E°_(Zn²⁺/Zn) = – 0.76 V,
E°_(Ag²⁺/Ag) = 0.80 V, 1 F = 96,500 C mol^-1]
Answer:
E°_cell = E°_(Ag²⁺/Ag) – E°_(Zn²⁺/Zn)
= 0.80 – (-0.76) = 1.56V
ΔG° = -nFE°_cell          (Here n = 2)
= – 2 × 96500 × 1.56
= – 301080 J mol^-1
or = – 301.080 kJ mol^-1

(b) How can you determine limiting molar conductivity (∧_m°,) for strong electrolyte and weak electrolyte? (CBSE Al 2019)
Answer:
For strong electrolytes ∧_m° can be obtained as intercept from the plot of Am vs C^-1/2 graph. For weak electrolytes ∧_m° can be obtained from Kohlrausch law.

Question 2.
(i) Explain the following:
(a) CO₂ is always present in natural water. Explain its effect (increases, stops or no effect) on rusting of iron.
(b) Rusting of iron is quicker in saline water than in ordinary water. Explain.
Answer:
(i) (a) Presence of CO₂ in natural water increases rusting of iron. It dissolves in water to form H₂CO₃ which gives H⁺ ions. The H⁺ ions accelerate the process of corrosion.
In rusting of iron, Fe oxidises to Fe²⁺ ions
Fe (s) → Fe²⁺ (aq) + 2e^– (anode)
The released electrons go to the cathode and reduce oxygen in the presence of H⁺ ions (obtained from H₂CO₃). The reaction occurs at cathode. Thus, CO₂ increases rusting.
O₂ (g) + 4H⁺(aq) + 4e^– → 2H₂O (l)

(b) Rusting of iron is quicker in saline water (salt water) than in ordinary water. This is mainly due to the fact that saline water increases the electrical conduction of electrolyte solution formed on the metal surface. Therefore, rusting becomes more serious problem where salt water is present.

(ii) Discuss electrical protection for preventing rusting of iron pipes in underground water.
Answer:
In this method iron articles which are in contact with water such as underground water pipes are protected from rusting. The article of iron is connected with more active metals like magnesium or zinc. This prevents its tendency to lose electrons and therefore, corrosion is prevented. The cathodes of magnesium of zinc can be fixed to the surface of iron or burned in sub-soil water near by the pipes.

Question 3.
Give the construction and working of hydrogen standard electrode potential?
Answer:
The standard hydrogen electrode consists of platinum wire sealed in a glass tube and has a platinum foil attached to it. The foil is coated with finely divided platinum and acts as platinum electrode. It is dipped into an acid solution containing H⁺ ions in 1 M concentration (1 M HCl). Pure hydrogen gas at 1 atmosphere pressure is constantly bubbled into the solution at constant temperature of 298 K. The surface of the foil acts as a site for the reaction. This is shown in figure.
The following reactions occur in this half cell depending upon whether it acts as an anode or as a cathode:

If S.H.E. acts as anode:
H₂(g) → 2H⁺ + 2e^–
If S.H.E. acts as cathode:
2H⁺ + 2e^– → H₂(g)
The electrode potential of an electrode can be determined by connecting this half cell with a standard hydrogen electrode. The electrode potential of the standard hydrogen electrode is taken as zero.

Measurement of the standard electrode potential (E°): The standard electrode potential of a metal electrode is measured with respect to a standard hydrogen electrode. A cell is prepared in which the metal electrode constitutes one half cell and the S.H.E. as the other half cell (anode). The electrons released by the metal in the oxidation half cell are accepted by the H+ ions of the acid in the reduction half cell.
The cell may be represented as:
Pt(s) | H₂(g, 1 atm) | H⁺ (aq, 1 M) || Mⁿ⁺ (aq, 1M) | M
Now, e.m.f. of cell
e.m.f. = E_R – E_L
Since the potential of S.H.E. has been fixed to be zero, i.e. E_L = 0 so that
e.m.f. = E_R – 0
or E_R = e.m.f.
From the knowledge of e.m.f. of the cell, the electrode potential of the electrode can be calculated. For example, if we wish to determine the electrode potential of zinc electrode in 1 M solution of ZnSO₄, it is combined with S.H.E. The e.m.f. of the cell is found to be – 0.76 V so that
e.m.f. = E_L – E_L
– 0.76 = E_R – 0
or E_R = – 0.76 V.

Question 4.
What are fuel cells? Discuss briefly hydrogen-oxygen fuel cell?
Answer:
Fuel cells. These are voltaic cells in which the reactants are fed continuously to the electrodes. These are designed to convert the energy from the combustion of fuel such as H₂, CO, CH₄, etc. directly into electrical energy. The common example is hydrogen-oxygen fuel cell as described below:

Fig. A simple H₂ – O₂ fuel cell.

In this cell, hydrogen and oxygen are bubbled through a porous carbon electrode into concentrated aqueous sodium hydroxide as shown in figure. The diffusion rates of the gases into the cell are carefully regulated to get maximum efficiency. In the anode compartment hydrogen is oxidised while oxygen in the cathode compartment is reduced. The net reaction is the same as burning of hydrogen and oxygen to form water.
The reactions are given below:
At anode:
2 [H₂ (g) + 2OH^–(aq) → 2H₂O (l) + 2e^–]

At cathode:
O₂ (g) + 2H₂O (l) + 4e^– → 4OH^–(aq)

Overall reaction:
2H₂ (g) + O₂(g) → 2H₂O (l)
The catalysts (Pt, Ag or Co) are also added.
This cell runs continuously as long as the reactants are fed. These fuel cells are more efficient than conventional methods of generating electricity on a large scale by burning hydrogen, carbon fuels because these fuel cells convert the energy of the fuel directly into electricity.
The cell has been used for electric power in the Apollo space programme.

Question 5.
What is corrosion? What are the factors which affect corrosion?
Answer:
Corrosion. It is a process of eating away of metals when exposed to the atmosphere surrounding it. Many metals when exposed to the atmosphere, react with air or water in the environment to form undesirable compound on their surface. In case of iron, the corrosion is called rusting. The red or orange coating that forms on the surface of iron when exposed to air and moisture is called rust. Chemically, rust is a hydrated form of ferric oxide, Fe₂O₃ . xH₂O.
Factors which affect corrosion. The main factors which affect corrosion are:
1. Position of metals in e.m.f. series: The reactivity of a metal depends upon its position in the electrochemical series. More the reactivity of the metal, more will be the possibility of the metal getting corroded.

2. Presence of impurities in metals: The impurities help in setting up a voltaic cell, which increases the speed of corrosion.

3. Presence of electrolytes: Presence of electrolytes in water also increases the rate of corrosion. For example, corrosion of iron in sea water takes place to larger extent than in distilled water.

4. Presence of CO₂ in water. Presence of CO₂ in natural water increases rusting of iron. Water containing CO₂ acts as an electrolyte and increases the flow of electrons from one place to another.

5. Presence of protective coatings. When the iron surface is coated with layers of metals more active than iron, then the rate of corrosion is retarded. For example, coating of zinc on iron prevents rustings.

Question 6.
What type of battery is lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery. (CBSE 2011)
Answer:
It is a secondary cell.
Anode reaction:
Pb (s) + SO₄^2- (aq) → PbSO₄ (s) + 2e^–

Cathode reaction:
PbO₂ (s) + 4H⁺ (aq) + SO₄^2- (aq) + 2e^– → PbSO₄ (s) + 2H₂O (l)

Overall reaction:
Pb (s) + PbO₂ (s) + 2H₂S0₄ → 2PbSO₄ (s) + 2H₂O (l)

Question 7.
The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere. (CBSE 2011)
Answer:
The chemistry of corrosion of iron is an electrochemical theory which involves oxidation and reduction reactions. According to this theory it is believed that non-uniform surface of metal or impurities present in iron behave like small electric cells (called corrosion couples) in the presence of water containing dissolved oxygen or carbon dioxide. A film of moisture with dissolved CO₂ acts as electrolytic solution covering the metal surface at various places. This is shown in Fig. In these small electrolytic cells, pure iron acts as anode while cathodes are impure portions. The overall rusting involves the following steps:
Oxidation occurs at the anode of each electrochemical cell. Therefore, at each anode iron is oxidised to Fe²⁺ ions.

At anode:
Fe (s) → Fe²⁺ (aq) + 2e^– …… (i)
Thus, the metal atoms in the lattice pass into the solution as ions, leaving electrons on the metal itself. These electrons move towards the cathode region through the metal.

At the cathode of each cell, the electrons are taken up by hydrogen ions (reduction takes place). The H⁺ ions are obtained either from water or from acidic substance in water:
H₂O ⇌ H⁺ + OH^– …… (ii)
or CO₂ + H₂O → H⁺ + H⁺CO₃^– …… (iii)

At cathode:
H⁺ + e^– → H+ …… (iV)
Thus, hydrogen atoms on the iron surface reduce dissolved oxygen.
4H + O₂ → 2H₂O ……..(v)
Therefore, the overall reaction at cathode of different electrochemical cells may be written as:
4H⁺(aq) + O₂ (g) + 4e^– → 2H₂O (l) ……. (vi)
The overall redox reaction may be written by multiplying reaction at anode Eq. (i) by 2 and adding reaction at cathode Eq. (iv) to equalise number of electrons lost and gained, i.e. Oxidation half reaction:
Fe (s) → Fe²⁺ (aq) + 2e^– × 2

Reduction half reaction:
4H⁺(aq) + O₂ (g) + 4e^– → 2H₂O (l)

Overall cell reaction:
2Fe (s) + 4H⁺(aq) + O₂ (g) → 2Fe²⁺(aq) + 2H₂O (l)
The ferrous ions are oxidised further by atmospheric oxygen to Fe³⁺ (as Fe₂O₃) and form rust
4Fe²⁺ + O₂ (g) + 4H₂O → 2Fe₂O₃ + 8H⁺
and Fe₂O₃ + xH₂O → Fe₂O₃. xH₂O
The H⁺ ions produced above are also used for reaction (iv).

Mechanism of rusting of air.

Question 8.
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution? (CBSE 2014)
Answer:
Kohlrausch law of independent migration of ions states that at infinite dilution when the dissociation is complete, each ion makes a definite contribution towards molar conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated. Kohlrausch law also means that the limiting molar conductivity of an electrolyte is sum of the individual contributions of the ions of the electrolyte.

Conductivity (κ) of the electrolyte solution decreases with dilution because the number of ions per unit volume furnished by an electrolyte decreases with dilution.

Question 9.
(i) Following reactions occur at cathode during the electrolysis of aqueous silver chloride
solution:
Ag⁺(aq) + e^– → Ag(s)        E° = + 0.80 V
H⁺(aq) + e^– → \(\frac{1}{2}\)H₂(g)         E ° = 0.00 V
On the basis of their standard electrode potential (E°) values which reaction is feasible at the cathode and why?
(ii) Define limiting molar conductivity. Why does conductivity of an electrolyte solution decrease with the decrease in concentration? (CBSE Delhi 2015)
Answer:
(i) The reaction, Ag⁺(aq) + e^– → Ag(s) (E° = + 0.80 V) is feasible at cathode because its reduction potential is higher than other reaction (H⁺ + e^– → \(\frac{1}{2}\) H₂; E° = 0.0 V)

(ii) The molar conductivity of the solution when the concentration approaches zero (infinite dilution) is called limiting molar conductivity. Conductivity is the conductance of one centimeter cube of the solution. Upon diluting the solution, the concentration of ions per centimeter cube decreases and hence conductivity decreases with dilution.

Question 10.
(i) Following reactions occur at cathode during the electrolysis of aqueous sodium chloride
solution:
Na⁺(aq) + e^– → Na(s)        E° = -2.71V
H⁺(aq) + e^– → H₂(g)         E ° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
Answer:
(i) The reaction: H^–(aq) + e^– → H₂(g) (E° = 0.00 V) is feasible at cathode because its reduction
potential is higher than the other electrode reaction.

(ii) Why does the cell potential of mercury cell remain constant throughout its life? (CBSE 2015)
Answer:
The cell potential of mercury cell remains constant because the overall reaction does not involve any ion in the solution whose concentration changes during its life time.

Question 11.
Zinc rod is dipped in 0.01 M solution of zinc sulphate when temperature is 298 K. Calculate the electrode potential of zinc.
(Given: E°_zn²⁺/Zn = – 0.76 V; log 10 = 1) (CBSE 2019 C)
Answer:

Question 12.
Write Nernst equation and calculate e.m.f. of the following cells at 298 K:
(i) Mg (s) | Mg²⁺ (0.001 M) | | Cu²⁺ (0.0001 M) | Cu (s)
Given: E^⊖_Mg²⁺/Mg = -2.37 V, E^⊖_Cu²⁺/Cu = 0.34V
Answer:
(i) The electrode reactions and cell reactions are:

Since the reaction involves 2 moles of electrons and therefore, n = 2 and the Nernst equation for the cell at 298 K is:

(ii) Fe (s) | Fe²⁺ (0.001 M) | | H⁺ (1M) | H₂ (1 atm) | Pt
Given: E^⊖_Fe²⁺/Fe = -0.44V (CBSE Delhi 2013)
Answer:
Fe_s | Fe²⁺ (0.001 M) || H⁺ (1 M) | H₂ (1 atm) | Pt
The electrode reactions and overall cell reactions are:

Since the reaction involves 2 moles of electrons, therefore, n = 2 and the Nernst equation at 298 K

(iii) Sn (s) | Sn²⁺ (0.050 M) | | H⁺ (0.020 M) | H₂ (1 atm) | Pt
Given: E^⊖_Sn²⁺/Cn = – 0.14V (CBSE Al 2018)
Answer:
The cell is:
Sn(s) | Sn²⁺(0.050M) | | H⁺(0.020 M) | H₂(1 atm) | Pt
The electrode reactions and cell reactions are:

The reaction involves 2 moles of electrons, therefore, n = 2 and the Nernst equation is:

Question 13.
(a) The e.m.f. of the following cell at 298 K is 0.1745 V:
Fe (s) | Fe²⁺ (0.1 M) | | H⁺ (x M) | H₂ (g) (1 bar) | Pt (s)
Given: E°_Fe²⁺Fe = – 0.44 V
Calculate the H⁺ ions concentration of the solution at the electrode where hydrogen is being produced.
Answer:
(a)

(b) Aqueous solutions of copper sulphate and silver nitrate are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different? Explain your answer.
Answer:
The mass of copper and silver deposited at the cathode will be different.
Faraday’s second law of electrolysis: It states that when same quantity of electricity is passed through different electrolytic solutions connected in series, the weights of the substances produced at the electrodes are directly proportional to their chemical equivalent weights.

For example, when same current is passed through two electrolytic solutions, containing copper sulphate (CuSO₄) and silver nitrate (AgNO₃) connected in series, the weights of copper and silver deposited are:

OR

(a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 × 10^-5 S cm^-1.
Given λ°_H⁺ = 349.6 S cm² mol^-1; λ°_CH3COO^– = 40.9 S ² mol^-1
Answer:
(a)

Electrolyte B is a strong electrolyte.

(b) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases to a smaller extent while that of ‘A’ increases to a much larger extent comparatively. Which of the two is a strong electrolyte? Justify your answer. (CBSE Sample Paper 2019)
Answer:
Limiting molar conductivity increases only to a smaller extent for a strong electrolyte, as on dilution the interionic interactions are overcome. Limiting molar conductivity increases to a larger extent for a weak electrolyte, as on dilution the degree of dissociation increases, therefore the number of ions in total volume of solution increases. Therefore, ‘B’ is a strong electrolyte.

Question 14.
(i) Calculate E°_cell for the following reaction at 298 K:
2Cr(s) + 3Fe²⁺ (0.01 M) → 2Cr³⁺ (0.01 M) + 3Fe(s)
Given: E_cell = 0.261 V
Answer:

(ii) Using the E° values of A and B, predict which one is better for coating the surface of iron [Ee°_Fe²⁺/Fe) = – 0.44 V] to prevent corrosion and why?
Given: E°_(A²⁺/A) = -2.37 V: E°_(B²⁺/B) = – 0.14 V
Answer:
‘A’ will prevent iron from corrosion. So, we can coat the iron surface with metal A because it has more negative E° value.

OR

(i) The conductivity of 0.001 mol L^-1 solution of CH₃COOH Is 3.905 × 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation (α).
Given λ°(H⁺) = 349.65 cm² mol^-1 and λ°(CH₃COO^–) = 40.9 S cm² mol^-1.
Answer:

(ii) Define electrochemical cell. What happens if external potential applied becomes greater than E°cell of electrochemical cell? (CBSE 2016)
Answer:
Electrochemical cell is a device used for the production of electricity from energy released during spontaneous chemical reaction. Electrochemical cell converts chemical energy into electrical energy.
If E°_cell (external) > E°_cell, the cell starts acting as an electrolytic cell. In this case, electrical energy is used to carry out non-spontaneous chemical reaction.

Question 15.
(i) Define the following terms:
(a) Limiting molar conductivity
(b) Fuel cell
(ii) Resistance of a conductivity cell filled with 0.1 mol L^-1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L^-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L^-1 KCl solution. The conductivity of 0.1 mol L^-1 KCl solution is 1.29 × 10^-2 Ω^-1 cm^-1.

OR

(i) State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu²⁺ to Cu.
(ii) Calculate emf of the following cell at 298 K:
Mg(s) | Mg²⁺(0.1 M) | | Cu²⁺(0.01)|Cu(s) [Given E°_cell = +2.71 V, 1 F = 96500 C mol^-1] (CBSE Delhi 2014)
Answer:
(a) The molar conductivity of a solution when concentration approaches zero is called limiting molar conductivity and is expressed as ∧°_m.

(b) Fuel cells are voltaic cells in which the reactants are continously supplied to the electrodes and are designed to convert energy from the combustion of fuels such as H₂, CO, CH₄, etc. directly into electrical energy.

(ii) Let us first calculate cell constant
Cell constant, G* = Conductivity (κ) × Resistance (R)
Resistance of 0.1 M KCl solution = 100 Ω
Conductivity of 0.1 M KCl solution = 1.29 Sm^-1
Cell constant = 1.29 (Sm^-1) × 100 Ω
= 129 m^-1 = 1.29 cm^-1
Calculation of conductivity of 0.02 M KCl solution,
Resistance of solution = 520 Ω
Cell constant (G*) = 1.29 cm^-1

OR

(i) Faraday’s first law of electrolysis states that the amount of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolytic solution.
Cu²⁺ + 2e^– → Cu(s)
Reduction of 1 mol of Cu²⁺ ion requires 2F of electricity.

(ii) The electrode reactions and cell reaction are
Mg(s) → Mg²⁺(aq) + 2e^– (at anode)
Cu²⁺(aq) + 2e^– → Cu(s) at cathode
Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)

= 2.71 – 0.0295
= 2.6805 v.

Question 16.
(i) How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO₃)₂ solution with a current of 2.00 A for 3 hours? [Hg(NO₃)₂ = 200.6 g mol^-1].
Answer:
Hg²⁺ + 2e^– → Hg
Quantity of electricity passed = 1 × t(sec)
= 2.0 A × 3.0 × 60 × 60 = 21 600C
2 × 96500 C of electricity produces mercury = 1 mol
21,600 C of electricity will produce mercury
= \(\frac{1}{2 \times 96500}\) × 21600 = 0.112 mol

(ii) A voltaic cell is set up at 25° C with the following half cells:
Al³⁺ (0.001 M) and Ni²⁺ (0.50 M).
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given: E°_Ni²⁺/Ni = – 0.25 V, E°_Ni²⁺/Ni = – 1.66V) (CBSE 2011)
Answer:
Cell reaction is:

Question 17.
(i) What type of battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.
Answer:
It is secondary cell.
Anode reaction:
Pb (s) + SO₄^2-(aq) → PbSO₄(s) + 2e^–

Cathode reaction:
PbO₂ (s) + 4H⁺(aq) + SO₄^2-(aq) + 2e^– → PbSO₄(s) + 2H₂O (l)

Overall reaction:
Pb (s) + PbO₂ (s) + 2H₂SO₄ → PbSO₄(s) + 2H₂O (l)

(ii) Calculate the potential for half cell containing 0.1 M K₂Cr₂O₇(aq), 0.20 M Cr³⁺(aq) and 1.0 × 10^-4 M H⁺(aq). The half cell reaction is
Cr₂O₇^2-(aq) + 14 H⁺(aq) + 6e^– → 2Cr³⁺(aq) + 7H₂O(l),
and the standard electrode potential is given as E° = 1.33 V. (CBSE 2011)
Answer:

Question  18.
(i) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law.
Answer:
Kohlrausch law states that at infinite dilution, each ion makes a definite contribution towards molar conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated. This means that the molar conductivity at infinite dilution for a given salt can be expressed as the sum of the individual contributions from the ions of the electrolyte.

For acetic acid, limiting molar conductvity at infinite dilution can be written as:
∧°(CH₃COOH) = λ° (CH₃COO^–) + λ° (H⁺) ……. (i)
This equation can be obtained by the knowledge of molar conductivity at infinite dilution for some electrolytes. For example, consider the strong electrolytes HCl, NaCl and CH₃COONa. From Kohlrausch’s law:
∧°(CH₃COONa) = λ°(CH₃COO^–) + A°(Na⁺) …… (i)
∧°(HCl) = λ°(H⁺) + λ°(Cl^–) …….. (ii)
∧° (NaCl) = λ° (Na⁺) + λ° (Cl^–) …….. (iii)
It is clear that
λ° (CH₃COO^–) + λ° (H⁺) = [λ°(CH₃COO^–) + A°(Na⁺)] + [λ°(H⁺) + λ°(Cl^–)] – λ° (Na⁺) + λ° (Cl^–)
∧°(CH₃COOH) = ∧°(CH₃COONa) + ∧°(HCl) – ∧° (NaCl)

(ii) Calculate ∧°_m for acetic acid
Given that ∧°_m(HCl) = 426 S cm² mol^-1
∧°_m(NaCl) = 126 S cm² mol^-1
∧°_m(CH₃COONa) = 91 S cm² mol^-1 (CBSE Delhi 2010)
Answer:
We know
∧°(CH₃COOH) = ∧°(CH₃COONa) + ∧°(HCl) – ∧° (NaCl)
= 91 + 426 – 126 = 391 S cm² mol^-1

Question 19.
E°_cell for the given redox reaction is 2.71 V.
Mg(s) + Cu²⁺(0.01 M) → Mg²⁺(0.001 M) + Cu(s)
Calculate E_cell for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and
(ii) greater than 2.71 V
Answer:
Mg(s) + Cu²⁺(0.01 M) → Mg²⁺(0.001M) + Cu(s)

= 2.71 – 0.0295 log 10^-1
= 2.71 + 0.0295
= 2.7395 V or approximately 2.74 V
(i) When E_ext < 2.71, current flows from Cu to Mg
(ii) When E_ext > 2.71, current flows from Mg to Cu.

OR

(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeS04 and ZnS04 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y. (Molar mass: Fe = 56 g mol^-1, Zn = 65.3 g mol^-1, 1F = 96500 C mol^-1)
(b) In the plot of molar conductlvtty (∧_m) vs square root of concentration (c^1/2), the following curves are obtained for two electrolytes A and B:

Answer the following:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation 0f ∧_m to concentration approaching zero for electrolytes A and B? (CBSE Delhi 2019)
Answer:
(a) Cell-X contains FeSO₄
FeSO₄(aq) → Fe²⁺ + SO₄^2-
Fe²⁺ + 2e^– → Fe (at cathode)
1 mole or 56g of Fe is deposited by 2 × 96500C
2.8g of Fe with be deposited by= 2 × 96500 × 2.8
= 9650 C
Q = I × t
9650 = 2 × t
or t = \(\frac{9650}{2}\) = 4825s
In cell Y
1 mol or 65.3 g of Zn is deposited by 2 × 96500 C
or 2 × 96500 C of electricity will deposit Zn = 65.3 g
9650 C of electricity will deposit Zn = \(\frac{65.3 \times 9650}{2 \times 96500}\) = 3.265 g

(b) (i) Electrolyte A is strong electrolyte.
(ii) Electrolyte B is weak electrolyte.
(iii) For A, the intercept will give ∧°_m.
For B, we cannot obtain ∧°_m on extrapolation.

Question 20.
(i) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte.
Answer:
Molar conductivity is the conducting power of all the ions produced by dissolving one gram mole of an electrolyte in solution. It is expressed as ∧_m and is defined as:
∧_m = \(\frac{\kappa \times 1000}{M}\)
where M is the molarity of the solution. Its units are ohm^-1 cm² mol^-1.
The molar conductance of an electrolytic solution decreases with increase in concentration. For weak electrolytes, molar conductivity increases sharply with dilution. On the other hand, for strong electrolytes, molar conductivity increases slowly with dilution.

For weak electrolytes, the increase in molar conductivity is due to increase in degree of ionisation with dilution. For strong electrolytes, the increase in molar conductance with dilution is because of decrease in interactions between ions with dilution. The decrease in molar conductivity for weak electrolyte (CH₃COOH) and strong electrolyte (KCl) is shown below:

(ii) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1 ? (CBSE Delhi 2012)
Answer:
R = 1500 Ω
Molarity = 0.001 M
K = 0.146 × 10^-3S cm^-1
\(\frac{1}{A}\) = ?
Conductivity (κ) = Conductance (G) × Cell Constant \(\left(\frac{1}{A}\right)\)
or κ = \(\frac{1}{R} \times \frac{l}{A}\)
or Cell constant \(\frac{1}{A}\) = R × κ
= 1500 Ω × 0.146 × 10^-3S cm^-1
\(\frac{1}{A}\) = 0.219 cm^-1

Question 21.
(i) What type of a battery is the lead storage battery ? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it.
Answer:
It is secondary cell.
Anode reaction:
Pb (s) + SO₄^2-(aq) → PbSO₄(s) + 2e^–

Cathode reaction:
PbO₂ (s) + 4H⁺(aq) + SO₄^2-(aq) + 2e^– → PbSO₄(s) + 2H₂O (l)

Overall reaction:
Pb (s) + PbO₂ (s) + 2H₂SO₄ → PbSO₄(s) + 2H₂O (l)

(ii) In the button cell, widely used in watches, the following reaction takes place
Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH^–(aq)
Determine E° and ∆G° for the reaction.
(given: E°_Ag²⁺/Ag = + 0.80V, E°_Zn²⁺/Zn = – 0.76V)
Answer:
E°_Ag²⁺/Ag = + 0.80V, E°_Zn²⁺/Zn = – 0.76V
E°_cell = E°_Ag²⁺/Ag – E°_Zn²⁺/Zn
E°_cell = + 0.80 – (- 0.76) = 1.56 V
ΔG° = – nFE°
n = 2, F = 96500 C
∴ ΔG° = – 2 × 96500 × 1.56
= – 301080 J = – 301.08 kJ

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 2 Solutions. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 2 Important Extra Questions Solutions

Solutions Important Extra Questions Very Short Answer Type

Question 1.
What is meant by reverse osmosis? (CBSE 2011)
Answer:
The direction of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution side through the semi-permeable membrane.

Question 2.
Define an ideal solution and write one of its characteristics. (CBSE Delhi 2014)
Answer:
An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of temperature and pressure. For ideal solution

• Heat of mixing is zero
• Volume change of mixing is zero.

Therefore first find the number of moles of each compound present and then use the mole fraction equation.

Question 3.
(i) Write the colligative property which is used to find the molecular mass of macromolecules.
(ii) In non-ideal solution, what type of deviation shows the formation of minimum boiling azeotropes? (CBSE Delhi 2016)
Answer:
(i) Osmotic pressure
(ii) Minimum boiling azeotropes show positive deviation from Raoult’s law.

How to find molar concentration with density how to calculate molarity of liquid solutions tank mixing problem dilution of two solutes

Question 4.
What is the relation between normality and molarity of a given solution of sulphuric acid?
Answer:
Normality = 2 × Molarity.

Question 5.
Given below is the sketch of a plant for carrying out a process.

(i) Name the process occurring in the above plant.
(ii) To which container does the net flow of solvent take place?
(iii) Name one SPM which can be used in this plant.
(iv) Give one practical use of the plant. (CBSE Sample Paper 2007)
Answer:
(i) Reverse osmosis
(ii) To fresh water container
(iii) Film of cellulose acetate
(iv) This can be used as a desalination plant to meet potable water requirements.

Free Molarity Calculator Chemistry determines the molarity of the chemical solution in a short span of time.

Question 6.
A and B liquids on mixing produce a warm solution. Which type of deviation from Raoult’s law is there? (CBSE Sample Paper 2011)
Answer:
Negative deviation.

Question 7.
Define ebullioscopic constant. (CBSE AI 2011)
Answer:
Ebullioscopic constant is the elevation in boiling point of a solution containing 1 mole of solute dissolved in 1000 g of the solvent.

This colloid osmotic pressure calculator determines the pressure induced by proteins in blood plasma.

Question 8.
What are isotonic solutions?
Answer:
The solutions having same osmotic pressure at the same temperature are called isotonic solutions. These have equimolar concentrations at same temperature.

Question 9.
Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’?
Answer:
The solutions (liquid mixtures) which boil at constant temperature and can distil unchanged in composition are called azeotropes or azeotropic mixtures.

Question 10.
Define‘colligative properties’
Answer:
Colligative properties are those properties of solutions which depend only on the number of solute particles and not on the nature of the solute.

Question 11.
Define ‘Molality (m)’.
Answer:
Molality is the number of moles of solute dissolved per 1000 g (or 1 kg) of the solvent.

Question 12.
Define ‘Ideal solution’.
Answer:
The solutions which obey Raoult’s law over the entire range of concentration are called ideal solutions.

Question 13.
Define ‘Abnormal molar mass’.
Answer:
Molar mass of a substance, calculated based on its colligative properties, may not be correct if the solute particles undergo dissociation or association in the solution. Molar mass thus calculated is called abnormal molar mass.

Solutions Important Extra Questions Short Answer Type

Question 1.
State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations. (CBSE Delhi 2019)
Answer:
Raoult’s law states that at a given temperature, the partial vapour pressure of any volatile component of a solution is equal to the product of the vapour pressure of the pure component and its mole fraction in the solution.

• No heat is evolved or absorbed when these components are mixed to form a solution.
• Volume of the solution is exactly same as the sum of the volumes of the components.

Question 2.
Define the terms ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions? (CBSE 2010)
Answer:
The flow of solvent from solution of low concentration to higher concentration through a semipermeable membrane is called osmosis.
The excess pressure which must be applied to a solution to prevent the passage of solvent through a semi-permeable membrane is called osmotic pressure. Osmotic pressure measurement is preferred over all other colligative properties because
1. even in dilute solutions, the osmotic pressure values are appreciably high and can be measured accurately.
2. osmotic pressure can be measured at room temperature. On the other hand, elevation in boiling point is measured at high temperature where the solute may decompose. The depression in freezing point is measured at low temperatures.

Question 3.
State the following:
(i) Raoult’s law in its general form in reference to solutions.
(ii) Henry’s law about partial pressure of a gas in a mixture. (CBSE 2011)
Answer:
(i) For a solution of volatile liquids, at a given temperature, the partial vapour pressure of each component in solution is equal to the product of vapour pressure of the pure component and its mole fraction.

(ii) Henry’s law states that the mass of a gas dissolved per unit volume of the solvent at a constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution.

Question 4.
The experimentally determined molar mass for what type of substances is always lower than the true value when water is used as solvent? Explain. Give one example of such a substance and one example of a substance which does not show a large variation from the true value. (CBSE Sample Paper 2019)
Answer:
Ionic compounds when dissolved in water dissociate into cations and anions. When there is dissociation of solute into ions, in dilute solutions, the number of particles increases if we ignore the interionic interactions.

As the value of the colligative properties depends on the number of particles of the solute, the experimentally observed value of colligative property will be higher than the true value. Therefore the experimentally determined molar mass is always lower than the true value.

If we dissolve KCl in water, the experimentally determined molar mass is always lower than the true value. Glucose (non-electrolyte) does not show a large variation from the true value.

Question 5.
What mass of ethylene glycol (molar mass = 62.0 g mol^-1) must be added to 5.50 kg of water to lower the freezing point of water from 0 to – 10.0°0 (k_f for water = 1.86 K kg mol^-1). (CBSE 2010)
Answer:
ΔT_f = \(\frac{\mathrm{K}_{f} \times 1000 \times \mathrm{w}_{\mathrm{B}}}{\mathrm{w}_{\mathrm{A}} \times \mathrm{M}_{\mathrm{B}}}\)
K_f = 1.86 K kg mol^-1, w_A = 5.50 kg = 5500 g
W_B = ?
M_B = 62.0 g/mol, ΔT_f = 0 – (- 10) = 10°C
10 = \(\frac{1.86 \times 1000 \times w_{B}}{5500 \times 62}\)
W_B = \(\frac{10 \times 5500 \times 62}{1.86 \times 1000}\) = 1833.3 g = 1.833 kg

Question 6.
15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at – 0.34°C. What is the molar mass of this material? (k_f for water = 1.86 K kg mol^-1) (CBSE Delhi 2012)
Answer:
Molecular mass, M_B is
M_B = \(\frac{\mathrm{K}_{f} \times \mathrm{w}_{\mathrm{B}} \times 1000}{\mathrm{w}_{\mathrm{A}} \times \Delta \mathrm{T}_{f}}\)
ΔT_f = 0 – (- 0.34) = 0.34°,
K_f = 1.86 K Kg mol^-1, w_B = 15.0 g, W_A = 450 g
M_B = \(\frac{1.86 \times 15.0 \times 1000}{450 \times 0.34}\) = 182.3 g/mol

Question 7.
State Henry’s law and write its two applications. (CBSE Delhi 2019)
Answer:
Henry’s law states that the mole fraction of the gas in the solution is directly proportional to the partial pressure of the gas over the solution.
P = K_H.X
Applications of Henry’s law

• In the production of carbonated beverages.
• In deep sea diving (scuba diving)

Question 8.
A solution of glycerol (C₃H₈O₃) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C while pure water boils at 100°C. What mass of glycerol was dissolved to make the solution?
(Kb for water = 0.512 K kg mol^-1)         (CBSE Delhi 2012, CBSE AI 2012)
Answer:
Elevation in boiling point
T_b = \(\frac{\mathrm{K}_{b} \times 1000 \times \mathrm{w}_{\mathrm{B}}}{\mathrm{w}_{\mathrm{A}} \times \mathrm{M}_{\mathrm{B}}}\)
ΔT_b = 100.42 – 100 = 0.42°, w_A = 500 g
w_A = ?, K_b = 0.512 k Kg mol^-1
M_B = 3 × 12 + 8 × 1 + 3 × 16 = 92 g/mol
0.42 = \(\frac{0.512 \times 1000 \times w_{B}}{92 \times 500}\)
or w_B = \(\frac{0.42 \times 500 \times 92}{0.512 \times 1000}\)
= 37.33 g
Mass of glycerol to be added = 37.33 g

Question 9.
A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10°C and K_b for benzene = 2.53 °C kg mol^-1) (CBSE Delhi 2010)
Answer:

k_b = 2.53°C kg mol^-1
w_A = 99.0 g, w_B = 1.25 g
M_B = \(\frac{2.53 \times 1000 \times 1.25}{0.21 \times 99.0}\)
= 152.1 g mol^-1

Question 10.
18 g of glucose, C₆H₁₂O₆ (molar mass = 180 g mol^-1), is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil?
(K_b for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K). (CBSE Delhi 2013)
Answer:
ΔT_b = \(\frac{\mathrm{K}_{b} \times 1000 \times \mathrm{w}_{\mathrm{B}}}{\mathrm{w}_{\mathrm{A}} \times \mathrm{M}_{\mathrm{B}}}\)
w_B = 18 g, w_A = 1000 g, M_B = 180 g/mol,
K_b = 0.52 K kg mol^-1
∴ ΔT_b = \(\frac{0.52 \times 1000 \times 18}{1000 \times 180}\) = 0.052 K
Boiling point of solution
= 373.15 + 0.052 = 373.202 K.

Question 11.
A solution containing 15 g urea (molar mass = 60 g mol^-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol^-1) in water. Calculate the mass of glucose present in one litre of its solution. (CBSE Al 2014)
Answer:
For isotonic solutions,
π (urea) = π (glucose)

Question 12.
A 1.00 molal solution of trichloroacetic acid (CCl₃COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid (K_b for water = 0.512 K kg mol^-1). (CBSE Delhi 2012)
Answer:
Observed boiling point elevation,
ΔT_b = 100.18 – 100.0 = 0.18°C
Molality of solution = 1.00 m
Calculated boiling point elevation,
ΔT_b(calc.) = K_b × m
= 0.512 × 1 = 0.512
van’t Hoff factor,

Question 13.
Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution? (CBSE AI 2016)
Answer:
(i) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or blood cells. When blood cells are placed in this solution, water flows out of the cells and they shrink due to loss of water by osmosis.

(ii) 0.4% sodium chloride solution is hypotonic with respect to 0.9% sodium chloride solution or blood cells. When blood cells are placed in this solution, water flows into the cells and they swell.

Question 14.
1.00 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute.
(K_f for benzene = 5.12 K kg mol^-1). (CBSE AI 2013)
Answer:

Question 15.
(a) Out of 0.1 molal aqueous solution of glucose and 0.1 molal aqueous solution of KCI, which one will have higher boiling point and why?
Answer:
(a) 0.1 m KCI solution will have higher boiling point. This is because KCl dissociates in water to give two ions (K⁺ and Cl^–) whereas glucose does not dissociate. Therefore, number of solute particles is greater in 0.1 m KCl as compared to 0.1 m glucose.

(b) Predict whether van’t Hoff factor
(i) is less than one or greater than one in the following:
(i) CH₃COOH dissolved in water
(ii) CH₃COOH dissolved in benzene (CBSE AI 2019)
Answer:
(i) i > 1 because CH₃COOH dissociates in water.
(ii) i < 1 because CH₃COOH associates in benzene.

Question 16.
Urea forms an ideal solution in water. Determine the vapour pressure of an aqueous solution containing 10% by mass of urea at 40°C. (Vapour pressure of water at 40°C = 55.3 mm of Hg) (CBSE AI 2006)
Answer:
w_A = 90 g. w_B = 10 g

55.3 – p_A = 1.84
∴ p_A = 53.46 mm Hg

Question 17.
A solution of chloroform and acetone is an example of maximum boiling azeotrope. Why? (CBSE Sample Paper 2012)
Answer:
The solution of chloroform and acetone has lower vapour pressure than ideal solution because of stronger interactions between chloroform and acetone molecules. As a result, total vapour pressure becomes less than the corresponding ideal solution of same composition (i.e. negative deviations). Therefore, the boiling points of solutions are increased and form maximum boiling azeotropes.

Question 18.
A solution of glucose (C₆H₁₂O₆) in water is labelled as 10% by weight. What would be the molality of the solution? (CBSEAI 2013)
Answer:
10% by weight means that 10 g of glucose is present in 100 g of solution.
Mass of solvent = 100 – 10 = 90 g 10
Moles of glucose = \(\frac{10}{180}\) = 0.0556 moles
Molality = \(\frac{0.0556}{90}\) × 1000
= 0.618 m.

Question 19.
What is meant by positive deviations from Raoult’s law? Give an example. What is the sign of Δ_mix H for positive deviation?
OR
Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult’s law? Give an example. (Delhi 2015)
Answer:
When the observed vapour pressure of a liquid solution is higher than the value expected from Raoult’s law, it is called positive deviation from Raoult’s law.
P_total > P°_A X_A + P°_B X_B
for the two components A and B.
Example: Ethyl alcohol and cyclohexane
ΔH_mix is positive.
OR
Liquid mixtures which boil at constant temperature and can distil unchanged in composition are called azeotropes.
The mixture which shows positive deviation from Raoult’s law forms minimum boiling azeotrope.
Example: A mixture of ethanol and water containing 95.4% ethanol forms minimum boiling azeotrope.

Question 20.
Write two differences between an ideal solution and a non-ideal solution. (CBSE Delhi 2019)
Answer:

Ideal solution	Non-ideal solution
(i) Each component of solution obeys Raoult’s Law at all temperatures and concentrations.	(i) Their components do not obey Raoult’s law.
(ii) There is no enthalpy change on mixing. Δmixing H = 0	(ii) There is enthalpy change on mixing. Δmixing H ≠ 0

Question 21.
Define an ideal solution and write one of its characteristics. (CBSE 2014)
Answer:
An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of temperature and pressure. For ideal solution

• Heat of mixing is zero
• Volume change of mixing is zero.

Question 22.
State Henry’s law. What is the effect of temperature on the solubility of a gas in a liquid? (CBSE 2014)
Answer:
Henry’s law states that the mass of a gas dissolved per unit volume of the solvent at a constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution.

The solubility of a gas decreases with increase in temperature. However, it may be noted that there are certain gases like hydrogen and inert gases whose solubility increases slightly with increase in temperature especially in non-aqueous solvents such as alcohols, acetone, etc.

Question 23.
State Raoult’s law for the solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law? (CBSE 2014)
Answer:
Raoult’s law states that at a given temperature for a solution of volatile liquids, the partial vapour pressure of each component in solution is equal to the product of the vapour pressure of pure component and its mole fraction.
Similarity between Raoult’s law and Henry’s law

• Both Raoult’s law and Henry’s law apply to volatile component in solution. .
• Both laws state that the vapour pressure of one component is proportional to the mole fraction of that component.

Solutions Important Extra Questions Long Answer Type

Question 1.
Calculate the freezing point of a solution containing 0.5 g KCI (Molar mass = 74.5 g/ mol) dissolved in 100 g water, assuming KCI to be 92% ionised. K_f of water = 1.86 K kg / mol. (CBSE Sample Paper 2019)
Answer:
Let us consider one mole of KCI whose degree of dissociation is α. The dissociation of KCI can be represented as:
KCI → K⁺ + Cl^–
1 – α       α      α
Total number of moles after dissociation
= 1 – α + α + α
= 1 + α
Hence i = \(\frac{1+\alpha}{1}\)
If the solute dissolves in the solvent giving n ions, then here n = 2
i = 1 + (n -1) α
= 1 + (2 – 1) α = 1 + α
Now, ∆T_f = i K_fm
= (1 + 0.92) × 1.86 × \(\frac{0.5 \times 1000}{74.5 \times 100}\)
∆T_f = 0.24
∆T_f = T°_f – T’_f = 0 – 0.24
T’_f = -0.24°C

Question 2.
State Henry’s law. Why do gases always tend to be less soluble in liquids as the temperature is raised?
OR
State Raoult’s taw for the solution containing volatile components. Write two differences between an ideal solution and a non-ideal solution. (CBSE Delhi 2015)
Answer:
Henry’s law states that the mass of a gas dissolved per unit volume of the solvent is directly proportional to the pressure of the gas in equilibrium with the solution.

If m is the mass of the gas dissolved in a unit volume of the solvent and p is the pressure of the gas in equilibrium with the solution, then
m ∝ p
or m = K.p
(where K is the proportionality constant) or “partial pressure of the gas in its vapour phase (p) is directly proportional to the mole fraction of the gas (x) in the solution”.
P = K_H.X
The dissolution of a gas in a liquid is exothermic process. Therefore, in accordance with Le Chatelier’s principle, with increase in temperature, the equilibrium shifts in the backward direction. As a result, solubility decreases with increase in temperature.
OR
Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component in solution is equal to the product of the vapour pressure of the pure component and its mole fraction. For a binary solution of two components A and B,
P_A = P_A° × X_A
P_B = P_B° × X_B
Differences between ideal and non-ideal solutions

Ideal solution Non-ideal solution

Ideal solution	Non-ideal solution
1. It obeys Raoult’s law over the entire range of concentration of solution.	1. It does not obey Raoult’s law.
2. Solute-solvent interactions are nearly same as in pure solvent.	2. Solute-solvent interactions are not same as solute-solute or solvent-solvent interactions.

Question 3.
Calculate the amount of KCI which must be added to 1 kg of water so that the freezing point is depressed by 2 K. (K_f for water = 1.86 K kg mol^-1) (CBSE Delhi 2012)
Answer:
ΔT_f = 2 K
KCl ⇌ K⁺ + Cl^–
i = 2
K_f = 1.86 K kg mol^-1
w_A = 1 kg = 1000 g
M_B = (39 + 35.5) = 74.5 g
w_B = ?

Question 4.
What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50°C? The freezing point depression constant (K_f) for water is 1.86°C/m. Assume van’t Hoff factor for NaCl is 1.87 (Molar mass of NaCI = 58.5 g mol^-1). (CBSE 2011)
Answer:
Lowering in freezing point,
ΔT_f = 7.50°C
K_f = 1.86 °C/m,
Mass of water, W_A = 65.0 g
Molar mass of NaCl, M_B = 58.5 g
Mass of NaCl, W_B = ?
van’t Hoff factor, i = 1.87

Question 5.
Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10^-2 g of K₂SO₄ in 2 L of water at 25°C, assuming that it is completely dissociated. (CBSE Delhi 2013)
(R = 0.0821 L atm K^-1 mol^-1, Molar mass of K₂SO₄ = 174 g mol^-1)
Answer:
Since K₂SO₄ dissociates completely,
K₂SO₄ → 2K⁺ + SO₄^2-
One mole of K₂SO₄ will give 3 mole particles and therefore, the value of ‘i’ is 3.
Osmotic pressure, π = iCRT
= i\(\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{V}}\)RT
W_B = 2.5 × 10^-2 g, V = 2.0 L, M_B = 174 g/mol
R = 0.821 L atm mol^-1 K^-1
∴ π = \(\frac{3 \times 2.5 \times 10^{-2} \times 0.0821 \times 298}{174 \times 2.0}\)
= 5.27 × 10^-3 atm

Question 6.
Calculate the mass of compound (molar mass = 256 g mol^-1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (K_f = 5.12 K kg mol^-1). (CBSE Delhi 2014)
Answer:
ΔT_f = 0.48 K, K_f = 5.12 K kg mol^-1
M = 256 g mol^-1, w_A = 75g, w_B = ?

Question 7.
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van’t Hoff factor and predict the nature of solute (associated or dissociated).
(Given: Molar mass of benzoic acid = 122 g mol^-1, K_f for benzene = 4.9 K kg mol^-1) (CBSE Delhi 2015)
Answer:
ΔT_f = i K_f × m
= \(\frac{i \times K_{f} \times W_{2} \times 1000}{M_{2} \times W_{1}}\)
W₂ = 3.9 g, W₁ = 49 g, ΔT_f = 1.62 K,
M₂ = 122 g mol^-1
K_f = 4.9 K kg mol^-1
1.62 = \(\frac{i \times 4.9 \times 3.9 \times 1000}{122 \times 49}\)
or i = \(\frac{1.62 \times 122 \times 49}{4.9 \times 3.9 \times 1000}\) = 0.506
Since i is less than 1, the solute is associated.

Question 8.
A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% by mass. What would be the molality and molarity of the solution? (Density of solution = 1.2 g mL^-1) (Delhi Al 2014)
Answer:
10% (by mass) solution of glucose means that 10 g of glucose is present in 100 g of solution or in 90 g of water.

(i) Calculation of molality
Mass of glucose = 10 g
Moles of glucose = \(\frac{10}{180}\) = 0.0556
(Molar mass of glucose = 180 g/mol)
Mass of water = 90 g

= \(\frac{0.0556}{90}\) × 1000
= 0.618 m

(ii) Calculation of molarity
Moles of glucose = 0.0556 Mass

Question 9.
A 4% solution (w/w) of sucrose (M = 342 g mol^-1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol^-1) in water. (Given: Freezing point of pure water = 273.15 K)
Answer:
For Sucrose solution,
W_B = 4 g, W_A = 100 – 4 = 96 g,
ΔT_f = 273.15 – 271.5 = 2°C

Freezing point of solution = 273.15 – 4.8
= 268.35 K

Question 10.
What would be the molar mass of a compound if 6.21 g of it dissolved in 24.0 g of chloroform form a solution that has a boiling point of 68.04°C. The boiling point of pure chloroform is 61.7°C and the boiling point elevation constant, K_b, for chloroform is 3.63 °C/m. (CBSE Delhi 2011)
Answer:
Elevation in boiling point,
ΔT_b = 68.04 – 61.7 = 6.34 °C
Mass of substance,
w_B = 6.21 g,
Mass of chloroform,
w_A = 24.0g
K_b = 3.63 °C/m

Question 11.
Two elements A and B form compounds having molecular formula AB₂ and AB₄. When dissolved in 20 g of benzene, 1 g of AB₂ lowers the freezing point by 2.3 K, whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate the atomic mass of A and B. (Delhi Al 2004)
Answer:
Let us first calculate molar masses of AB₂ and AB₄.
For AB₂ compound
M_B = \(\frac{\mathrm{K}_{f} \times \mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{w}_{\mathrm{A}} \times \Delta \mathrm{T}_{f}}\)
ΔT_f = 2.3 K, w_B = 1.0 g,
w_A = 20.0 g
K_f = 5.1 K kg mol^-1
M_AB2 = \(\frac{5.1 \times 1.0 \times 1000}{20.0 \times 2.3}\)
∴ M_AB2 = 110.87
For AB₄ compound
ΔT_f= 1.3 K, w_B = 1.0 g,
w_A = 20.0 g
M_AB4 = \(\frac{5.1 \times 1.0 \times 1000}{20.0 \times 1.3}\)
∴ M_AB4 = 196.15
Let a be the atomic mass of A and b be the atomic mass of B, then
M_AB2 = a + 2b = 110.87 …… (i)
M_AB4 = a + 4b = 196.15 ……. (ii)
Subtracting eqn. (ii) from eqn. (i)
– 2b = – 85.28
∴ b = 42.64
Substituting the value of b in eqn (i)
a + 2 × 42.64 = 110.87
a = 110.87 – 85.28 = 25.59
Atomic mass of A = 25.59 g
Atomic mass of B = 42.64 g

Question 12.
A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene fragment is a non-electrolyte, determine its molar mass. (CBSE Delhi 2011, Delhi Al 2011)
Answer:
Mass of gene fragment = 8.95 mg
= 8.95 × 10^-3 g
Volume of water = 35.0 mL
= 35.0 × 10^-3 L
Osmotic pressure,
π = 0.335 torr
= 0.335/760 atm
Temperature = 25°C
= 273 + 25 = 298 K

= 14193.3 g mol^-1 or
1.42 × 1o⁴g mol^-1.

Question 13.
Calculate the boiling point of solution when 2 g of Na₂SO₄ (M = 142 g mol^-1) was dissolved in 50 g of water, assuming Na₂SO₄ undergoes complete ionization. (K_b for water = 0.52 K kg mol^-1) [Delhi Al 2016)
Answer:
ΔT_b = \(\frac{i \times K_{b} \times w_{B} \times 1000}{M_{B} \times w_{A}}\)
Weight of solute, w_B = 2 g
Molar mass = 142 g mol^-1
Weight of solvent = 50 g
K_b = 0.52 K kg mol^-1
Na₂SO₄ undergoes complete ionisation as:
Na₂SO₄ ⇌ 2Na⁺ + SO₄^2-
One mole of Na₂SO₄ gives 3 mole particles and therefore,
i = 3
∴ ΔT_b = \(\frac{3 \times 0.52 \times 2 \times 1000}{142 \times 50}\) = 0.439
Boiling point of solution = 373 + 0.439
= 373.439 K

Question 14.
Calculate the amount of CaCl₂ (molar mass = 111 g mol^-1) which must be added to 500 g of water to lower its freezing point by 2 K, assuming CaCl₂ is completely dissociated, (K_f for water = 1.86 K kg mol^-1) (Delhi Al 2015)
Answer:
CaCl₂ undergoes complete dissociation as:
CaCl₂ → Ca²⁺ + 2Cl^–
One mole of CaCl₂ will give 3 mole particles and therefore, the value of T will be equal to 3.
ΔT_f = i K_f × m
= \(\frac{i \times K_{f} \times w_{B} \times 100}{M_{B} \times W_{A}}\)
K_f = 1.86 K kg mol^-1, w_A = 500 g, w_B = ?, ΔT_f = 2 K, i = 3, M_B = 111 g mol^-1
2 = \(\frac{3 \times 1.86 \times w_{B} \times 1000}{111 \times 500}\)
∴ W_B = \(\frac{2 \times 111 \times 500}{3 \times 1.86 \times 1000}\) = 19.89 g

Question 15.
A solution 0.1 M of Na₂SO₄ is dissolved to the extent of 95%. What would be its osmotic pressure at 27°C? (R = 0.0821 L atm K^-1 mol^-1) (CBSE AI 2019)
Answer:
Na₂SO₄ ⇌ 2Na⁺ + SO₄^2-
n = 3
If α is the degree of dissociation, then
α = \(\frac{i-1}{n-1}\)
0.95 = \(\frac{i-1}{3-1}\)
0.95 × 2 = i – 1 or i = 1.90 + 1 = 2.90
π = iCRT
= 2.90 × 0.1 × 0.0821 × 300
= 7.143 atm

Question 16.
(i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution? (Delhi Al 2016)
Answer:
(i) The elevation in boiling point is a colligative property and depends upon the number of moles of solute added. Higher the concentration of solute added, higher will be the elevation in boiling point. Thus, 2M glucose solution has higher boiling point than 1 M glucose solution.

(ii) When the external pressure applied becomes more than the osmotic pressure of the solution, then the solvent molecules from the solution pass through the semipermeable membrane to the solvent side. This process is called reverse osmosis.

Question 17.
Which of the following solutions has higher freezing point?
0.05 M Al₂(SO₄)₃, 0.1 M K₃[Fe(CN)₆] Justify. (CBSE Sample Paper 2017-18)
Answer:
0.05 M Al₂(SO₄)₃ has higher freezing point.
ΔT_f ∝ i × concentration
For 0.05 M Al₂(SO₄)₃ i = 5
ΔT_f ∝ 5 × 0.05 = 0.25 moles of ions
For 0.1 M K₃[Fe(CN)₆], i = 4
ΔT_f ∝ 4 × 0.1 = 0.40 moles of ions
∴ Depression in freezing point for 0.05 M Al₂(SO₄)₃ will be less and hence freezing point will be higher.

Question 18.
A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K. Given: (Molar mass of sucrose = 342 g mol^-1) (Molar mass of glucose = 180 g mol^-1) (CBSE AI 2017, CBSE Delhi 2017)
Answer:
For sucrose solution,
K_f = \(\frac{\Delta T_{f} \times w_{A} \times M_{B}}{w_{B} \times 1000}\)
ΔT_f = 273.15 – 269.15 = 4.0 k, w_B = 10 g
W_A = 100 – 10 = 90 g
k_f = \(\frac{4.0 \times 90 \times 342}{10 \times 1000}\) = 12.31 km^-1
For glucose solution,
w_B = 10 g, w_A = 100 – 10 = 90 g
M_B = 180 g
ΔT_f = \(\frac{12.31 \times 10 \times 1000}{90 \times 180}\) = 7.6 k
Freezing point of glucose solution
= 273.15 – 7.6
= 265.55 K

Question 19.
30 g of urea (M = 60 g mol^-1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg. (CBSE Al 2017)
Answer:

Question 20.
The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the composition of liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition of the mixture in vapour phase.
(CBSE Sample Paper 2010)
Answer:
Let mole fraction of liquid A in solution = x_A
Mole fraction of liquid B in solution, x_B = 1 – x_A
P = P_A°X_A + P_B°X_B or = P_A°X_A + P_B°(1 – x_A)
p = 600 mm Hg
600 = 450 x_A + 700 (1 – x_A)
Solving, X_A = \(\frac{100}{250}\) = 0.4
Mole fraction of liquid A = 0.4
Mole fraction of liquid B = 1 – 0.4 = 0.6
Calculation of composition of vapour phase
p_A = P_A°X_A = 450 mm × 0.4 = 180 mm
p_B = P_B°X_B = 700 × 0.6 = 420 mm
P_total = p_A + p_B = 180 + 420 = 600 mm

Question 21.
Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to
250.0 g of water. (K_b for water = 0.512 K kg mol^-1, molar mass of NaCl = 58.44 g) (CBSE Delhi 2011)
Answer:
ΔT_b = \(\frac{\mathrm{iK}_{\mathrm{b}} \times 1000 \times \mathrm{W}_{2}}{\mathrm{~W}_{1} \times \mathrm{M}_{2}}\)
NaCl dissociates as:
NaCl → Na⁺ + Cl^–
∴ i = 2
W₂ = 15.0 g, W₁ = 250.0 g, M₂ = 58.44 g mol^-1
∴ K_b = 0.512 K kg mol^-1
∴ ΔT_b = \(\frac{2 \times 0.512 \times 1000 \times 15.0}{250.0 \times 58.44}\)
∴ Boiling point of solution = 100 + 1.05
= 101.5°C

Question 22.
(i) Define the following terms:
(a) Mole fraction
(b) van’t Hoff factor
(ii) 100 mg of a protein is dissolved in enough water to make 10.0 mL of a solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of protein? (R = 0.0821 L atm. mol^-1 K^-1 and 760 mm Hg = 1 atm.) (CBSE Delhi 2010)
Answer:
(i) (a) Mole fraction is the ratio of number of moles of one component to the total number of motes In a mixture. For example, in a binary mixture containing n₁ and n₂ moles of two components, Mole fraction of one component,
x₁ = \(\frac{n_{1}}{n_{1}+n_{2}}\)
Mote fraction of second component,
x₂ = \(\frac{n_{2}}{n_{1}+n_{2}}\)
(b) van’t Hoff factor is the ratio of the normal molar mass to the observed or abnormal molar mass of a solute in a solution due to association or dissociation.

(ii) Osmotic pressure,

Question 23.
(i) Differentiate between molarity and molality of a solution. How does a change in temperature influence their values?
(ii) Calculate the freezing point of an aqueous solution contaning 10.50 g of MgBr2 in 200 g of water (Molar mass of MgBr₂ = 184 g mol^-1, = for water = 1.86 K kg mol^-1). (CBSE Delhi 2011)
Answer:
(i) Molality is the number of moles of solute per 1000 g of solvent, whereas molarity is the number of moles of solute per 1000 ml of the solution. Molality is represented as m, whereas molarity is represented as M.

Molarity changes with change in temperature because of change in volume. On the other hand, there is no effect of temperature on the molality of the solution.

(if) Moles of MgBr₂
= \(\frac{10.50}{184}\)
= 0.0571 mol
Mass of water = 200 g
Molality = \(\frac{0.0571}{200}\) × 1000
= 0.2855 m
MgBr₂ ionises as:
MgBr₂ → Mg²⁺ + 2Br^–
Assuming complete dissociation of MgBr₂,
i = 3
Freezing point depression
ΔT_f = i × K_f × m
= 3 × 1.86 × 0.2855
= 1.59
Freezing point = 0 – 1.59°C
= – 1.59°C

Question 24.
(i) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
(ii) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (K_b for water = 0.512 K kg mol^-1, molar mass of NaCl = 58.44 g mol^-1) (CBSE Delhi 2011)
Answer:
(i) Osmosis is the flow of solvent from solution of lower concentration to higher concentration through a semi- permeable membrane.

Osmotic pressure is the excess pressure which must be applied to a solution to prevent the passage of solvent through a semipermeable memberane.

It has been found experimentally that for n moles of the solute dissolved in V litres of the solution, the osmotic pressure (π) at temperature T is
πV = nRT
where R is a gas constant.
or π = \(\frac{n}{V}\) RT
= C RT
where C is the molar concentration of the solution.
For a solution at given tempeature, both R and T are constant so that
π ∝ C
Thus, osmotic pressure depends upon the molar concentration of solution and therefore, is a colligative property.

(ii) ΔT_b = \(\frac{i \mathrm{~K}_{b} \times 1000 \times \mathrm{W}_{2}}{\mathrm{~W}_{1} \times \mathrm{M}_{2}}\)
NaCl dissociates as:
NaCl → Na⁺ + Cl^–
∴ i = 2
W₂ = 15.0 g, W₁ = 250.0 g,
M₂ = 58.44 g mol^-1
K_b = 0.512 K kg mol^-1 .
∴ ΔT_b = \(\frac{2 \times 0.512 \times 1000 \times 15.0}{250.0 \times 58.44}\)
= 1.05°C
∴ Boiling point of solution = 100 + 1.05
= 101.5°C

Question 25.
(a) A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water. The freezing point of pure water is 273.15 K.
(b) Why is osmotic pressure of 1 M KCl higher than 1 M urea solution?
(c) What type of liquids form ideal solutions?
OR
(a) 1.0 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^-1. Find the molar mass of the solute.
(b) What is the significance of Henry’s law constant, K_H?
(c) What leads to anoxia? (CBSE 2019C)
Answer:
(a)

(b) On dissolving in water KCl dissociates into K⁺ and Cl^– ions but urea does not dissociate into ions.

(c) Ideal solutions: An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of concentration.
Such solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecular forces. In these solutions, the intermolecular interactions between the components (A – B attractions) are of same magnitude as the intermolecular interactions in pure components (A – A and B – B attractions).

The ideal solutions have also the following characteristics:
1. Heat change on mixing is zero. Since there is no change in magnitude of the attractive forces in the two components present, the heat change on mixing, i.e. Δ_mixingH in such solutions must be zero.

(ii) Volume change on mixing is zero. In ideal solutions, the volume of the solution is the sum of the volumes of the components before mixing, i.e. there is no change in volume on mixing or Δ_mixingV is zero.

For example, when we mix 100 cm³ of benzene with 100 cm³ of toluene, the volume of the solution is found to be exactly 200 cm³. Therefore, there is no change in volume on mixing, i.e. Δ_mixingV = 0. It has been noticed that the solutions generally tend to become ideal when they are dilute.

Examples of ideal solutions: In fact, ideal solutions are quite rare but some solutions are nearly ideal in behaviour at least when they are very dilute. A few examples of ideal solutions are:

• Benzene and toluene
• n-hexane and n-heptane
• Bromoethane and iodoethane
• Chlorobenzene and bromobenzene.

OR

(a)

(b) (i) Henry’s law constant, K_H depends upon the nature of the gas.
(ii) Higher the value of K_H at a particular pressure, the lower is the solubility of the gas in the liquid. (∵ x = \(\frac{1}{\mathrm{~K}_{\mathrm{H}}}\) . p)

(iii) The value of K_H increases with increase in temperature indicating that the solubility of gases decreases with increase of temperature. This is the reason that aquatic species are more comfortable in cold water rather than warm water.

(c) At high altitudes, the partial pressure of oxygen is less than that at the ground level. This leads to low concentration of oxygen in the blood and the tissues of the people living at high altitudes. As a result of low oxygen in the blood, the people become weak and unable to think clearly. These are the symptoms of a condition known as anoxia.

Question 26.
(i) State Raoutt’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law?
(ii) 1.00 g of non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (K_f for benzene = 512 K kg mol^-1)
OR
(i) Define the following terms:
(a) Ideal solution
(b) Azeotrope
(c) Osmotic pressure
(ii) A solution of glucose (C₆H₁₂O₆) in
water is labelled as 10% by weight. What would be the molality of the solution? (CBSE 2013)
(Molar mass of glucose = 180 g mol^-1)
Answer:
(i) Raoult’s law states that at a given temperature, for a solution of volatile liquids, the partial pressure of each component in solution is equal to the product of the vapour pressure of the pure component and its mole fraction. For example, for a binary solution of two volatile liquids A and B having mole fractions x_A and x_B,
P_A = P°_A X_A and P_B = P°_B X_B
where p_A and p_B are the vapour pressures of the components in solutions and P°_A and P°_B are vapour pressure of pure components.
According to Henry’s law for a gas dissolved in a liquid, the pressure of the gas is directly proportional to mole fraction, i.e.
p = K_Hx
where K_H is a proportionality constant known as Henry’s constant.
But Raoult’s law states that
p = p°x
∴ K_H = p°
This means that Raoult’s law is a special case of Henry’s law.

(ii) Molar mass of solute,
M_B = \(\frac{K_{f} \times W_{B} \times 1000}{W_{A} \times \Delta T_{f}}\)
W_B = 1.0g
W_A = 50.0 g,
ΔT_f = 0.40 K
K_f = 5.12 K kg mol^-1
M_B = \(\frac{5.12 \times 1.0 \times 1000}{50 \times 0.40}\)
= 256 g/mol

OR

(i) (a) Ideal solution. A solution which obeys Raoult’s law exactly over the entire range of concentration is called ideal solution.

(b) Azeotrope. The solutions or liquid mixtures which boil at constant temperature and can distil unchanged in composition are called azeotropes.

(c) Osmotic pressure. The excess pressure which must be applied to a solution to prevent the passage of solvent into it through a semipermeable membrane is called osmotic pressure.

(ii) 10% by weight means that 10 g of glucose is present in 100 g of solution.
Mass of solvent = 100 – 10 = 90 g
Moles of glucose = \(\frac{10}{180}\) = 0.0556
Molality = \(\frac{0.0556}{90}\) × 1000
= 0.618 m

Question 27.
(i) Define the following terms:
(a) Molarity
(b) Molal elevation constant (K_b)
(ii) A solution containing 15 g urea
(molar mass = 60 g mol^-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol^-1) in water. Calculate the mass of glucose present in one litre of its solution. (CBSE 2014)
OR
(i) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(ii) A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL^-1)
Answer:
(i) (a) Molarity is defined as number of moles of solute dissolved per litre of solution.

Its unit is mol L^-1 or M.

(b) Molal elevation constant K_b is the elevation in boiling point for 1 molal solution.
(ii) For isotonic solutions,
π (urea) = π (glucose)

OR

(i) Mixture of ethanol and acetone shows positive deviations from Raoult’s law. This is because in ethanol, the molecules are held together due to hydrogen bonding as:

When acetone is added to ethanol, there are weaker interactions between acetone and ethanol than ethanol-ethanol interactions. Some molecules of acetone occupy spaces between ethanol molecules and consequently, some hydrogen bonds in alcohol molecules break and attractive forces between ethanol molecules are weakened.

Therefore, the escaping tendency of ethanol and acetone molecules from solution increases. Thus, the vapour pressure of the solution is greater than the vapour pressure as expected according to Raoult’s law.

(ii) W_B = 10 g, wt. of solvent = 90 g
M_B = 180 g mol^-1

Question 28.
(i) Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol-1) was dissolved in 50 g of water, assuming MgCl₂ undergoes complete ionisation.
(K_f for water = 1.86 K kg mol^-1)
(ii) (a) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(b) What happens when the external pressure applied becomes more than the osmotic pressure of solution? (CBSE Delhi 2016)
OR
(i) When 2.56 g of sulphur was dissolved in 100 g of CS₂, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_x).
(K_f for CS₂ = 3.83 K kg mol^-1, Atomic mass of sulphur = 32 g mol^-1)
(ii) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(a) 1.2% sodium chloride solution?
(b) 0.4% sodium chloride solution?
Answer:
(i) ΔT_f = \(\frac{i \mathrm{~K}_{f} \times \mathrm{w}_{\mathrm{B}} \times 1000}{M_{\mathrm{B}} \times \mathrm{w}_{\mathrm{A}}}\)
w_B = 1.9 g, w_A = 50 g, M_B = 95 g mol^-1
K_f = 1.86 K kg mol^-1
MgCl₂ ⇌ Mg²⁺ + 2Cl^–
i = 3
ΔT_f = \(\frac{3 \times 1.86 \times 1.9 \times 1000}{95 \times 50}\)
= 2.232 K
Freezing point of solution
= 273 – 2.232 = 270.768 K

(ii) (a) The elevation in boiling point is a colligative property and depends upon the number of moles of solute added. Higher the concentration of solute added, higher will be the elevation in boiling point. Thus, 2 M glucose solution has higher boiling point than 1 M glucose solution.

(b) When the external pressure applied becomes more than the osmotic pressure of the solution, then the solvent molecules from the solution pass through the semipermeable membrane to the solvent side. This process is called reverse osmosis.
OR
(i) M_B = \(\frac{\mathrm{K}_{f} \times 1000 \times \mathrm{W}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{A}} \times \Delta \mathrm{T}_{f}}\)
W_B = 2.56 g, W_A = 100 g, ΔT_f = 0.383 K_f = 3.83 K kg mol^-1
Let the molecular formula of sulphur
= S_x
32 × x = 256
x = \(\frac{256}{32}\) = 8
∴ Molecular formula = S₈

(ii) (a) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or blood cells. When red blood cells are placed in this solution, water flows out of the cell and they shrink due to loss of water by osmosis.

(b) 0.4% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or blood cells. When red blood cells are placed in this solution, water flows into the cells and they swell.

Question 29.
A 4% solutioin (w/w) of sucrose (M = 342 g mol^-1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol^-1) in water.
(Given: Freezing point of pure water = 273.15 K)
Answer:
In case of sucrose:
ΔT_f = (273.15 – 271.15) K = 2.00 K
Molar mass of sucrose (C₁₂H₂₂O₁₁)
= (12 × 12) + (22 × 1) + (11 × 16) = 342 g mol^-1
4% solution (w/w) of sucrose in water means 4 g of cane sugar in (100 – 4) g = 96 g of water.
Number of moles in 4 g sucrose in water
= 4/342 mol or 0.01169 mol
Therefore, molality of the solution,
m = 0.011696 mol / 0.096 kg
Or
m = 0.1217 mol kg^-1
Now applying the relation,
ΔT_f = K_f × m
⇒ K_f = ΔT_f / m
⇒ 2.00 K/0.1217 mol kg^-1
= 16.4338 K kg mol^-1
Molar mass of glucose (C₆H₁₂O₆) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g mol^-1
5% glucose in water means 5 g of glucose is present in (100 – 5) g = 95 g of water.
∴ Number of moles of glucose = 5/180 mol
= 0.0278 mol
Therefore, molality of the solution, m = 0.0278 mol / 0.095 kg
= 0.2926 mol kg^-1
Applying the relation, ΔT_f = K_f × m
ΔT_f = (16.4338 K kg mol^-1) × (0.2926 mol kg^-1 )
= 4.81 K (approx.)
Hence, the freezing point of 5% glucose solution is (273.15 – 4.81) K = 268.34 K.

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 13
Chapter Name	Nuclei
Number of Questions Solved	31
Category	NCERT Solutions

Question 1.
(a) Two stable isotopes of lithium \(_{ 3 }^{ 6 }{ Li } \) and \(_{ 3 }^{ 7 }{ Li } \) have respective abundance of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u respectively. Find the atomic weight of lithium.
(b) Boron has two stable isotopes \(_{ 5}^{ 10 }{ Li } \) and \(_{5 }^{ 11 }{ Li } \) ^.Their respective masses are 10.01294 u and 11.00931 u and the atomic weight of boron is 10.811 u. Find the abundances of \(_{ 5 }^{ 10 }{ Li } \) and
\(_{5 }^{ 11 }{ Li } \)
Answer:
(a) Atomic weight of lithium


Question 2.
The three stable isotopes of neon :\(_{ 20}^{ 10 }{ Ne } \) and \(_{ 22}^{ 10 }{ Ne } \)  have respective abundance of 90.51%, 0.27% and 9.22%. The atomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.
Answer:

nuclear equation calculator is available in my webiste.

Question 3.
Obtain the binding energy of a nitrogen nucleus (\(_{ 7}^{ 14 }{ N } \)) from the following data :
m_H = 1.00783 u
m_n = 1.00867 u
_mn = 14.00307 u
Give your answer in MeV.
Answer:

Question 4.
Obtain the binding energy of the nuclei \(_{ 26 }^{ 56 }{ Fe } \) and
in units of \(_{ 83 }^{ 209 }{ Bi} \) from the following data:
m_H =1007825u
m_n =1008665u
m (\(_{ 26 }^{ 56 }{ Fe } \))= 55.934939 u
m (\(_{83}^{209 }{ Bi} \))
Which nucleus has greater binding energy per nucleon?
Answer:

Question 5.
A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of \(_{ 29 }^{ 63 }{ Cu } \) atoms (of mass 62.92960 u). The masses of proton and neutron are 1.00783 u and 1.00867 u, respectively.
Answer:

Question 6.
Write nuclear equations for :
(a) the α-decay of \(_{ 86 }^{226 }{ Ra} \)
(b) the β^–-decay of \(_{ 15 }^{ 32 }{ p } \)
(c) the β⁺-decay of \(_{ 6 }^{ 11 }{ p } \)
Answer:

Question 7.
A radioactive isotope has a half-life of T years. After how much time is its activity reduced to 3.125% of its original activity (b) 1% of original value ?
Answer:

Question 8.
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive \(_{ 6 }^{ 14 }{ C } \) present with the stable carbon isotope
\(_{ 12 }^{ 6 }{ C } \) When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of \(_{ 6 }^{ 14 }{ C } \) , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of \(_{ 6 }^{ 14 }{ C } \) dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Answer:

Question 9.
Obtain the amount of \(_{ 27 }^{ 60 }{ Co } \) necessary to provide a radioactive source of 8.0 mCi strength. The half­ life of \(_{ 27 }^{ 60 }{ Co } \) is 5.3 years.
Answer:

Question 10.
The half-life of \(_{ 38 }^{ 90 }{ Sr } \) is 28 years. What is the disintegration rate of 15 mg of this isotope?
Answer:

Question 11.
Obtain approximately the ratio of the nuclear radii of the gold isotope \(_{ 79 }^{ 197}{ Au } \) and silver isotope \(_{47}^{107 }{ Au } \).
Answer:

Question 12.

Answer:

Question 13.

Answer:

Here m_N stands for the nuclear mass of the element or particle. In order to express the Q value in terms of the atomic masses, 6 m_e mass has to be subtracted from the atomic mass of \(_{6}^{11 }{ Au } \) and 5 m_e mass has to be

Question 14.
The nucleus \(_{ 10 }^{ 23 }{ Ne} \) decays by β~ emission. Write down the p-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
m(\(_{ 10 }^{ 23 }{ Sr } \)) = 22.994466 u
m(\(_{ 11 }^{ 23 }{ Sr } \)) = 22.989770 u.
Answer:

Question 15.
The Q value of a nuclear reaction A + b ⇒ C + d is defined by [Q = m_A + m_b-m_c– m_d] c² where the masses refer to nuclear rest masses. Determine from the given data whether the following reactions are exothermic or endothermic.

Answer:

Question 16.
Suppose, we think of fission of a \(_{ 26}^{ 56 }{ Fe} \) nucleus into two equal fragments, if \(_{ 13}^{ 28 }{ Al } \). Is the fission energetically possible ? Argue by working out Q of the process. Given, m (\(_{ 26}^{ 56 }{ Fe} \)) = 55.93494 u and m (\(_{ 13}^{ 28 }{ Al } \))= 27.98191
Answer:


Question 17.
The fission properties of \(_{ 94}^{ 239 }{ Pu} \) are very similar to those of \(_{ 92}^{ 235 }{ u} \)u. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure
\(_{94}^{ 239 }{ Pu} \) undergo fission ?
Answer:

Question 18.
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much
\(_{ 92}^{ 235 }{ u} \) did it contain initially ? Assume that all the energy generated arises from the fission of \(_{92}^{ 235 }{ u} \) and that this nuclide is consumed by the fission process.
Answer:

Question 19.
How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium ? The fusion reaction can be taken as

Answer:

Question 20.
Calculate the height of Coulomb barrier for the head on collision of two deuterons. The effective radius of deuteron can be taken to be 2.0 fm.
Answer:
The initial mechanical energy E of the two deutrons before collision is given by
E = 2 K.E.

Question 21.
From the relation R = R₀ A^1/3, where R₀  is a constant and A is the mass number of a nucleus, show that nuclear matter density is nearly constant (i.e. independent of A)
Answer:

Question 22.
For the β⁺ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K- shell, is captured by the nucleus and a neutrino is emitted).

Show that if β⁺ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa
Answer:

Question 23.
In a Periodic Table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on Earth. The three isotopes and their masses are \(_{ 12}^{ 24 }{ Mg} \) (23.98504u), ? \(_{12}^{ 25 }{ Mg} \) (24.98584) and \(_{ 12}^{ 26 }{ Mg} \) (25.98259u). The natural abundance of \(_{ 12}^{ 24 }{ Mg} \) is 78.99% by mass. Calculate the abundances of the other two isotopes.
Answer:

Question 24.
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei \(_{ 12}^{ 24 }{ Ca} \) and \(_{ 13}^{ 27 }{ Al} \) from the following data :

Answer:

Question 25.

Answer:

Question 26.
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes :

(a) Calculate the Q values for these decays and determine that both are energetically possible.
(b) The Coulomb barrier height for α-particle

Answer:

Question 27.
Consider the fission of \(_{ 92}^{ 239}{u} \) by fast neutrons. In one fission event, no neutrons are emitted and the final stable end products, after the beta-decay of the primary fragments, are \(_{58}^{ 140}{Ce} \) and \(_{44}^{ 99}{Ru} \). Calculate Q for this fission process. The relevant atomic and particle masses

Answer:

Question 28.
Consider the D-T reaction (deuterium-tritium-fusion) given in eqn. :

(b) Consider the radius of both deuterium and tritium to be approximately 1.5 fm. What is the kinetic energy needed to overcome the Coulomb repulsion? To what temperature must the gases be heated to initiate the reaction?
Answer:
From the equation given in the question,

m_N refers to the nuclear mass of the element given in the brackets and m_n = mass of the neutron. If in represents the atomic mass, then


Question 29.
Obtain the maximum kinetic energy of p-particles and the radiation frequencies to y decay in the following decay scheme. You are given that
m (¹⁹⁸Au) = 197.968233 u
m (¹⁹⁸Hg) = 197.966760 u
Answer:
The total energy released for the transformation of \(_{79}^{ 198}{Au} \) to \(_{80}^{ 198}{u} \) can be found by considering the energies of ϒ-rays. We first find the frequencies of the ϒ-rays emitted.




Question 30.
Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within the sun and (b) the fission of 1.0 kg of ²³⁵U in a fission reactor.
Answer:

Question 31.
Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plant. Suppose we are given that, on average, the efficiency of utilisation (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium did our country need per year by 2000 ? Take the heat energy per fission of ²³⁵U to be about 200 MeV. Avogadro’s number = 6.023 x 10²³ mol^-1.
Answer:

We hope the We hope theNCERT Solutions for Class 12 Physics Chapter 13 Nuclei, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 13 Nuclei, drop a comment below and we will get back to you at the earliest.

By going through these CBSE Class 12 Maths Notes Chapter 12 Linear Programming, students can recall all the concepts quickly.

Linear Programming Notes Class 12 Maths Chapter 12

Linear Programming Problems: Problems that minimize or maximize a linear function Z subject to certain conditions, determined by a set of linear inequalities with non-negative variables, are known as Linear Programming Problems.

Objective Function: A linear function Z = ax + by, where a and b are constants, which has to be maximized or minimized according to a set of given conditions, is called a linear objective function.

Decision Variables: In the objective function Z = ax + by, the variables x, y are said to be decision variables.

Constraints: The restrictions in the form of inequalities on the variables of linear programming problems are called constraints. The conditions x > 0, y > 0 are known as non-negative restrictions.

Example: A dealer deals in only two items: wall hangings and artificial plants. He has ₹ 5,000 to invest in and a space to store at the most 80 pieces. A wall hanging costs him ₹300 and an artificial plant ₹150. He can sell a wall hanging at a profit of ₹50 and an artificial plant at a profit of ₹18. Assuming he can sell all the items that he buys, formulate a linear programming problem in order to maximize his profit.

Let x be the number of wall hanging and y be the number of artificial plants.
Profit on one wall-hanging = ₹50.
∴ Profit by selling x wall hangings = ₹50x
Profit on one artificial plant = ₹18

∴ Profit by selling y artificial plants = ₹18y
The total profit = 50x + 18y
∴ Z = 50x + 18y is the objective function.

A wall hanging costs = ₹300.
∴ Cost of x wall-hangings = ₹300x
The cost of one artificial plant = ₹150

∴ Cost of y artificial plants = ₹150y
The total investment = ₹(300x + 150y)

But the maximum investment = ₹15000
⇒ 300x + 150y ≤ 15000
or
2x + y ≤ 100.

Further, storage capacity to a maximum is 80 pieces.
⇒ x + y ≤ 80.

We know that x ≥ 0, y ≥ 0
Thus, the constraints are 2x + y ≤ 100, x + y ≤ 80,
x ≥ 0, y ≥ 0

R Programming MCQ questions are useful for the campus placement for all freshers including experience professionals and engineers.

Graphical Method of solving a linear programming Problem:
In the above example, the constraints are 2x + y ≤ 100, x + y ≤ 80, x ≥ 0, y ≥ 0.

Now, we draw the lines 2x + y = 100, x + y = 80, x = 0, y = 0 and we find the regions which represent these inequalities.
(i) Consider the inequality
2x + y ≤ 100 .
The line 2x + y = 100 passes through A(50, 0) and B(0,100).

Putting x = 0, y = 0 in 2x + y ≤ 100, we get 0 ≤ 100, which is true.

This shows origin lies in this region, i.e., the region 2x + y ≤ 100 lies on and below AB as shown in the figure.
1. Next inequality is x + y ≤ 80.
The line x + y = 80 passes through C(80,0) and D(0, 80).
Putting x = 0, y = 0 in x + y < 80, we get 0 ≤ 80, which is true.
⇒ Origin lies in this region, i.e., the region x + y ≤ 80 is on and below CD.
(iii) x > 0 is the region on and to the right of the y-axis.
(iv) y > 0 is the region on and above the x-axis.

→ Feasible Region: The common region determined by all the constraints including non-negative constraints x, y > 0 of linear programming problem is known as feasible region.
In the figure drawn, the shaded area OAPD is the feasible region, which is the common area of the regions drawn under the given constraints.

→ Feasible Solution: Points within and on the boundary of the feasible region represents feasible solutions of constraints.

In the above figure, every point of the shaded area (feasible region) is the feasible solution.
In the feasible region, there are infinitely many points (solutions) that satisfy the given conditions. But we would like to know the points, where Z is maximum or minimum.

→ Theorem 1: Let R be the feasible region for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimum value (maximum or minimum), where variables x and y are subject to constraints described by linear inequalities, the optimal value must occur at a corner point (vertex) of the feasible region.

→ Theorem 2: Let R be the feasible region for a linear programming problem and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both maximum and minimum value on R and each of these occurs at a corner point (vertex) of R.

Solving the equation 2x + y = 100 and x + y = 80, we get the coordinates of the point P(20, 60). •
In the example given above,
The values of Z at corner points are
At D (0, 80), Z = 50x + 18y = 18 × 80 – 1440
At P(20, 60), Z = 50 × 20 + 18 × 60 = 1000 + 1080 = 2080
At A(50, 0), Z = 50 × 50 + 0 = 2500
At 0(0, 0), Z = O.
Thus, the maximum value of Z is ₹2080 at P(20, 60) and the minimum value of Z is ₹1440 at D(0, 80). In fact, the minimum value is ₹0 at O.

However, if the feasible region is unbounded, the optimal value obtained may not be maximum or minimum.

To determine the optimal point, we proceed as follows:

1. Let M be the maximum value found as explained above, if the open half-plane determined by ax + by > M has no point in common with the feasible region, then M is the maximum value of Z. Otherwise Z has no maximum value.
2. Similarly, if m is the minimum value of Z and if the open half-plane determined by ax + by < m, has no point in common with the feasible region, then m is the minimum value of Z, otherwise, Z has no minimum value.

Different Types of Linear Programming Problems:
A few important linear programming problems are as follows:
1. Manufacturing Problem: In such a problem, we determine

1. a number of units of different products to be produced and sold.
2. Manpower required, Machine hours needed, warehouse space available, etc. The objective function is to maximize profit.

2. Diet Problem: Here we determine the number of different types of constituents or nutrients which should be included in the diet. The objective function is to minimize the cost of production.

3. Transportation Problems: These problems deal with the costs of transportation which are to be minimized under given constraints.

1. LINEAR PROGRAMMING

Linear programming is a too! which is used in decision making in business for obtaining maximum and minimum values of quantities subject to certain constraints.

2. MATHEMATICAL FORMULATION

Let the linear function Z be defined by :
(i) Z = c₁ x₁ + c₂x₂ + … + c_nx_n, where c’s are constants.
Let a_ij. be (mn) constants and b_j. be a set of m constants such that:
a₁₁x₁ + a₁₂ x₂ + … + a_1n x_n (≤ , = ,≥) b₁
a₂₁x₁ + a₂₂ x₂ + … + a_2n x_n (≤ , = ,≥) b₂
………………………………..
………………………………..
a_m1x₁ + a_m2 x₂ + … + a_mn x_n (≤ , = ,≥) b_m
and let (iii) x₁ ≥ 0, x₁ ≥ 0, …, x ≥ 0.

The problem of determining the values of x , x^, …, x which maximizes (or minimizes) Z and satisfying above (ii) is called General Linear Programming Problem (LPP).
Note : In (ii), there may be any sign <. =, or >.

3. SOME DEFINITIONS

A. (i) Objective Function. The linear function Z = ax + by, where a, b are constants, which is to be maximized or minimized is called a linear objective function of LPP.
The variables x and y are called decision variables.

(ii) Constraints. The linear inequalities or inequations or restrictions on the variables of a linear programming problem are called constraints of LPP.
The conditions ,r > 0, y > 0 are called non-negative restrictions of LPP.

(iii) Optimization Problem. A problem, in which it is required to maximize or minimize a linear function (say of two variables x and y) subject to certain constraints, determined by a set of linear inequalities, is called the optimizations problem.

(iv) Solution. The values of x, y which satisfy the constraints of LPP is called the solution of the LPP.

(v) Feasible Solution. Any solution of LPP, which satisfies the non-negative restrictions of
the problem, is called a feasible solution to LPP.

(vi) Optimum Solution. Any feasible solution, which optimizes (minimizes or maximizes) the objective function of LPP is called an optimum solution of the general LPP.

B. (i) Feasible Region. The common region, which is determined by all the constraints including non-negative constraints x, y > 0 of a LPP is called feasible region (or solution region.) The region, other than the feasible one, is called an infeasible region.

(ii) Feasible Solution. The points, which lie on the boundary and within the feasible region, represent the feasible solution of the constraints.
Any part outside the feasible solution is called infeasible solution.

(iii) Optimal (Feasible) Solution. Any point in the feasible region, which gives the optimal value (maximum or minimum) of the objective function, is called optimal solution. These points are infinitely many in number.

Theorem I. Let R be the feasible region (convex polygon) for LPP and Z = ax + by, the objective function.

When Z has an optimal value (max. or min.) where x, y are subject to constraints, this optimal value will occur at a comer point (vertex).

Theorem II. Let R be the feasible region for LPP and Z = ax + by, the objecti ve function. If R is bounded , then the objective function Z has both maximum and minimum values of R and each occurs at a corner point (vertex).

Key Point
When R is unbounded, max. (or min.) value of objective function may not exist. In case it exists, it occurs at a comer point of R.

4. CORNER POINT METHOD

In order to solve a Linear Programming Problem we use Corner Point Method which is as below

Step I. Obtain the feasible region of LPP and determine its comer points (vertices).
Step II. Evaluate the objective function Z = ax + by at each comer point (vertex).
Let M and m be the largest and smallest values at these points.
Step III. (a) When the feasible region is bounded,
then M and m are the maximum and minimum values of Z.

(b) When the feasible region is unbounded, then
(i) M is the maximum value of Z if the open half-plane determined by ax + by > M has no common point with the feasible region; otherwise Z has no maximum value
(ii) m is the minimum value of Z if the open half-plane determined by ax + by < m has no common point with the feasible region; otherwise Z has no minimum value.

Here we are providing Online Education for Class 12 Hindi Important Extra Questions and Answers Aroh Chapter 1 आत्म-परिचय, एक गीत. Important Questions for Class 12 Hindi are the best resource for students which helps in class 12 board exams.

Online Education for आत्म-परिचय, एक गीत Class 12 Important Extra Questions Hindi Aroh Chapter 1

प्रश्न 1.
‘आत्म-परिचय’ कविता का मूल भाव स्पष्ट कीजिए।
उत्तर
श्री हरिवंश राय बच्चन’ द्वारा रचित कविता ‘आत्म-परिचय’ ‘बुद्ध और नाचघर’ संग्रह से संकलित है जिसमें कवि ने यह चित्रण किया है कि मनुष्य द्वारा अपने को जानना या आत्मबोध दुनिया को जानने से अत्यंत कठिन है। समाज से मनुष्य का नाता खट्टा-मीठा होता है। इस संसार से निरपेक्ष रहना असंभव है। मनुष्य चाहकर भी जग से विमुख नहीं हो सकता। मनुष्य एक सामाजिक प्राणी है, अतः मनुष्य का इस जग से अटूट संबंध है।

संसार अपने व्यंग्य-बाणों तथा शासन-प्रशासन से उसे चाहे कितने ही कष्ट एवं पीड़ाएं। क्यों न दे, पर मनुष्य इस जगह से अलग नहीं रह सकता। ये दुनिया ही उसकी पहचान है। जहाँ पर वह अपना परिचय देते हुए इस संसार से द्विविधात्मक एवं वंद्वात्मक संबंधों का मर्म उद्घाटित करता हुआ जीवन जीता है। इस दुनिया में मनुष्य का जीवन द्वंद्व
एवं विरुधों का सामंजस्य है। सुख-दुख का समन्वय है।

प्रश्न 2.
प्रस्तुत काव्यांश का भाव-सौंदर्य एवं काव्य-सौंदर्य स्पष्ट कीजिए।

मैं स्नेह-सुरा का पान किया करता हूँ,
मैं कभी न जग का ध्यान किया करता हूँ,
जग पूछ रहा उनको, जो जग की गाते,
मैं अपने मन का गान किया करता हूँ।
उत्तर
भाव-सौंदर्य प्रस्तुत काव्यांश श्री हरिवंश राय बच्चन द्वारा रचित कविता ‘आत्म-परिचय’ से अवतरित है। इस काव्यांश में कवि ने संसार
की स्वार्थपरता का चित्रांकन किया है। कवि कहता है कि मैं तो प्रेम रूपी मदिरा का पान कर उसकी मस्ती में मस्त रहता हूँ। मुझे इस संसार की बिल्कुल भी चिंता नहीं है। यह संसार मेरे बारे में क्या कहता है, क्या सोचता है मुझे इसकी कोई चिंता नहीं। यह तो स्वार्थी है, केवल उनको पूछता है जो इसका गान करते हैं अर्थात इसकी चापलूसी करते हैं। मैं तो इस जग से दूर केवल अपनी मनोभावनाओं को गाता रहता हूँ।

काव्य-सौंदर्य
(i) प्रस्तुत काव्यांश हालावाद के प्रवर्तक कवि हरिवंश राय बच्चन द्वारा रचित ‘आत्म-परिचय’ कविता से अवतरित है। इसमें कवि ने संसार की स्वार्थपरता का चित्रण किया है कि यह संसार केवल उनको पूछता है, जो इसकी चापलूसी करते हैं।
(ii) प्रस्तुत काव्यांश में खड़ी बोली भाषा का प्रयोग है जो सरल, सरस एवं प्रवाहमयी है।
(iii)स्नेह-सुरा में रूपक अलंकार की छटा शोभनीय है।
(iv) अनुप्रास, स्वरमैत्री तथा पदमैत्री अलंकारों की सुंदर प्रयोग है।
(v) संस्कृत के तत्सम और तद्भव शब्दों का सफल प्रयोग है।
(vi) गीति शैली का प्रयोग है।
(vii) मुक्तक छंद है।
(vi) माधुर्य गुण है।
(ix) श्रृंगार एवं शांत रस है।
(x) अभिधा शब्द-शक्ति का प्रयोग है।
(xi) बिंब योजना अत्यंत सुंदर है।

प्रश्न 3.
काव्यांश का भाव-सौंदर्य एवं काव्य-सौंदर्य स्पष्ट कीजिए।
‘मैं और, और जग और, कहाँ का नाता
मैं बना बना कितने जग रोज मिटाता’
उत्तर
भाव-सौंदर्य-श्री हरिवंश राय बच्चन ने प्रस्तुत काव्यांश में संसार को स्वयं से भिन्न स्थापित किया है।
(i) प्रस्तुत अवतरण श्री हरिवंश राय बच्चन द्वारा रचित ‘आत्म-परिचय’ नामक कविता से अवतरित है। इसमें कवि ने संसार को स्वयं से बिल्कुल भिन्न स्थापित किया है।
(ii) खड़ी बोली की भाषा सरल, सरस एवं भावमयी है।
(iii) शैली अत्यंत गंभीर एवं रोचक है।
(iv) तत्सम, तद्भव एवं उर्दू व फ़ारसी के शब्दों का समन्वय है।
(v) ‘और-और’ में भिन्नर्थक आपूर्ति होने से यमक अलंकार की शोभा है।
(vi) ‘बना-बना’ में पुनरुक्ति प्रकाश की छटा है।
(vii) इसके साथ इसमें अनुप्रास, पदमैत्री अलंकारों का प्रयोग भी हुआ है।
(viii) शैली गीतिमयता से परिपूर्ण है।
(ix) माधुर्य गुण का समावेश है।
(x) शांत रस है।
(xi) बिंब योजना सार्थक एवं सारगर्भित है।

प्रश्न 4.
‘दिन जल्दी-जल्दी ढलता है!
हो जाए न पत्र में रात कहीं
मंजिल भी तो है दूर नहीं
यह सोच थका दिन का पंथी
भी जल्दी-जल्दी चलता है।
उपर्युक्त पंक्तियों का भाव-सौंदर्य
एवं काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर :
भाव-सौंदर्य- प्रस्तुत अवतरण श्री हरिवंश राय बच्चन विरचित ‘एक गीत’ नामक कविता से अवतरित है। इसमें कवि ने समय की परिवर्तनशीलता एवं जीवन की क्षणभंगुरता का स्पष्ट चित्रांकन किया है। समय निरंतर चलायमान एवं परिवर्तनशील है। वह प्रतिपल परिवर्तित होता रहता है। यही धारणा लेकर थका हुआ यात्री शीघ्रता से अपनी मंजिल की तरफ़ चलता है। हालाँकि उसकी मंजिल ज्यादा दूर नहीं है किंतु वह यह सोचकर बेचैन होता है कि कहीं लक्ष्य पर पहुँचने से पहले ही रात न हो जाए।

काव्य-सौंदर्य
(i) इस काव्यांश में श्री हरिवंश राय बच्चन ने समय की चिर परिवर्तनशीलता एवं जीवन की क्षणभंगुरता का वर्णन किया है।
(ii) भाषा सरल, सरस खड़ी बोली है।
(iii) तत्सम, तद्भव एवं उर्दू-फ़ारसी शब्दावली है।
(iv) ‘जल्दी-जल्दी’ में पुनरुक्ति प्रकाश अलंकार है।
(v) अनुप्रास-पदमैत्री की छटा भी है।
(vi) मुक्तक छंद का प्रयोग है।
(vii) शांत रस है।
(viii) प्रसाद गुण है।
(ix) बिंब-योजना सार्थक एवं सटीक है।

प्रश्न 5.
कवि जग-जीवन को भार स्वरूप क्यों समझता है?
उत्तर
कवि जग-जीवन को भार स्वरूप इसलिए समझता है क्योंकि वह जीवन को प्रेम, मस्ती, आनंद एवं सौंदर्य से परिपूर्ण मानता है जबकि संसार इनसे दूर कहीं कोरी कल्पनाओं में ही डूबा हुआ है। संसार को प्रेम, मस्ती एवं आनंद से कोई मतलब नहीं है। वह तो ईर्ष्या, द्वेष आदि की भावनाओं में ही जकड़ा रहता है।

प्रश्न 6.
कवि कैसे जीवन की कामना करता है?
उत्तर
कवि ऐसे जीवन की कामना करता है जो प्रेम, मस्ती, आनंद और सौंदर्य से भरपूर हो, जिसमें चारों तरफ सुंदरता हो। प्रेम रूपी मदिरा का मौसम हो। जहाँ वह केवल प्रेम रूपी मदिरा पीकर उसी में डूबकर आनंद विभोर हो जाए। कहीं कोई ईर्ष्या-द्वेष जैसे कुविचार न हों, जिसमें योवन की मदमस्त करने वाली उन्माद हो।

कवि का मंतव्य है कि यह संसार उससे बिल्कुल भिन्न है। उसके विचार, उसकी भावनाएँ इस क्रूर संसार से समन्वय स्थापित नहीं कर सकतीं। इसका दृष्टिकोण उससे बिल्कुल अलग है अत: इस संसार के साथ उसका कोई संबंध नहीं हो सकता। यह संसार तो केबल एक परिपाटी पर अपना जीवनयापन कर रहा है, जबकि वह प्रतिदिन ऐसे अनेक लोक बनाकर उन्हें नष्ट कर देता है। काव्य-सौंदर्य

प्रश्न 7.
आत्म-परिचय कविता में कवि संसार को क्या संदेश देना चाहता है?
उत्तर
आत्म-परिचय कविता श्री हरिवंश राय बच्चन द्वारा रचित है। इसमें कवि संसार को मस्ती का संदेश देना चाहता है। ऐसी मस्ती जिसमें संपूर्ण संसार मदमस्त होकर आनंद विभोर हो उठे। वह आनंदित होकर नृत्य करने लगे और इसी मस्ती में सदैव लहराता रहे।

प्रश्न 8.
आत्म-परिचय कविता के माध्यम से स्पष्ट कीजिए कि कवि क्यों रोया होगा?
उत्तर :
कवि प्रेम एवं मस्ती की कोमल भावनाओं से ओतप्रोत है। उसे अपने जीवन में किसी प्रिया से असीम प्रेमभाव हुआ, किंतु दुर्भाग्य से वह प्रेम पूर्ण नहीं हो सका जिसके बाद कवि को अगाध विरह-पीड़ा को सहन करना पड़ा। यही पीड़ा उसके रोदन के माध्यम से प्रस्फुटित हुई। इसी कारण है कि कवि रोया होगा।

प्रश्न 9.
कवि और संसार में क्या भिन्नता है?
उत्तर :
कवि प्रेम एवं मस्ती में मदमस्त रहता है। वह प्रेम रूपी मदिरा पीकर उसी के आनंद में डूबा हुआ है। वह चारों तरफ प्रेम, मस्ती, आनंद एवं सौंदर्य का वातावरण फैलाना चाहता है जबकि यह संसार परस्पर ईर्ष्या, द्वेष, अहं की भावनाओं में डूबा है। इसे किसी के प्रेम व आनंद से कोई मतलब नहीं है। यह तो प्रेम की मस्ती को भी व्यर्थ समझता है। यह सदैव दूसरों की कोमल भावनाओं से खिलवाड़ करता है और उनका मज़ाक उड़ाता है।

प्रश्न 10.
कवि ज्ञान की बजाय किसे अपनाना चाहता है और क्यों?
उत्तर :
कवि ज्ञान की अपेक्षा प्रेम को अपनाना चाहता है क्योंकि वह ज्ञान के रास्ते को अत्यंत कठिन एवं दुष्प्राप्य मानता है, जिसे अनादि काल से संसार के बड़े-बड़े महात्मा, मुनि एवं संतजन भी कोटि-कोटि प्रयास करने पर भी नहीं जान सके।

प्रश्न 11.
पथिक जल्दी-जल्दी क्यों चलता है?
उत्तर:
पथिक जल्दी-जल्दी इसलिए चलता है क्योंकि उसे यह भय रहता है कि कहीं पथ में ही रात न हो जाए। उसकी मंज़िल उससे छूट न जाए।

प्रश्न 12.
चिड़िया के पंखों में चंचलता कैसे भरती है?
उत्तर
चिड़िया जब सोचती है कि उसके बच्चे घोंसले में उसकी राह देख रह होंगे। वे अकेले होंगे और अपनी माँ के आने की प्रतीक्षा में नीड़ों से निरंतर रास्ते की ओर झोंक रहे होंगे, तो उसके पंखों में चंचलता भर जाती है। पनि जल्दी-जल्दी डालता जीत काया राय बच्चन का वा निम पाल का संकलित है।

प्रश्न 13.
दिन जल्दी-जल्दी ढलता है’ कविता का मूल भाव अथवा प्रतिपाद्य स्पष्ट कीजिए।
उत्तर
‘दिन जल्दी-जल्दी ढलता है’ गीत हरिवंश राय बच्चन के ‘निशा-निमंत्रण’ गीत संग्रह से संकलित है। इस गीत में कवि ने प्रकृति थी दैनिक परिवर्तनशीलता के संदर्भ में प्राणी के धड़कते हृदय की भावनाओं को सुनने का प्रयास किया है। समय चिर परिवर्तनशील है। किसी प्रिय के आलंबन या विषय से भावी साक्षात्कार का आश्वासन ही हमारे प्रयास के चरणों की गति में और अधिक गतिशीलता एवं साहस पैदा कर देता है।

प्रश्न 14.
‘आत्म-परिचय’ कविता कहाँ से ली गई है और इसमें किस चीज़ का वर्णन किया गया है?
उत्तर
‘आत्म-परिचय’ कवित हरिवंश राय बच्चन के काव्य संग्रह ‘बुद्ध और नाचघर’ से ली गई है। इस कविता में कवि ने मानव के आत्म-परिचय का चित्रण किया है।

प्रश्न 15.
‘आत्म-बोध’ का अर्थ बताते हुए बतलाइए कि कवि ने कविता में किसका परिचय दिया है ?
उत्तर
‘आत्म-बोध’ का अर्थ है-स्वयं को जानना, जो संसार को जानने से अधिक कठिन है। व्यक्ति का समाज से घनिष्ठ संबंध है। कविता में कवि ने अपना परिचय देते हुए लगातार दुनिया से अपनी दुविधा और वंद्वपूर्ण संबंधों का रहस्य प्रकट किया है।

प्रश्न 16.
कवि किस चीज़ का भार लेकर अपने कंधों पर घूम रहा था तथा किस चीज़ ने उसके मन को झंकृत कर दिया?
उत्तर
कवि सांसारिक जीवन का संपूर्ण भार अपने कंधों पर लेकर घूम रहा था। इस सारे भार को उठाने के बाद भी वह संसार में अपने जीवन में प्यार के लिए घूम रहा था। कवि की हृदय रूपी वीणा को किसी ने प्रेम से छूकर झंकृत कर दिया। वह उसी झंकृत वीणा की साँसों को लिए दुनिया में घूम रहा है।

प्रश्न 17.
कवि का किससे कोई लेना-देना नहीं है और क्यों?
उत्तर
कवि का इस संसार से कोई लेना-देना नहीं है। उसकी दृष्टि में यह संसार एकदम अपूर्ण है, अतः उसे अच्छा नहीं लगता। वह अपने हृदय के भाव एवं उपहार लिए घूम रहा है। उसका अपना एक स्वप्निल संसार है। उसी संसार को लिए वह घूम रहा है।

प्रश्न 18.
कवि ने संसार को क्या माना है और उसे किस चीज़ की आवश्यकता नहीं है?
उत्तर
कवि सुख-दुख दोनों अवस्थाओं में मान रहता है। उसने संसार को एक सागर के समान माना है। उसके अनुसार संसार के लोग इस संसार रूपी सागर को पार करने हेतु नाव बना सकते हैं, किंतु उसे इस नाव की कोई आवश्यकता नहीं है। वह सांसारिक खुशियों में डूबकर यूं ही बहना चाहता है।

प्रश्न 19.
कवि आंतरिक रूप से क्यों रोता रहता है?
उत्तर
कवि का कहना है कि उसके पास जवानी का जोश है तथा इस जोश में छिपा दुख है। इसी कारण वह बाहय रूप से तो हँसता हुआ
दिखता है, लेकिन आंतरिक रूप से निरंतर रोता रहता है।

प्रश्न 20.
कवि और संसार में किस प्रकार का संबंध है?
उत्तर
कवि और संसार का कोई संबंध नहीं है। उसकी राह कोई और है तथा संसार की कोई और। वह न जाने प्रतिदिन कितने जग बना-बना कर मिटा देता है। कवि के अनुसार, यह संसार जिस पृथ्वी पर रहकर अपना वैभव जोड़ना चाहता है, वह प्रतिपग इस पृथ्वी के वैभव को तुकरा देता है।

प्रश्न 21.
कवि के सदन में क्या छिपा है तथा वह अपने संग क्या लेकर घूमता रहता है?
उत्तर
कवि के रूदन में एक राग छिपा है तथा उसकी शीतल वाणी में क्रोध की आग समाहित है। वह एक विराट खंडहर का अंश अपने साथ लेकर घूमता रहता है। जिस पर बड़े-बड़े राजाओं के महल भी अर्पित हो जाते हैं।

प्रश्न 22.
कवि ने संसार को अजीब क्यों कहा है?
उत्तर
कवि ने संसार को अजीब इसलिए कहा है, क्योंकि वह उसके रोने को भी गीत समझता है। दुखों की अपार वेदना के कारण जब वह फूट-फूट कर रोया तो संसार ने उसका छंद समझा।

प्रश्न 23.
कवि ने स्वयं को दीवाना क्यों कहा है?
उत्तर
कवि ने स्वयं को दीवाना इसलिए कहा है क्योंकि वह तो दीवानों का वेश धारण कर अपनी मस्ती में मस्त होकर घूम रहा है। साथ ही वह अपने संग मस्ती का संदेश लेकर घूम रहा है जिसे सुनकर ये सारा संसार झूम उठेगा, झुक जाएगा तथा लहराने लगेगा।

प्रश्न 24.
कवि मस्ती में डूबकर मन के गीत क्यों गाता रहता है?
उत्तर
प्रेम रूपी मदिरा को पी लेने के कारण कवि इसी में मग्न रहता है। उसे इस संसार का बिल्कुल भी ध्यान नहीं है। कवि के अनुसार यह संसार मात्र उन्हीं की पूछ करता है जो उसका गान करते हैं। यह स्वार्थ के नशे में डूबकर औरों की अनदेखी कर देता है, लेकिन कवि अपनी मस्ती में डूब मन के गीत गाता रहता है।

प्रश्न 25.
कविता में कवि अपने हृदय में क्या जलाकर रखता है?
उत्तर
कविता में कवि अपने हृदय में अग्नि जलाकर रखता है तथा स्वयं भी उसी में जलता रहता है। वह अपने जीवन को समभाव होकर जीता है। वह सुख-दुख दोनों अवस्थाओं में मग्न रहता है।

प्रश्न 26.
कवि को संसार अपूर्ण क्यों लगता है?
उत्तर
कवि को संसार इसलिए अपूर्ण लगता है, क्योंकि यह समस्त संसार स्वार्थी है। यहाँ हर कोई अपनी स्वार्थ पूर्ति में डूबा हुआ है। संसार केवल उन्हीं को पूछता है तो उसकी जय-जयकार करते हैं।

प्रश्न 27.
कवि किसकी यादों का अपने दिल में संजोकर घूमता है?
उत्तर
कवि कहता है कि उसने अपनी जवानी में किसी से प्रेम किया था और उसकी यादों को अपने हृदय में संजोया था। आज उसी की यादों का अपने हृदय में संजोकर घूम रहा है। प्रश्न

प्रश्न 28.
कवि ने संसार को मूर्ख क्यों कहा है?
उत्तर
कवि संसार को मूर्ख इसलिए कहता है, क्योंकि यह संसार सत्य की खोज में मिटे असंख्य महापुरुषों को देखकर भी सचेत नहीं हुआ। सत्य को जानने के लिए असंख्य महापुरुष लोग यत्न कर मिट गए। वह बात जानकर भी यह संसार कुछ भी सीखना नहीं चाहता।

प्रश्न 29.
कवि संसार में किस चीज़ से प्रभावित नहीं होता?
उत्तर
कवि पृथ्वी पर व्याप्त धन-ऐश्वर्य तथा शान-ओ-शौकत से तनिक भी प्रभावित नहीं होता। ऐसी शान-ओ-शौकत को वह पग-पग रा पर ठुकराता चलता है। वह कहता है कि जिस पृथ्वी पर यह संसार झूठे धन-ऐश्वर्य तथा शान और शौकत खड़ा करता है, वह ऐसी ऐश्वर्य परिपूर्ण पृथ्वी को पग-पग पर ठुकरा देता है। यह धन-वैभव कवि को तनिक भी विचलित नहीं कर सकता।

प्रश्न 30.
कवि संसार से उसे कवि न मानने के लिए क्यों कहता है?
उत्तर
कवि संसार को संबोधन करते हुए कहता है कि यह संसार उसे एक कवि मानकर क्यों अपनाना चाहता है। वह एक कवि नहीं, अपितु इस संसार का एक नया प्रेमी है। एक नया दीवाना है, जो अपनी प्रेमवाणी का बखान कर रहा है।

प्रश्न 31.
कवि संसार में किसका रूप धारण करके जीवनयापन कर रहा है?
उत्तर
कवि का कथन है कि वह इस संसार में प्रेम में पागल प्रेमियों का रूप लेकर जीवनयापन कर रहा है। वह अपने हृदय में एक प्रेमी को बिठाकर उसी का वेश धारण किए हुए है। कवि के हृदय में थोड़ी-सी मादकता व नशा बाकी है और उसी मादकता में डूबकर वह जी रहा है।

प्रश्न 32.
‘एक गीत’ कविता किसकी है और कहाँ से ली गई है?
उत्तर
‘एक गीत’ कविता श्री हरिवंश राय बच्चन द्वारा रचित उनके काव्य संग्रह ‘बुद्ध और नाचघर’ से संकलित है।

प्रश्न 33.
प्रस्तुत कविता में कवि ने क्या कहता चाहा है?
उत्तर
प्रस्तुत कविता में कवि ने समय के व्यतीत होने के एहसास के साथ-साथ लक्ष्य तक पहुँचने के लिए प्राणी द्वारा कुछ कर गुजरने के जज़्बे का चित्रण किया है। इस कविता में कवि की रहस्यवादी चेतना भी मुखरित हुई है।

प्रश्न 34.
कवि ने कविता में समय के बारे में क्या कहा है?
उत्तर
कवि का कथन है कि समय परिवर्तनशील है जो निरंतर चलायमान अवस्था में रहता है। वह कभी भी नहीं रुकता। समय के इसी परिवर्तन और अभाव को देखकर दिन में चलने वाला पाथिक भी यह सोचकर अत्यंत शीघ्रता से चलता है कि कहीं जीवन रूपी रास्ते में ही रात न हो जाए, जबकि उसकी मंज़िल भी अधिक दूर नहीं है।

प्रश्न 35.
‘मानव-जीवन क्षणभंगुर है’ पंक्ति का आशय स्पष्ट कीजिए।
उत्तर
कवि के कहने का अभिप्राय यह है कि मानव-जीवन क्षणभंगुर है। अर्थात समय बहुत कम है जो अत्यंत तेज़ी से गुज़रता है। मनुष्य रूपी यात्री को यह चिंता रहती है कि उसके लक्ष्य को प्राप्त करने से पहले ही उसके रास्ते में रात न हो जाए। कवि कहता है कि दिन अत्यंत शीघ्रता से व्यतीत होता है।

प्रश्न 36.
‘मंज़िल भी तो दूर नहीं’ पंक्ति में निहित लाक्षणिक अर्थ को स्पष्ट कीजिए।
उत्तर
इस पंक्ति का आशय है कि मनुष्य की वह मंजिल भी अधिक दूर नहीं है जहाँ उसे मृत्यु के बाद पहुँचना है।

प्रश्न 37.
कौन-सा ध्यान चिड़ियों के पंखों में न जाने कितनी चंचलता भर देता है?
उत्तर
चिड़ियों के पंखों में यह ध्यान चंचलता उत्पन्न करता है कि संध्या के समय घोंसलों के उनके बच्चे उनके आने की प्रतीक्षा करते हुए घोंसले से झाँक रहे होंगे। इससे उनके पंखों में वात्सल्य भाव के कारण अत्यधिक स्फूर्ति आ जाती है और वे शीघ्र अतिशीघ्र अपने घोंसलों तक पहुंचना चाहते हैं।

प्रश्न 38.
कविता में जीवन रूपी रथ पर बैठते हुए राही क्या सोचता है?
उत्तर
कवि का कथन है कि अपने जीवन रूपी पथ पर चलता हुआ प्राणी चिंतन करता है। वह निरंतर सोचता रहता है कि इस समय उससे मिलने के लिए कौन व्याकुल होगा।

प्रश्न 39.
पथिक के पाँवों को कौन-सा प्रश्न कमजोर कर देता है?
उत्तर
कवि कहता है कि जब-जब पथिक के हृदय में यह प्रश्न उठता है कि कौन उसके लिए व्याकुल है अर्थात कौन उसकी प्रतीक्षा कर रहा है तो यह प्रश्न उसके पाँवों को कमजोर कर देता है और यह उसे सुस्त बना देता है और इससे राही के हृदय में अपार व्याकुलता भर जाती है क्योंकि उसे लगता है कि किसी को उसकी प्रतीक्षा नहीं है।

प्रश्न 40.
घोंसलों में किसे न पाकर कौन परेशान हो उठता है?पक्षियों के बच्चे अपने घोंसले में अपने माँ-बाप को न पाकर परेशानी से भर उठते हैं। घोंसलों में अपने बच्चों को अकेला छोड़कर गए
पक्षी भी इसी चिंता में रहते हैं कि उनके बच्चे भी उनके आने की आशा में अपने-अपने घोंसलों से झाँक रहे होंगे।

प्रश्न 41.
कविवर ‘बच्चन’ किस प्रकार के साहित्यकार हैं?
उत्तर
‘बच्चन जी हालावाद के प्रवर्तक कवि हैं तथा आधुनिक हिंदी साहित्य के महान साहित्यकार माने जाते हैं। इनका एक कहानीकार के रूप में भी उदय हुआ था, किंतु काव्य के क्षेत्र में इन्होंने अद्भुत साहित्य की रचना की। ये एक कवि के रूप अधिक प्रसिद्ध हुए। इन्होंने हालावाद का प्रवर्तन कर साहित्य को एक नया मोड़ दिया।

प्रश्न 42.
आशय स्पष्ट कीजिए-“मैं और, और जग और, कहाँ का नाता।” (A.I. C.B.S.E 2016)
उत्तर
‘मैं और, और जग और, कहाँ का नाता’-हरिवंशराय बच्चन के द्वारा रचित है जिसमें कवि मानता है कि उसका समाज से कोई नाता नहीं है। समाज और उसकी प्रकृति में बहुत बड़ा अंतर है क्योंकि उन दोनों के लक्ष्यों में बहुत बड़ा अंतर है। कवि तो अपने कल्पना के लोक में प्रतिदिन अनेक संसार बना-बनाकर नष्ट कर देता है। वह नित नई कल्पना करता है और फिर उसे मिटा देता है। वह धन-ऐश्वर्य और शान-ओ-शौकत से प्रभावित नहीं होता जबकि संसार धन-दौलत और झूठी शान-ओ-शौकत पाना चाहता है।

प्रश्न 43.
‘बच्चन’ के संकलित गीत में दिन ढलते समय पथिकों और पक्षियों की गति में तीव्रता और कवि की गति में ष्ठिाथिलता के कारण लिखिए। (Outside Delhi 2017)
उत्तर
पथिक मंजिल से पूर्व रात होने के कारण तीव्रता से चलता है। पक्षी घोंसलों में अपने बच्चों द्वारा प्रतीक्षा के कारण गतिष्ठील बन जाते हैं। कवि की किसी के द्वारा प्रतीक्षा न करने के कारण गति में ष्ठिाथिलता आ जाती है।

प्रश्न 44.
आत्म परिचय कविता में परस्पर विपरीत कथनों से कवि क्या कहना चाहता है? (Outside Delhi 2017 Set-III)
उत्तर
आत्म परिचय कविता में परस्पर विपरीत कथनों से कवि यह कहना चाहता है कि मनुज्य एक सामाजिक प्राणी है। समाज से मनुज्य का नाता खट्टा-मीठा होता है। उसके जीवन में दुःख-सुख दोनों ही आते हैं। दुनिया अपने व्यंग्य-बाण और टासन-प्रष्ठासन से मनुज्य को अनेक कज्टों के रूप में जीवन-भार प्रदान करती है। चाहकर भी मनुज्य इस जीवन-भार से अलग नहीं हो सकता। इस जीवन-भार को उसे आजीवन ढोना ही पड़ता है।

लेकिन दूसरी ओर कवि का यह कहना है कि जीवन इस जग को देकर नहीं जीता क्योंकि वह इसे हृदयहीन और स्वार्थी मानता है। वह तो केवल संसार के वैभव से अलग अपनी मस्ती में मस्त होकर जीवन जीना चाहता है। इसलिए वह कहता है कि मैं कभी भी जग का ध्यान नहीं करता।

सप्रसंग व्याख्या, अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

1. मैं जग-जीवन का भार लिए फिरता हूँ,
फिर भी जीवन में प्यार लिए फिरता हूँ;
कर दिया किसी ने झंकृत जिनको छूकर
कि मैं साँसों के दो तार लिए फिरता हूँ। (C.B.S.E. Sample Paper-I, C.B.S.E. 2014 set I, III)

शब्दार्थ : जग-जीवन-संसार का जीवन, सांसारिक जीवन । छूकर-स्पर्श करके। झंकृत-बजाना।

प्रसंग : प्रस्तुत काव्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित ‘आत्म परिचय’ नामक कविता से अवतरित है जिसके रचयिता ‘हरिवंशराय बच्चन’ हैं। इसमें कवि ने आत्म-परिचय का चित्रण किया है।

व्याख्या : कवि बच्चन अपने जीवन का बोध कराते हुए कहते हैं कि मैं इस सांसारिक जीवन के भार को निरंतर वहन करता हुआ
जीवन-यापन कर रहा हूँ। मेरे जीवन पर इस जग का बहुत भार है लेकिन मैं इस भार को देख दुखी नहीं होता और न ही कभी विचलित होता हूँ। मैं अपने जीवन में असीम प्यार लिए घूम रहा हूँ। अपने विगत जीवन का स्मरण करते हुए कवि कहता है कि मेरे जीवन में किसी ने पदार्पण किया था तथा प्रेम-भरे हाथों से मेरी हृदय रूपी वीणा को झंकृत कर दिया था।

आज मैं उनकी यादों के रूप में अपने साँसों के दो तार लिए जी रहा हूँ। कवि का कहने का अभिप्राय यह है कि एक प्रिया का मेरे जीवन में आगमन हुआ था। उसने मुझे प्यार से छुआ था लेकिन उसका साथ नहीं रहा। बस यादों के रूप में उसके कोमल हाथों से झंकृत साँसों के तार लिए जीवन जी रहा हूँ।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. उपर्युक्त काव्यांश के कवि तथा कविता का नाम लिखिए।
2. कवि किसका भार लिए फिरता है?
3. कवि अपने जीवन में क्या लेकर घूमता है?
4. साँसों के तारों को किसने छुआ होगा?
5. काव्यांश का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. उपर्युक्त काव्यांश के कवि श्री हरिवंश राय बच्चन हैं तथा कविता का नाम ‘आत्मपरिचय’ है।
2. कवि जग-जीवन का भार लिए फिरता है। वह भार दुख और दर्द के कारण जीवन के लिए अति दुखदायी है।
3. कवि अपने जीवन में प्यार लेकर घूमता है।
4. साँसों के तारों को कवि की प्रियतमा ने छुआ होगा।
5. काव्य-सौंदर्य

• कवि ने विगत जीवन का यथार्थ बोध कराया है।
• भाषा सहज, सरल, सरस खड़ी बोली है।
• संस्कृत के तत्सम और तद्भव शब्दों का प्रयोग है।
• गेय-मुक्तक शैली का प्रयोग है।
• अनुप्रास, पदमैत्री तथा स्वरमैत्री अलंकार की शोभा दर्शनीय है।
• अभिधा शब्द-शक्ति का प्रयोग है।
• बिंब-योजना सार्थक एवं सटीक है।
• शृंगार रस का प्रयोग किया गया है।
• माधुर्य गुण विद्यमान है।

2. मैं स्नेह-सुरा का पान किया करता हूँ,
मैं कभी न जग का ध्यान किया करता हूँ,
जग पूछ रहा उनको, जो जग की गाते,
मैं अपने मन का गान किया करता हूँ !
(C.B.S.E. Sample Paper-I, A.I. C.B.S.E. 2011 Set-I), (C.B.S.E. Delhi 2013, Set-I, II, III, C.B.S.E. Outside Delhi 2013, Set-II, A.L.C.B.S.E, 2014)

शब्दार्थ : स्नेह-सुरा-स्नेह रूपी मदिरा, प्रेम रूपी शराब। गान-बखान, कहना। जग-संसार।

प्रसंग : प्रस्तुत काव्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित कवि श्री हरिवंशराय बच्चन द्वारा रचित है। इसमें कवि ने आत्म-परिचय का बोध कराया है।

व्याख्या : कवि कहता है कि मैं प्रेम रूपी मदिरा को पीने वाला हूँ। मैं इसी प्रेम रूपी मदिरा को पीकर इसी की मस्ती में डूबा रहता हूँ। मैं केवल अपने में मग्न रहता हूँ। मैं कभी भी संसार का ध्यान नहीं करता। कवि कहता है कि यह संसार स्वार्थी है। यह केवल उसको पूछता है, जो इसका बखान करते हैं और उसके अनुकूल कार्य करते हैं। अपने प्रतिकूल कार्य करने वालों को यह संसार कभी नहीं पूछता। कवि कहता है कि मैं तो अपनी मस्ती में डूबकर अपने मन का बखान करता हूँ, अपने मन की भावनाओं तथा संवेदनाओं को सुनाता रहता हूँ।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. कवि किसका पान किए विचरण करता है?
2. कवि जग का ध्यान क्यों नहीं करता?
3. कवि किसका गान करता है?
4. जग कवि को क्यों नहीं पूछता?
5. काव्यांश का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. कवि स्नेह रूपी मदिरा का पान किए विचरण करता है।
2. कवि जग का ध्यान इसलिए नहीं करता, क्योंकि वह स्नेह रूपी सुरा को पीकर उसी में मस्त रहता है।
3. कवि अपने मन का गान करता है।
4. जग कवि को इसलिए नहीं पूछता क्योंकि वह जग का गान नहीं करता; उसकी जय-जयकार नहीं करता।
5. काव्य-सौंदर्य

• कवि ने संसार की स्वार्थपरता का चित्रांकन किया है।
• भाषा सहज, सरल, सरस खड़ी बोली है।
• संस्कृत के तत्सम और तद्भव शब्दों का प्रयोग है।
• स्नेह-सुरा में रूपक अलंकार की छटा शोभनीय है।
• अनुप्रास, पदमैत्री, स्वरमैत्री अलंकार द्रष्टव्य हैं।
• गेय-मुक्तक शैली का सुंदर प्रयोग है।
• बिंब योजना सार्थक एवं सटीक है।
• माधुर्य गुण है।
• शृंगार रस का चित्रण है।

3. मैं निज उर के उद्गार लिए फिरता हूँ, (C.B.S.E. Delhi 2013, Set-II, Set-III, A.I.C.B.S.E, 2014)
मैं निज उर के उपहार लिए फिरता हूँ,
है यह अपूर्ण संसार न मुझको भाता
मैं स्वप्नों का संसार लिए फिरता हूँ। (C.B.S.E. Sample Paper)

शब्दार्थ : निज-अपना। भाता-अच्छा लगना। उद्गार-भाव, अधीरता।

प्रसंग : प्रस्तुत अवतरण हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित आत्म-परिचय नामक कविता से अवतरित है। इसके रचयिता श्री हरिवंश राय बच्चन हैं। इस काव्यांश में कवि ने अपने हृदय के उद्गार व्यक्त किए हैं।

व्याख्या : कवि कहता है कि मैं अपने हृदय के भाव तथा उपहार लिए जीवन-यापन कर रहा हूँ। कवि का कहने का अभिप्राय यह है कि मैंने अपने हदय में अनेक भाव उपहार स्वरूप सँजो रखे हैं। इन्हीं भावों तथा उपहारों को लिए विचरण कर रहा हूँ। कवि कहता है कि यह संसार तो अपूर्ण है इसलिए इसकी अपूर्णता मेरे हृदय को अच्छी नहीं लगती। मैं तो इस संसार की उपेक्षा करके अपने ही सपनों का संसार लेकर जी रहा हूँ। मैं तो अपने ही स्वप्निल संसार में डूबा रहता हूँ।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. कवि किसके उद्गार और उपहार लेकर फिरता है?
2. कवि को यह संसार अपूर्ण क्यों लगता है?
3. कवि जीवन में किसका संसार लिए फिरता है?
4. उपर्युक्त काव्यांश का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. कवि अपने हृदय के उद्गार और उपहार लेकर फिरता है।
2. कवि को यह संसार इसलिए अपूर्ण लगता है क्योंकि यह समस्त संसार स्वार्थी है। यहाँ हर कोई अपनी स्वार्थ-पूर्ति में डूबा हुआ है। यह संसार केवल उन्हीं को पूछता है जो इसकी जय-जयकार करते हैं।
3. कवि अपने जीवन में सपनों का संसार लिए फिरता है।
4. काव्य-सौंदर्य

• कवि ने संसार को अपूर्ण बताया है।
• भाषा सरस, प्रवाहमयी है।
• गीति शैली का प्रयोग है।
• अनुप्रास, पदमैत्री तथा स्वरमैत्री अलंकारों की छटा शोभनीय है।
• संस्कृत के तत्सम और तद्भव शब्दों का सार्थक प्रयोग है।
• शांत रस है।
• प्रसाद गुण है।

4. मैं जला हृदय में अग्नि, दहा करता हूँ,
सुख-दुख दोनों में मग्न रहा करता हूँ,
जग भव-सागर तरने को नाव बनाए,
मैं भव मौजों पर मस्त बहा करता हूँ।  (C.B.S.E. Model Question Paper, 2008, A.I.C.B.S.E, 2014)

शब्दार्थ : अग्नि-आग। जग-संसार। नाव-किश्ती। दहा-जला। भव-सागर-संसार रूपी सागर। मौजों पर-तरंगों पर, हिलोरों या लहरों पर।

प्रसंग : प्रस्तुत काव्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित कवि श्री हरिवंश राय बच्चन द्वारा रचित ‘आत्म-परिचय’ नामक कविता से अवतरित है। इसमें कवि ने आत्म-परिचय का चित्रण करते हुए सुख-दुखावस्था में एक समान रहने की प्रेरणा दी है।

व्याख्या : कवि कहता है कि मैंने अपने हृदय में वियोग रूपी अग्नि को जला रखा है जिसमें मैं निरंतर जलता रहता हूँ। मैं सुख और दुःख दोनों अवस्थाओं में मग्न रहता हूँ। मैं न तो सुख आने पर अत्यधिक खुश होता हूँ और न ही दुःख की प्रतिकूल परिस्थितियों में अधिक दुखी। दोनों भावों को अपने जीवन में एकसमान ग्रहण करता हूँ। कवि संसार को संबोधन करते हुए कहता है कि अन्य लोग इस संसार रूपी सागर को पार करने के लिए भले ही नाव का निर्माण करे, पर मेरी संसार रूपी सागर को पार करने की कोई इच्छा नहीं है।

मैं तो इस संसार की लहरों पर ही मस्ती में बहना चाहता हूँ। कवि का अभिप्राय यह है कि इस संसार रूपी सागर को पार करने अर्थात मोक्ष की कामना हेतु भले ही औरों को किसी अन्य सहारे रूपी नाव की आवश्यकता हो, पर उसे किसी अन्य की कोई आवश्यकता नहीं और न ही मोक्ष प्राप्ति की कोई कामना है। वह तो इसी संसार की लहरों पर बहना चाहता है।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. उपर्युक्त अवतरण के रचयिता का नाम लिखिए।
2. कवि किसमें मग्न रहता है?
3. कवि किन पर मस्ती में बहता है?
4. कवि अपने हृदय में किस अग्नि को जलाने की बात करता है?
5. इस अवतरण का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. उपर्युक्त अवतरण के रचयिता श्री हरिवंश राय बच्चन हैं।
2. कवि सुख और दुख दोनों में मग्न रहता है।
3. कवि इस संसार की तरह-तरह की मौजों पर मस्ती में भरकर बहता है।
4. कवि अपने हृदय में वियोग रूपी अग्नि को जलाने की बात करता है।
5. काव्य-सौंदर्य

• कवि ने सुख-दुःख में सम रहने की प्रेरणा दी है।
• भाषा सरल, साधारण, सरस खड़ी बोली है।
• भव-सागर में रूपक अलंकार की छटा है।
• अनुप्रास, पदमैत्री, स्वरमैत्री अलंकारों का प्रयोग है।
• प्रतीकात्मक शैली का प्रयोग है।
• बिंब योजना सार्थक है।
• वियोग-शृंगार की अद्भुत छटा है।
• माधुर्य गुण का सार्थक प्रयोग है।

5. मैं यौवन का उन्माद लिए फिरता हूँ,
उन्मादों में अवसाद लिए फिरता हूँ,
जो मुझको बाहर हँसा, रुलाती भीतर,
मैं, हाय, किसी की याद लिए फिरता हूँ, (C.B.S.E. Model Question Paper, 2008)

शब्दार्थ: यौवन-जवानी। अवसाद-दुख, पीड़ा। भीतर-आंतरिक रूप। उन्माद-जोश, मस्ती। बाहर-बाह्य रूप।

प्रसंग : प्रस्तुत पद्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित ‘आत्म-परिचय’ नामक कविता से अवतरित किया गया है, जिसके रचयिता श्री हरिवंश राय बच्चन हैं। ये हालावाद के प्रवर्तक कवि माने जाते हैं। इस काव्यांश में कवि ने अपने जीवन की आंतरिक पीड़ा का चित्रण किया है।

व्याख्या : कवि कहता है कि मैं तो अपनी जवानी का जोश लिए जी रहा हूँ। मेरे हृदय में आज भी वह जवानी की मस्ती और जोश है जिसमें मस्त होकर मैं जीवन-यापन कर रहा हूँ। इसी जवानी के पागलपन में अनेक दुख समाए हुए हैं। कवि कहता है कि मैंने जवानी में किसी से प्रेम करके उसकी यादों को अपने हृदय में संजोया था। आज उसी की यादों को अपने हृदय में सजाकर फिर रहा हूँ। ये । यादें मुझे बाह्य रूप से हँसा देती हैं लेकिन आंतरिक रूप से रुलाती हैं। मैं अपनी प्रिया की यादों को लेकर जीवन जी रहा हूँ। इस जहाँ को मैं बाहर से भले ही हँसता हुआ दिखाई दूं, लेकिन मैं हृदय से रोता रहता हूँ। उसकी यादें आज भी मुझे सताती रहती हैं।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. कवि किसका उन्माद लिए फिरता है?
2. कवि को किसकी याद रुलाती है? कैसे?
3. कवि के जीवन में जवानी के उन्माद के साथ-साथ और क्या है?
4. उपर्युक्त पंक्तियों का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. कवि यौवन का उन्माद लिए फिरता है।
2. कवि को उसकी प्रियतमा की याद रुलाती है। वह कवि को बाहर से हँसाती है तथा अंदर से रुलाती है।
3. कवि के जीवन में जवानी के उन्माद के साथ-साथ अनेक अवसाद हैं।
4. काव्य-सौंदर्य

• आंतरिक पीड़ा का प्रतिपादन किया है।
• भाषा खड़ी बोली है जो अत्यंत सरस एवं प्रवाहमयी है।
• प्रतीकात्मक शैली का सार्थक प्रयोग है। यहाँ बाहर से ‘बाह्य स्वरूप’ का तथा भीतर से आंतरिक रूप का परिचय होता है।
• अनुप्रास, पदमैत्री अलंकारों की शोभा है।
• संस्कृत के तत्सम और तद्भव शब्दों का प्रयोग है।

6. कर यल मिटे सब, सत्य किसी ने जाना ?
नादान वहीं है, हाय, जहाँ पर दाना !
फिर मूढ़ न क्या जग, जो इस पर भी सीखे ?
मैं सीख रहा हूँ, सीखा ज्ञान भुलाना !

शब्दार्थ : यल – प्रयास। दाना – अक्लमंद, समझदार। जग = संसार। नादान – नासमझ। मूढ़ – मूर्ख।

प्रसंग : प्रस्तुत पद्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2′ में संकलित कवि श्री हरिवंशराय बच्चन’ द्वारा रचित ‘आत्म-परिचय’ नामक कविता से अवतरित किया गया है। इसमें कवि ने ‘शाश्वत सत्य’ का वर्णन किया है।

व्याख्या : कवि प्रश्न करते हुए पूछता है कि इस संसार में बड़े-बड़े महापुरुषों ने सत्य को जानने का प्रयास किया, लेकिन आज तक कोई भी सत्य को नहीं जान पाया। सत्य (ब्रह्म) की खोज करते-करते सब समाप्त हो गए, फिर भी उसे नहीं जान पाए। कवि का अभिप्राय यह है कि इस संसार में आज तक कोई भी सत्य को पहचान नहीं सका। वह कहता है कि समाज में जहाँ पर अक्लमंद या समझदार लोग रहते हैं, वहीं पर नासमझ या मूर्ख भी निवास करते हैं। संसार को संबोधन कर कवि कहता है कि यह बात जानकर भी यह संसार मूर्ख है जो इसके बावजूद भी सीखना चाहता है। मैं तो भूले हुए ज्ञान को सीखने का प्रयास कर रहा हूँ।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. ‘कर यत्न मिटे सब, सत्य किसी ने जाना?’ पंक्ति में कवि किस सत्य की बात कहता है?
2. ‘नादान वहीं है, हाय, जहाँ पर दाना।’ पंक्ति का भाव स्पष्ट कीजिए।
3. कवि संसार को मूर्ख क्यों कहता है ?
4. काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. इस पंक्ति में कवि अंतिम सत्य अर्थात परमात्मा की बात करता है। इस सत्य को जानने के लिए असंख्य महापुरुष लोग और उनके अनेक यत्न कर मिट गए।
2. इस पंक्ति का भाव यह है कि सृष्टि में जहाँ समझदार और विद्वान लोग रहते हैं, वहीं नासमझ और मूर्ख लोग भी निवास करते हैं।
3. कवि संसार को मूर्ख इसलिए कहता है क्योंकि यह संसार सत्य की खोज में मिटे असंख्य महापुरुषों को देखकर भी सचेत नहीं हुआ।
4. काव्य-सौंदर्य

• कवि ने शाश्वत सत्य अर्थात ब्रह्म की ओर संकेत किया है।
• संस्कृत के तत्सम और तद्भव शब्दों का प्रचुर प्रयोग हुआ है।
• कवि की रहस्यवादी चेतना का चित्रण हुआ है।
• अनुप्रास, पदमैत्री, प्रश्नवाचक अलंकारों की शोभा है।
• प्रतीकात्मक शैली का प्रयोग है।
• शांत रस है।
• प्रसाद गुण है।

7. मैं और, और जग और, कहाँ का नाता,
मैं बना-बना कितने जग रोज़ मिटाता;
जग जिस पृथ्वी पर जोड़ा करता वैभव,
मैं प्रति पग से उस पृथ्वी को ठुकराता! (C.B.S.E. Outside Delhi 2013, Set-1)

शब्दार्थ : वैभव-धन, ऐश्वर्य। प्रति-पग-प्रत्येक चरण।

प्रसंग : प्रस्तुत काव्यांश हमारी हिंदी का पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित कवि ‘श्री हरिवंश राय बच्चन’ द्वारा रचित ‘आत्म-परिचय’ नामक कविता से अवतरित किया गया है। इसमें कवि ने बताया है कि उसका इस संसार से कोई नाता नहीं है। कवि कहता है कि मेरा इस संसार के साथ कोई संबंध नहीं है। मेरी और संसार की प्रकृति में बहुत अंतर है। मेरा स्वभाव कुछ और है तथा संसार का कुछ और। मेरी कोई और मंजिल है तथा इस स्वार्थी संसार की कोई और।

व्याख्या : कवि कहता है कि मैं तो प्रतिदिन ऐसे अनेक संसार बना-बना कर खत्म कर देता हूँ। मैं हर दिन ऐसे अनेक संसार की कल्पना करता हूँ और फिर उसे मिटा देता हूँ। ये संसार धन-ऐश्वर्य से प्रेरित होकर जिस पृथ्वी पर धन-ऐश्वर्य तथा शान-ओ-शौकत जोड़ता है, मैं उससे तनिक भी प्रभावित नहीं होता। ऐसी शान-ओ-शौकत को मैं पग-पग पर ठुकराता चलता हूँ। जिस पृथ्वी पर यह संसार झूठे धन-ऐश्वर्य खड़ा करता है मैं ऐसी ऐश्वर्य से परिपूर्ण पृथ्वी को पग-पग पर ठुकरा देता हूँ। ये धन-वैभव मुझे बिल्कुल भी विचलित नहीं कर सकते।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. कवि राग किसमें लिए हुए हैं?
2. ‘शीतल वाणी में आग लिए फिरता हूँ’ पंक्ति में विरोधाभास स्पष्ट कीजिए।
3. कवि किस खंडहर का अंग लिए हुए हैं?
4. उपर्युक्त अवतरण का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. कवि अपने रोदन में राग लिए हुए है।
2. शीतल वाणी अर्थात हृदय के उद्गार, आग अर्थात क्रोध रूपी आग। इसका आशय है कि कवि अपने हृदय की शीतल वाणी में भी क्रोध एवं व्यंग्य रूपी आग बरसाता है।
3. कवि उस खंडहर का अंग लिए हुए हैं, जिस पर अनेक राजाओं के महल न्योछावर हो जाते हैं।
4. काव्य-सौंदर्य

• कवि ने अपने हृदय की पीड़ा का प्रतिपादन किया है।
• खड़ी बोली भाषा सरस एवं प्रवाहमयी है।
• तत्सम शब्दावली का प्रचुर प्रयोग है।
• प्रतीकात्मक शैली का प्रयोग है।
• अनुप्रास, स्वरमैत्री तथा पदमैत्री की छटा है।
• माधुर्य गुण है।
• शृंगार रस की वियोगावस्था का चित्रण हुआ है।
• लक्षणा शब्द-शक्ति का प्रयोग भावानुरूप हुआ है।

8. मैं निज रोदन में राग लिए फिरता हूँ,
शीतल वाणी में आग लिए फिरता हूँ,
हों जिस पर भूपों के प्रासाद निछावर,
मैं वह खंडहर का भाग लिए फिरता हूँ। (C.B.S.E. Model Q. Paper, 2008)

शब्दार्थ : निज-अपना। भूपों के राजाओं के। निछावर-अर्पण, न्योछावर। भाग-अंश। रोदन-रोना, दुख। प्रासाद-महल। खंडहर-टूटा-फूटा महल।

प्रसंग : प्रस्तुत काव्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित कवि ‘श्री हरिवंश राय बच्चन’ द्वारा रचित ‘आत्म-परिचय’ नामक कविता से अवतरित किया गया है। इसमें कवि ने आत्मीय पीड़ा का चित्रण किया है।

व्याख्या : कवि कहता है कि मैं अपने आँसुओं में भी प्रेम-भरा गीत छिपाए फिरता हूँ। जिसे संसार मेरा रोदन समझता है, उसमें मेरे अनेक गीत छिपे हुए हैं। मैं अपनी अश्रुओं की धारा के माध्यम से अपने प्रेमगीतों का बखान करता चलता हूँ। मेरे हृदय में वाणी अत्यंत शीतल और कोमल है लेकिन मैं इस शीतलता में भी क्रोध रूपी आग छिपाए हूँ। कवि जग को संबोधित करते हुए कहता है कि मैं उस खंडहर का अंश लिए हूँ जिस पर महान प्रतापी राजा अपने महलों को न्योछावर कर देते हैं।

मेरे पास उस खंडहर का अंशमात्र है जिसके सामने बड़े-बड़े राजाओं के आलीशान महलों का भी कोई मूल्य नहीं है अर्थात कवि का प्रिया से वियोग होने पर हृदय विच्छिन हो गया, जो खंडहर की भाँति पड़ा है लेकिन फिर भी उसका मूल्य अनमोल है। वह इतना सुंदर है कि उसकी शोभा के समक्ष बड़े से बड़े राजा का आलीशान महल भी नगण्य है।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. संसार किसको गाना कहता है?
2. कवि के फूट-फूटकर रोने को संसार ने क्या नाम दिया?
3. कवि संसार से क्या अपेक्षा करता है?
4. उपर्युक्त अवतरण का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. संसार कवि के रोने को गाना कहता है।
2. कवि के फूट-फूट कर रोने को संसार ने छंद बनाना नाम दिया।
3. कवि संसार से यह अपेक्षा करता है कि संसार उसे एक कवि कहकर न बुलाए, बल्कि उसे एक नया दीवाना कहकर संबोधित करे।
4. काव्य-सौंदर्य

• कवि ने अपनी अंत:पीड़ा का प्रतिपादन किया है।
• संस्कृत के तत्सम और तद्भव शब्दों का प्रयोग है।
• भाषा खड़ी बोली है।
• अनुप्रास, स्वरमैत्री, पदमैत्री अलंकारों की शोभा दर्शनीय है।

9. मैं रोया, इसको तुम कहते हो गाना,
मैं फूट पड़ा, तुम कहते, छंद बनाना,
क्यों कवि कहकर संसार मुझे अपनाए,
मैं दुनिया का हूँ एक नया दीवाना !

शब्दार्थ : फूट पड़ा-जोर से रोया। दीवाना-प्रेम करने वाला, आसक्त। छंद बनाना-कविता लिखना या कहना।

प्रसंग : प्रस्तुत पद्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित तथा कवि ‘हरिवंश राय बच्चन’ द्वारा रचित कविता
‘आत्म-परिचय’ से अवतरित किया गया है। इसमें कवि ने हृदय की पीड़ा का चित्रण किया है।

व्याख्या : कवि संसार को संबोधन करते हुए कहता है कि मैं दुख में अत्यंत दुखी होकर रोया था लेकिन तुम मेरे रोने को भी गीत समझ रहे हो। हृदय में अपार वेदना के कारण मैं तो जोर-जोर से रोया लेकिन इसे भी तुम कविता कहना समझते रहे। कवि का कहने का अभिप्राय यह है कि यह संसार तो बिल्कुल अजीब है।

यह किसी की आंतरिक भावनाओं को नहीं समझ सकता। मैं जब साधारण रूप से रोया था तो उसे यह मेरा गीत गाना समझ रहे थे, लेकिन जब असीम पीड़ा के कारण मेरा हृदय जोर-जोर से चिल्लाकर रोने लगा तो उसे इसने मेरा कविता करना मान लिया। इस प्रकार यह हृदयहीन संसार मेरी आंतरिक पीड़ा को नहीं समझ रहा है। कवि पुनः इस जग को संबोधित करते हुए कहता है कि यह संसार मुझे एक कवि मानकर क्यों अपनाना चाहता है। मैं एक कवि नहीं हूँ बल्कि मैं तो इस जहाँ का एक नया प्रेमी हूँ। एक नया दीवाना हूँ जो अपनी प्रेमवाणी का बखान कर

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. संसार किसको गाना कहता है?
2. कवि के फूट-फूटकर रोने को संसार ने क्या नाम दिया?
3. कवि संसार से क्या अपेक्षा करता है?
4. उपर्युक्त अवतरण का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. संसार कवि के रोने को गाना कहता है।
2. कवि के फूट-फूट कर रोने को संसार ने छंद बनाना नाम दिया।
3. कवि संसार से यह अपेक्षा करता है कि संसार उसे एक कवि कहकर न बुलाए, बल्कि उसे एक नया दीवाना कहकर संबोधित करे।
4. काव्य-सौंदर्य

• कवि ने अपनी अंत:पीड़ा का प्रतिपादन किया है।
• संस्कृत के तत्सम और तद्भव शब्दों का प्रयोग है।
• भाषा खड़ी बोली है।
• अनुप्रास, स्वरमैत्री, पदमैत्री अलंकारों की शोभा दर्शनीय है।

10. मैं दीवानों का वेश लिए फिरता हूँ,
मैं मादकता निःशेष लिए फिरता हूँ,
जिसको सुनकर जग झूमे, झुके, लहराए,
मैं मस्ती का संदेश लिए फिरता हूँ।

शब्दार्थ : दीवानों का-प्रेम में पागल व्यक्तियों का, प्रेमियों का। मादकता-नशा, उन्माद । वेश-पहनावा, रूप। निःशेष-बिल्कुल थोड़ा-सा, समाप्त।

प्रसंग : प्रस्तुत काव्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित तथा श्री हरिवंश राय बच्चन द्वारा रचित ‘आत्म-परिचय’ नामक कविता से अवतरित किया गया है। इसमें कवि ने संसार को प्रेम और मस्ती का संदेश दिया है जिसमें यह संसार झूम उठे तथा लहराने लगे।

व्याख्या : कवि का कथन है कि मैं इस संसार में प्रेम में पागल प्रेमियों का रूप लेकर जीवनयापन कर रहा हूँ अर्थात मैंने अपने हृदय में एक प्रेमी को बिठाकर उसी का वेश धारण कर लिया है। मेरे हृदय में अभी भी थोड़ी-सी मादकता का नशा बाकी है और उसी मादकता में डूबकर मैं जी रहा हूँ। कवि कहता है कि मैं इस हृदयहीन और दुखी संसार को एक ऐसा मस्ती का संदेश देना चाहता हूँ, जिसको सुनकर यह संपूर्ण दुखी संसार झूम उठे और मस्ती में डूब कर लहराने लगे तथा इस मस्ती के आगे झुक जाएं।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. कवि किसका वेश लिए फिरता है?
2. कवि किसका संदेश देता है?
3. कवि के संदेश को सुनकर संसार क्या-क्या प्रतिक्रियाएँ करता है?
4. उपर्युक्त अवतरण का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. कवि दीवानों का वेश लिए फिरता है।
2. कवि संसार को मस्ती का संदेश देता है।
3. कवि का संदेश सुनकर यह संसार झूम उठता है, झुक जाता है तथा लहराने लगता है।
4. काव्य-सौंदर्य

• कवि दुखी हृदयहीन संसार को मस्ती का संदेश देना चाहता है।
• खड़ी बोली, भाषा सहज, सरस एवं प्रवाहमयी है।
• गेय मुक्तक शैली का प्रयोग है।
• माधुर्य गुण है। शृंगार रस की छटा है।
• अनुप्रास, स्वरमैत्री तथा पदमैत्री अलंकारों की शोभा दर्शनीय है।
• अभिधा शब्द-शक्ति का सटीक प्रयोग हुआ है।
•  तत्सम शब्दावली का प्रचुर प्रयोग है।।

11. दिन जल्दी-जल्दी ढलता है!
हो जाए न पथ में रात कहीं,
मंजिल भी तो है दूर नहींयह
सोच थका दिन का पंथी भी जल्दी-जल्दी चलता है!
दिन जल्दी-जल्दी ढलता है!

शब्दार्थ : ढलता है-अस्त होता है। मंजिल-लक्ष्य, जहाँ पहुँचना है। पथ-रास्ता। दिन का पंथी-सूर्य।

प्रसंग : प्रस्तुत काव्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित कवि श्री हरिवंश राय बच्चन द्वारा रचित ‘एक गीत’ नामक कविता से अवतरित किया गया है। बच्चन जी हालावाद के प्रवर्तक तथा आधुनिक हिंदी साहित्य के प्रमुख कवि माने जाते हैं। इस काव्यांश में कवि ने समय के व्यतीत होने के साथ-साथ पथिक के मंजिल पर पहुँचने तथा उसे प्राप्त करने के जज्बे का चित्रण किया है।

व्याख्या : कवि का कथन है कि समय परिवर्तनशील है जो निरंतर चलायमान अवस्था में रहता है। वह कभी भी नहीं रुकता। समय के इसी परिवर्तन और अभाव को देखकर दिन में चलने वाला पथिक अर्थात सूर्य भी यह सोचकर अत्यंत शीघ्रता से चलता है कि कहीं रास्ते में ही रात न हो जाए, जबकि उसकी मंजिल भी अधिक दूर नहीं है। कवि का कहने का अभिप्राय यह है कि मानव जीवन क्षणभंगुर है। अतः समय बहुत कम है जो अत्यंत तेजी से गुजरता हुआ चलता है। मनुष्य रूपी यात्री को यह चिंता रहती है कि कहीं उसके लक्ष्य को प्राप्त करने से पहले ही उसके रास्ते में रात न हो जाए। यह सोच वह अत्यंत शीघ्रता से चलता है। कवि कहता है कि दिन अत्यंत शीघ्रता से व्यतीत होता है।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. उपर्युक्त काव्यांश के कवि तथा कविता का नाम बताएँ।
2. ‘हो जाए न पथ में रात कहीं’ यहाँ कवि किस पथ और रात की बात करता है?
3. ‘मंजिल भी तो है दूर नहीं’ पंक्ति में निहित लाक्षणिक अर्थ को स्पष्ट कीजिए।
4. इस काव्यांश का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. उपर्युक्त काव्यांश के कवि का नाम श्री हरिवंश राय बच्चन है तथा कविता का नाम ‘एक गीत’ है।
2. यहाँ कवि जीवन रूपी पथ और मृत्यु रूपी रात की बात करता है।
3. इस पंक्ति का आशय है कि मनुष्य की वह मंज़िल भी अधिक दूर नहीं है जहाँ उसे मृत्यु के बाद पहुँचना है।
4. काव्य-सौंदर्य

• कवि ने अपनी रहस्यवादी चेतना का चित्रण किया है।
• ‘जल्दी-जल्दी’ में पुनरुक्ति प्रकाश अलंकार की छटा दर्शनीय है।
• भाषा खड़ी बोली, सरल, सरस तथा प्रवाहमयी है।
• तत्सम और तद्भव शब्दावली का प्रयोग है। (५) प्रसाद गुण है।
• शांत रस है।
• बिंब योजना अत्यंत सटीक एवं सार्थक है।

12. बच्चे प्रत्याशा में होंगे,
नीड़ों से झाँक रहे होंगे
यह ध्यान परों में चिड़ियों के भरता कितनी चंचलता है।
दिन जल्दी-जल्दी ढलता है। (A.I. C.B.S.E. 2012, Set-I, 2018)

शब्दार्थ : प्रत्याशा-आशा। झाँकना-देखना। नीड़ों से-घोंसलों से। परों में-पंखों में।

प्रसंग : प्रस्तुत काव्यांश हमारी हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित ‘एक गीत’ नामक कविता से अवतरित किया गया है जिसके रचयिता ‘श्री हरिवंश राय बच्चन’ जी हैं। इस काव्यांश में कवि ने दाने की खोज में गए पक्षियों (प्राणियों) का अपने बच्चों के प्रति वात्सल्य भाव का चित्रण किया है।

व्याख्या : कवि का कथन है कि लौटते पक्षियों को जब यह महसूस होता है कि उनके बच्चे घोंसलों में उनकी राह देख रहे होंगे और उनके आने की आशा मन में लिए घोंसलों से झाँक रहे होंगे। यह ध्यान उन चिड़ियों के पंखों में न जाने कितनी चंचलता भर देता है अर्थात जब भी चिड़ियों को अपने बच्चों की याद आती है तो उनके पंखों में अपने बच्चों की वात्सल्य भाव के कारण और अधिक स्फूर्ति छा जाती है। कवि कहता है कि दिन अत्यंत शीघ्रता से व्यतीत हो रहा है।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. बच्चे किसकी प्रत्याशा में होंगे?
2. नीड़ों से कौन झाँक रहे होंगे और क्यों?
3. चिड़ियों के पंखों में कौन-सा ध्यान चंचलता उत्पन्न करता है?
4. इस अवतरण का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. बच्चे अपनी माताओं की प्रत्याशा में होंगे।
2. नीडों से चिडियों के बच्चे झाँक रहे होंगे, क्योंकि संध्या के समय उनकी माताएँ उनके पास नहीं हैं।
3. चिड़ियों के पंखों में यह ध्यान चंचलता उत्पन्न करता है कि संध्या के समय घोंसलों में उनके बच्चे उनके आने की प्रतीक्षा करते हुए घोंसलों से झाँक रहे होंगे।
4. काव्य-सौंदर्य

• चिड़ियों का बच्चों के प्रति वात्सल्य भाव का चित्रण है।
• खड़ी बोली की भाषा सरल एवं सरस है।
• मुक्तक छंद है।
• अनुप्रास, पुनरुक्ति प्रकाश, पदमैत्री, स्वरमैत्री अलंकारों की शोभा है।
• तत्सम और तद्भव शब्दावली का प्रयोग है।
• प्रसाद गुण है।
• अभिधा शब्द-शक्ति का प्रयोग है।

13. मुझसे मिलने को कौन विकल?
मैं होऊँ किसके हित चंचल?
यह प्रश्न शिथिल करता पद को, भरता उर में विहवलता है!
दिन जल्दी-जल्दी ढलता है।

शब्दार्थ : विकल-याकुल। शिथिल करना-सुस्त करना, मंद करना। विह्वलता-व्याकुलता। हित-के लिए। उर-हृदय।

प्रसंग : प्रस्तुत काव्यांश हिंदी की पाठ्य-पुस्तक ‘आरोह-भाग-2’ में संकलित ‘एक गीत’ से अवतरित है जिसके रचयिता हरिवंश राय बच्चन हैं। इस काव्यांश में कवि ने प्राणी की व्याकुलता का चित्रण किया है।

व्याख्या : कवि का कथन है कि अपने जीवन रूपी पथ पर चलता हुआ प्राणी चिंतन करता है। वह अपने मन ही मन में सोच रहा है कि इस जीवन मार्ग में अपनी मंजिल की ओर आगे बढ़ते हुए मुझसे मिलने के लिए कौन व्याकुल हो रहा है तथा मैं भी किसके लिए दुखी हो रहा हूँ। कवि कहता है कि जब-जब पथिक के हृदय में यह प्रश्न उठता है तो यह प्रश्न

उसके पाँवों को कमजोर कर देता है, उन्हें सुस्त बना देता है तथा राही के हृदय में अपार व्याकुलता भर देता है अर्थात जीवन रूपी मार्ग पर अग्रसर होते हुए जब भी पथिक को किसी का ध्यान आता है तो वह उसके पैरों को सुस्त कर उसके हृदय में व्याकुलता भर देता है क्योंकि उसे लगता है कि ऐसा कोई भी नहीं है जो उससे मिलने को व्याकुल हो। कवि कहता है कि दिन अत्यंत शीघ्रता से व्यतीत हो रहा है। समय शीघ्र गुजर रहा है।

अर्थग्रहण एवं सौंदर्य-सराहना संबंधी प्रश्नोत्तर

प्रश्न
1. ‘मुझसे मिलने को कौन विकल?’ पंक्ति में निहित भाव स्पष्ट कीजिए।
2. कौन-सा प्रश्न पद को शिथिल करता है?
3. हृदय में व्याकुलता कौन भरता है?
4. काव्यांश का काव्य-सौंदर्य स्पष्ट कीजिए।
उत्तर
1. इस पंक्ति का भाव यह है कि जीवन रूपी पथ पर बैठते हुए राही सोचता है कि उससे इस राह में मिलने के लिए कौन व्याकुल हो रहा है।
2. जीवन रूपी राह पर चलते हुए मुझसे मिलने के लिए कौन व्याकुल है तथा मैं किसके लिए रोमांचित हो जाऊँ, यह प्रश्न पद को शिथिल करता है।
3. जीवन में उठने वाले ये प्रश्न कि उससे मिलने के लिए व्याकुल कौन है तथा वह किसके लिए रोमांचित है, यह प्रश्न हृदय में व्याकुलता भरता है।
4. काव्य-सौंदर्य

• कवि ने राही के पथ से ध्यान भटक जाने की ओर संकेत किया है।
• प्रश्न अलंकार, पुनरुक्ति प्रकाश, पदमैत्री, स्वरमैत्री तथा अनुप्रास अलंकारों की छटा शोभनीय है।
• भाषा सहज, सरल एवं सरस है।
• तत्सम शब्दावली का प्रचुरता से प्रयोग है। (५) मुक्तक छंद है।
• अभिधात्मक शैली का प्रयोग है।
• प्रसाद गुण है।
• बिंब योजना सार्थक एवं सटीक है।