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In this page, we are providing Life Processes Class 10 Extra Questions and Answers Science Chapter 6 pdf download. NCERT Extra Questions for Class 10 Science Chapter 6 Life Processes with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 6 Extra Questions and Answers Life Processes

Extra Questions for Class 10 Science Chapter 6 Life Processes with Answers Solutions

Extra Questions for Class 10 Science Chapter 6 Very Short Answer Type

Life Processes Class 10 Extra Questions Question 1.
What will happen to a plant if its xylem is removed?  [CBSE 2009]
Answer:
Xylem helps in the transport of water and minerals to the various parts of the plant. If xylem is removed it would ultimately lead to the death of the plant.

Life Processes Class 10 Extra Questions With Answers Question 2.
Name the green dot like structures in some cells observed by a student when a leaf peel was viewed under a microscope. What is the green colour due to?  [CBSE 2010]
Answer:
The green dots like structures seen are the chloroplasts. The green colour is due to the pigment called chlorophyll.

Life Processes Extra Questions Question 3.
Give one reason why multicellular organisms require special organs for exchange of gases between their body and their environment.  [CBSE 2010]
Answer:
Simple diffusion is not sufficient for the exchange of gases in multicellular organisms as all their cells are not in direct contact with the environment. So, they require special organs for exchange of gases between their body and their environment.

Extra Questions Of Life Processes Class 10 Question 4.
What process in plants is known as transpiration?  [CBSE 2008]
Answer:
The loss of water in the form of vapour from the aerial parts of the plant is known as transpiration.

Class 10 Science Chapter 6 Extra Questions Question 5.
What is osmoregulation?  [CBSE 2006]
Answer:
The maintenance of optimum concentration of water and salts (electrolytes) in the body fluids is called as osmoregulation.

Life Process Extra Question Answer Question 6.
Why is carbon dioxide mostly transported in dissolved form?
Answer:
Carbon dioxide is mostly transported in the dissolved form as it is more soluble in water.

Life Processes Class 10 Extra Questions Pdf Question 7.
When we breathe out, why does the air passage not collapse?  [CBSE 2014]
Answer:
Rings of cartilage present on trachea prevent it from collapsing during the passage of air.

Class 10 Life Processes Extra Questions Question 8.
Herbivores have longer small intestine while carnivores have shorter small intestine. Give reason.  [CBSE 2014]
Answer:
Herbivores have a longer small intestine compared to the carnivores to allow time for the cellulose present in the grass to get digested.

Extra Questions On Life Processes Class 10 Question 9.
Mention the respiratory unit of lungs and excretory unit of kidneys.  [CBSE 2014]
Answer:
The respiratory unit of lungs are alveoli and excretory unit of kidneys are nephrons.

Class 10 Science Ch 6 Extra Questions Question 10.
Some organisms derive nutrition from plants or animals without killing them. What are these organisms called? Write one example.  [CBSE 2014]
Answer:
They are called parasites, e.g. Cuscuta and tapeworm are the parasites of plants and animals respectively.

Important Questions For Class 10 Science Chapter 6 Life Processes Question 11.
What are the major constituents of urine?  [CBSE 2008]
Answer:
Urine is an aqueous solution of water, urea, chloride, sodium, potassium, creatinine and other dissolved ions, inorganic and organic compounds (proteins, hormones, and metabolites).

Chapter 6 Science Class 10 Extra Questions Question 12.
Where does the urine produced by the kidneys get stored?  [CBSE 2006]
Answer:
Urinary bladder.

Extra Questions Life Processes Class 10 Question 13.
How does transpiration help in upward transport of substances?  [CBSE 2008]
Answer:
Transpiration creates a suction pressure which pulls up water along with the minerals through the xylem.

Life Processes Class 10 Extra Questions And Answers Question 14.
When the right atrium contract, blood flows from it to which part of the heart?  [CBSE 2007]
Answer:
Right ventricle.

Ch 6 Science Class 10 Extra Question Question 15.
State the functions of the following:
(i) Blood
(ii) WBC  [CBSE 2011, 2012]
Answer:
Function of RBC – To carry oxygen to various parts of the body.
Function of WBC – To protect the body against both infectious diseases and foreign invaders.

Extra Questions For Class 10 Science Chapter 6 Question 16.
Write the functions of the two upper chambers of the human heart.  [CBSE 2011]
Answer:
The right atrium receives deoxygenated blood from various parts of the body through the vena cava.
The left atrium receives the oxygenated blood from the lungs through the pulmonary artery.

Ncert Class 10 Science Chapter 6 Extra Questions Question 17.
Leakage of blood from vessels reduces the efficiency of pumping system. How is the leakage prevented? [CBSE 2010, 2011]
Answer:
Leakage is prevented by the blood platelets present in the blood which help in clotting the blood at the site of injury.

Class 10 Life Process Extra Questions Question 18.
Which mechanism plays an important role in transportation of water in plants
(i) During daytime
(ii) At night?  [CBSE 2011, 2012]
Answer:
During daytime – Transpiration; At Night – Root pressure.

Extra Questions Of Chapter 6 Science Class 10 Question 19.
Why are valves present in heart and the veins?   [CBSE 2010,2011]
Answer:
Valves present in the heart does not allow the blood to flow backwards when the atria or ventricles contracts. Valves are present in the veins to prevent the back flow of blood in the veins as it travels at very slow rate in the veins.

Extra Questions for Class 10 Science Chapter 6 Short Answer Type I

Class 10 Science Life Process Extra Questions Question 1.
How are the fats digested in our bodies? Where does this process take place?  [CBSE 2011]
Answer:
Bile juice produced by the liver breaks down the large fat globules into smaller globules by the process of emulsification. These small globules are then digested by the fat digesting enzymes. This process takes place in the small intestine.

Extra Questions From Life Processes Class 10 Question 2.
State the function of the epiglottis.  [CBSE 2004]
Answer:
Epiglottis covers the opening of the wind pipe (the glottis) and prevents the entry of food into the wind pipe during swallowing.

Extra Questions For Class 10 Science Life Processes Question 3.
The breathing cycle is rhythmic, whereas exchange of gases is a continuous process. Comment upon this statement.
Answer:
Some volume of air called as residual volume is left behind in the lungs even after forceful breathing out of air. This helps to provide sufficient time for oxygen to be absorbed and for carbon dioxide to be released. Even in the absence of continuous breathing, the exchange of these gases is continuous. Hence, breathing cycle is rhythmic, whereas exchange of gases is a continuous process.

Class 10 Chapter 6 Science Extra Questions Question 4.
What are the end products formed during fermentation in yeast? Under what condition a similar process takes place in our body that lead to muscle cramps?  [CBSE 2010]
Answer:
The end products formed during fermentation in yeast are ethanol and carbon dioxide. A similar process occurs in the muscles and produces lactic acid during anaerobic respiration in the muscles. Accumulation of lactic acid in the muscle cells lead to muscular cramps.

Extra Questions Of Chapter Life Processes Class 10 Question 5.
Give Reasons:
(a) Rings of cartilage are present in the trachea.
(b) Lungs always contain a residual volume of air.  [CBSE 2013]
Answer:
(a) The walls of trachea have rings of cartilage on them which prevent it from collapsing.
(b) The volume of air left behind in the lungs even after forceful breathing out of air is called as residual volume. This helps to provide sufficient time for oxygen to be absorbed and for the carbon dioxide to be released.

Question 6.
State in brief the role of lungs in the exchange of gases.  [CBSE 2012]
Answer:
Lungs have alveoli which provide a larger surface for exchange of gases and are richly supplied with blood vessels to enable faster exchange. So, lungs help in providing oxygen to various tissues of the body and removal of carbon dioxide from the body.

Question 7.
What is the basic unit of kidney called? Why is it composed of very thin blood capillaries?  [CBSE 2015]
Answer:
The basic unit of kidney is called nephron. It is composed of a cluster of very thin blood capillaries as they help in filtration of blood and remove the nitrogenous wastes from the body in the form of urine.

Question 8.
How does the plant get rid of excretory products?  [CBSE 2009]
Answer:
Excess oxygen and carbon dioxide removed through stomata.
Plant waste products are also removed by:

• Storage in cellular vacuoles
• Storage in leaves that fall off
• Storing as resins and gums in old xylem
• By excreting into the soil around them.

Question 9.
Tabulate two differences between renal artery and renal vein. [CBSE 2009]
Answer:
Renal Artery

1. Blood in renal artery contains glucose, oxygen and cellular waste products.
2. It takes blood towards the kidney.

Renal Vein:

1. Blood in renal vein is filtered, and is free from cellular waste and any other impurities.
2. It takes blood away from the kidney towards the heart.

Question 10.
(a) What is the main toxic waste that kidney filters from the blood?
(b) Name any two substances which are selectively reabsorbed from the tubules of a nephron.  [CBSE 2010, 2012]
Answer:
(a) Urea is the main excretory product removed by the kidneys of human beings.
(b) The substances selectively reabsorbed by the kidneys are water, glucose, electrolytes, etc.

Question 11.
What is excretion? How do unicellular organisms remove their wastes?  [CBSE 2012]
Answer:
Removal of metabolic wastes from the body is called as excretion. Many unicellular organisms remove metabolic wastes from the body surface into the surrounding water by simple diffusion.

Question 12.
Write a function of (a) blood vessels (b) blood platelets.  [CBSE 2008]
Answer:
(a) Blood vessels help in carrying blood to various parts of the body.
(b) Blood platelets help in the clotting of blood at the point of injury to prevent non-stop bleeding.

Question 13.
How are water and minerals absorbed by the plant?  [CBSE 2010]
Answer:
The water and minerals in the soil are absorbed by plants with the help of root hairs present on their roots. Root hairs provide a larger surface area for absorption.

Question 14.
What are capillaries? Sate the function performed by them.  [CBSE 2012]
Answer:
The capillaries are one-cell thick, small blood vessels which help in the exchange of materials between the blood and the surrounding tissues.

Question 15.
Mention the two main components of the transport system in plants. State one function of each one of these components.  [CBSE 2010, 2011]
Answer:
The two main components of the transport system in the plants are xylem and phloem. Xylem helps to transport water and minerals to various parts of the plant. Phloem helps to carry food from leaves to the various parts of the plant.

Question 16.
During one cycle how many times does blood go to the heart of fish and why?  [CBSE 2010]
Answer:
The blood passes only once through the heart in one cycle in fishes because the two-chambered heart of the fishes pump the blood to gills for oxygenation. The blood from gills is then directly passed to the various parts of the body in the fishes.

Question 17.
What would be the consequences of deficiency of haemoglobin in our bodies?  [CBSE 2012]
Answer:
Haemoglobin helps in transport of oxygen to the body parts. Deficiency of haemoglobin will affect transport of oxygen and the person will suffer from improper metabolism, weakness, fatigue and pain.

Question 18.
Name the following:
(а) The process in plants that links light energy with chemical energy.
(b) Organisms that can prepare their own food.
(c) The cell organelle where photosynthesis occurs.
(d) Cells that surround a stomatal pore.
(e) Organisms that cannot prepare their own food.
(f) An enzyme secreted from gastric glands in stomach that acts on proteins. [NCERT Exemplar]
Answer:
(a) Photosynthesis
(b) Autotrophs
(c) Chloroplast
(d) Guard cells
(e) Heterotrophs
(f) Pepsin

Question 19.
“All plants give out oxygen during the day and carbon dioxide during night”. Do you agree with this statement? Give reason.  [NCERT Exemplar]
Answer:
The rate of photosynthesis is higher than the rate of respiration during the daytime, so the net result is the evolution of oxygen. In the absence of photosynthesis at night, only respiration occurs in the plants so carbon dioxide is released at night.

Question 20.
How do the guard cells regulate opening and closing of stomatal pores?  [NCERT Exemplar]
Answer:
The entry of water into the guard cells of the stomata causes an increase in turgor pressure in the guard cells which leads to opening of the stomata. The loss of water from the guard cells results in their shrinking and closes the stomata.

Question 21.
Two green plants are kept separately in oxygen free containers, one in the dark and the other in continuous light. Which one will live longer? Give reasons.  [NCERT Exemplar]
Answer:
Plant kept in continuous light will perform photosynthesis and release oxygen for its respiration. Hence, it will live longer than the plant kept in the dark.

Question 22.
If a plant is releasing carbon dioxide and taking in oxygen during the day, does it mean that there is no photosynthesis occurring? Justify your answer.  [NCERT Exemplar]
Answer:
During the day time the plants take in carbon dioxide and release oxygen as a by product of photosynthesis. Release of carbon dioxide and taking in air during the daytime means that either the rate of photosynthesis is too low or its not occurring at all.

Question 23.
Why do fishes die when taken out of water? [NCERT Exemplar]
Answer:
Fishes take water from mouth and send it to the gills which are richly supplied with blood capillaries for absorbing the oxygen dissolved in water. But the fishes cannot absorb gaseous oxygen, so they die soon after they are taken out of water.

Question 24.
Is ‘nutrition’ a necessity for an organism? Discuss.  [NCERT Exemplar]
Answer:
Nutrition (food) is a necessity for an organism as

• It provides energy for the various metabolic processes in the body.
• It is essential for the growth and repair of various cells and tissues.
• It helps to provide resistance against various diseases.

Question 25.
What would happen if green plants disappear from the Earth?  [NCERT Exemplar]
Answer:
The green plants are the source of energy for the entire organisms on the Earth. Herbivores depend directly on the plants while the carnivores and omnivores depend either directly or indirectly on plants. So, all the organisms will die due to starvation if all the green plants disappear from the Earth.

Question 26.
Leaves of a healthy potted plant were coated with vaseline. Will this plant remain healthy for long? Give reasons for your answer.  [NCERT Exemplar]
Answer:
This plant will not remain healthy for a long time because the stomata will get blocked, so plant

• Will not get carbon dioxide for photosynthesis.
• Will not get oxygen for respiration.
• Will not be able to do transpiration which will in turn affect the upward transport of water and minerals.

Extra Questions for Class 10 Science Chapter 6 Short Answer Type II

Question 1.
(a) Explain why the rate of photosynthesis is low both at lower and higher temperatures.
(b) Is green light most or least useful in photosynthesis and why?  [CBSE 2005]
Answer:
(a) The reactions occurring during the process of photosynthesis is under the control of enzymes, which work under an optimum range of temperature only. The very high temperatures as well as very low temperatures decrease the activity of the enzymes. So, the rate of photosynthesis is low both at lower and higher temperatures.
(b) Chlorophyll does not absorb the green light, so the green light is least useful in photosynthesis.

Question 2.
Draw a well labelled diagram of stomata. List two functions of stomata.  [CBSE 2011]
Answer:
The function of stomata are:
(a) Help in the exchange of gases like carbon dioxide and oxygen from the leaves of the plants.
(b) Help in the transport of water, minerals and food materials in plants by transpiration.
(c) Transpiration occurring through stomata on leaves helps in cooling of leaf surface.

Question 3.
Explain how the products of photosynthesis and other substances are translocated in plants?  [CBSE 2015]
Answer:
Translocation is the transport of soluble products of photosynthesis through phloem. Sucrose is transferred into sieve tubes of phloem via the companion cells using energy from ATP. This increases the osmotic pressure inside the sieve tubes which causes movement of water into the sieve tubes from the adjacent xylem. This pressure helps in translocation of material in the phloem to tissues which have less pressure.

Question 4.
Write three events which occur during the process of photosynthesis.  [CBSE 2015]
Answer:
The three events which occur during photosynthesis are:

1. Absorption of light energy of chlorophyll.
2. Conversion of light energy to chemical energy + splitting of water molecules into hydrogen and oxygen.
3. Reduction of carbon dioxide to carbohydrates.

Question 5.
Explain why the transportation of materials is necessary in animals?  [CBSE 2014]
Answer:
The transportation system is necessary to transport the various nutrients and gases to and from the various parts of the body. It also helps in removing the wastes from the body of the organisms. The various life processes are maintained and carried out due to an efficient transport system in animals.

Question 6.
Plants absorb water from the soil. Explain how does the water reach the tree top?  [CBSE 2014]
Answer:
There are two ways for the transport of water in plants:
(а) By root pressure: The cells of root in contact with soil actively take up ions which creates a difference in ion concentration between the root and the soil. Water moves into the root from the soil to eliminate this difference, creating a column of water that is steadily pushed upwards.

(b) By transpiration pull: Loss of water from stomata by transpiration gets replaced by the xylem vessels in the leaf which creates a suction to pull water from the xylem cells of the roots. This strategy is used during day time and helps to transport water to the highest points of the plant body.

Question 7.
State the function of the following in the alimentary canal.
(a) Liver
(b) Gallbladder
(c) Villi   [CBSE2014]
Answer:
(a) Liver: Helps in detoxification of harmful chemicals. Produces bile juice which helps in digestion of fats.
(b) Gall bladder: Helps in the storage of bile juice released from the liver.
(c) Villi: Helps to increase the surface area of the small intestine and aid in absorption of the digested nutrients.

Question 8.
(a) How does exchange of respiratory gases—oxygen and carbon dioxide take place between tissues and blood in human beings?
(b) Name the respiratory pigment in humans. Where is it found?  [CBSE 2014]
Answer:
(a) The arteries have oxygenated blood having oxygen at higher pressure to that present in the tissues. Carbon dioxide is present at a higher pressure inside the tissues. The oxygen is thus exchanged at the tissue surface with carbon dioxide in order to supply oxygen to the tissues of the body.

(b) Haemoglobin is the respiratory pigment in humans which is present in the red blood cells.

Question 9.
Explain giving any three reasons the significance of transpiration in plants.   [CBSE 2014]
Answer:
The significance of transpiration is:

1. Absorption and upward movement of water and minerals.
2. It also helps in temperature regulation by cooling the leaf surface.
3. It helps to maintain the shape and size of the cells.

Question 10.
Mention the pathway of urine starting from the organ of its formation. Name four substances which are reabsorbed from the initial filtrate in the tubular part of the nephron.  [CBSE 2014]
Answer:
The urine formed in the kidney moves through the ureters to the urinary bladder. Ureters takes urine into the urinary bladder where it is stored until it is released through the urethra. Release of urine is under nervous control.

Glucose, amino acids, salts and water are reabsorbed from the initial filtrate in the tubular part of the nephron.

Question 11.
Major amount of water is selectively reabsorbed by the tubular part of nephron. On what factor does the amount of water reabsorbed depends?  [CBSE 2010, 2011, 2012]
Answer:
The amount of water reabsorbed by the tubular part of nephron depends on two factors:

1. The amount of nitrogenous and other excretory wastes present in the body.
2. The amount of excess water present in the body.

Question 12.
State the functions of
(a) Renal artery
(b) kidney
(c) urinary bladder.  [CBSE 2012]
Answer:
(a) Renal artery: It brings blood having nitrogenous wastes to nephrons for filtration.
(b) Kidney: It help to filter the nitrogenous and other wastes from the blood.
(c) Urinary bladder: Stores urine until it is eliminated.

Question 13.
(a) Write the important functions of the structural and functional unit of kidney.
(b) Write any one function of an artificial kidney.  [CBSE 2011]
Answer:
(a) The structural and functional units of kidney are the nephrons which help in filtration, reabsorption and secretion. They help to eliminate the wastes from the body.

(b) Artificial kidneys help to remove the harmful nitrogenous wastes from the body of a patient whose kidneys are not functioning properly.

Question 14.
How is urine produced?  [CBSE 2011]
Answer:
The urine formation involves three steps:
(a) Glomerular filtration: Nitrogenous wastes, glucose water, amino acid are filtered from the blood in blood capillaries into Bowman Capsule of the nephrons.

(b) Selective reasbsorption: Some substances in the initial filtrate, such as glucose, amino acids, salts and a major amount of water are selectively reabsorbed back by capillaries surrounding the nephrons.

(c) Tubular secretion: Some ions like K⁺, H⁺, etc. are secreted into the tubule which opens up into the collecting duct.

Question 15.
Blood does not clot in the blood vessels. Give reasons.  [CBSE 2009]
Answer:
Blood does not clot inside the blood vessels because the chemical which causes the blood to clot in the cut area gets activated only when it comes in contact with air (oxygen actually) to form the clotting substance. The chemical for clotting of blood is released by the blood platelets.

Question 16.
What is translocation? How does it take place in plants?  [CBSE 2011]
Answer:
Translocation is the movement of materials from leaves to other tissues throughout the plant by the phloem. During translocation, sucrose is transferred into sieve tubes of phloem via the companion cells using energy from ATP. This increases the osmotic pressure inside the sieve tubes which causes movement of water into the sieve tubes from the adjacent xylem. This pressure helps in translocation of material in the phloem to tissues which have less pressure.

Question 17.
(a) Transport of food in plants require living tissues and energy. Justify this statement.
(b) Name the components of food that are transported by the living tissues.  [CBSE 2012]
Answer:
(a) The energy in the form of ATP is required both during loading of sucrose in the sieve tubes at the leaf surface and during uploading of the sucrose at the site which needs sucrose for its metabolic activities. So, it can be said that transport of food in plants require living tissues i.e., phloem and energy in the form of ATP.

(b) The components of food transported by the phloem are sugars, hormones, and minerals elements dissolved in water.

Question 18.
What are the adaptations of leaf for photosynthesis?  [NCERT Exemplar]
Answer:
The various adaptations in leaf for photosynthesis are:

• Leaves have expanded portion to provide large surface area for maximum light absorption.
• Leaves are arranged at right angles to the light source in a way that causes overlapping.
• The extensive network of veins helps in quick transport of substances in leaf.
• Easy gaseous exchange is facilitated by the presence of stomata on the leaves.
• The chloroplasts are more in number on the upper surface of leaves.

Question 19.
Why is small intestine in herbivores longer than in carnivores?  [NCERT Exemplar]
Answer:
Herbivores have a longer small intestine compared to the carnivores to allow time for the cellulose present in the grass to get digested. Cellulose takes longer time to get digested. Animals have shorter small intestine as they are not able to digest cellulose.

Question 20.
What will happen if mucus is not secreted by the gastric glands?  [NCERT Exemplar]
Answer:
Mucus forms the lining of the stomach and protects it from the action of hydrochloric acid released by the gastric glands of stomach. If mucus is not secreted then acidity and ulcers may be caused due to erosion of the inner lining of the stomach by the acid.

Question 21.
What is the significance of emulsification of fats?  [NCERT Exemplar]
Answer:
The large globules of fats are broken down into smaller fat globules by the action of bile salts present in the bile juice. This is called emulsification and it increases the efficiency of the fat digesting enzymes that can digest the smaller fat globules.

Question 22.
Why does absorption of digested food occur mainly in the small intestine?  [NCERT Exemplar]
Answer:
Maximum absorption occurs in small intestine because

• The process of digestion gets completed in the small intestine by intestinal juice.
• Finger-like projections called villi provide larger surface area for absorption.
• The villi are richly supplied with blood vessels to enable easy absorption into the bloodstream.

Question 23.

Group (A)	Group (B)
(a) Autotrophic nutrition	(i) Leech
(b) Heterotrophic nutrition	(ii) Paramecium
(c) Parasitic nutrition	(iii) Deer
(d) Digestion in food vaceolus	(iv) Green plants

Answer:
(a) (iv)
(b) (iii)
(c) (i)
(d) (ii)

Question 24.
Why is the rate of breathing in aquatic organisms much faster than in terrestrial organisms?  [NCERT Exemplar]
Answer:
Aquatic animals use the oxygen dissolved in water. They breathe at a faster rate since the amount of dissolved oxygen is fairly low compared to the amount of oxygen in the air.

Question 25.
In each of the following situations what happens to the rate of photosynthesis?  [NCERT Exemplar]
(a) Cloudy days
(b) No rainfall in the area
(c) Good manuring in the area
(d) Stomata get blocked due to dust
Answer:
(a) Decreases
(b) Decreases
(c) Increases
(d) Decreases

Question 26.
Name the energy currency in the living organisms. When and where is it produced?  [NCERT Exemplar]
Answer:
The energy currency in the cells is Adenosine triphosphate (ATP). It is produced in mitochondria during respiration and in chloroplasts during photosynthesis in plants.

Question 27.
What is common for Cuscuta, ticks and leeches?  [NCERT Exemplar]
Answer:
Cuscuta, ticks and leeches are parasites which obtain their nutrition from plants and animals without killing them.

Question 28.
Explain the role of mouth in digestion of food.  [NCERT Exemplar]
Answer:
The mouth plays an important role in digestion because

• The teeth present in the mouth crush the food into small pieces.
• Tongue helps in thorough mixing of food with saliva and to swallow the food.
• Saliva secreted in mouth contains an enzyme called salivary amylase which helps to break down the starch into maltose.

Question 29.
What are the functions of gastric glands present in the wall of the stomach?
Answer:
The functions of gastric glands present in the wall of the stomach are:

1. Produce pepsin enzyme that digests proteins.
2. Secrete mucus which protects the inner lining of stomach from the action of acids.

Extra Questions for Class 10 Science Chapter 6 Long Answer Type

Question 1.
How is oxygen and carbon dioxide transported in the human beings?  [CBSE All India 2008]
Answer:
At the alveolar surface, the oxygen diffuses out from the alveoli to the capillaries which surround the alveoli. The oxygenated blood is transported from the lungs to the heart which pumps it to the various parts of the body to supply it to various tissues.

At the tissue surface, the oxygen from the blood diffuses out into the tissues and carbon dioxide which is at higher concentration in the tissues, diffuses out into the capillaries. The capillaries transport the carbon dioxide rich blood from tissues to the alveoli where carbon dioxide diffuses out into the alveoli and oxygen enters into the capillaries to be transported to the tissues.

Question 2.
(a) State reason for the following:
(i) Herbivores need a longer small intestine while carnivores have shorter small intestine.
(ii) The lungs are designed in human beings to maximise the area for exchange of gases.
(b) The rate of breathing in aquatic organisms is much faster than that seen in terrestrial organisms.  [CBSE 2014]
Answer:
(a) (i) The cellulose present in the grasses takes a longer time to digest. So, the herbivores have a longer small intestine than the carnivores.
(ii) The alveoli present in the lungs provide a larger surface area for the exchange of gases in the lungs.

(b) Aquatic organisms breathe at a faster rate since the amount of dissolved oxygen is fairly low compared to the amount of oxygen in the air.

Question 3.
Draw a diagram of excretory unit of human kidney and label the following:
(a) Bowman’s capsule
(b) Glomerulus
(c) Collecting duct
(d) Renal artery  [CBSE 2011, 2012]
Answer:

Question 4.
Explain how deoxygenated blood travels from body to lung for purification. Draw well labelled diagram in support of your answer.  [CBSE 2011]
Answer:
The deoxygenated blood from the various parts of the body is collected by the veins which transport the blood to the heart through the vena cava. Vena cava pours the deoxygenated blood in the right atrium of the heart. The right atrium contracts and the blood moves into the right ventricle. On contraction of the right ventricle the deoxygenated blood is transported to the lungs through the pulmonary artery for purification.

Schematic representation of transport and exchange of oxygen and carbon dioxide

Question 5.
Draw the sectional view of human heart and label the following parts given below:
(a) Chamber where oxygenated blood from lungs is collected
(b) Largest blood vessel in the body
(c) Muscular wall separating right and left chambers
(d) Blood vessel that carries blood from heart to the lungs  [CBSE 2010]
OR
(a) Part which receives deoxygenated blood from vena cava
(b) Part which sends deoxygenated blood to lung through pulmonary artery
(c) Part which receives oxygenated blood from lungs
(d) Part which sends oxygenated blood to all parts of the body through aorta  [CBSE 2012]
OR
(a) the chamber of heart that pumps out deoxygenated blood
(b) the blood vessel that carries away oxygenated blood from the heart
(c) the blood vessel that receives deoxygenated blood from the lower part of our body  [CBSE 2010, 2011, 2012]
Answer:

(а) Chamber where oxygenated blood from lungs is collected – Left atrium
(b) Largest blood vessel in the body – Aorta
(c) Muscular wall separating right and left chambers – Septum
(d) Blood vessel that carries blood from heart to the lungs – Pulmonary artery
OR
(а) Part which receives deoxygenated blood from vena cava – Right atrium
(b) Part which sends deoxygenated blood to lung through pulmonary artery – Right Ventricle
(c) Part which receives oxygenated blood from lungs – Left atrium
(d) Part which sends oxygenated blood to all parts of the body through aorta – Left Ventricle
OR
(а) the chamber of heart that pumps out deoxygenated blood – Right Ventricle
(b) the blood vessel that carries away oxygenated blood from the heart – Aorta
(c) the blood vessel that receives deoxygenated blood from the lower part of our body – Inferior Vena cava

Question 6.
Explain the process of nutrition in Amoeba.  [NCERT Exemplar]
Answer:
Nutrition in Amoeba:
Temporary finger-like extensions of the cell surface called pseudopodia are used by Amoeba to engulf food. Pseudopodia fuse over the food particle forming a food-vacuole in which complex substances are broken down into simpler ones and diffuse into the cytoplasm. The remaining undigested material moves to the surface of the cell and gets thrown out.

Question 7.
Describe the alimentary canal of man.  [NCERT Exemplar]
Answer:
Mouth: Helps in intake of whole food.
Teeth: Helps in chewing and grinding of food.
Tongue: Helps in tasting food + rolling food + swallowing food.

Salivary glands: Secrete saliva and mucus. The enzyme called salivary amylase is present in saliva which breaks down the complex starch into sugar.

Oesophagus (food pipe): Food moves towards stomach through oesophagus by rhythmic contraction of its muscles called as peristaltic movements or peristalsis.

Stomach: Muscular walls of stomach help in mixing food thoroughly with digestive juices. Stomach has gastric glands which secrete gastric juice containing pepsin for digestion of proteins, hydrochloric acid for creating acidic medium and mucus to protect inner lining of stomach from acid.

Small Intestine: Digestive juices like pancreatic juice, bile juice and the intestinal juices are secreted in the small intestine to help to complete the process of digestion. After digestion, the nutrients are absorbed by the villi present in the walls of the small intestine.

Large intestine: The unabsorbed food is sent into the large intestine where more villi absorb water from this material and remove the wastes through the anus by egestion. The exit of this waste material is regulated by the anal sphincter.

Question 8.
Explain the process of breathing in man.  [NCERT Exemplar]
Answer:
Air enters the body after getting filtered by fine hairs and mucus in the nostrils.
The air then passes through trachea (present in throat) into the lungs. The trachea divide into bronchi which enter the lungs and divide further into bronchioles which finally terminate in balloon-like structures called alveoli which have a rich supply of blood vessels and help in exchange of gases.

During inhalation (breathing in), the volume of the chest cavity becomes larger as the ribs get lifted and diaphragm gets flattened. Air gets sucked into the lungs and fills the expanded alveoli. The blood brings carbon dioxide from the rest of the body to the alveoli and exchanges it for oxygen to be transported to all the cells in the body.

During exhalation (breathing out), the volume of the chest cavity becomes smaller as the ribs get relaxed and diaphragm moves upward (relaxes). Air rich in carbon dioxide gets pushed out of the lungs to come out through the nostrils.

Question 9.
Explain the importance of soil for plant growth.
Answer:
Soil is important for the plant growth as it helps in the

• Anchoring the plant.
• Acts as a source of water and minerals for the plants.
• Ensures availability of oxygen for respiration of root cells.
• Microbes living in symbiotic association are found in the soil which helps to provide nitrogen for the plants.

Life Processes HOTS Questions With Answers

Question 1.
Which part of visible the spectrum is mostly ineffective in the process of photosynthesis? Why?
Answer:
The green part of the visible spectrum is mostly ineffective in the process of photosynthesis because chlorophyll involved in the process does not absorb green light. The green component of the visible spectrum gets reflected by the chlorophyll due to which the leaves appear green.

Question 2.
What difference would be seen in the small intestine of a grass eating animal and a flesh eating animal?
Answer:
The grass eating animal would have a longer small intestine than a flesh eating animal as the cellulose present in the grass takes a longer time to get digested.

Question 3.
The gall bladder of a patient is removed when a stone was observed in the gall bladder. Which kind of nutrient in the diet should be absent from the diet given to the patient?
Answer:
The gall bladder release bile juice which helps in the digestion of fats in the human beings. So, the patient should be advised a diet free from fats during the treatment process, as the digestion of fats will be most affected due to removal of gall bladder.

Question 4.
Stomata of desert plants remain closed during the daytime. Then how do they take up carbon dioxide and perform photosynthesis?  [CBSE 2010, 2012]
Answer:
The stomata of the desert plants open during the night time and absorb carbon dioxide to use it during the daytime for performing photosynthesis.

Question 5.
How do carbohydrates, proteins and fats get digested in human beings?
Answer:
An enzyme called salivary amylase is present in the saliva produced in the mouth. Salivary amylase helps in the breakdown of carbohydrates into maltose in the mouth.

The proteins get digested initially by the pepsin enzyme in the stomach and then by the enzymes present in the pancreatic juice secreted into the small intestine. The fats are broken down into small globules by the bile salts present in the bile juice secreted into the small intestine. This helps in increasing the efficiency of fat digesting enzymes.

The intestinal juice secreted by the walls of the small intestine help to complete the digestion of carbohydrates into glucose/monosaccharide, digestion of proteins into amino acids and the digestion of fats into fatty acids and glycerol.

Extra Questions for Class 10 Science Chapter 6 Value Based Questions

Question 1.
Raksha planted many plants in her home garden as she knew that the plants help to purify air. To get purified air in her room she kept some of the plants in a closed room, it leaves turned pale in colour after some days.
(a) Define photosynthesis. What are the conditions necessary for photosynthesis?
(b) What are the values shown by Raksha?
(c) What can be the most probable reason which resulted in the pale leaves?
Answer:
(a) The process by which the autotrophs synthesise their own food is called as photosynthesis.
Photosynthesis,occurs in the presence of carbon dioxide, water, chlorophyll and sunlight.
(b) The values shown by Raksha are intelligence, environment friendly, scientific temper, creative thinking.

Question 2.
Ayush experienced muscular cramps during the training session for his upcoming cricket match. His coach advised him a schedule of aerobic exercises to overcome this problem. Ayush followed his coach’s advice and did not experience any muscular cramps during the game. Based on this, answer the following questions:
(а) What was the reason for the muscular cramps?
(b) What was the effect of aerobic exercises?
(c) What are the values shown by Ayush?
Answer:
(a) The muscular cramps were caused as the muscles produced lactic acid during anaerobic respiration in the muscle cells. The accumulation of lactic acid caused the cramps.

(b) Aerobic exercises helped in providing enough oxygen for the various muscle cells so that they had a more supply of oxygen.

(c) Values shown by Ayush are respect for his coach, responsibility, hard work and patience.

Question 3.
Anshika noticed that one of the plants in her garden had wrinkled and drooping leaves. She went to the garden and felt the soil below the plant which was too dry. She immediately made provisions for watering the plant and made a daily schedule for watering it every day. Within a few days the plant revived and became green and healthy. On the basis of this text, answer the following questions:
(a) How is water transported up into the plants?
(b) What is the role of the leaves and the roots in such transport?
(c) What are the values shown by Anshika?
Answer:
(a) Water is transported up into the plants with the help of xylem. The movement of water in the xylem occurs due to transpiration and root pressure.

(b) Roots absorb water through their root hairs. Leaves have stomata through which transpiration occurs and pulls up the water through the xylem to the top of the plant.

(c) The values shown by Anshika are: Love for nature, scientific aptitude, awareness, care

Question 4.
Sameer went to a forest and observed the villagers collecting reins and gums from the plants. He asked the villagers about the use of the resins and gums. They told him that they will be used in making the paints and varnishes for furniture. He learnt the technique of collecting resins from the villagers and enjoyed the activity with the villagers by helping them in collection.
On the basis of the text, answer the following questions:
(а) Why are resins and gums released by plants?
(b) What are the ways in which plants get rid of their waste products?
(c) What are the values shown by Sameer?
Answer:
(a) Resins and gums are the excretory products of the plants.
(b) Plants remove excess oxygen and carbon dioxide through stomata, excess water by transpiration through stomata and other waste products by storage in cellular vacuoles, leaves that fall off, storage as resins and gums in old xylem or by excreting into the soil around them.
(c) Values shown by Sameer are: Care of the environment, scientific attitude, helpfulness, curiosity.

Here we are providing Areas Related to Circles Class 10 Extra Questions Maths Chapter 12 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-maths/

Extra Questions for Class 10 Maths Areas Related to Circles with Answers Solutions

Extra Questions for Class 10 Maths Chapter 12 Areas Related to Circles with Solutions Answers

Areas Related to Circles Class 10 Extra Questions Very Short Answer Type

Area Related To Circle Class 10 Extra Questions With Solutions Question 1.
Find the area of a square inscribed in a circle of diameter p cm.

Solution:
Diagonal of the square = p cm
∴ p² = side² + side²
⇒ p² = 2side²
or side² = \(\frac{p^{2}}{2}\) cm² = area of the square

Areas Related To Circles Class 10 Extra Questions Question 2.
Find the area of the circle inscribed in a square of side a cm.

Solution:
Diameter of the circle = a

Areas Related To Circles Extra Questions Question 3.
Find the area of a sector of a circle whose radius is and length of the arc is l.
Solution:
Area ola sector ola circle with radius r

Area Related To Circle Class 10 Extra Questions Question 4.
Find the ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal.
Solution:

Areas Related To Circles Class 10 Extra Questions With Answers Question 5.
A square inscribed in a circle of diameter d and another square is circumscribing the circle. Show that the area of the outer square is twice the area of the inner square.

Solution:
Side of outer square = d {Fig. 12.5]
∴ Its area = d
Diagonal of inner square = d
∴ Side = \(\frac{d}{\sqrt{2}}\)
⇒ Area = \(\frac{d^{2}}{2}\)
Area of outer square = 2 × Area of inner square.

Class 10 Maths Chapter 12 Extra Questions Question 6.
If circumference and the area of a circle are numerically equal, find the diameter of the circle.
Solution:
Given, 2πr = πr²
⇒ 2r = r²
⇒ r(r – 2) = 0 or r = 2
i.e. d = 4 units

Area Related To Circles Extra Questions Question 7.
The radius of a wheel is 0.25 m. Find the number of revolutions it will make to travel a distance of 11 km.
Solution:

Area Related To Circle Class 10 Extra Questions With Solutions Pdf Question 8.
If the perimeter of a semi-circular protractor is 36 cm, find its diameter.
Solution:
Perimeter of a semicircular protractor = Perimeter of a semicircle
= (2r + πr) cm
⇒ 2r + πr = 36
⇒ r\(\left(2+\frac{22}{7}\right)\) = 36
⇒ r = 7cm
Diameter 2r = 2 × 7 = 14 cm.

Area Related To Circle Class 10 Extra Questions Pdf Question 9.
If the diameter of a semicircular protractor is 14 cm, then find its perimeter.
Solution:
Perimeter of a semicircle = πr + 2r
= \(\frac{22}{7}\) × 7 + 2 × 7 = 22 + 14 = 36cm

Areas Related to Circles Class 10 Extra Questions Short Answer Type 1

Area Related To Circle Class 10 Important Questions With Solutions Question 1.
If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square?

Solution:
Let radius of the circle be r units.
Then, diagonal of the square = 2r

Area Related To Circle Difficult Questions Question 2.
What is the area of the largest triangle that is inscribed in a semi circle of radius r unit?
Solution:

Area of largest ∆ABC = \(\frac{1}{2}\) × AB × CD
\(\frac{1}{2}\) × 2r × r = r² sq. units

Extra Questions Of Area Related To Circles Class 10 Question 3.
What is the angle subtended at the centre of a circle of radius 10 cm by an arc of length 5π cm?
Solution:

Areas Related To Circles Class 10 Important Questions Question 4.
What is the area of the largest circle that can be drawn inside a 4 rectangle of length a cm and breadth b cm (a > b)?

Solution:
Diameter of the largest circle that can be inscribed in the given b
rectangle = b cm
∴ Radius = \(\frac{b}{2}\) cm

Ch 12 Maths Class 10 Extra Questions Question 5.
Difference between the circumference and radius of a circle is 37 cm. Find the area of circle.
Solution:
Given 2π r – r = 37
or r (2π – 1) = 37

Extra Questions For Class 10 Maths Areas Related To Circles Question 6.
The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution:
Let r be the radius of required circle. Then, we have
πr² = π(8)² + π(6)²
⇒ πr² = 64π + 36π
⇒ πr² = 100π
∴ r² = \(\frac{100 \pi}{\pi}\) = 100
⇒ r = 10cm
Hence, radius of required circle is 10 cm.

Class 10 Areas Related To Circles Extra Questions Question 7.
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Solution:
Let R be the radius of required circle. Then, we have
2πR = 2π(19) + 2π (9)
⇒ 2πR = 2π (19 + 9)
⇒ R = \(\frac{2 \pi \times 28}{2 \pi}\) = 28
⇒ R = 28 cm
Hence, the radius of required circle is 28 cm.

Area Related To Circle Extra Questions Question 8.
Find the area of a circle whose circumference is 22 cm.
Solution:
Let r be the radius of the circle. Then,
Circumference = 22 cm

Class 10 Maths Ch 12 Extra Questions Question 9.
The area of a circular playground is 22176 m². Find the cost of fencing this ground at the rate of ₹50 per m.
Solution:
Area of circular playground = 22176 m²
πr² = 22176
⇒ \(\frac{22}{7}\) r² = 2176
⇒ \(\frac{22176 \times 7}{22}\)
⇒ r = 84 m
∴ Circumference of the playground =2πr = 2 × \(\frac{22}{7}\) × 84 = 44 × 12 = 528 m .
∴ Cost of fencing this ground = ₹ 528 × 50 = ₹ 26400.

Class 10 Area Related To Circles Extra Questions Question 10.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
We know that

Important Questions Of Area Related To Circle Class 10 Question 11.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Let r be the radius of circle, then circumference = 22 cm

Chapter 12 Maths Class 10 Extra Questions Question 12.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Since the minute hand rotates through 6° in one minute, therefore, area swept by the minute hand in one minute is the area of a sector of angle 6° in a circle of radius 14 cm.

Extra Questions On Areas Related To Circles Question 13.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Solution:
We have, r = 16.5 km and 0 = 80°
∴ Area of the sea over which the ships are warned =

Areas Related to Circles Class 10 Extra Questions Short Answer Type 2

Area Related To Circles Class 10 Extra Questions Question 1.
If the perimeter of a semicircular protractor is 66 cm, find the diameter of the protractor
(Take π = \(\frac{22}{7}\)).
Solution:
Let the radius of the protractor be r сm. Then,
Perimeter = 66 cm
= πr + 2r = 66 [∴ Perimeter of a semicircle = πr + 2r]

Extra Questions Areas Related To Circles Class 10 Question 2.
The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle.
Solution:
Let the radius of the circle be r сm. Then,
Diameter = 2r cm and Circumference = 2πr cm
According to question,
Circumference = Diameter + 16.8
⇒ 2πr = 2r + 16.8
⇒ 2 × \(\frac{22}{7}\) × r = 2r + 16.8
⇒ 44r = 14r + 16.8 × 7
⇒ 44r – 14r = 117.6 or 30r = 117.6
⇒ r = \(\frac{117.6}{30}\) = 3.92
Hence, radius = 3.92 cm.

Question 3.
A race track is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track.
Solution:
Let the outer and inner radii of the ring be R m and r m respectively. Then,
2πR = 396 and 2πr = 352

Hence, width of the track = (R – r) m = (63 – 56) m = 7 m

Question 4.
The inner circumference of a circular track [Fig. 12.10] is 220 m. The track is 7 m wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of ₹2 per metre.
Solution:

Let the inner and outer radii of the circular track berm and R m respectively. Then,
Inner circumference = 2πr = 220 m

Since the track is 7 m wide everywhere. Therefore,
R = Outer radius = r + 7 = (35 + 7)m = 42 m
∴ Outer circumference = 2πR = 2 × \(\frac{22}{7}\) × 42m = 264m
Rate of fencing = ₹ 2 per metre
∴ Total cost of fencing = (Circumference × Rate) = ₹(264 × 2) = ₹ 528

Question 5.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Solution:
The diameter of a wheel = 80 cm.
radius of the wheel = 40 cm.
Now, distance travelled in one complete revolution of wheel = 2π × 40 = 80π
Since, speed of the car is 66 km/h
So, distance travelled in 10 minutes = \(\frac{66 \times 100000 \times 10}{60}\)
= 11 × 100000 cm = 1100000 cm.
So, Number of complete revolutions in 10 minutes

Question 6.
An umbrella has 8 ribs which are equally spaced (Fig. 12.11). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:
We have, r = 45 cm

Question 7.
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (Fig. 12.12). Find (i) the area of that part of the field in which the horse can graze;
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Solution:
Let the horse be tied at point O and the length of the rope is OH (Fig. 12.13).
Thus, (i) the area of the part of the field in which the horse can graze
= Area of the quadrant of a circle (OAHB)

(ii) Now r = 10 m and


Increase in the grazing area
= (78.5 – 19.625) m²
= 58.875 m²

Question 8.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115o. Find the total area cleaned at each sweep of the blades.
Solution:
We have, r = 25 cm and θ = 115°.
∴ Total area cleaned at each sweep of the blades
= 2 × (Area of the sector having radius 25 cm and angle θ = 115°).

Question 9.
In Fig. 12.15, sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area
of the shaded region.
Solution:
Let A₁ and A₂ be the areas of sectors OAB and OCD respectively. Then, A, = Area of a sector of angle 30° in a circle of radius 7 cm.

Question 10.
The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 9 AM and 9.35 AM.
Solution:
We have,
Angle described by the minute hand in one minute = 6°
∴ Angle described by the minute hand in 35 minutes = (6 × 35)° = 210°
Area swept by the minute hand in 35 minutes = Area of a sector of a circle of radius 10 cm

Question 11.
Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector. (Use π = 3.14)
Solution:

Area of the sector OAPB = \(\frac{\theta}{360^{\circ}} \times \pi r^{2}\)
\(\frac{30^{\circ}}{360^{\circ}}\) × 3.14 × 4 × 4 cm²
= \(\frac{12.56}{3}\)cm²
= 4.19 cm² (approx.)
Area of the corresponding major sector = πr² – Area of sector OAPB
= (3.14 × 4 × 4 – 4.19)cm² = (50.24 – 4.19) cm²
= 46.05 cm² = 46.1 cm² (approx.)

Question 12.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)
Solution:
We have, r = 15 cm and θ = 60°
Given segment is APB

Now, area of major segment = Area of circle – Area of minor segment
= π(15)² – 20.44 = 3.14 × 225 – 20.44
= 706.5 – 20.44 = 686.06 cm

Question 13.
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)
Solution:
We have, r = 12 cm and θ = 120°
Given segment is APB
Now, area of the corresponding segment of circle
= Area of the minor segment

Question 14.
A round table cover has six equal designs as shown in Fig. 12.19. If the radius of the cover is 28 cm?, find the cost of making the designs at the rate of ₹ = 0.35 per cm². (Use √3 = 1.7)
Solution:
Area of one design = Area of the sector OAPB – Area of ΔAOB

Area of 6 such designs = 77.47 × 6 = 464.8 cm²
Hence, cost of making such designs = ₹ 162.69

Question 15.
Find the area of the shaded region in Fig. 12.20, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Solution:

Here, ROQ is the diameter of given circle, therefore ∠RPQ = 90°
Now, in right angled ΔPRQ, we have
RQ² = RP² + PQ² (by Pythagoras Theorem)
RQ² = (7)² + (24)² = 49 + 576 = 625
RQ = \(\sqrt{625}\) = 25 cm
Therefore, radius r = \(\frac{25}{2}\) cm
Now, area of shaded region
= Area of the semi-circle – Area of ARPQ

Question 16.
Find the area of the shaded region in Fig. 12.21, if radii of the two concentric circles with centre 0 are 7 cm and 14 cm respectively and ∠AOC = 40°.
Solution:
Area of shaded region
= Area of sector AOC – Area of sector OBD

Question 17.
Find the area of the shaded region in Fig. 12.22, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Solution:
We have, radius of semicircles = 7 cm
∴ Area of shaded region
= Area of square ABCD – Area of semi-circles (APD +BPC)

Question 18.
Find the area of the shaded region in Fig. 12.23, where a circular arc of radius 6 cm has been drawn with vertex 0 of an equilateral triangle OAB of side 12 cm as centre.
Solution:
We have, radius of circular region
= 6 cm and each side of ΔOAB = 12 cm.
∴ Area of the circular portion
= Area of circle – Area of the sector

Quesiton 19.
From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.24. Find the area of the remaining portion of the square.
Solution:
We have, the side of the square ABCD = 4 cm
Area of the square ABCD = (4)² = 16 cm²
Since, each quadrant of a circle has radius 1 cm.
∴ The sum of the areas of four quadrants

Question 20.
In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.

Solution:
We have, each side of square ABCD = 14 cm
∴ Area of square ABCD = (14²)cm² = 196 cm²
Now, radius of each quadrant of circle,
r = \(\frac{14}{2}\) = 7 cm
∴ The sum of the area of the four quadrants at the four corners of the square

Now, area of shaded portion
= Area of square ABCD – The sum of the areas of four quadrants at the four corners of the square
= (196 – 154) cm² = 42 cm²

Question 21.
On a square handkerchief, nine circular designs, each of radius 7 cm are made (see Fig. 12.26). Find the area of the remaining portion of the handkerchief.

Solution:
Total area of circular design = 9 × Area of one circular design
= 9 × π × (7)²
= 9 × \(\frac{22}{7}\) × 7 × 7 = 1386 cm²
Now, each side of square ABCD = 3 × diameter of circular design
= 3 × 14 = 42 cm
∴ Area of square ABCD = (42)² = 1764 cm²
∴ Area of the remaining portion of handkerchief
= Area of square ABCD – Total area of circular design
= (1764 – 1386) cm = 378 cm²

Question 22.
In Fig. 12.27, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region.
OR
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.
Solution:

Question 23.
In Fig. 12.28, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14).
Solution:
We have,
Radius of quadrant

= 200 × 3.14 – 400 = 628 – 400 = 228 cm²

Question 24.
Calculate the area of the designed region in Fig. 12.29, which is common between the two quadrants of circles of radius, 8 cm each.
Solution:
Here, radius of each quadrant ABPD and BQDC = 8 cm

Question 25.
In the given Fig. 12.30, find the area of the shaded region.
Solution:

Clearly, diameter of the circle
= Diagonal BD of rectangle ABCD
Now, BD

Let r be the radius of the circle. Then, r = \(\left(\frac{10}{2}\right)\)cm = 5 cm
Hence, area of the shaded region = Area of the circle – Area of rectangle ABCD
= πr² – l × b = (3.14 × 5 × 5) – (8 × 6)
= (78.50 – 48) cm² = 30.50 cm²

Question 26.
A square park has each side of 100 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius 14 m as shown in Fig. 12.31. Find the area of the remaining part of the park.
Solution:

We have,
Area of 4 quadrants of a circle of radius 14 m

Area of square park having side 100 m long
= (100 × 100) m² = 10,000 m².
Hence, area of the remaining part of the park
= Area of square – Area of 4 quadrants at each corner
= (10,000 – 616) m² = 9384 m²

Question 27.
Find the area of the shaded region in Fig. 12.32, where ABCD is a square of side 14 cm each.

Solution:
Area of square ABCD = 14 × 14 cm² = 196 cm²
Diameter of each circle = \(\frac{14}{2}\) cm = 7 cm
So, radius of each circle = \(\frac{7}{2}\) cm

∴ Area of shaded region = Area of square – Area of 4 circles
= (196 – 154) cm² = 42 cm²

Question 28.
In Fig. 12.33, ABCD is a trapezium of area 24.5 sq. cm. In it, AD || BC, ∠DAB = 90°, AD = 10 cm and BC = 4 cm. If ABE is a quadrant of a circle, find the area of the shaded region. [Take π = \(\frac{22}{7}\)
Solution:

Area of trapezium = 24.5 cm²
\(\frac{1}{2}\)[AD + BC] AB = 24.5
\(\frac{1}{2}\)[10 + 4] × AB = 24.5
AB = 3.5 cm ⇒ r = 3.5 cm
Area of quadrant = \(\frac{1}{4}\)πr²
.025 × \(\frac{22}{7}\) × 3.5 × 3.5 = 9.625 cm²
The area of shaded region = 24.5 – 9.625 = 14.875 cm²

Question 29.
In Fig. 12.34, O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)
Solution:

In ΔABC, ∠ACB = 90° (Angle in the semicircle)
∴BC² + AC² = AB²
∴ BC² = AB² – AC²
= 169 – 144 = 25
∴ BC = 5 cm
Area of the shaded region = Area of semicircle – Area of right ΔABC

Question 30.
In Fig. 12.35, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre 0 and radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M.
If OP = PQ = 10 cm show that area of shaded region is \(25\left(\sqrt{3}-\frac{\pi}{6}\right)\)cm².
Solution:
Since OP = PQ = QO
⇒ ΔPOQ is an equilateral triangle
∴ ∠POQ = 60°
Area of segment PAQM

Areas Related to Circles Class 10 Extra Questions Long Answer Type

Question 1.
PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semicircles are drawn on PQ and QS as diameters is shown in Fig. 12.36. Find the perimeter and area of the shaded region.

Solution:
We have,
PS = Diameter of a circle of radius 6 cm = 12 cm
∴ PQ = QR = RS = \(\frac{12}{3}\) = 4 cm
Fig. 12.36 QS = QR + RS = (4 + 4) cm = 8 cm
Hence, required perimeter
= Arc of semicircle of radius 6 cm + Arc of semi circle of radius 4 cm + Arc of semi-circle of radius 2 cm
= (π × 6 + π × 4 + π × 2) cm = 12π cm = 12 × = \(\frac{22}{7}\) = \(\frac{264}{7}\) = 37.71 cm.
Required area = Area of semicircle with PS as diameter + Area of semicircle with PQ as diameter – Area of semi-circle with QS as diameter

Question 2.
Figure 12.37 depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Solution:

Question 3.
The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days.
Solution:
In 2 days, the short hand will complete 4 rounds.
Distance moved by its tip = 4 (Circumference of a circle of radius 4 cm)

Question 4.
Fig. 12.38, depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
(i) the distance around the track along its inner edge.
(ii) the area of the track.
Solution:

Here, we have
OE = O’G = 30 m
AE = CG = 10 m
OA = O’C = (30 + 10) m = 40 m
AC = EG = FH = BD = 106 m

(i) The distance around the track along its inner edge
= EG + FH + 2 × (circumference of the semicircle of radius OE = 30cm)

(ii) Area of the track = Area of the shaded region
= Area of rectangle AEGC + Area of rectangle BFHD + 2 (Area of the semicircle of radius 40 m – Area of the semicircle with radius 30 m)

Question 5.
The area of an equilateral triangle ABC is 17320.5 cm. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.39). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)
Solution:

Let each side of the equilateral triangle be x cm. Then,
Area of equilateral triangle ABC = 17320.5 cm (Given)

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.40. Find the area of the design.
Solution:
Here, ∆ABC is an equilateral triangle. Let O be the circumcentre of circumcircle.
Radius, r = 32 cm.
Now, area of circle = πr²

Question 7.
In Fig. 12.41, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
Solution:

Question 8.
In Fig. 12.42 ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Solution:
In ∆ABC, we have

Hence, area of the shaded region = Area of the semi-circle BQC – Area of the segment BPC
= (154 – 56)cm² = 98cm²

Question 9.
In Fig 12.43, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region. [Use π = 3.14 and √3 = 1.73]

Solution:
Construction:
Join OA, OB and OC
Draw OZ ⊥ BC, OX ⊥ AB and OY ⊥ AC.
Let the radius of the circle be r сm.
Area of ∆ABC = Area of ∆AOB + Area of ∆BOC + Area of ∆AOC


= 1.73 × 3 × 12 – 3.14 × 4 × 3
= 62.28 – 37.68 = 24.6 cm²

Question 10.
In Fig. 12.45, PSR, RTQ and PAQ are three semicircles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region. [Use π = 3.14]
Solution:

Radius of semicircle PSR = \(\frac{1}{2}\) × 10 cm = 5 cm
Radius of semicircle RTQ = \(\frac{1}{2}\) × 3 cm = 1.5 cm
Radius of semicircle PAQ = \(\frac{1}{2}\) × 7 em = 3.5 cm
Perimeter of shaded region = Circumference of semicircle PSR + Circumference of semicircle RTQ + Circumference of semicircle PAQ.

= π[5 + 1.5 + 3.5]= 3.14 × 10 = 31.4cm

Question 11.
An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 12.46). From one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length of the belt that is still in contact with the pulley. Also find the shaded area. (Use π = 3.14 and √3 = 1.73)
Solution:

PA = 5√3 cm = BP [Tangents from an external point are equal]

Question 12.
In Fig. 12.47, a sector OAP of a circle with centre O, containing ∠θ. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is

Solution:

Question 13.
Find the area of the shaded region in Fig. 12.48, where APD, AQB, BRC , CSD are semi-circles of diameter 14 cm, 3.5 cm, 7 cm and 3.5 cm respectively. (Use π = \(\frac{22}{7}\) )
Solution:

Question 14.
A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the 14 cmcentre of circle. Find the area of major and minor segments of the Fig. 12.48 circle.
Solution:
Area of minor segment
= Area of minor sector having angle 60° at centre – area of equilateral ∆OPQ

Question 15.
In the given figure, ∆ABC is a right-angled triangle in which ∠A is 90°. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region.
Solution:
∵ ABC is right angled triangle

Question 16.
In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.
Solution:
In right angle triangle ABC

Areas Related to Circles Class 10 Extra Questions HOTS

Question 1.
Two circles touch internally. The sum of their areas is 116 ncmo and distance between their centres is 6 cm. Find the radii of the circles.
Solution:

Let R and r be the radii of the circles [Fig. 12.52].
Then, according to question,
⇒ πR² +πr² = 116π
⇒ R² + r² = 116 …..(i)
Distance between the centres = 6 cm
⇒ OO’ = 6 cm
⇒ R – s = 6 …(ii)
Now, (R + r)² + (R – 1)² = 2(R² + m²)
Using the equation (i) and (ii), we get
(R + r)² + 36 = 2 × 116
= (R + r)² = (2 x 116 – 36) = 196
= R + r = 14 …..(iii)
Solving (ii) and (iii), we get R = 10 and r = 4
Hence, radii of the given circles are 10 cm and 4 cm respectively.

Question 2.
A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
Solution:
Let the radius of the wheel be r сm.

∴ Diameter = 2r cm = (2 × 35) cm = 70 cm
Hence, the diameter of the wheel is 70 cm.

Question 3.
Find the area of the shaded design of Fig. 12.53, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter (use π = 3.14).

Solution:

Let us mark the four unshaded regions as I, II, III and IV (Fig. 12.53).
Area of I + Area of III
= Area of ABCD – Areas of two semicircles of radius 5 cm each

= (100 – 3.14 × 25) cm² = (100 – 78.5) cm² = 21.5 cm²
Similarly, Area of II + Area of IV = 21.5 cm²
So, Area of the shaded design
= Area of ABCD – Area of (I + II + III + IV)
= (100 – 2 × 21.5) cm²
= (100 – 43) cm² = 57 cm²

Question 4.
A copper wire, when bent in the form of a square, encloses an area of 484 cm². If the same wire is bent in the form of a circle, find the area enclosed by it.
Solution:
We have, Area of the square = a² = 484 cm²
∴ Side of the square = √1484 cm = 22 cm
So, Perimeter of the square = 4 (side) = (4 × 22) cm = 88 cm
Let r be the radius of the circle. Then, according to question,
Circumference of the circle = Perimeter of the square

Question 5.
Two circles touch externally. The sum of their areas is 130 r sq. cm and the distance between their centres is 14 cm. Find the radii of the circles.
Solution:

If two circles touch externally, then the distance between their centres is equal to the sum of their radii.
Let the radii of the two circles be r, cm and r² cm respectively [Fig. 12.55).
Let C, and C, be the centres of the given circles. Then,
C₁C₂ = r₁ +r₂
= 14 = r₁ +r₂
[∵ C₁C₂ = 14 cm given]
= r₁ +r₂ = 14
It is given that the sum of the areas of two circles is equal to 130 n cm²

Solving (i) and (iv), we get r₁ = 11 cm and r₂ = 3 cm.
Hence, the radii of the two circles are 11 cm and 3 cm.

Question 6.
In Fig. 12.56, from a rectangular region ABCD with A
AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. [Use π = 3.14]

Solution:
Area of shaded region
= Area of rectangle – Area of triangle + Area of semicircle. In right ∆ADE.
AD² = AE² + DE²

Area of rectangle = AB × BC = 20 × 15 = 300 cm²
Area of shaded region = 300 + 88.31 – 54 = 334.31 cm²

Question 7.
In the given Fig. 12.58, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O’ are centres of the circles. Find the area of shaded region.

Solution:
Area of shaded region
= Area of square + Area of 2 major sectors having angle 270° at centre

Question 8.
In the given Fig. 12.59, ABCD is a rectangle of dimensions 21 cm × 14 cm. A semicircle is drawn with BC as diameter. Find the area and the perimeter of the shaded region in the figure.
Solution:

Area of shaded region
Area of shaded region
= Area of rectangle – Area of semicircle
=(l × b) – \(\frac{1}{2}\) × π × r²
= (21 × 14) – \(\frac{1}{2}\) × π × 7 × 7
= 294 – \(\frac{1}{2}\) × \(\frac{22}{7}\) × 7 × 7
= 294 – 77 = 217 cm²
Now, perimeter of shaded region = 2l + b + circumference of semicircle i.e.; πr
= 2 × 21 + 14 + \(\frac{22}{7}\) × 7
= 56 + 22 = 78 cm

In this page, we are providing Carbon and its Compounds Class 10 Extra Questions and Answers Science Chapter 4 pdf download. NCERT Extra Questions for Class 10 Science Chapter 4 Carbon and its Compounds with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 4 Extra Questions and Answers Carbon and its Compounds

Extra Questions for Class 10 Science Chapter 4 Carbon and its Compounds with Answers Solutions

Extra Questions for Class 10 Science Chapter 4 Very Short Answer Type

Carbon And Its Compounds Class 10 Extra Questions With Answers 2021 Question 1.
Draw the electron dot notation of O₂ molecule.
Answer:

Carbon And Its Compounds Class 10 Extra Questions With Answers Question 2.
Name a molecule that has triple bond.
Answer:
Nitrogen (N₂).

Carbon And Its Compounds Extra Questions Question 3.
Name the hardest substance which is an allotrope of carbon.
Answer:
Diamond.

Class 10 Science Chapter 4 Extra Questions Question 4.
Name the allotrope of carbon which have the structure of C-60.
Answer:
Fullerenes.

Chapter 4 Science Class 10 Extra Questions Question 5.
Name the unique ability of carbon to form bonds with other atoms of carbon.
Answer:
Catenation.

Carbon And Its Compounds Class 10 Questions With Answers Question 6.
Mention the two characteristic features seen in carbon.
Answer:
Tetravalency and catenation.

Class 10 Science Ch 4 Extra Questions Question 7.
Name the first organic compound synthesised by Wohler.
Answer:
Urea.

Carbon And Its Compounds Class 10 Extra Questions With Answers Pdf Question 8.
Write the general molecular formula of alkane series.
Answer:
C_nH_2n+2

Carbon And Its Compounds Class 10 Extra Questions Question 9.
Write the IUPAC name of the following compound:
CH₃CH₂CH₂CH₂—C ≡ C—H
Answer:
Hex-1-yne.

Class 10 Carbon And Its Compounds Extra Questions Question 10.
Identify the functional group present in the following compound:

Answer:
Aldehyde.

Extra Questions For Class 10 Science Chapter 4 Question 11.
How many covalent bonds are there in a molecule of ethane, C₂H₆?
Answer:
Seven (7)

Class 10 Chapter 4 Science Extra Questions Question 12.
Draw the structure of the hexanal molecule, C₅H₁₁CHO.
Answer:

Extra Questions Of Carbon And Its Compounds Question 13.
Write the name and formula of the 2^nd member of homologous series having general formula C_nH_2n.  [CBSE 2015]
Answer:
C_nH_2n : Alkene
2^nd member = C₃H₆ (propene)

Class 10 Science Chapter 4 Extra Question Answer Question 14.
Write the molecular formula of an alkyne containing 10 atoms of hydrogen.
Answer:
C₆H₁₀.

Extra Questions Of Ch 4 Science Class 10 Question 15.
Why is ethanoic acid known as glacial acetic acid?
Answer:
Acetic acid freezes at 290 K to form crystals which look like glaciers, so pure ethanoic acid is known as glacial acetic acid.

Question 16.
Which property of ethanol makes it suitable for preparing medicines such as tincture iodine, cough syrup and other tonics?
Answer:
Ethanol is a good solvent.

Question 17.
What is the function of cone. H₂SO₄ in the formation of ethene from ethanol?
Answer:
Dehydrating agent

Question 18.
Name the alcohol which is an active ingredient of all alcoholic drinks.
Answer:
Ethanol or ethyl alcohol (C₂H₅OH)

Question 19.
Which two of the following compounds could belong to the same homologous series?
C₂H₆O₂, C₂H₆O, C₃H₂₈, CH₄O
Answer:
CH₄O and C₂H₆O (General formula C_nH_2n+1. OH)

Question 20.
Which of the following molecule is called buckminsterfullerene?
C₉₀, C₆₀, C₇₀, C₁₂₀
Answer:
C₆₀.

Question 21.
Name the gas evolved when sodium carbonate and bicarbonate is added to ethanoic acid.
Answer:
Carbon dioxide (CO₂)

Question 22.
Among CH₄, C₂H₆ and C₄H₁₀ which is expected to show isomerism?
Answer:
C₄H₁₀.

Question 23.
Write the structural formula of a saturated hydrocarbon whose molecule contains three atoms of carbon.
Answer:
C₃H₈.

Question 24.
A neutral organic compound is warmed with some ethanoic acid and a little cone. H₂SO₄. Vapours having sweet smell or fruity smell are observed. Identify the functional group present in the organic compound.
Answer:

Question 25.
Name the oxidising agent which can oxidise ethanol to ethanoic acid.
Answer:
Alkaline potassium permanganate (KMnO₄/KOH) or acidified potassium dichromate (K₂Cr₂O₇/H₂SO₄).

Question 26.
Write the formula and name of next homologue of CH₃COCH₃.
Answer:
CH₃CH₂COCH₃, Butanone.

Question 27.
Why do alkanes burn with a blue flame?
Answer:
Alkanes generally burn with a blue flame or clean flame because the combustion is complete and no unbumt carbon particles are released.

Question 28.
Draw the structure of an unsaturated cyclic compound having six carbon atoms. Also draw its electron dot structure. (Cyclohexene)
Answer:

Question 29.
How do the melting and boiling points of the hydrocarbons change with increase in molar mass?
Answer:
Intermolecular forces of attraction increases due to increase in molar mass, hence the melting and boiling points increase.

Extra Questions for Class 10 Science Chapter 4 Short Answer Type I

Question 1.
Draw the electron dot structure of ethyne and also draw its structural formula.   [NCERT Exemplar]
Answer:
Ethyne, C₂H₂

Question 2.
Name the functional groups present in the following compounds:
(a) CH₃ CO CH₂ CH₂ CH₂ CH₃
(b) CH₃ CH₂ CH₂ COOH
(c) CH₃ CH₂ OH   (NCERT Exemplar)
Answer:
(a) Ketone
(b) Carboxylic acid
(c) Aldehyde
(d) Alcohol

Question 3.
Which of the following hydrocarbons can undergo addition reactions:
C₂H₆, C₄H₁₀, C₃H₆, C₃H₄, CH₄, C₂H₂, C₄H₈
Answer:
C₃H₆, C₃H₄, C₂H₂ and C₄H₈ because these compounds are unsaturated organic compounds and hence can undergo addition reactions.

Question 4.
Draw the electron dot structure of O₂ and N₂ molecules.
Answer:

Question 5.
Give the general formula of alkanes. Write the name, structural formula and physical state of the compound containing:
(i) 3-carbon atoms
(ii) 8-carbon atoms.
Answer:
(a) General formula of alkanes is C_nH_2n+2
n = 1, 2, 3…
(i) Propane, CH₃—CH₂—CH₃
or

Propane is a gas.

(ii) CH₃—CH₂—CH₂—CH₂—CH₂—CH₂—CH₂—CH₃
or

Octane is a liquid

Question 6.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Carbon atoms have 4 valence electrons in their valence shell, it needs to gain or lose 4 electrons to attain the noble gas configuration.

(i) It could gain four electrons forming C^4- anion. But it would be difficult for the nucleus with six protons to hold on to ten electrons.

(ii) It could lose four electrons forming C⁴⁺ cation. But it would require a large amount of energy to remove four electrons from its outermost shell.
Therefore, carbon shares its valence electrons to complete its octet with other atoms to form covalent bonds.

Question 7.
List the common physical properties of carbon compounds.
Answer:

• They have covalent bonds between their atoms therefore they do not form ions. So they are poor conductors of electric current.
• These compounds have low melting and low boiling points.
• They are generally insoluble in water but soluble in the organic solvents like ether, carbon- tetrachloride, etc.

Question 8.
Draw the structures of diamond and graphite.
Answer:
In diamond, each carbon atom is bonded to four other carbon atoms forming a rigid three dimensional structure.

In graphite, each carbon atom is bonded to three other carbon atoms in the same plane giving a hexagonal array. One of these bonds is a double bond.

Question 9.
Write the general IUPAC names of alcohol, carboxylic acid, aldehyde and ketone.
Answer:

Compound	General IUPAC name
Alcohol	Alkanol
Carboxylic acid	Alkanoic acid
Aldetjyde	Alkanal
Ketone	Alkanone

Question 10.
Ethane , Ethene, Ethanoic acid, Ethyne, Ethanol
From the box given above, name:
(i) The compound with —OH as a part of its structure.
(ii) The compound with —COOH as a part of its structure.
(iii) Gas used in welding.
(iv) Homologue of the homologous series with general formula C_nH_2n+2.
Answer:
(i) Ethanol
(ii) Ethanoic acid
(iii) Ethyne
(iv) Ethane

Question 11.
Give two uses each of methyl alcohol and ethyl alcohol.
Answer:
Uses of ethyl alcohol:

• It is used in the manufacture of dyes, perfumes, antiseptics, etc.
• It is used in alcoholic drinks.

Uses of methyl alcohol:

• It is used as a solvent
• It is used as an antifreeze

Question 12.
List two main points of difference between organic and inorganic compound.
Answer:
Organic compounds:

1. They are made up of few elements (C, H, O, N, S) through covalent bonds.
2. They are combustible.

Inorganic compounds:

1. They are made up of all the known elements, which involve ionic bond.
2. They are generally non-combustible.

Question 13.
What are substitution reactions? Justify your answer with a suitable example.
Answer:
A chemical reaction in which atom(s) or group of atoms of an organic compound is/are replaced by other atom(s) or group of atoms without any change in the rest of the molecule is called a substitution reaction.
For example,
CH₄ + Cl₂ → CH₃Cl + HCl
In this chemical reaction Cl atom substitutes one hydrogen atom from methane.

Question 14.
Give any four uses of ethanoic acid.
Answer:

1. Ethanoic acid is used in making synthetic vinegar.
2. Ethanoic acid is used as a reagent in chemistry laboratory.
3. Ethanoic acid is used for making dyes, perfumes and esters.
4. Ehanoic acid is used for coagulating rubber from latex and casein from milk.

Question 15.
List four differences between soaps and detergents.
Answer:
Soaps:

1. Soaps are sodium salts of higher fatty acids.
2. Biodegradable.
3. Soaps cannot be used in acidic medium.
4. Soaps cannot be used in hard water.

Synthetic Detergents:

1. Detergents are sodium alkyl sulphates or sodium alkyl benzene sulphonates with alkyl group having more than ten carbon atoms.
2. Non-biodegradable.
3. They can be used in acidic medium.
4. Detergents can be used even in hard water.

Extra Questions for Class 10 Science Chapter 4 Short Answer Type II

Question 1.
What is the role of metal or reagents written on arrow’s in the given chemical reactions?

[NCERT Exemplar]
Answer:
(a) Ni acts as a catalyst.
(b) Concentrated H₂SO₄ acts as a catalyst.
(c) Alkaline KMnO₄ acts as an oxidising agent.

Question 2.
The molecular formula of an organic compound X is C₂H₄O₂ which has vinegar like smell.
(i) Identify the compound.
(ii) Write its chemical formula and name.
(iii) What happens when sodium bicarbonate is added into it?
Answer:
(i) The organic compound X is acetic acid.
(ii) Chemical formula: CH₃COOH
IUPAC name: Ethanoic acid
(iii) Ethanoic acid produces effervescence with sodium bicarbonate liberating carbon dioxide gas.

Question 3.
Write one chemical equation to represent each of the following types of reactions of organic substances.
(a) Esterification
(b) Saponification
(c) Substitution   [CBSE 2011]
Answer:
(a) Esterification. Ethanol reacts with ethanoic acid on warming in presence of a few drops of conc. H₂SO₄ to form a sweet smelling ester, ethyl ethanoate.

(b) Saponification. The alkaline hydrolysis of esters is known as saponification as this reaction is used for the preparation of soaps.

(c) Substitution. The reaction, in which one (or more) hydrogen atoms of a hydrocarbon are replaced by some other atoms, is called substitution reaction.

4. Write chemical equations for what happens when
(a) Sodium metal is added to ethanoic acid
(b) Solid sodium carbonate is added to ethanoic acid
(c) Ethanoic acid reacts with a dilute solution of sodium hydroxide.
Answer:
(a) 2Na + 2CH₃COOH → 2CH₃COONa + H₂(g)
(b) Na₂CO₃ + 2CH₃COOH → 2CH₃COONa + CO₂ (g)+ H₂O
(c) CH₃COOH + NaOH → CH₃COONa + H₂O

Question 5.
Name the oxidising agent used for the conversion of ethanol to ethanoic acid. Distinguish between ethanol and ethanoic acid on the basis of (i) litmus test, (ii) reaction with sodium hydrogen carbonate.
Answer:
Alkaline KMnO₄ or acidified potassium dichromate (K₂Cr₂O₇) are used for the conversion of ethanol to ethanoic acid.
Ethanol:

• Ethanol has no effect on red or blue litmus solution.
• No gas evolved when ethanol is treated with sodium hydrogen carbonate.

Ethanoic acid:

• Ethanoic acid turns blue litmus solution red.
• Ethanoic acid reacts with sodium hydrogen carbonate to evolve carbon dioxide along with the formation of salt and water.
  CH₃COOH + NaHCO₃ → CH₃COONa + CO₂ + H₂O

Question 6.
(a) Differentiate between alkanes and alkenes. Name and draw the structure of one member of each.  [CBSE 2013]
(b) Alkanes generally burn with clean flame. Why?
Answer:
Alkanes:

1. An alkane is a hydrocarbon in which the carbon atoms are connected by only single covalent bond.
2. General formula of alkane is C_nH_2n+2.
3. The simplest alkane is methane (CH₄).
   
4. Alkanes generally burn in air with a blue and non-sooty flame.
5. Alkanes undergo substitution reactions.
6. Alkanes do not decolourise red brown colour of bromine water.

alkenes:

1. An alkene is an unsaturated hydrocarbon in which the two carbon atoms are connected by a double bond.
2. General formula of alkene is C_nH_2n.
3. The simplest alkene is ethene (C₂H₄).
   
4. Alkenes burn in air with a yellow and sooty flame.
5. Alkenes undergo addition reactions.
6. Alkenes decolourise bromine water.

(b) Alkanes burn in air with a blue and non-sooty flame because the percentage of carbon in the alkane is comparatively low which gets oxidised completely by oxygen present in air.

Question 7.
What happens when
(а) ethanol is burnt in air?
(b) ethanol is heated with excess cone. H₂SO₄ at 443 K?
(c) a piece of sodium is dropped into ethanol?  [CBSE 2013]
Answer:
(a) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O + Heat + Light

(c) 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂(g)

Question 8.
What is meant by homologous series of carbon compounds? Write the general formula of
(i) alkenes, and
(ii) alkynes. Draw the structures of the first member of each series to show the bonding between the two carbon atoms.   [CBSE 2014]
Answer:
Homologous series:
A series of carbon compounds in which the same functional group substitutes for hydrogen on a carbon chain is called a homologous series. There is a difference of –CH₂ in the molecular formulae of two nearest compounds of a homologous series. Each such series has same general molecular formula and has a general scientific name. There is a difference of 14 u (unified mass) in the molecular masses of two nearest compounds of a series.

Members of homologous series of aldehydes:
H – CHO Methanal
CH₃ – CHO Ethanal
C₂H₅ – CHO Propanal

General formula:
(i) Alkenes, C_nH_2n
(ii) Alkynes, CnH_2n-2

Structure:

• The first member of alkenes is ethene and its structure is
  
• The first member of alkynes is ethyne and its structure is H—C ≡ C—H

Question 9.
With the help of an example, explain the process of hydrogenation.
Mention the essential conditions for the reaction and state the change in physical property with the formation of the product.   [CBSE 2015]
Answer:
Process of hydrogenation:

The addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydrocarbon is called hydrogenation.

Essential conditions for the reaction are:

• Presence of an unsaturated hydrocarbon.
• Presence of a catalyst such as nickel (Ni) or palladium.

Changes observed:

• Change observed in the physical property is the change of unsaturated compound from the liquid state to saturated compound in the solid state.
• The boiling or melting points of a product is increased.

Question 10.
The list of some organic compounds is given below:
Ethanol, ethane, methanol, methane, ethyne, ethene
From the above list, name a compound:
(i) formed by the dehydration of ethanol by conc. H₂SO₄.
(ii) which will give red precipitate with ammoniacal cuprous chloride solutions,
(iii) which forms ethanoic acid on oxidation with KMnO₄.
(iv ) which has vapour density 14 and decolourises pink alkaline potassium permanganate.
(v) which forms chloroform on halogenation in the presence of sunlight.
(vi) which decolourises bromine solution in carbon tetrachloride.
Answer:
(i) Ethene
(ii) Ethyne
(iii) Ethanol
(iv) Ethene
(v) Methane
(vi) Ethene

Question 11.
The molecule of alkene family are represented by a general formula C_nH_2n. Now answer the following:
(i) What do n and 2n signify?
(ii) What is the name of alkene when n = 4?
(iii) What is the molecular formula of alkene when n = 6?
(iv) What is the molecular formula of the alkene if there are six H-atoms in it?
(v) What is the molecular formula and structural formula of the first member of the alkene family?
(vi) Write the molecular formulae of lower and higher homologues of an alkene which contains four carbon atoms.
Answer:
(i) n indicates number of carbon atoms and 2n indicates number of hydrogen atoms.
(ii) Butene
(iii) C₆H₁₂
(iv) C₃H₆

(vi) Lower homologue —C₃H₁₆, Higher homologue – C₅H₁₀

Question 12.
Copy and complete the following table which relates to three homologous series of hydrocarbons:

Answer:

Extra Questions for Class 10 Science Chapter 4 Long Answer Type

Question 1.
Write the names of the following compounds:(a) Pentanoic acid (c) Heptanal (e) Methyl ethanoate

Answer:
(a) Pentanoic acid
(b) But-l-yne or Butyne
(c) Heptanal
(d) Pentanol
(e) Methyl ethanoate

Question 2.
(а) What are hydrocarbons? Give examples.
(b) Give the structural differences between saturated and unsaturated hydrocarbons with two examples
(c) What is a functional group? Give examples of four different functional groups. [NCERT Exemplar]
Answer:
(a) Hydrocarbons are compounds of carbon and hydrogen. For example, ethane, ethene, benzene, etc.

(b) Saturated compounds are saturated by means of number of bonds or they contains only single bonds. For example,
CH₃—CH₃ (ethane), CH₃CH₂CH₃ (propane)
Unsaturated hydrocarbon contains multiple bonds (C=C or C≡C)
For example, CH≡CH (ethyne) and CH₂=CH₂ (ethene).

(c) Functional group is an atom or group of atoms which imparts certain characteristic properties to the organic compound.

Question 3.
(a) Distinguish between esterification and saponification reactions of organic compounds.
(b) With a labelled diagram describe an activity to show the formation of an ester.
Answer:
(a) Esterification:
When carboxylic acid reacts with alcohol in the presence of a little concentrated sulphuric acid to form ester, the reaction is called esterification.

Saponification:
When an ester is heated with sodium hydroxide solution then the esters get hydrolysed to form alcohol and sodium salt of carboxylic acid. This alkaline hydrolysis of esters is known as saponification as it is used in the preparation of soap.

(b) Activity to show the formation of an ester:

1. 1 ml ethanol and 1 mL glacial acetic acid along with a few drops of concentrated sulphuric acid are taken in a test tube.
2. The mixture is allowed to warm in a water-bath for at least five minutes.
3. The hot mixture of the test tube is poured into a beaker containing 20-50 ml of water.
4. A sweet smelling substance called ester is formed.

 

Question 4.
List in tabular form three physical and two chemical properties on the basis of which ethanol and ethanoic acid can be differentiated.   (CBSE 2012)
Answer:

Question 5.
Give one example each of:
1. Cracking
2. Hydrogenation
3. Dehydration
4. Substitution reaction
5. Addition reaction.
Answer:
1. Cracking

2. Hydrogenation

3. Dehydration

4. Substitution reaction

5. Addition reaction

Carbon and its Compounds HOTS Questions With Answers

Question 1.
A compound X is formed by the reaction of a carboxylic acid C₂H₄O₂ and an alcohol in presence of a few drops of H₂SO₄. The alcohol on oxidation with alkaline KMnO₄ followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of (a) carboxylic acid, (b) alcohol and (c) the compound X. Also write the reaction.   (NCERT Exemplar)
Answer:
The available information suggests that the alcohol which gives the same carboxylic acid upon oxidation has two carbon atoms. It is therefore ethanol (C₂H₅OH). The structures of the different compounds are:

Question 2.
Carbon, Group (14) element in the Periodic Table, is known to form compounds with many elements.
Write an example of a compound formed with
(a) chlorine (Group 17 of Periodic Table)
(b) oxygen (Group 16 of Periodic Table)  (NCERT Exemplar)
Answer:
(a) Carbon tetrachloride (CCl₄)
(b) Carbon dioxide (CO₂)

Question 3.
A salt X is formed and a gas is evolved when ethanoic acid reacts with sodium hydrogen carbonate. Name the salt X and the gas evolved. Describe an activity and draw the diagram of the apparatus to prove that the evolved gas is the one which you have named. Also, write chemical equation of the reaction involved.   (NCERT Exemplar)
Answer:

X is sodium ethanoate. Gas evolved is carbon dioxide (CO₂).

Activity:

1. Take sodium hydrogen carbonate in a test tube and add 2 ml ethanoic acid in it.
2. Carbon dioxide gas is evolved with brisk effervescence.
3. Pass the gas through lime water, it will turn milky. This shows that the gas evolved is carbon dioxide (CO₂).

Question 4.
A compound C (molecular formula, C₂H₄O₂) reacts with Na metal to form a compound R and evolves a gas which bums with a pop up sound. Compound C on treatment with an alcohol A in presence of an acid forms a sweet smelling compound S (molecular formula, C₃H₆O₂). On addition of NaOH to C, it also gives R and water. S on treatment with NaOH solution gives back R and A.
Identify C, R, A, S and write down the reactions involved.   (NCERT Exemplar)
Answer:
C – Ethanoic acid
R – Sodium salt of ethanoic acid (sodium acetate) and gas evolved is hydrogen
A – Methanol
S – Ester (Methyl acetate)

Question 5.
Look at figure given on the next page and answer the following questions:
(a) What change would you observe in calcium hydroxide solution taken in test tube B?
(b) Write the reaction involved in test tubes A and B respectively.
(c) If ethanol is given instead of ethanoic acid, would you expect the same change?
(d) How can a solution of lime water be prepared in the laboratory?   (NCERT Exemplar)

Answer:
(a) It will become milky.

(b) In test tube A, ethanoic acid react with sodium carbonate to form sodium ethanoate along with carbon dioxide. The gas is evolved accompanied by brisk effervescence.
CH₃COOH + Na₂CO₃ → CH₃COONa + CO₂ + H₂O

In test tube B, calcium hydroxide reacts with carbon dioxide to form a milky solution of calcium carbonate.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

(c) No, it would be different. No chemical reaction is possible between ethanol and sodium carbonate.

(d) Lime water is prepared by keeping a suspension of calcium hydroxide overnight in a beaker. The solution is decanted and is transferred to another beaker. It contains traces of calcium hydroxide and is called lime water.

Question 6.
An organic compound A on heating with concentrated H₂SO₄ forms a compound B which on addition of one mole of hydrogen in presence of Ni forms a compound C. One mole of compound C on combustion forms two moles of CO₂ and 3 moles of H₂O. Identify the compounds A, B and C and write the chemical equations of the reactions involved.  (NCERT Exemplar)
Answer:
Since compound C gives 2 moles of CO₂ and 3 moles of H₂O, it shows that it has the molecular formula C₂H₆ (ethane). C is obtained by the addition of one mole of hydrogen to compound B, so the molecular formula of B should be C₂H₄ (ethene). Compound B is obtained by heating compound A with concentrated H₂SO₄ which shows it to be an alcohol. So compound A could be C₂H₅OH (Ethanol).

Extra Questions for Class 10 Science Chapter 4 Value Based Questions

Question 1.
Ethanol is one of the most important industrial chemicals. It is used in medicine, to synthesise many important compounds and as an excellent solvent.
However, inspite of its benefits it causes many social problems. If a person drinks alcohol regularly, he becomes an alcoholic. Alcohol is non-toxic but it produces physiological effects disturbing brain activity. These persons are also a threat to the lives of others.
(a) Give three reasons in favour and three reasons against ‘alcohol-free world’.
(b) ‘Alcohol drinking should not be portrayed on media’. Give valid reasons to justify.
(c) As a student what initiative would you take in the concern of “We should condemn drinking alcohol”,
Answer:
(a) In favour of‘Alcohol-free world’:

• Alcohol drinking lowers inhibitions which leads to increased violence and crime in the society.
• A liver disease ‘cirrhosis’ caused by alcohol can lead to death.
• Drunken driving leads to increased road accidents.

Against ‘Alcohol-free world’:

• Alcohol is used for making some medicines like cough syrups, tincture iodine, some tonics, etc.
• Mixed with petrol, it is now being used as a fuel for light vehicles.
• It is used for making antifreeze material for cooling engines of vehicles.

(b) ‘Alcohol drinking should not be portrayed on media’ because young people and children are greatly influenced by the media.

(c) Initiatives taken by a student to create awareness about drinking alcohol could be:

• By writing slogan
• Through debates
• By writing articles
• By role plays/skits

Question 2.
Intake of small quantity of methanol can be lethal. Comment.  (NCERT Exemplar)
Answer:
Methanol is oxidised to methanal in the liver. Methanal reacts with the component of the cells. It causes the protoplasm to coagulate. It also affect the optic nerve, due to which it causes blindness.

Here we are providing How I Taught My Grandmother to Read Extra Questions and Answers Class 9 English Literature Reader, Extra Questions for Class 9 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-english/

How I Taught My Grandmother to Read Extra Questions and Answers Class 9 English Literature

How I Taught My Grandmother to Read Extra Questions and Answers Short Answer Type

Answer the following questions briefly.

How I Taught My Grandmother To Read Question And Answers Pdf Question 1.
Why did the narrator and the other people in the village wait eagerly for the bus?
Answer:
The narrator and the other people waited eagerly because it brought the papers, the weekly magazine, and post.

How I Taught My Grandmother To Read Question Answers Question 2.
How do we know that her grandparents’ village was very remote?
Answer:
The morning paper was received in the afternoon, the weekly magazine came a day late, and the transport system was not good.

How I Taught My Grandmother Question Answer Question 3.
Who speaks the line: “The happiness Kashi” (para 3). What does the line show about the lady’s
character?
Answer:
These lines are spoken by the old lady in the novel Kashi Yatre. It shows how wise, magnanimous, and compassionate the lady was. She sacrifices her deepest desire to go to Kashi for the happiness of an orphan girl. This act of hers expresses her selflessness and her humane quality which is that the highest ideal of love does not expect anything in return.

How I Taught My Grandmother To Read Questions And Answers Question 4.
Why had the grandmother not gone to school?
Answer:
The grandmother had lost her mother as a young girl and her father had married her off at a very young age. Very soon, she had children and there had been no time to go to school. Also, girls were rarely sent to school in those days.

How I Taught My Grandmother To Read Summary Questions And Answers Question 5.
Why was the grandmother so upset when her granddaughter went away to attend a wedding?
Answer:
The grandmother felt frustrated and was upset because she could not read the next episode of her favourite story that was serialized in the magazine, thus, she had to wait for her granddaughter to return and read it out.

Question Answers Of How I Taught My Grandmother To Read Question 6.
How do we know that the grandmother was determined to read?
Answer:
We know that the grandmother was determined to read as it was her who approached her granddaughter to teach her, keeping her embarrassment aside. She also set a deadline by which she would learn to read, and started the process from the very next day.

Question 7.
How did the granddaughter react to her grandmother’s request to teach her to read?
Answer:
At first, she made fun of her but seeing her grandmother’s determination to overcome all obstacles, she started teaching her in earnest.

Question 8.
What were the gifts exchanged by the granddaughter and grandmother?
Answer:
The grandmother gifted the narrator frock material while the granddaughter gifted her the novel “Kashi Yatre” which had been published in the form of a book.

Question 9.
Why did the grandmother touch the feet of her granddaughter?
Answer:
The grandmother wanted to show her respect for her granddaughter who had acted as her teacher and taught ‘ her to read and write.

Question 10.
Explain the statement “student had passed with flying colours”.
Answer:
When the grandmother read the title of the book effortlessly, it proved that she had succeeded in her quest to learn to read and write thus proving she had passed with flying colours.

How I Taught My Grandmother to Read Extra Questions and Answers Long Answer Type

Question 1.
You are the grandmother from the story. Write a diary entry expressing your feelings on the day that you finally learn to read and write.
Answer:
8th January, 20xx, Wednesday
Today I am very happy as I have finally learnt to read and write. All the credit for this goes to my darling granddaughter. What a loving and thoughtful girl she is! She is only twelve but so wise! She taught me with so much love and affection that within a short time, I achieved my goal.

Today I have been able to read “Kashi Yatre” myself. I just can’t believe it. I, an old lady of 62, am finally independent! I can read whatever I wapt, and never feel embarrassed about being uneducated.

Over the last few weeks, as I practiced reading and writing under the guidance of my granddaughter, I often recalled my childhood. If only I had been able to go to school then, I would not have lost so many valuable years of reading. Even though I have been happy and busy all these years, with my marriage and children and household duties, the happiness and satisfaction that I have experienced during the last few weeks and especially today, are unmatched. Today I have received wealth that is duly invaluable—thanks to my little teacher!

Question 2.
You are a neighbour of the grandmother from the story. You have just witnessed the touching ceremony where the grandmother touched her young teacher’s feet, and have decided to write to your friend in another village, sharing what happened and how you feel about it.
Answer:
12 March 20xx
XYZ
Dear Krisna
How are you? Hope you are fine with your family. We are all fine here too.

I have written to give you a startling piece of news. Do you remember my neighbour Krishtakka? Well she has recently learnt how to read and write! it seems her granddaughter who is hardly 12 years old taught her how to read and write in these last three months. They had a very moving ceremony on Saraswati Puja during Dasara.The grandmother touched the feet of the granddaughter and gave her a gift! We were all shocked but later on we realised that it was an appropriate gesture as the girl had become her teacher and to show respect to your teacher is the right thing to do! I wish I had the same determination and the enthusiasm, she has been encouraging all of us to study as well.

According to her, it is never too late and she says, “for a good cause, if you are determined you can overcome any obstacle!” I have decided to join the classes that the grandmother and granddaughter are holding for anyone in the village who wants to study. What do you think? Do write and let me know.
Your friend always,
Janaki

Question 3.
Describe the character of the grandmother.
Answer:
The grandmother was a very cheerful woman and totally involved in bringing up her children and grandchildren. Though she had not been educated herself, she knew the value of education and made sure that all her children were well-educated. She was also very determined. She wanted to learn how to read and write. She did not feel embarrassed to approach her granddaughter and ask her to teach her.

She was very hard-working as we see that she worked very hard to learn how to read and write, often redoing her work to learn within the deadline that she had set for herself. We also see that she was not self-conscious and did not feel any embarrassment in touching her granddaughter’s feet as she had assumed the role of her teacher. She was very wise and understood the value of education thus she said, “We are well off, but what use is money when I cannot be independent.”

Question 4.
After reading the story, what do you think is the role of a teacher and what do you think is the duty of a student?
(Encourage students to think creatively and formulate their own answers.)
Answer:
In the story, we find that each person has a role and a duty to accomplish. A teacher’s goal is to educate his students and impart his knowledge whereas the duty the students hold is to acquire the skills and knowledge imparted. We learn from the story that the narrator has taken up the responsibility of teaching her grandmother. Although she finds it amusing in the beginning, she begins to see her grandmother’s determination.

Thus, her sole responsibility became imparting her knowledge of alphabets to her grandmother while showing compassion to understand her student’s problem. The duty of the student is however best described in the act the grandmother displays. She bends down to wash the feet of her granddaughter. Not only was she a humble student, she was diligent and determined.

How I Taught My Grandmother to Read Extra Questions and Answers Reference to Context

Read the extracts given below and answer the questions that follow.

Question 1.
“Her style was easy to, read and very convincing. Her stories usually dealt with complex psychological problems in the lives of ordinary people and were always interesting. ”
(a) Whom does “her” in the above lines refers to
Answer:
Triveni, the popular writer.

(b) What does the narrator mean when she says, ‘Her style’? What is she referring to?
Answer:
The narrator is talking about the style of writing that the writer Triveni uses. Triveni wrote stories that were convincing, easy to read and talked about complex psychological problems in the lives of ordinary people.

(c) Find a word from the extract that means to make someone believe in something.
Answer:
Convincing

Question 2.
“My grandmother too never went to Kashi, and she identified herself with the novel’s protagonist. So more than anybody else she was the one most interested in knowing what happened next in the story and used to insist that I read the serial out to her. ”
(a) In what way did the grandmother relate to the central character of the story “Kashi Yatre”?
Answer:
The grandmother could relate to the central character of the story “Kashi Yatre” because they both had a – strong desire to visit Kashi.

(b) Why did the grandmother not read the serial herself?
Answer:
The grandmother could not read the serial by herself as she could not read or write; grandmother was illiterate.

(c) The leading character in a drama, film or novel is called a
Answer:
Protagonist.

Question 3.
“After hearing what happened next in Kashi Yatre, she wouldjoin her friends at the temple courtyard where we children would also gather to play hide and seek. She would discuss the latest episode with her friends. At that time, I never understood why there was so much of debate about the story. ”
(a) What is “Kashi Yatre”?
Answer:
“Kashi Yatre” is a novel written by the Kannada writer, Triveni, which appeared as a serial in the Kannada weekly, Karmaveera.

(b) Why did the women at the temple discuss the latest episode of “Kashi Yatre”?
Answer:
The women at the temple would discuss the latest episode of “Kashi Yatre” as they could relate with the protagonist of the serial.

(c) Can you give different words that have the same meaning as ‘debate’
Answer:
discussion, argument.

Question 4.
“When I came back to my village, I saw my grandmother in tears. I was surprised, for I had never seen her cry even in the most difficult situations. What had happened? I was worried. ”
(a) Where had the narrator gone?
Answer:
The narrator had gone for a wedding to the neighbouring village.

(b) Why did the granddaughter find her grandmother in tears on her return?
Answer:
The granddaughter found her grandmother in tears on her return as the grandmother had been unable to read the story “Kashi Yatre” on her own.

(c) What kinds of emotion were expressed in this extract? What is the word/phrase used to describe sadness?
Answer:
Sadness, surprise/shock, anxiety were expressed in this extract. The phrase ‘in tears’ and the word ‘cry’ describes sadness.

Question 5.
“My grandmother never went to school, so she could not read. ”
(a) How did the grandmother know the details of the story published in her favourite magazine?
Answer:
The grandmother got to know the details of the story published in the favourite magazine as her granddaughter used to read it out to her.

(b) Why had the grandmother not gone to school?
Answer:
The grandmother had not gone to school because in her times people did not think it was important to educate girls.

(c) Can you think of a Word that has the same meaning as ‘not being able to read’?
Answer:
illiterate.

Question 6.
“Awa came and sat next to me. Her affectionate hands touched my forehead. I realised she wanted to speak. ”
(a) What does the term Avva mean? What did Awa tell the narrator?
Answer:
The term Awa means mother. Awa told her about her childhood, why she never went to school and asked her to teach her how to read and write.

(b) What kind of relationship did Awa and the narrator share?
Answer:
The relationship between the narrator and her grandmother was very close, warm, and loving.”

(c) Can you explain the phrase “affectionate hands”?
Answer:
The phrase affectionate hands means that the grandmother was showing her affection and care with her hands as she gently touched her granddaughter’s head.

Question 7.
“She was a good looking lady who was usually smiling. Even today I cannot forget the worried expression on her face. ”
(a) Why was “she” upset?
Answer:
The grandmother was upset because she realised that she was helpless and dependent on others to read and write.

(b) What quality in “her” character is brought out by this episode?
Answer:
Her determination becomes evident through this episode.

(c) What does the word ‘expression’ mean in this context?
Answer:
In this context, the word expression means ‘a look on someone’s face’

Question 8.
“For a good cause if you are determined you can overcome any obstacle. ”
(a) What is the “good cause” mentioned here?
Answer:
The good cause mentioned here is learning how to read and write.

(b) What light does this remark throw on the character of the speaker?
Answer:
This remark brings out the fact that she is determined and hardworking.

(c) Can you find a word that means a ‘hindrance’ or ‘hurdle’ in the extract?
Answer:
Obstacle

Question 9.
“Then she did something unusual. She bent down and touched my feet. I was surprised and taken aback”
(a) What was unusual about the gesture?
Answer:
It was unusual that a grandmother was touching the feet of a granddaughter, as the custom is that the younger ones touch the feet of the elders as a mark of respect.

(b) Why did she behave in this manner?
Answer:
This gesture on her part was to honour and thank the person who had taught her to read and”write.

(c) What does the narrator mean when she said she was taken aback?
Answer:
When the narrator said she was taken aback, she meant that she was startled.

Question 10.
“Now I am independent. It is my duty to respect a teacher. ”
(a) Who is the speaker?
Answer:
The speaker of the extract is the grandmother.

(b) How had she become independent?
Answer:
The grandmother had become independent as she no longer had to depend on others to read or write.

(c) How did she show her respect for her teacher?
Answer:
The grandmother expressed her respect for her teacher by touching her feet.

Question 11.
“Secretly I bought Kashi Yatre which had been published as a novel by that time. ”
(a) What was Kashi Yatre?
Answer:
“Kashi Yatre” was a novel.

(b) Why had the narrator bought it “secretly”?
Answer:
The narrator bought it “secretly” as she wanted it to be a surprise.

(c) Why did she buy “Kashi Yatre” for her?
Answer:
The narrator bought “Kashi Yatre” for her grandmother as her grandmother loved the story and she wanted to give her a gift for completing her studies.

Question 12.
“It is a great tradition but today the reverse has happened. ”
(a) What does the term “great tradition” refer to?
Answer:
The “great tradition” refers to the tradition of touching the feet of elders.

(b) What does the narrator mean by “reverse”?
Answer:
In the given extract the narrator meant that the opposite had happened as the grandmother touched the feet of the granddaughter.

(c) Why had this happened?
Answer:
The “reverse” had happened as the granddaughter had taught her grandmother to read and write.

Here we are providing Polynomials Class 10 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-maths/

Extra Questions for Class 10 Maths Polynomials with Answers Solutions

Extra Questions for Class 10 Maths Chapter 2 Polynomials with Solutions Answers

Polynomials Class 10 Extra Questions Very Short Answer Type

The graphs of y = p(x) for some polynomials (for questions 1 to 4) are given below. Find the number of zeros in each case.

Polynomials Class 10 Extra Questions Question 1.

Answer:
There is no zero as the graph does not intersect the X-axis.

Polynomial Class 10 Extra Questions Question 2.

Answer:
The number of zeros is four as the graph intersects the X-axis at four points.

Polynomials Class 10 Extra Questions With Answers Question 3.

Answer:
The number of zeros is three as the graph intersects the X-axis at three points.

Class 10 Polynomials Extra Questions Question 4.

Answer:
The number of zeros is three as the graph intersects the X-axis at three points.

Class 10 Maths Chapter 2 Extra Questions Question 5.
What will the quotient and remainder be on division of ax² + bx + c by px² + qx² + rx + 5, p ≠ 0?
Answer:
0, ax² + bx + C.

Polynomials Class 10 Questions With Answers Question 6.
If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p(x) and g(x)?
Answer:
Since the quotient is zero, therefore
deg p(x) < deg g(x)

Polynomials Extra Questions Class 10 Question 7.
Can x – 2 be the remainder on division of a polynomial p(x) by x + 3?
Answer:
No, as degree (x – 2) = degree (x + 3)

Extra Questions Of Polynomials Class 10 Question 8.
Find the quadratic polynomial whose zeros are -3 and 4.
[NCERT Exemplar]
Answer:
Sum of zeros = -3 + 4 = 1,
Product of zeros = – 3 x 4 = -12
∴ Required polynomial = x² – x – 12

Extra Questions On Polynomials Class 10 Question 9.
If one zero of the quadratic polynomial x² – 5x – 6 is 6 then find the other zero.
Answer:
Let α,6 be the zeros of given polynomial.
Then α + 6 = 5 3 ⇒ α = -1

Extra Questions For Class 10 Maths Chapter 2 Question 10.
If both the zeros of the quadratic polynomial ax² + bx + c are equal and opposite in sign, then find the value of b.
Answer:
Let α and -α be the roots of given polynomial.
Then α + (-α) = 0 ⇒ \(-\frac{b}{a}=0\) ⇒ b = 0.

Polynomial Extra Questions Class 10 Question 11.
What number should be added to the polynomial x² – 5x + 4, so that 3 is the zero of the polynomial?
Answer:
Let f(x) = x² – 5x + 4
Then f(3) = 3² – 5 x 3 + 4 = -2
For f(3) = 0, 2 must be added to f(x).

Polynomials Extra Questions Question 12.
Can a quadratic polynomial x² + kx + k have equal zeros for some odd integer k > 1?
Answer:
No, for equal zeros, k = 0,4 ⇒ k should be even.

Class 10 Maths Polynomials Extra Questions Question 13.
If the zeros of a quadratic polynomial ax² + bx + c are both negative, then can we say a, b and c all have the same sign? Justify your answer.
Answer:
Yes, because \(-\frac{b}{a}\) = sum of zeros < 0, so that \(\frac{b}{a}=0\) > 0. Also the product of the zeros = \(\frac{c}{a}=0\) > 0.

Chapter 2 Maths Class 10 Extra Questions Question 14.
If the graph of a polynomial intersects the x-axis at only one point, can it be a quadratic polynomial?
Answer:
Yes, because every quadratic polynomial has at the most two zeros.

Extra Questions Of Polynomials Class 10 With Solutions Question 15.
If the graph of a polynomial intersects the x-axis at exactly two points, is it necessarily a quadratic polynomial?
Answer:
No, x⁴ – 1 is a polynomial intersecting the x-axis at exactly two points.

Polynomials Class 10 Extra Questions Short Answer Type 1

Question 1.
If one of the zeros of the quadratic polynomial f(x) = 4x² – 8kx – 9 is equal in magnitude but opposite in sign of the other, find the value of k.
Answer:
Let one root of the given polynomial be α.
Then the other root = -α
Sum of the roots = (-α) + α = 0
⇒ \(-\frac{b}{a}\) = 0 or \(-\frac{8k}{4}\) = 0 or k = 0

Question 2.
If one of the zeros of the quadratic polynomial (k – 1)x² + kx + 1 is -3 then find the value of k.
Answer:
Since – 3 is a zero of the given polynomial
∴ (k – 1)(-3)² + k(-3) + 1 = 0 :
⇒ 9k – 9 – 3k + 1 = 0 ⇒ k = 4/3.

Question 3.
If 1 is a zero of the polynomial p(x) = ax² – 3(a – 1)x -1, then find the value of a.
Answer:
Put x = 1 in p(x)
∴ p(1) = a(1)² – 3(a – 1) x 1 – 1 = 0
⇒ a – 3a + 3 – 1 = 0 ⇒ 2a = -2 ⇒ a = 1

Question 4.
If α and β are zeros of polynomial p(x) = x² – 5x + 6, then find the value of α + B – 3aß.
Answer:
Here, α + β = 5, αβ = 6
= α + β – 3αβ = 5 – 3 x 6 = -13

Question 5.
Find the zeros of the polynomial p(x) = 4x² – 12x + 9.
Answer:
p(x) = 4x² – 12x + 9 = (2x – 3)²
For zeros, p(x) = 0
⇒ (2x – 3)(2x – 3) = 0 ⇒ x = \(\frac{3}{2}, \frac{3}{2}\)

Question 6.
If one root of the polynomial p(y) = 5y² + 13y + m is reciprocal of other, then find the value of m.
Answer:

Question 7.
If α and β are zeros of p(x) = x² + x – 1, then find \(\frac{1}{\alpha}+\frac{1}{\beta}\)
Answer:
Here, α + β = -1, αβ = -1,
So \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-1}{-1}=1\)

Question 8.
Given that one of the zeros of the cubic polynomial ax³ + bx² + cx + d is zero, find the product of the other two zeros.
Answer:
Let α, β, γ be the roots of the given polynomial and α = 0.
Then αβ + βγ + γα = c/a ⇒ βγ = c/a

Question 9.
If the product of two zeros of the polynomial p(x) = 2x³ + 6x² – 4x + 9 is 3, then find its third zero.
Answer:
Let α, β, γ be the roots of the given polynomial and αβ = 3
Then αβγ = \(-\frac{9}{2}\)
⇒ 3 x γ = \(\frac{-9}{2}\) or γ = \(\frac{-3}{2}\)

Question 10.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.
(i) \(-\frac{1}{4}, \frac{1}{4}\)
(ii) \(\sqrt{2}, \frac{1}{3}\)
Answer:
Let α, β be the zeros of polynomial.
(i) We have, α + β = \(-\frac{1}{4}\) and αβ = \(\frac{1}{4}\)
Thus, polynomial is
p(x) = x2 – (a + B) x + aß

Quadratic polynomial 4x² + x + 1

(ii) We have, α + β = √2 and  αβ = \(\frac{1}{3}\)
Thus, polynomial is p(x) = x² – (α + β) x + αβ
= x² – √2x + \(\frac{1}{3}\) = \(\frac{1}{3}\) (3x² – 3√2x + 1)
Quadratic polynomial = 3x² – 3√2x + 1

Polynomials Class 10 Extra Questions Short Answer Type 2

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients (Q. 1 – 2).

Question 1.
6x² – 3 – 7x
Answer:
We have, p(x) = 6x² – 3 – 7x
p(x) = 6x² – 7x – 3
(In general form)
= 6x² – 9x + 2x – 3 = 3x (2x – 3) + 1 (2x – 3)
= (2x – 3) (3x + 1)
The zeros of polynomial p(x) is given by
p(x) = 0) = (2x – 3) (3x + 1) = 0 ⇒ \(x=\frac{3}{2},-\frac{1}{3}\)
Thus, the zeros of 6x² – 7x – 3 are α = \(-\frac{3}{2}\) and β = \(-\frac{1}{3}\)
Now, sum of the zeros = α + β = \(\frac{3}{2}-\frac{1}{3}\) = \(\frac{9-2}{6}=\frac{7}{6}\)

Question 2.
4u² + 8u
Answer:
We have, p(u) = 4u² + 8u = p(u) = 4u (u + 2)
The zeros of polynomial p(u) is given by
p(u) = 0 ⇒ 4u (u + 2) = 0 .
∴ u = 0, -2
Thus, the zeros of 4u² + 8u are α = 0 and β = -2
Now, sum of the zeros = α + β = 0 – 2 = -2

Question 3.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2 (ii) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
Answer:
(i) We have,

Clearly, remainder is zero, so x2 + 3x + 1 is a factor of polynomial 3x⁴ + 5x³ – 7x² + 2x + 2

(ii) We have,

Clearly, remainder is zero, so t’ – 3 is a factor of polynomial 2t⁴ + 3t³ – 2t² – 9t – 12.

Question 4.
If α and β are the zeros of the quadratic polynomial f(x) = 2x² – 5x + 7, find a polynomial whose zeros are 2α + 3β and 3α + 2β.
Answer:
Since α and β are the zeros of the quadratic polynomial f(x) = 2x² – 5x + 7

Let S and P denote respectively the sum and product of the zeros of the required polynomial.

Question 5.
What must be subtracted from p(x) = 8x⁴ + 14x³ – 2x² + 7x – 8 so that the resulting polynomial
is exactly divisible by g(x) = 4x² + 3x – 2?
Answer:
Let y be subtracted from polynomial p(x)
: 8x⁴ + 14x³ – 2x² + 7x – 8 – y is exactly divisible by g(x)
Now,

∵ Remainder should be 0.
∴ 14x – 10 – y = 0 or 14x – 10 = y or y = 14x – 10
∴ (14x – 10) should be subtracted from p(x) so that it will be exactly divisible by g(x)

Question 6.
What must be added to f(x) = 4x⁴ + 2x³ – 2x² + x – 1 so that the resulting polynomial is divisible
by g(x) = x² + 2x – 3?
Answer:
By division algorithm, we have
f(x) = g(x) × q(x) + r(x)
= f(x) – r(x) = g(x) × q(x) ⇒ f(x) + {-r(x)} = g(x) × q(x)
Clearly, RHS is divisible by g(x). Therefore, LHS is also divisible by g(x). Thus, if we add –r(x) to f(x), then the resulting polynomial is divisible by g(x). Let us now find the remainder when f(x) is divided by g(x).

∴ r(x) = -61x + 65 or -r(x) = 61x – 65
Hence, we should add –r(x) = 61x – 65 to f(x) so that the resulting polynomial is divisible by g(x).

Question 7.
Obtain the zeros of quadratic polynomial 3x² – 8x + 4√3 and verify the relation between its zeros and coefficients.
Answer:
We have,

Question 8.
If α and β are the zeros of the polynomial 6y² – 7y + 2, find a quadratic polynomial whose zeros are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\)
Answer:
Let p(y) = 6y² – 7y + 2

Question 9.
If one zero of the polynomial 3x² – 8x + 2k + 1 is seven times the other, find the value of k.
Answer:
Let α and β be the zeros of the polynomial. Then as per question β = 7α

Question 10.
If one zero of the polynomial 2x² + 3x + λ is 1/2 find the value of and other zero.
Answer:
Let P(x) = 2x² + 3x + λ

Question 11.
If one zero of polynomial (a² + 9)x² + 13x + 6a is reciprocal of the other, find the value of a.
Answer:
Let one zero of the given polynomial be α.
Then, the other zero is 1/α
∴ Product of zeros = α × \(\frac{1}{\alpha}\) = 1
But, as per the given polynomial product of zeros = \(\frac{6 a}{a^{2}+9}\)
∴ \(\frac{6 a}{a^{2}+9}\) = 1 ⇒ a² + 9 = 6a
⇒ a² – 6a + 9 = 0) ⇒ (a – 3)² = 0
⇒ a – 3 = 0 ⇒ a = 3
Hence, a = 3.

Question 12.
If the polynomial (x⁴ + 2x³ + 8x² + 12x + 18) is divided by another polynomial (x² + 5), the remainder comes out to be (px +q). Find values of p and q.
Answer:
Let f(x) = (x⁴ + 2x³ + 8x² + 12x + 18) and g(x) = (x² + 5)
On dividing f(x) by g(x), we get

Now, px + 9 = 2x + 3 ⇒ p = 2,q = 3 (By comparing the coefficient of x and constant term).

Polynomials Class 10 Extra Questions Long Answer Type 1

Question 1.
Verify that the numbers given alongside the cubic polynomial below are their zeros. Also verify the relationship between the zeros and the coefficients.
x³ – 4x² + 5x – 2; 2,1,1
Solution:
Let p(x) = x³ – 4 x² + 5x – 2
On comparing with general polynomial px) ax³ + bx² + cx + d, we get a = 1, b = -4, c = 5 and d = -2
Given zeros 2, 1, 1.
∴ p(2) = (2)³ – 4(2)² + 5(2) – 2 = 8 – 16 + 10 – 2 = 0
and p(1) = (1)³ – 4(1)² + 5(1) – 2 = 1 – 4 + 5 – 2 = 0
Hence, 2, 1 and I are the zeros of the given cubic polynomial.
Again, consider α = 2, β = 1, γ = 1
∴ α + 13 + y = 2 + 1 + 1 = 4

Question 2.
Find a cubic polynomial with the sum of the zeros, sum of the products of its zeros taken two at a time, and the product of its zeros as 2, -7, -14 respectively.
Solution:
Let the cubic polynomial be p(x) = ax³ + bx² + cx + d. Then

p(x) = a[x³ + (-2)x² + (-7)x + 14] ⇒ p(x) = a[x3 – 2x² – 7x + 14]
For real value of a = 1, p(x) = x³ – 2x² – 7x + 14

Question 3.
Find the zeros of the polynomial f(x) = x3 – 5x² – 2x + 24, if it is given that the product of its two zeros is 12.
Solution:
Let α, β and γ be the zeros of polynomial (fx) such that αβ = 12.

Now, α + β + γ = 5 α + β – 2 = 5
= α + β = 7 a = 7 – β
= (7 – β) β =12 ⇒ 7β – β² – 12
= β² + 7β + 12 = 0 ⇒ β² – 3β – 4β + 12 = O
= β = 4 or β = 3
β = 4 or β = 3
∴ α = 3 or α = 4

Question 4.
If the remainder on division of x3 – kx² + 13x – 21 by 2x – 1 is -21, find the quotient and the value of k. Hence, find the zeros of the cubic poIyncmia1 x³ – kx² + 13x.
Solution:
Let f(x) = x³ – kx² + 13x – 21

Question 5.
Obtain all other zeros of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeros are \(\sqrt{\frac{5}{3}}\) and \(\sqrt{\frac{5}{3}}\).
Solution:

Question 6.
Given that √2 is a zero of the cubic polynomial 6x³ + √2x² – 10x – 4√2, find its other zeros.
Solution:
The given polynomial is f(x) = (6x³ +√2x² – 10x – 4√2). Since √2 is the zero of f(x), it follows that (x – √2) is a factor of f(x).
On dividing f(x) by (x – √2), we get

Polynomials Class 10 Extra Questions HOTS

Question 1.
If α, β, γ bezerosofpo1ynomia1 6x³ + 3x² – 5x + 1, then find die value of α^-1 + β^-1 + γ^-1.
Solution:
∵ p(x) = 6x³ + 3x² – 5x + 1 so a = 6, b = 3, c = -5, d = 1
∴ α, β and γ are zeros of the polynomial p(x).

Question 2.
Find the zeros of the polynomial f(x) = – 12x² + 39x – 28, if it is given that the zeros are in AP.
Solution:
If α, β, γ are in AP., then,
β – α = γ + β ⇒ 2β = α + γ
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-(-12)}{1}\) = 12 ⇒ α + γ = 12 – β …….. (i)
From (i) and (ii)
2β = 12 – β or 3β = 12 or β = 4
Putting the value of β in (i), we have
8 = a + γ
αβγ = – \(\frac{d}{a}\) = \(\frac{-(-28)}{1}\) = 28 …….. (iii)
(αγ) 4 = 28 or αγ = 7 or γ = \(\frac{7}{α}\) ….. (iv)
Putting the value of γ = \(\frac{7}{α}\) in (iii), we get
⇒ 8 = α + \(\frac{7}{α}\) ⇒ 8α = α2 + 7
⇒ α² – 8α + 7 = 0 ⇒ α² – 7α – 1α + 7 = 0
⇒ α(α – 7)-1 (α – 7) = 0 ⇒ (α – 1)(α – 7) = 0
⇒ α = 1 or α = 7

Question 3.
If the polynomial f(x) = x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a. Find k and a.
Solution:
By division algorithm, we have Dividend = Divisor × Quotient + Remainder
⇒ Dividend – Remainder = Divisor × Quotient
⇒ Dividend – Remainder is always divisible by the divisor.
When f(x) = – 6x³ + 16x² – 25x + 10 is divided by x² – 2x + k the remainder comes out to be x + a.
∴ f(x) – (x + a) = x⁴ – 6x³+ 16x² – 25x + 10 – (x + a)
= x⁴ – 6x³ + 16x² – 25x + 10 – x – a x⁴ – 6x³ + 16x² – 26x + 10 – a
is exactly divisible by x² – 2x + k
Let us now divide x⁴ – 6x³ + 16x² – 26x + 10 – a by x² – 2x + k.

For f(x) – (x + a) = x⁴ – 6x³ + 16x² – 26x + 10 – a to be exactly divisible by x² – 2x + k, we must
have (-10 + 2k)x + (10 – a – 8k + k²) = 0 for all x
= – 10 + 2k = 0 and 10 – a – 8k + k² = 0
⇒ k = 5 and 10 – a – 40 + 25 = 0
⇒ k = 5 and a – 5

In this page, we are providing Sources of Energy Class 10 Extra Questions and Answers Science Chapter 14 pdf download. NCERT Extra Questions for Class 10 Science Chapter 14 Sources of Energy with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 14 Extra Questions and Answers Sources of Energy

Extra Questions for Class 10 Science Chapter 14 Sources of Energy with Answers Solutions

Extra Questions for Class 10 Science Chapter 14 Very Short Answer Type

Sources Of Energy Class 10 Extra Questions And Answers Question 1.
Define fossil fuel.
Answer:
Fossil fuels were formed millions of years ago, when plants and animal remains got buried under the earth and were subjected to high temperature and pressure conditions. For example, Coal, Petroleum, etc.

Sources Of Energy Class 10 Questions And Answers Question 2.
Write down the disadvantages of fossil fuels.
Answer:
These fossil fuels are non-renewable sources of energy and cause environmental problems due to pollution.

Sources Of Energy Class 10 Important Questions And Answers Question 3.
Define Nuclear energy.
Answer:
Nuclear energy: Energy released when some changes take place in the nucleus of the atom of a substance, is called Nuclear energy.

Questions On Sources Of Energy Class 10 Question 4.
Define solar cell.
Answer:
Solar cell is a device that converts solar energy into electricity.

Source Of Energy Class 10 Questions And Answers Question 5.
Define non-conventional sources of energy.
Answer:
Non-conventional sources of energy are those which are not used as the conventional ones and meet our energy requirement only on a limited scale.
Examples:

• Tidal energy
• Geothermal energy.

Sources Of Energy Extra Questions Question 6.
Define conventional sources of energy.
Answer:
Conventional sources of energy are those which are used extensively and meet a major portion of our energy requirement.
Examples:

• Fossil fuels
• Hydropower plants.

Sources Of Energy Questions And Answers Question 7.
How is charcoal produced and what is the advantage of charcoal in comparison to wood?
Answer:
When wood is burnt in a limited supply of oxygen, the volatile materials present in it get removed, and charcoal is produced. Charcoal bums without flames and is comparatively less smoky.

Extra Questions for Class 10 Science Chapter 14 Short Answer Type I

Extra Questions Of Chapter 14 Class 10 Science Question 1.
Write down the characteristics of a good fuel.
Answer:
Characteristics of a good fuel:

• Has high calorific value
• Produces less smoke
• Produces less residue after burning
• Easy availability
• Inexpensive
• Easy to store and transport

Class 10 Sources Of Energy Question Answer Question 2.
What is a thermal power plant?
Answer:
In a thermal power plant

• Coal, petroleum and natural gas is used to produce thermal electricity.
• Electricity transmission is very efficient.
• The steam produced by burning fossil fuels runs the turbine to produce electricity.

Source Of Energy Class 10 Important Questions Question 3.
Write down the advantages and disadvantages of a nuclear power plant.
Answer:
Advantage of Nuclear energy:

• Alternative source of energy due to depletion of fossil fuels.
• From a small amount of fuel, a large amount of energy is released.

Disadvantages of nuclear energy:

• Difficult to store and dispose nuclear waste which may cause environmental contamination.
• High cost of setting up a nuclear plant.
• Limited availability of raw material (uranium).

Ch 14 Science Class 10 Extra Questions Question 4.
Describe energy harnessed from wind and write down its advantages and disadvantages.
Answer:
Wind energy:

• It can converted into mechanical and electrical energy.
• Kinetic energy of the wind is used for running windmills, which can be used to lift water, grind grains, etc.

Advantages:

• Eco friendly
• Renewable

Disadvantages:

• Wind speed not uniform always.
• Needs a large area to erect series of windmills.
• Big amount of investment is needed.
• Output is less as compared to investment.

Extra Questions for Class 10 Science Chapter 14 Short Answer Type II

Sources Of Energy Class 10 Questions And Answers Pdf Question 1.
How can solar energy be harnessed? Mention any two limitations of using solar energy. Write down the advantages of solar cell.
Answer:
Solar energy:
Solar radiations can be converted into electricity through solar cells (photovoltaic cells). Photovoltaic cells convert solar radiations directly into electricity through silicon solar cells. Solar cells are arranged on large flat sheets to form a solar panel.

Advantages of solar cell:

• Solar cell have no moving parts, require little maintenance and work quite satisfactorily.
• They can be setup in remote and inaccessible hamlets or very sparsely inhabited areas.

Limitations:

• Solar cells are expensive.
• Solar devices are only useful during day time and on a sunny day.

Sources Of Energy Questions Question 2.
Write two different ways of harnessing energy from the ocean.
Answer:
Energy from the sea:
1. Tidal energy: Locations in India – Gulf of Kutch, Gujarat and West Bengal

• Depends upon harnessing the rise and fall of sea level due to tidal action.
• Dams are constructed across a narrow part of sea the and a turbine converts tidal energy into electrical energy.

Disadvantage: Uniform tidal action is not seen.

2. Wave energy:

• Kinetic energy of the sea waves are used to rotate turbines.
• These turbines generate electrical energy.

Question 3.
Write a short note on geothermal energy.
Answer:

• Energy harnessed from the heat of the Earth is called geothermal energy.
• Magma is formed when this heat melts the rocks. The molten rocks and hot gases are called magma.
• The magma gets collected at some depths below the earth’s surface. These places are called “hot spots’.
• When underground water comes in contact with these hot spots, it changes into steam, which can be used to generate electricity.

Advantages of geothermal energy:

• Renewable.
• Cost of production is not much.

Disadvantages of geothermal energy:

• Only few sites available for harnessing energy.
• Expensive to set up.

Extra Questions for Class 10 Science Chapter 14 Long Answer Type

Question 1.
What is biomass? Explain the principle and working of a biogas plant using a labelled schematic diagram. Write down its advantages.
Answer:
Biomass is a source of conventionally used fuels that are used in our country, e.g., cow dung cakes, fire-wood, coal, charcoal.

Biogas: It is a mixture of gases produced during the decomposition of biomass in the absence of oxygen.
Methane is the major component of biogas. (Biogas contains 75% methane, carbon dioxide, hydrogen and hydrogen sulfide).

Biogas plant: Animal dung, sewage, crop residues, vegetable wastes, poultry droppings, etc. are used to produce biogas in biogas plants.

Construction and working:
The plant has a dome-like structure built with bricks. A slurry of cow dung and water is made in the mixing tank from where it is fed into the digester. The digester is a sealed chamber in which there is no oxygen. Anaerobic microorganisms, that do not require oxygen, decompose or breakdown complex compound of cow dung slurry and produce methane, carbon dioxide, hydrogen and hydrogen sulfide.

Advantages of biogas

• It bums without smoke and leaves no residue.
• Its heating capacity is high.
• The slurry of biogas plant is used as an excellent manure, rich in nitrogen and phosphorus.

Question 2.
Write down construction, advantages and limitations of a solar cooker.
Answer:
Solar cooker:
Construction:
Outer surface of the solar cooker is painted with black colour and a glass plate is used as cover.
Black colour absorbs more heat and the glass plate traps solar radiation by greenhouse effect. Mirror reflects the light.

Advantages of solar cookers

• Eco friendly
• Renewable
• Used in rural areas.
• Retains all the nutrients in food due to slow cooking.

Disadvantages of solar cooker

• Silicon cells are expensive.
• Solar radiations are not uniform over the Earth’s surface.
• Cannot be used at night or on cloudy days.
• Cannot be used to make chapattis for frying as these require a temperature of 140°C or more. (Maximum temperature of 100°C only can be achieved in a solar cooker.)

Question 3.
What is hydroelectric power plant? Write down its advantages and disadvantages.
Answer:
Hydroelectric power plant: A conventional renewable source of energy is obtained from water falling from a great height. Dams are constructed to collect water flowing in high altitude rivers. The stored water has a lot of potential energy. When water is allowed to fall from a height, potential energy changes to kinetic energy, which rotates the turbines to produce electricity.

Advantages:

• It is clean and non polluting source of energy.
• Hydropower is renewable source of energy.

Disadvantages:

• Highly expensive to construct.
• Dams cannot be constructed on all river sites.
• Large areas of human habitation and agricultural fields get submerged.
  People face social and environmental problems.

Sources of Energy HOTS Questions With Answers

Question 1.
What is the role of a plane mirror and a glass sheet in a solar cooker?
Answer:
Plane mirror reflects sunlight so that maximum sunlight can enter the box.
Glass sheet traps solar radiation by greenhouse effect.

Question 2.
What kind of mirror is used in concentrating type-solar cooker?
Answer:
A concave mirror is used in concentrating type solar cooker so that maximum heat can be concentrated at a given point.

Question 3.
Name the process that produces a large amount of energy in the sun.
Answer:
Nuclear fusion

Question 4.
Name the materials used for making solar cells.
Answer:
Silicon, Germanium and Selenium

Extra Questions for Class 10 Science Chapter 14 Value Based Questions

Question 1.
A. Reddy visited his grandfather’s village Koodankular with his younger sister. They saw people protesting against building a nuclear power plant there. Reddy’s sister asked the reason for the protests. Reddy said that it could be due to the risks involved in setting up the plant near people’s habitation.
(i) Write down the advantages and disadvantages of a nuclear power plant. ‘
(ii) Values shown by A. Reddy.
(iii) Which process is used to harness nuclear energy? Explain briefly.
Answer:
(i) Advantages of Nuclear energy:

• Alternative source of energy due to depletion of fossil fuels.
• From a small amount of fuel, a large amount of energy is released.

Disadvantages of Nuclear energy:

• Hazardous nature of nuclear waste and risk of nuclear waste leakage.
• High cost of setting up a nuclear plant.
• Displacement of people from their habitat.

(ii) A. Reddy is a very intelligent person.

(iii) Nuclear energy is produced by a process called nuclear fission. During this process, the nucleus of a heavy atom (such as uranium, plutonium or thorium) when bombarded with low-energy neutrons, can be split apart into lighter nuclei. The process releases a tremendous amount of heat energy. The released energy can be used to produce steam which is used to produce electricity.

Question 2.
During summer vacations Bhaskar visited his uncle’s village. He saw a biogas plant installed by villagers for their basic energy needs.
(i) What is biogas and biogas plant?
(ii) What values are shown by villagers?
Answer:
(i) Biogas: It is a mixture of gases produced during decomposition of biomass in the absence of oxygen. Methane is the major component of biogas.
Biogas plants: Animal dung, sewage, crop residues, vegetable wastes and, poultry droppings are used to produce biogas in biogas plants.

(ii) Villagers are intelligent and care for the environment.

Question 3.
Iatin aggarwal took admission in DTO (Delhi Technological University). He saw solar devices were installed in the university.
(i) Write down a short note on solar energy and solar cooker.
(ii) Write down the values shown by the university management.
Answer:
(i) (a) Solar energy:

• Solar radiations can be converted to electricity through solar cells (photovoltaic cells).
• Photovoltaic cells convert solar radiations directly into electricity through silicon solar cells.
• Solar cells arranged on a large flat sheet to form a solar panel.

(b) Solar cooker:
Solar cookers are painted black from outside and covered with a large glass plate to trap solar radiations by greenhouse effect.

(ii) University management is concerned about energy conservation.

In this page, we are providing Human Eye and Colourful World Class 10 Extra Questions and Answers Science Chapter 11 pdf download. NCERT Extra Questions for Class 10 Science Chapter 11 Human Eye and Colourful World with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 11 Extra Questions and Answers Human Eye and Colourful World

Extra Questions for Class 10 Science Chapter 10 Human Eye and Colourful World with Answers Solutions

Extra Questions for Class 10 Science Chapter 11 Very Short Answer Type

Class 10 Science Chapter 11 Extra Questions Question 1.
Why is the colour of the clear sky blue? (NCERT Exemplar)
Answer:
Colour of the clear sky is blue: The molecules of air and other fine particles in the atmosphere have size smaller than the wavelength of visible light.

When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour more strongly than red.

Human Eye And The Colourful World Class 10 Extra Questions And Answers Question 2.
Why do stars appear to twinkle?
Answer:
Stars appears to twinkle due to atmospheric refraction.

Human Eye Class 10 Extra Questions Question 3.
Define farthest point of an eye.
Answer:
Farthest point of an eye: The farthest point upto which the eye can see objects clearly is called far point of the eye. It is infinity for normal eye.

Human Eye And Colourful World Extra Questions Question 4.
Define power of accommodation.
Answer:
Power of accommodation: The ability of the eye lens to adjust its focal length is called accommodation.

Human Eye And The Colourful World Extra Questions Question 5.
Define least distance of distinct vision.
Answer:
Least distance of distinct vision: Minimum distance at which an object can be seen distinctly without any strain from the normal eye, i.e., 25 cm for normal vision.

Human Eye And Colourful World Class 10 Extra Questions Question 6.
Define Tyndall effect.
Answer:
Tyndall effect: The phenomenon of scattering of light by colloidal particles gives rise to Tyndall effect.
Tyndall effect can be observed when sunlight passes through a canopy of a dense forest. Here tiny droplets in mist scatters light.

Class 10 Science Ch 11 Extra Questions Question 7.
Define atmospheric refraction.
Answer:
Atmospheric refraction: If physical conditions of the refracting medium (air) are not stationary, the apparent position of the object fluctuates.

Extra Questions Of Human Eye And The Colourful World Question 8.
Why are danger signal lights red in colour?
Answer:
Danger signal lights are red in colour because red colour is least scattered by fog or smoke.

Extra Questions for Class 10 Science Chapter 11 Short Answer Type I

Cbse Class 10 Physics Human Eye And Colourful World Extra Questions Question 1.
What is meant by advance sunrise and delayed sunset? Draw a labelled diagram to explain these phenomena.
Answer:
Advance sunrise and delayed sunset is due to atmospheric refraction.
When the sun is slightly below the horizon, the sunlight coming from the less dense (vacuum) to more dense (air) medium is refracted downwards. So the sun appears to be above the horizon.

Similarly, even after actual sunset, the sun can be seen for a few minutes due to refraction of sunlight.

Class 10 Human Eye Extra Questions Question 2.
Explain formation of rainbow.
Answer:

Rainbow formation: A rainbow is a natural spectrum appearing in the sky after rain shower. It is caused by dispersion of sunlight by tiny water droplets, present in the atmosphere. The water droplets act like small prism. They refract and disperse the incident sunlight, then reflect it internally and finally refract it again.
Due to dispersion of light and internal reflection different colours appear.

Chapter 11 Science Class 10 Extra Questions Question 3.
Explain the refraction of light through a triangular glass prism using a labelled ray diagram. Hence define the angle of deviation. (NCERT Exemplar)
Answer:
Refraction of light through prism

PE – Incident ray
EF – Refracted ray
FS – Emergent ray
∠A – Angle of the prism
∠i – Angle of incidence
∠r – Angle of refraction
∠e – Angle of emergence
∠D – Angle of deviation

Refraction of light through a triangular glass prism

1. The refraction of light takes place at two surfaces firstly when light enters from air to prism and secondly when light emerges from prism.
2. Angle of prism: The angle between the two lateral faces of the prism is called angle of prism.
3. Angle of deviation: The angle between incident ray (produced forward) and emergent ray (produced backward).

Extra Questions for Class 10 Science Chapter 11 Short Answer Type II

Human Eye And The Colourful World Class 10 Questions And Answers Question 1.
What is the difference in colours of the sun observed during sunrise/sunset and noon? Give explanation for each.
Answer:
In the morning and evening, the sun lies near the horizon. Sunlight travels through a larger distance in the atmosphere and most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelength. This gives rise to the reddish appearance of the sun.
At noon sun appears white as only a little of blue and voilet colours are scattered.

Extra Questions Of Chapter 11 Science Class 10 Question 2.
Define the term dispersion of white light. Name the colour of light which bends (i) the most, (ii) the least, while passing through a glass prism. Draw a ray diagram to justify your answer.
Answer:
Dispersion of white light by a glass prism
Dispersion: The splitting of light into its component colours is called dispersion.
The red light bends the least while violet bends the most.
Spectrum: The band of the coloured components of a light beam is called spectrum.
i.e., VIBGYOR

The Human Eye And The Colourful World Extra Questions Question 3.
Explain twinkling of stars.
Answer:
Twinkling of stars:

• The twinkling of stars is due to atmospheric refraction of starlight.
• When starlight enters the earth’s atmosphere, it suffers refraction continuously. Since the physical conditions of the earth’s atmosphere are not stationary the stars appear twinkling.

Extra Questions for Class 10 Science Chapter 11 Long Answer Type

Human Eye And Colourful World Class 10 Questions With Answers Question 1.
List four common refraction defects of vision. Suggest the way of correcting these defects. (CBSE 2014)
Answer:
Defects of vision:
(i) Cataract: Crystalline lens of people at old age becomes milky and cloudy. This condition is called cataract.
It is possible to restore vision through cataract surgery.

(ii) Myopia: (Near sightedness)
A person with myopia can see nearby objects clearly but cannot see distant objects clearly.

Cause:

• Due to excessive curvature of the eye lens.
• Elongation of the eyeball.

Correction:
Concave lens of suitable power.
(a) Far point of myopic eye

(b) Myopic eye

(c) Correction for myopia

(a), (b) The myopic eye, and (c) correction for myopia with a concave lens

(iii) Hypermetropia (far-sightedness)
A person with hypermetropia can see distant objects clearly but cannot see nearby objects distinctly.

Cause:

• The focal length of the eye lens is too long.
• The eyeball has become too small.

Correction:
Convex lens of suitable power.



(a), (b) The hypermetropic eye, and (c) correction for hypermetropia

(iv) Presbyopia
The power of accommodation of the eye usually decreases with ageing. In this eye defect it is difficult to see nearby objects comfortably and distinctly without corrective eye glasses.

Cause:
Weakening of ciliary muscles and diminishing flexibility of eye lens.

Correction:
By using Bifocal lens: Upper portion consists of concave lens and lower part is convex lens.

Question 2.
Explain the structure and functioning of the human eye. How are we able to see nearby as well as distant objects?
Answer:

• Cornea: A thin membrane through which light enters the eye, maximum refraction occurs at the outer surface of cornea.
• Iris: A dark muscular membrane which controls size of pupil.
• Pupil: Regulates and controls the amount of light entering the eye.
• Eye lens: Composed of fibrous, jelly like material, with adjustable curvature, forms an inverted and real image of object at retina.
• Retina: It is a light sensitive screen on which image is formed.

The power of accommodation, that is, the ability of the eye lens to adjust its focal length, help us to see near and far objects clearly.

Human Eye and Colourful World HOTS Questions With Answers

Question 1.
How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.
Answer:
When an inverted prism is kept a little distance away from the prism causing dispersion or basically in the path of splitted beam, the spectrum recombines to form white light.

Recombination of the spectrum of white light

Question 2.
Is the position of a star as seen by us in its exact position? Justify your answer. (NCERT Exemplar)
Answer:
No, the starlight, on entering the earth’s atmosphere, undergoes refraction continuously before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position.

Question 3.
Why do we see a rainbow in the sky only after rainfall?
Answer:
The rainbow in the sky appears only after rainfall because the suspended water drops behave like prism, and refract, disperse and reflect the light rays internally.

Question 4.
A person needs a lens of power – 4.5 D for correction of his/her vision.
(a) What kind of defect in vision is he/she suffering from?
(b) What is the focal length of the corrective lens?
(c) What is the nature of the corrective lens?
Answer:
(a) Myopia
(b) f = \(\frac{1}{P}\) = \(\frac{100}{4.5}\) = 22.2 cm
(c) Concave lens

Question 5.
A narrow beam PQ of white light is passing through a glass prism ABC as shown in the diagram. (CBSE 2014)

Trace it on your answer sheet and show the path of the emergent beam as observed on the screen DE.
(i) Write the name and cause of the phenomenon observed.
(ii) Where else in nature is this phenomenon observed?
(iii) Based on this observation, state the conclusion which can be drawn about the constituents of white light.
Answer:
(i) Dispersion of light, because different colours of light bends through different angles.
(ii) Rainbow formation
(iii) White light contains seven colours i.e., VIBGYOR

Question 6.
On which factor does colour of scattered light depends?
Answer:
The colour of the scattered light depends on the size of the scattering particles. Very fine particles scatter mainly blue light while particle of larger size scatters light of longer wavelengths.

Extra Questions for Class 10 Science Chapter 11 Value Based Questions

Question 1.
Akshay, sitting in the last row in his class, could not see clearly the words written on the blackboard. When the teacher noticed it, he asked if any student sitting in the front row would volunteer to exchange his seat with Akshay. Salman immediately agreed to exchange his seat with Akshay. Akshay could now see the words written on the blackboard clearly. The teacher thought it fit to send message to Akshay’s parents advising them to get his eyesight checked.
In the context of the above event, answer the following questions:
(a) Which defect of vision is Akshay suffering from? Which type of lens is used to correct this defect?
(b) State the values displayed by the teacher and Salman.
Answer:
(a) Myopia, concave lens
(b) Teacher is very caring and knowledgeable. Salman has great concern for his friend, Akshay, and is very helpful.

Question 2.
On the rainy day, Ram reached his grandfather’s place in village. On the way to the house he saw a beautiful rainbow in the sky. In the night, he saw lots of twinkling stars in the clear sky. He was very excited to see these beautiful natural phenomenon, which he was not able to see in the city, where he lived with his father. Explain the phenomenon on the basis of science. Do you think that pollution in the atmosphere affects the formation of rainbow and twinkling of stars. Do you agree with the fact that pollution free environment will strengthen such natural phenomenon in the cities as well. Elaborate.
Answer:

• The twinkling of star is due to atmospheric refraction.
• The formation of rainbow is due to dispersion, refraction and internal reflection.
• Yes, pollution in atmosphere affects the formation of rainbow and twinkling of stars.

Question 3.
Vinay’s father cannot read a book placed 25 cm from his eye. But he feels a little comfortable when the book is placed 50 cm away. Vinay adviced his father for checkup of the eyes.
(a) From which defect of vision Vinay’s father may be suffering from? Give the proper correction.
(b) State the values of Vinay.
Answer:
(a) Hypermetropia, correction can be done by using suitable convex lens.
(b) Vinay is caring and intelligent.

Question 4.
Ankit’s grandma is facing a problem of clouded, blurred and dim vision. Ankit took her to the doctor.
(a) From which defect of vision Ankit’s grandma may be suffering from? Give proper correction.
(b) State the values of Ankit.
Answer:
(a) Cataract. It is possible to restore vision through cataract surgery.
(b) Ankit is caring and intelligent.

In this page, we are providing Periodic Classification of Elements Class 10 Extra Questions and Answers Science Chapter 5 pdf download. NCERT Extra Questions for Class 10 Science Chapter 5 Periodic Classification of Elements with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 5 Extra Questions and Answers Periodic Classification of Elements

Extra Questions for Class 10 Science Chapter 5 Periodic Classification of Elements with Answers Solutions

Extra Questions for Class 10 Science Chapter 5 Very Short Answer Type

Periodic Classification Of Elements Class 10 Extra Questions With Answers Question 1.
Name two elements whose properties were predicted on the basis of their positions in Mendeleev’s periodic table.
Answer:
Gallium (Eka-aluminium) and germanium (Eka-silicon).

Periodic Classification Of Elements Extra Questions Question 2.
Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the elements predicted by Mendeleev.  [NCERT Exemplar]
Answer:
Eka-aluminium represents gallium (Ga) with valency three and Eka-silicon is for germanium (Ge) with valency four. The formulae of their respective chlorides are GaCl₃ and GeCl₄.

Class 10 Science Chapter 5 Extra Questions Question 3.
How many elements are there in period 2?
Answer:
Eight (8).

Periodic Classification Of Elements Class 10 Extra Questions Question 4.
Name all the elements present in group-17 of the Modern Periodic Table.
Answer:
F, Cl, Br, I, At.

Periodic Classification Of Elements Class 10 Questions And Answers Question 5.
Name two elements whose atomic weight were corrected on the basis of their position in Mendeleev’s periodic table.
Answer:
Gold (Au) and platinum (Pt).

Extra Questions Of Periodic Classification Of Elements Question 6.
Arrange the following elements in the increasing order of their metallic character.
Mg, Ca, K, Ge, Ga     [NCERT Exemplar]
Answer:
Ge < Ga Mg < Ca < K.

Class 10 Periodic Classification Of Elements Extra Questions Question 7.
Name the scientist who proposed the Modern Periodic Law.
Answer:
Henry Moseley, a scientist, proposed the Modern Periodic Law.

Chapter 5 Science Class 10 Extra Questions Question 8.
What is meant by a group in the periodic table?
Answer:
Groups are the vertical columns/vertical lines in the periodic table.

Ch 5 Science Class 10 Extra Questions Question 9.
What is meant by a period in the periodic table?
Answer:
Periods are the horizontal rows in the periodic table.

Class 10 Science Ch 5 Extra Questions Question 10.
Name the elements present in period 1 of the modern periodic table.
Answer:
Hydrogen (H) and Helium (He).

Extra Questions Of Chapter 5 Science Class 10 Question 11.
An element A is in group II (group 2) of the periodic table:
(а) What will be the formula of its chloride?
(b) What will be the formula of its oxide?
Answer:
(a) ACl₂
(b) AO.

Extra Questions Of Periodic Classification Of Elements Class 10 Question 12.
Why are the elements calcium, strontium and barium named as alkaline earths?
Answer:
These elements are called alkaline earths because their oxides are alkaline in nature and exist in the earth.

Extra Question Of Periodic Classification Of Elements Question 13.
Give the name and electronic configuration of second alkali metal.
Answer:
The second alkali metal is sodium (Na). Its electronic configuration is 2, 8, 1.

Class 10 Chapter 5 Science Extra Questions Question 14.
What are Metalloids?
Answer:
Those elements which show some properties of both metals and non-metals are called metalloids. Examples: Boron, Silicon, Germanium, Arsenic, Antimony, Tellurium and Polonium.

Periodic Classification Extra Questions Question 15.
Name three elements whose atomic masses were correct on the basis of their positions in Mendeleev’s periodic table.
Answer:

1. Beryllium (Be, group II A)
2. gold (Au, group II B)
3. platinum (Pt, group VIII).

Question 16.
Name three elements which behave as metalloids.
Answer:
The elements are:

1. arsenic (As)
2. antimony (Sb)
3. germanium (Ge).

Question 17.
Name the inert gas which has two electrons in its valence shell.
Answer:
Helium.

Question 18.
Name the most metallic and most non-metallic element in the Periodic Table.
Answer:
The most metallic elements is francium (group 1) and most non-metallic element is fluorine (group 17).

Question 19.
In the Modern Periodic Table, which are the metals among the first ten elements?
Answer:
Lithium (Li), Beryllium (Be) and Boron (B) are the metal elements among first ten elements in the Modern Periodic Table.

Question 20.
Two elements A and B belong to the same period. What is common in them?
Answer:
They have the same number of shells.

Question 21.
The electronic configuration of an element is 2, 8, 6. Identify the element and name of the family to which it belongs.
Answer:
The element with configuration 2, 8, 6 (Z = 16) is sulphur. It belongs to the oxygen family.

Question 22.
Calcium, Strontium and Barium form a Dobereiner’s triad. The atomic masses of calcium and Barium are 40 and 137 respectively. Predict the atomic mass of strontium.
Answer:
According to Dobereiner’s triad, atomic mass of strontium is the arithmetic means of Ca and Ba as strontium lies between Ca and Ba.
Atomic mass of Sr = \(\frac{40+137}{2}=\frac{177}{2}=88.5\)

Question 23.
The formula of magnesium oxides is MgO. Write the formula of magnesium chloride.
Answer:
Oxygen is divalent in nature. The valency of magnesium in magnesium oxide is + 2. The formula of magnesium chloride is MgCl₂ since chlorine has valency equal to one.

Question 24.
State whether the following statement is true or false?
The valency of an element of group 17 is 7.
Answer:
This statement is wrong. The number of valence electrons in an element of group 17 = 17 – 10 = 7.
Therefore, the valency of the element = 8 – 7 = 1.

Question 25.
Name two elements whose atomic weight were corrected on the basis of their position in Mendeleev periodic table.
Answer:

1. Gold (Au)
2. Platinum (Pt).

Question 26.
The two isotopes of chlorine have atomic mass 35.4 and 37.4. Should they be placed in separate slots in the modern periodic table?
Answer:
No, they should be placed in the same slot because the modern periodic table is based on the atomic number of the elements.

Question 27.
List any two properties of the elements belonging to the first group of the modem periodic table.   [Delhi 2014]
Answer:
Elements of first group have one valence electron each in their atoms. All the elements of group 1 have the same valency of 1.

Question 28.
What is the basic difference in the electronic configuration of the elements belonging to group 1 and group 2?
Answer:
All elements belonging to group 1 have one electron in the valence shell while all elements belonging to group 2 have two electrons in their valence shell.

Question 29.
Element ‘Y’ with atomic number 3 combines with element ‘A’ with atomic number 17. What will be the formula of the compound?
Answer:
The electronic distribution in elements ‘A’ and ‘Y’ are 2, 1 and 2, 8, 7 respectively. Both have valency equal to 1. The formula of the compound is AY.

Question 30.
Name the property which remains unchanged on descending a group in the periodic table.
Answer:
Valence electrons.

Question 31.
An element X is in group 13 of the periodic table. What is the formula of its oxide?
Answer:
X₂O₃.

Question 32.
Elements ‘X’ and TP belong to groups 1 and 17 of the periodic table respectively. What will be the nature of the bond in the compound XY?
Answer:
Ionic bond.

Extra Questions for Class 10 Science Chapter 5 Short Answer Type I

Question 1.
The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called? Give one example of such a set of elements.   [NCERT Exemplar]
Answer:
The arrangement of these elements is known as Dobereiner triad. Example, Lithium, sodium and Potassium.

Question 2.
In group I of the periodic table, three elements X, Y and Z have atomic radii 1.33 A, 0.95 A and 0.60 A respectively. Arrange the elements (X, Y and Z) in the increasing order of atomic number and mention a suitable reason for it.
Answer:
In a given group, the atomic radii increases on moving down the group. Therefore, the arrangement of the given elements in the increasing order of their atomic numbers is as follows: Z, Y, X.

Question 3.
Elements have been arranged in the following sequence on the basis of their increasing atomic masses.
F, Na, Mg, Al, Si, P, S, Cl, Ar, K
(a) Pick two sets of elements which have similar properties.
(b) The given sequence represents which law of classification of elements?   [NCERT Exemplar]
Answer:
(a) F and Cl are first and eighth element in the above sequence, therefore, they have similar properties.
Please note that although Na and K have similar properties but they are not related as first and eighth element in the above sequence.

(b) This sequence represents Newland’s Law of Octaves.

Question 4.
State and explain Mendeleev’s periodic law.
Answer:
According to Mendeleev’s periodic law “Physical and chemical properties of elements are periodic function of their atomic masses”.

If elements are arranged in the increasing order of atomic masses, elements with similar properties lies after a certain interval in the periodic table.

Question 5.
“Hydrogen occupies a unique position in the Modern Periodic Table”. Justify the statement.   [NCERT Exemplar]
Answer:
The position of the element hydrogen is still not clear even in the Modern Periodic Table.

In electronic configuration, it resembles alkali metals of group 1. All of them have only one electron in the valence shell. Actually, hydrogen has only one shell (K-shell) which has one electron.

In characteristics, it resembles halogens of group 17. For example, like halogens hydrogen is a non-metal and diatomic as well. It has been therefore, decided to assign hydrogen a unique position in the Modern or Long Form Periodic Table. It is placed at the top in group 1 of alkali metals. However, it is not a member of that group.

Question 6.
If an element X is placed in group 14, what will be the formula and the nature of bonding of its chloride?  [NCERT Exemplar]
Answer:
Since element “X’ is placed in group 14, therefore, its valency is 14 – 10 = 4. Further, since it is difficult to either lose all the four valence electrons or gain four more electrons, therefore, it prefers to share these four electrons to acquire the stable electronic configuration of the nearest inert gas. Its formula will be XCl₄. Thus, the nature of the chloride of element ‘X’ is covalent.

Question 7.
Explain clearly, why atomic number of an element is most important to the chemist than its relative atomic mass.
Answer:
Atomic number corresponds to the number of electron in an atom or it reflects the electronic configuration of the element. The elements having similar electronic configurations can then be placed together in the same group, it helps in the systematic classification of elements.

On the other hand, the atomic mass of an element does not provide the electronic configuration of an element, so atomic number is more fundamental property for classification of elements.

Question 8.
What is the need of classification of the elements?
Answer:
The following reasons can be assigned for practical utility of classification of elements:

• To make the study of chemical elements easier.
• To correct atomic masses of various elements.
• To discover new elements.

Question 9.
The elements calcium, strontium and barium were put in one group or family on the basis of their similar properties.
(i) Mention the two similar properties?
(ii) What is the usual name of this group or family?
Answer:
(i) (a) All these elements are metals.
(b) All these elements have a valency of 2.

(ii) The usual name of this group or family is alkaline earth metals.

Question 10.
How does the electronic configuration of an atom of an element relate to its position in the modem periodic table? Explain with one example.   [CBSE 2011 (Delhi)]
Answer:
The electronic configuration of an atom of an element gives its position in the modern periodic table.
(i) The ‘period number’ of an element is equal to the number of electron shells in its atom.
(ii)

• The group number of an element having upto two valence electrons is equal to the number of valence electrons.
• The group number of an element having more than 2 valence electrons is equal to the number of valence electrons plus 10.

Example: If the electronic configuration of an element is 2, 8, 7.
Then its period number is 3 as it has three electrons shells.
Its group number is 17 as it has 7 valence electrons. (∵ Group no. = 7 + 10 = 17)

Question 11.
An element X (atomic number 17) combines with an element Y (atomic number 20) to form a compound.
(i) Write the positions of these elements in the modern periodic table.
(ii) Write the formula of the compound formed. Justify your answer in each case.   [CBSE 2013]
Answer:
(i) The electronic configurations of the two elements are:
X (Z = 17) 2, 8, 7 ; Y (Z = 20) 2, 8, 8, 2.
Element X is present in group 17 (halogen family) since it has 7 valence electrons. Element Y is placed in group 2 (alkaline earth family) since it has 2 valence electrons.

(ii) Element Y is a metal with valency 2 while element X is a non-metal with valency 1. Therefore, they two combine to form compound YX₂.

Question 12.
Explain why:
(i) All the elements in a group have similar chemical properties.
(ii) All the elements in a period have different chemical properties.
Answer:
(i) All the elements in a particular group have similar outer shell electronic configuration. Since chemical properties of an element are determined mainly by the outer shell configuration, all the elements in a group have similar chemical properties.

(ii) All the elements in a period have different valence shell electronic configuration because they have different number of electrons in the valence shell. Hence, the elements in a period have different chemical properties.

Question 13.
Given below are the atomic radii of three elements X, Y and Z of the periodic table, each having n electrons in the outermost shell of their atoms:

Element	X	Y	Z
Atomic radii	133 pm	157 pm	202 pm

Answer the following:
(a) Do these elements X, Y and Z belong to same group or to the same period?
(b) Which element will be least metallic?
Answer:
(a) Since the elements X, Y and Z contain same number of electrons (n) in the valence shell (outermost shell), they belong to the same group.
(b) In a group, the atomic radius increases on descending the group. Therefore, the element X with the smallest atomic radius is at the top of the group and the element Z having largest atomic radius is at the bottom.

Now, we know metallic character increases as we move from top to bottom in a group. Therefore, the least metallic element is X.

Question 14.
(a) The elements of the second period along with their atomic numbers in parentheses are given below:
B (5), Be (4), O (8), N (7), Li (3), C (6), F (9)
(i) Arrange them in the same order as they appear in the periodic table.
(ii) Which element has the largest and smallest atom?
(b) Why does atomic radius change as we move from left to right in a period?   [CBSE 2011]
Answer:
(a) (i) Li (3), Be (4), B (5), C (6), N (7), O (8), F (9)
(ii) The element Li has the largest atom. The element F has the smallest atom.

(b) Along a period, the nuclear charge increases and the electrons are attracted more towards the nucleus. Therefore, the atomic size or atomic radius decreases as we move from left to the right along a period.

Question 15.
Write two reasons responsible for the late discovery of noble gases.   [CBSE 2013]
Answer:

1. Noble gas elements were not present in earth crust as minerals like other elements and were present in air to a very small extent.
2. Their atoms have stable electronic configuration of their outermost shells also called valence shells. (2 in case of He and 8 for other elements). They do not combine with atoms of other elements.

That is why, noble gas elements were discovered at a later stage.

Question 16.
From the following elements:
₄Be; ₉F; ₁₉K; ₂₀Ca
(i) Select the element having one electron in the outermost shell.
(ii) two elements of the same group.
Write the formula of and mention the nature of the compound formed by the union of ₁₉K and element X (2, 8, 7). [Delhi 2015]
Answer:
(i) ₁₉K
(ii) ₄Be, ₂₀Ca belong to the same group.
Electronic configuration of K (19) = 2, 8, 8, 1   Valency of K = 1
Electronic configuration of X (2, 8, 7)   Valency of X = 1

The formula of the compound formed is KX. The compound KX is of ionic nature.
The bond is formed by transference of electrons.

Question 17.
Na, Mg and Al are the elements of the same period of Modern Periodic Table having one, two and three valence electrons respectively. Which of these elements (i) has the largest atomic radius, (ii) is least reactive? Justify your answer stating reason for each case.   [Delhi 2015]
Answer:

(1) Na has the largest atomic radius because on moving from left to right in the periodic table, the atomic radius decreases due to increase in positive charge on the nucleus which pulls the outermost electrons more close to the nucleus and the size of atom decreases.

(2) Al is least reactive because on moving from left to right in the periodic table the nuclear charge increases and the valence electrons are pulled in more close to the nucleus. Therefore, the tendency to lose electrons decreases and hence reactivity decreases.

Question 18.
How many groups and periods are there in the modern periodic table? How do the atomic size and metallic character of elements vary as we move: (a) down a group and (b) from left to right in a period. [Delhi 2015]
Answer:
Number of groups is 18 and number of periods in the modern periodic table is 7.
(a) All the elements in a group have the same valency. On going down in a group, the atomic size increases because a new shell of electrons is added to the atoms at every step.

(b) On moving from left to right in the periodic table, the valency of the elements first increases from 1 to 4 and then decreases to zero. On moving from left to right in a period, the metallic character of elements decrease because on moving from left to right in a period, the electropositive character of elements decrease.

Question 19.
The electronic configuration of an element X is:

(i) What is the group number of element X in the periodic table?
(ii) What is the period number of element X in the periodic table?
(iii) What is the number of valence electrons in an atom of X?
(iv) What is the valency of X?
Answer:
(i) From the above given electronic configuration we find that element X has 6 valence electrons in the outermost shell), so the group number of element X in the periodic table is 6 + 10 = 16.

(ii) Element X has 3 electron shells (K, L and M) in its atom, so the period number of X is 3. That is, X belongs to 3rd period of the periodic table.

(iii) Element X has 6 valence electrons.

(iv) Element X has 6 valence electrons so it needs 2 more electrons to complete its octet (8 electrons in valence shell) and become stable. Thus, the valency of element X is 2.

Extra Questions for Class 10 Science Chapter 5 Short Answer Type II

Question 1.
Identify the elements with the following property and arrange them in increasing order of their reactivity.
(a) An element which is a soft and reactive metal.
(b) The metal which is an important constituent of limestone.
(c) The metal which exists in liquid state at room temperature.  [NCERT Exemplar]
Answer:
(a) Alkali metals are soft and reactive. For example, Na, K, etc.
(b) Limestone is calcium carbonate, therefore, the important constituent of limestone is calcium.
(c) Metals which exists in the liquid state at room temperature are Mercury (Hg).

Question 2.
List three main anomalies of Mendeleev’s classification of elements.
Answer:
Anomalies or defects of Mendeleev’s periodic classification of elements are as follows:

1. Position of hydrogen.
2. Position of isotopes: Isotopes of an element were not assigned separate places.
3. Elements which are chemically similar such as gold and platinum have been placed in separate groups.

Question 3.
(a) State modern periodic law.
(b) What are the advantages of the long form of the periodic table over Mendeleev’s periodic table?
Answer:
(a) According to modern periodic law “physical and chemical properties of elements are periodic function of their atomic numbers”.

(b) (i) Classification of elements on the basis of atomic number is more closer to chemical properties.
(ii) No separate places for isotopes of an element are required.
(iii) In case of Ar and K, Ar has less atomic number so it should be placed before K according to the increasing order of atomic number.

Question 4.
Two elements X and Y have atomic number 12 and 17 respectively.
(i) Write the electronic configuration of both.
(ii) Which type of bond will they form?
(iii) Write the formula of the compound formed by their combination (in terms of X and Y).
Answer:
(i) The electronic configuration of element X (Z = 12) = 2, 8, 2
The electronic configuration of element Y (Z = 17) = 2, 8, 7

(ii) The bond formed will be of ionic nature. One atom of X will transfer 1 electron each to two atoms of Y.

(iii) The formula of the compound is:

Question 5.
Name:
(a) Three elements that have a single electron in their outermost shells.
(b) Two elements that have two electrons in their outermost shells.
(c) Three elements with filled outermost shells.
Answer:
(a) Lithium (Li), Sodium (Na), Potassium (K)
(b) Magnesium (Mg), Calcium (Ca)
(c) Helium (He), Neon (Ne), Argon (Ar)

Question 6.
Two elements X and Y belong to group 1 and 2 respectively in the same period of periodic table. Compare them with respect to:   [CBSE 2011]
(i) the number of valence electrons in their atoms
(ii) their valencies
(iii) metallic character
(iv) the sizes of their atoms
(v) the formulae of their oxides
(vi) the formulae of their chlorides.
Answer:
X and Y belong to same period.

• X belongs to group ‘1’.
• Y belongs to group ‘2’.

(i) Valence electron in X is 1 whereas valence electrons in Y are 2.
(ii) The valency of X is 1 whereas valency of Y is 2.
(iii) X is more metallic than Y because metallic character decreases on moving from left to right in a period.
(iv) The size of X is more than Y because size of the atom decreases on moving from left to right in a period.
(v) Oxide of X = X₂O,   Oxide of Y = YO
(vi) Chloride of X = XCl,   Chloride of Y = YCl₂

Question 7.
The element Li, Na and K, each having one valence electron, are in period 2, 3 and 4 respectively of the Modem Periodic Table.
(a) In which group of the periodic table should they be?
(b) Which one of them is least reactive?
(c) Which one of them has the largest atomic radius? Give reason to justify your answer in each case. [CBSE 2013]
Answer:
(a) Since the elements have one valence electron, they are placed in group 1 (Alkali metals) in the order Li (period = 2); Na (period = 3); K (period = 4).

(b) Since the reactivity of the elements increases down a group, the element Li is the least reactive chemically.

(c) Since the atomic size increases down a group, the element K has the large atomic size or atomic radius out of these elements.

Question 8.
Explain why are the following statements not correct:
(a) All groups contain metals and non-metals.
(b) Atoms of elements in the same group have the same number of electron(s).
(c) Non-metallic character decreases across a period with increase in atomic number.
(d) Reactivity increases with atomic number in a group as well as in a period.   [2010]
Answer:
(a) Because all groups do not contain metals and non-metals. For example, Alkali metal group contains metals only.
(b) Atom of the elements in the same group have same number of electrons in the valence shell.
(c) Non-metallic character increases across a period with increase in atomic number.
(d) Reactivity increases with atomic number in a group but not across a period.

Question 9.
This question refers to the elements of the periodic table with atomic numbers from 3 to 18. Some of the elements are shown by letters but the letters are not the usual symbols of the elements.

(a) Which of these
(i) are noble gases?
(ii) are halogens?
(iii) are alkali metals?
(b) If A combines with F, what would be the formula of the resulting compound?
(c) What is the electronic configuration of G?
Answer:
(a) (i) H, P;
(ii) G, O;
(iii) A, I
(b) F
(c) 2, 7

Question 10.
Name two other elements which are in the same group as:
(i) Carbon
(ii) Fluorine
(iii) Sodium, respectively.
Answer:
(i) The two other elements which are in the same group as carbon are: Silicon (Si) and germanium (Ge).
(ii) The two other elements which are in the same group as fluorine are : Chlorine (Cl) and bromine (Br).
(iii) The two other elements which are in the same group as sodium are : Lithium (Li) and potassium (K).

Question 11.
Atomic number of an element is 16. Predict
(i) the number of valence electrons in its atom
(ii) its valency
(iii) its group number
(iv) whether it is a metal or a non-metal
(v) the nature of oxide formed by it.
(vi) the formula of the chloride   [CBSE2011]
Answer:
The element with atomic number Z = 16 is sulphur. The electronic configuration of the element is 2, 8, 6.
(i) The number of valence electrons is 6
(ii) The valency of the element is 2 (8 – 6 = 2)
(iii) The group number of the element is 16.
(iv) The element is a non-metal.
(v) Since the element is non-metal, its oxide is acidic in nature. Actually, two oxides are formed. These are SO₂ and SO₃. Both are of acidic nature.
(vi) The formula of the chloride is SCl₂.

Question 12.
(a) How are the following related?
(i) Number of valence electrons of the different elements present in the same group.
(ii) Number of the shells of the elements in the same period.
(b) How do the following change:
(i) Number of shells of the elements as we go down a group.
(ii) Number of valence electrons of elements on moving from left to right in a period.
(iii) Atomic radius in moving from left to right along a period.
(iv) Atomic size down a group. [CBSE 2012]
Answer:
(a) (i) Number of valence electrons of the elements present in a group do not change.
(ii) The elements present in the same period have same number of shells.

(b) (i) The number of shells of elements in a group gradually increase downwards.
(ii) The number of valence electrons of the elements in moving from left to right along a period gradually increase.
(iii) Atomic radius decreases from left to the right.
(iv) Atomic size increases down a group.

Question 13.
(a) What is meant by periodicity in properties of elements with reference to the periodic table?
(b) Why do all the elements of the same group have similar properties?
(c) How will the tendency to gain electrons change as we go from left to right across a period? Why?
Answer:
(a) Periodicity in properties of elements with reference to the periodic table means the elements having similar properties are repeated after certain intervals or periods and the elements are arranged in the tabular form.

(b) All the elements of the same group have similar properties because all the elements belonging to the same group of the Periodic Table have same number of valence electrons.

(c) On moving from left to right in a period, the tendency of atoms to gain electrons increases because on moving from left to right in a period the nuclear charge (positive charge on nucleus) increases due to gradual increase in the number of protons so it becomes easier for the atoms to gain electrons.

Question 14.
What is meant by ‘group’ in the modern periodic table? How do the following change on moving from top to bottom in a group?
(i) Number of valence electrons
(ii) Number of occupied shells
(iii) Size of atoms
(iv) Metallic character of elements
(v) Effective nuclear change experienced by valence electrons.  [CBSE 2014]
Answer:
The vertical columns in a periodic table are called groups.
(i) All the elements of a group have the same number of valence electrons.
(ii) On moving down in a group the number of occupied (filled) shells increases gradually.
(iii) On going down in a group the size of atoms increases because a new shell of electrons is added to the atoms at every step.
(iv) On going down in a group the metallic character of elements increases.
(v) On moving down in a group, one more electron shell is added at every stage and size of the atom increases. Thus valence electrons move more and more away from the nucleus and hold of the nucleus or nuclear charge on valence electrons decreases.

Question 15.
The elements Be, Mg and Ca each having two electrons in their outermost shells are in periods 2, 3, and 4 respectively of the modern periodic table. Answer the following questions, giving justification in each case:  [CBSE 2014]
(i) Write the group to which these elements belong.
(ii) Name the least reactive element.
(iii) Name the element having largest atomic radius.
Answer:

Element	Valence electrons	Period	Electronic configuration
Be	2	2	2,2
Mg	2	3	2, 8, 2
Ca	2	4	2, 8, 8, 2

(i) All these elements belong to the 2nd group all have two electrons in their outermost shell.

(ii) Be is the least reactive metal because reactivity of metals increases in a period as the tendency to lose electrons in a group increases.

Therefore, Be being the smallest in all the given elements of a period has its valance electrons nearest to the nucleus. So the removal of electrons from its valance shell will be difficult.

(iii) Ca has the largest atomic radius because it has maximum number of shells, i.e., 4.

Question 16.
An element ‘M’ with electronic configuration (2, 8, 2) combines separately with (NO₃)^–, (SO₄)^2- and (PO₄)^3- radicals. Write the formula of three compounds so formed. To which group and period of the Modern Periodic Table does the elements ‘M’ belong? Will ‘M’ from covalent or ionic compounds? Give reason to justify your answer.
Answer:
Element M with electronic configuration =

Total number of electrons = 2 + 8 + 2 = 12, i.e., the given element is Magnesium (Mg). Formula of compound formed are:

Formula of compound = Mg (NO₃)₂

Formula of compound = MgSO₄

Formula of compound = Mg₃ (PO₄)₂

M will form ionic compound with radicals given. An ionic compound is a chemical compound comprising ions held together by electrostatic forces. In this compound no sharing of electrons takes place.

Extra Questions for Class 10 Science Chapter 5 Long Answer Type

Question 1.
An element X which is a yellow solid at room temperature shows catenation and allotropy. X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystals and are the major air pollutants.
(a) Identify the element X
(b) Write the electronic configuration of X
(c) Write the balanced chemical equation for the thermal decomposition of ferrous sulphate crystals.
(d) What would be the nature (acidic/basic) of oxides formed?
(e) Locate the position of the element in the Modern Periodic Table.  [NCERT Exemplar]
Answer:
(a) The element X which is a yellow solid at room temperature and shows catenation and allotropy is sulphur (S).

(b) The atomic number of sulphur is 16. Therefore, its electronic configuration is 2, 8, 6.

(c)

Both SO₂ and SO₃ are major air pollutants.

(d) Since sulphur is a non-metal, therefore, both SO₂ and SO₃ are acidic oxides since they dissolve in water to form the corresponding acids.

(e) Since sulphur contains six valence electrons, therefore, it lies in group 6 + 10 = 16. Further, since atomic number of sulphur (S) is 16, it lies in the 3rd period.

Question 2.
An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide.
(a) Where in the periodic table are elements X and Y placed?
(b) Classify X and Y as metal, non-metal or metalloid.
(c) What will be the nature of oxide of element Y? Identify the nature of bonding in the compound formed.
(d) Draw the electron dot structure of the divalent halide.   [NCERT Exemplar]
Answer:
(a) X belongs to Group 17 and 3^rd Period.
Y belongs to Group 2 and 4^th Period.
(b) X-non-metal and Y-metal.
(c) Basic oxide; Ionic bonding.
(d)

Question 3.
(a) Why do we classify elements?
(b) What were the two criteria used by Mendeleev in creating his Periodic Table?
(c) Why did Mendeleev leave some gaps in his Periodic Table?
(d) In Mendeleev’s Periodic Table, why was there no mention of Noble gases like Helium, Neon and Argon?
(e) Would you place the two isotopes of chlorine, Cl-35 and Cl-37 in different slots because of their different atomic masses or in the same slot because their chemical properties are the same? Justify your answer.
Answer;
(a) As different elements were being discovered, scientists gathered more information about the properties of these elements. It was observed that it was difficult to organise all the information or properties of these elements. So scientists started discovering some pattern in their properties to classify all the known elements to make their study easier.

(b) Atomic mass and similarity of chemical properties (compounds formed by elements with oxygen and hydrogen) were the two criteria used by Mendeleev in his Periodic Table.

(c) Mendeleev left some gaps in his Periodic Table as he predicted the existence of some elements that had not been discovered at that time.

(d) Noble gases like helium, neon, argon, etc. were not mentioned in Mendeleev’s Periodic Table because these gases were discovered later as they are very inert and present in extremely low concentrations in our atmosphere. After the discovery of noble gases they could be placed in a new group without disturbing the existing order of the Periodic Table.

(e) The Modem Periodic Table states that properties of elements are a periodic function of their atomic numbers, since Cl-35 and Cl-37 isotopes have the same atomic number (17), hence they will have same chemical properties even though their atomic masses are different. So, they should be placed in the same slot of the periodic table.

Question 4.
Atomic number of a few elements are give below:
10, 20, 7, 14
(a) Identify the elements.
(b) Identify the Group number of these elements in the Periodic Table.
(c) Identify the Periods of these elements in the Periodic Table.
(d) What would be the electronic configuration for each of these elements?
(e) Determine the valency of these elements.  [NCERT Exemplar]
Answer:
(a) The element are: Neon (Z = 10), Calcium (Z = 20), Nitrogen (Z = 7) and Silicon (Z = 14).
(b) Group number: Neon (18), Calcium (2), Nitrogen (15), Silicon (14).
(c) Periods: Neon (2), Calcium (4), Fluorine (2), Silicon (3).
(d) Electronic Configuration: Neon (2, 8); Calcium (2, 8, 8, 2); Nitrogen (2, 5); Silicon (2, 8, 4).
(e) Valency: Neon (zero); Calcium (2); Nitrogen (3); Silicon (4).

Question 5.
This question refers to the elements of the periodic table with atomic numbers from 3 to 18. Some of the elements are shown by letters, but the letters are not the usual symbols of the elements.

3	4	5	6	7	8	9	10
A					E		G
11	12	13	14	15	16	17	18
B	C		D			F

(a) Which of these:
(i) is a noble gas?
(ii) is a halogen?
(iii) is an alkali metal?
(iv) is an element with valency 4?
(b) If A combines with F, what would be the formula of the resulting compound?
(c) What is the electronic arrangement of G?
Answer:
(a) (i) G having 8 electrons in its valence shell is a noble gas.
₁₀G : 2, 8

(ii) F having 7 electrons in its valence shell is a halogen.
₁₇F : 2, 8, 7

(iii) B having 1 electron in its valence shell is an alkali metal.
₁₁B : 2, 8, 1
Element A is also an alkali metal.
₃A : 2, 1

(iv) The element D having 4 electrons in its valence shell will have valency 4.
₁₄D : 2, 8, 4

(b) A is an electropositive element with valency 1 while F is an electronegative element with valency 1.
Hence, the formula of the compound between A and F would be AF.

(c) The electronic configuration of G is:
₁₀G : 2, 8.

Question 6.
The elements of one short period of the periodic table are given below in the order from left to right.
Li, Be, B, C, O, F, Ne,
(a) To which period these elements belong?
(b) One element of this period is missing. Which is the missing element and where should it be placed?
(c) Which one of the element in this period shows the property of catenation?
(d) Place the three elements fluorine, beryllium and nitrogen in the order of increasing electronegativity.
(e) Which one of the above elements belong to the halogen series?
Answer:
(a) Second period (Atomic number: 3-10)
(b) Nitrogen (N), It should be placed between Carbon (Atomic number 6) and Oxygen (Atomic number 8)
(c) Carbon
(d) Be, N, F
(e) Fluorine.

Question 7.
Which element has
(a) two shells, both of which are completely filled with electrons?
(b) the electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?
Answer:
(a) Neon [Ne] (2, 8)
(b) Magnesium [Mg]
(c) Silicon [Si] (2, 8, 4)
(d) Boron [B] (2, 3)
(e) Carbon [C] (2, 4).

Periodic Classification of Elements HOTS Questions With Answers

Question 1.
(a) In this ladder symbols of elements are jumbled up. Rearrange these symbols of elements in the increasing order of their atomic number in the Periodic Table.
(b) Arrange them in the order of their group also.  [NCERT Exemplar]

Answer:
(a) H, He, Li, Be, B, C, N, O, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca
(b) Group 1 – H, Li, Na, K
Group 2 – Be, Mg, Ca
Group 13 – B, Al
Group 14 – C, Si
Group 15 – N, P
Group 16 – O, S
Group 17 – F, Cl
Group 18 – He, Ne, Ar.

Question 2.
The electronic configuration of an element T is 2, 8, 7.
(а) What is the group number of T?
(b) What is the period number of T?
(c) How many valence electrons are there in an atom of T?
(d) Is it a metal or non-metal?
Answer:
(a) Group-17
(b) Third period
(c) 7
(d) Non-metal.

Question 3.
Mendeleev predicted the existence of certain elements not known at that time and named two of them as Eka-silicon and Eka-aluminium.
(a) Name the elements which have taken the place of these elements.
(b) Mention the group and the period of these elements in the Modern Periodic Table.
(c) Classify these elements as metals, non-metals or metalloids.
(d) How many valence electrons are present in each one of them?  [NCERT Exemplar]
Answer:
(a) Eka-silicon for germanium (Ge) and Eka-aluminium for Gallium (Ga).
(b) Group number of Ga is 13 and its period is 4^th and group number of Ge is 14 and its period is also 4^th.
(c) Both Ga and Ge are metalloids.
(d) Ga lies in group 13, therefore, it has 13 – 10 = 3 valence electrons. Similarly, Ge lies in group 14 and hence it has 14 – 10 = 4 valence electron.

Question 4.
The atomic numbers of three elements A, B and C are given below:

Element	Atomic number
A	5
B	7
C	10

(i) Which element belongs to group-18?
(ii) Which element belongs to group-5?
(iii) Which element belongs to group-3?
(iv) To which period/periods do these elements belong?
Answer:
(i) C
(ii) B
(iii) A
(iv) Second period.

Question 5.
(a) Electropositive nature of the element(s) increases down the group and decreases across the period.
(b) Electronegativity of the element decreases down the group and increases across the period.
(c) Atomic size increases down the group and decreases across a period (left to right).
(d) Metallic character increases down the group and decreases across a period.
On the basis of the above trends of the Periodic Table, answer the following about the elements with atomic numbers 3 to 9.
(a) Name the most electropositive element among them
(b) Name the most electronegative element
(c) Name the element with smallest atomic size
(d) Name the element which is a metalloid
(e) Name the element which shows maximum valency.  [NCERT Exemplar]
Answer:
(a) Lithium (Z = 3) is the most electropositive element.
(b) Fluorine (Z = 9) is the most electronegative element.
(c) Fluorine (Z = 9) has the smallest atomic size.
(d) Boron (Z = 5) is a metalloid.
(e) Carbon (Z = 6) shows the maximum valency (4).
However, the element nitrogen (Z = 7) can show valency 5 in some compounds (example, N₂O₅).

Question 6.
The following table shows the position of six elements A, B, C, D, E and F in the period table.

Groups/periods	1	2	3 to 12	13	14	15	16	17	18
2.		A				B			C
3.		D			E				F

Using the above table answer the following questions:
(a) Which element will form only covalent compounds?
(b) Which element is a metal with valency of 2?
(c ) Which element is a non-metal with valency of 3?
(d) Out of D and E, which one has more atomic radius and why?
(e) Write a common name for the family of elements C and F.
Answer:
(a) The element ‘E’ present in group 14 is a non-metal. Its name is silicon (Si) and the compounds of the element are only covalent.
(b) The element ‘D’ present in group 2 is a metal known as magnesium (Mg). It exhibits valency 2 in its compounds.(c) The elements ‘B’ present in group 15 is a non-metal. It is nitrogen (N) and exhibits valency 3 in its compounds.
(d) The element ‘D’ has more atomic radius than the element ‘E’ as the atomic size decreases along a period.
(e) The elements ‘C’ and ‘F’ present in group 18 belong to a family known as noble gases.

Question 7.
From the part of the periodic table given, answer the following questions

1 Lithium	2	13	14 Carbon	15	16 Oxygen	17 L	18 Neon
X			S		P	Q
Y						R
Z						T

(a) Which is the most reactive metal?
(b) Name the family of L, Q, R and T.
(c) Name one element of group 2 and 15.
(d) Name one member of group 18 other than neon.
(e) Give the name of the element S placed below carbon in group 14.
Answer:
(a) Z
(b) L, Q, R and T belongs to halogen family.
(c) Group-2 element – Ca, Mg
Group-15 element – N, P
(d) The element argon (Ar) is also present in group 18.
(e) The element is silicon (Si).

Extra Questions for Class 10 Science Chapter 5 Value Based Questions

Question 1.
First systematic classification was done by Mendeleev. He arranged the elements in the increasing order of atomic masses. Elements with similar properties lies in a group, he left certain gaps for unknown elements.
(i) Name two elements for which blank spaces were left out by Mendeleev.
(ii) List two main drawbacks of the Mendeleev’s periodic classification.
(iii) Mention the values exhibited by Mendeleev.
Answer:
(i) Gallium (Ga), Germanium (Ge).
(ii) (a) Position of hydrogen.
(b) Position of isotopes.
(iii) Use of knowledge of Chemistry, Thinking ability.

Question 2.
Ria and Ram are students of class X. Ria is very organised and systematic. The teachers love her. She earns a great respect in the class whereas Ram is unorganised and always faces a lot of problem in handling life situations.
(i) In your opinion how does the organisation help in daily life?
(ii) How can you relate the above fact with the chapter classification of elements. How does classification of elements help us in studying them properly?
Answer:
(i) Organised approach always make the work easier and quicker. Hence, classification makes the things easier.

(ii) Organised approach in classification of elements make the study easier and we get a group of elements with similar properties.
By this approach we can study the elements easily which lies in a group with similar chemical properties.

Here we are providing Pair of Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-maths/

Extra Questions for Class 10 Maths Quadratic Equations with Answers Solutions

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers

Quadratic Equations Class 10 Extra Questions Very Short Answer Type

Quadratic Equation Class 10 Extra Questions Question 1.
What will be the nature of roots of quadratic equation 2x² + 4x – n = 0?
Solution:
D = b² – 4ac
⇒ 4² – 4 x 2 (-7)
⇒ 16 + 56 = 72 > 0
Hence, roots of quadratic equation are real and unequal.

Quadratic Equation Extra Questions Question 2.
If \(\frac{1}{2}\) is a root of the equation x² + kx – \(\frac{5}{4}\) = 0, then find the value of k.
Solution:
∴ \(\frac{1}{2}\) is a root of quadratic equation.
∴ It must satisfy the quadratic equation.

Quadratic Equations Class 10 Extra Questions Question 3.
If ax² + bx + c = 0 has equal roots, find the value of c.
Solution:
For equal roots D = 0
i.e., b² – 4ac = 0
⇒ b² = 4 ac
⇒ c = \(\frac{b^{2}}{4 a}\)

Quadratic Equation Class 10 Questions Question 4.
If a and b are the roots of the equation x² + ax – b = 0, then find a and b.
Solution:
Sum of the roots = a + b = – \(\frac{B}{A}\) = – a
Product of the roots = ab = \(\frac{B}{A}\) = – b
= a + b = – a and ab = -b
⇒ 2a = -b and a = -1
⇒ b = 2 and a = -1

Class 10 Maths Chapter 4 Extra Questions With Solutions Question 5.
Show that x = – 2 is a solution of 3x² + 13x + 14 = 0.
Solution:
Put the value of x in the quadratic equation,
⇒ LHS = 3x² + 13x + 14
⇒ 3(-2)² + 13(-2) + 14
⇒ 12 – 26 + 14 = 0
⇒ RHS Hence, x = -2 is a solution.

Quadratic Equation Class 10 Extra Questions With Answers Question 6.
Find the discriminant of the quadratic equation 4√2x² + 8x + 2√2 = 0).
Solution:
D = 62 – 4ac = (8)² – 4(4√2)(2√2)
⇒ 64 – 64 = 0

Quadratic Equations Class 10 Extra Questions Short Answer Type 1

Class 10 Quadratic Equation Extra Questions Question 1.
State whether the equation (x + 1)(x – 2) + x = 0 has two distinct real roots or not. Justify your answer.
Solution:
(x + 1)(x – 2) + x = 0
⇒ x² – x – 2 + x = 0
⇒ x² – 2 = 0
D = b² – 4ac
⇒ (-4(1)(-2) = 8 > 0
∴ Given equation has two distinct real roots.

Extra Questions Of Quadratic Equations Class 10 Question 2.
Is 0.3 a root of the equation x² – 0.9 = 0? Justify.
Solution:
∵ 0.3 is a root of the equation x² – 0.9 = 0
∴ x² – 0.9 = (0.3)² – 0.9 = 0.09 – 0.9 ≠ 0
Hence, 0.3 is not a root of given equation.

Quadratic Equations Extra Questions Question 3.
For what value of k, is 3 a root of the equation 2x² + x + k = 0?
Solution:
3 is a root of 2x² + x + k = 0, when
⇒ 2(3)² + 3 + k = 0
⇒ 18+ 3 + k = 0
⇒ k = – 21

Quadratic Equation Class 10 Important Questions With Solutions Question 4.
Find the values of k for which the quadratic equation 9x² – 3kx + k = 0 has equal roots.
Solution:
For equal roots:
D = 0
⇒ b² – 4ac = 0
⇒ (- 3k)² – 4 × 9 × k = 0
⇒ 9k² = 36k
⇒ k = 4

Class 10 Quadratic Equations Extra Questions Question 5.
Find the value of k for which the equation x² + k(2x + k – 1)+ 2 = 0 has real and equal roots.
Solution:
Given quadratic equation: x² + k(2x + k-1) + 2 = 0)
= x² + 2kx + (k² – k + 2) = 0
For equal roots, b² – 4ac = 0
⇒ 4k² – 4k² + 4k – 8 = 0
⇒ 4k = 8
⇒ k = 2

Quadratic Equation Questions For Class 10 Question 6.
If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then find the value of k.
Solution:
Since – 5 is a root of the equation 2x² + px – 15 = 0
∴ 2(-5)² + p(-5) – 15 = 0
⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p(x² + x) + k = 0 or 7x² + 7x + k = 0 has equal roots
∴ D = 0
i.e., b² – 4ac = 0 or 49- 4 × 7k = 0
⇒ k = \(\frac{49}{28}\) = \(\frac{7}{4}\)

Extra Questions On Quadratic Equations Class 10 Question 7.
Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.
Solution:
Yes, x² – 4x + 1 = 0 is a quadratic equation with rational co-efficients.

Questions On Quadratic Equations Class 10 Question 8.
Write the set of values of k for which the quadratic equation 2x² + kx + 8 = 0 has real roots.
Solution:
For real roots, D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ k² – 4(2)(8) ≥ 0
⇒ k² – 64 ≥ 0
⇒ k² ≥ 64
⇒ k ≤ -8 and k ≥ 8

Class 10 Maths Quadratic Equations Extra Questions Question 9.
Solve the quadratic equation 2x² + ax – a² = 0 for x.
Solution:
2x² + ax – a² = 0
Here, a = 2, b = a and c = -a².
Using the formula,

Class 10 Quadratic Equations Important Questions Question 10.
Find the roots of the quadratic equation
Solution:
The given quadratic equation is
√2x² + 7x + 5√2 = 0
By applying mid term splitting, we get
√2x² + 2x + 5x + 5√2 = 0
⇒ √2x(x + √2) + 5(x + √2)
⇒ (√2x + 5) + 5(x + √2) = 0
⇒ x = \(\frac{-5}{\sqrt{2}}/latex], -√2 or [latex]\frac{-5 \sqrt{2}}{2}\), -√2

Quadratic Equation Class 10 Extra Questions Pdf Question 11.
Find the values of p for which the quadratic equation 4x² + px + 3 = 0 has equal roots.
Solution:
For equal roots;
D = 0
⇒ b² – 4ac = 0
⇒ p² – 4 × 4 × 3 = 0
⇒ p² – 48 = 0
⇒ p² = 48
⇒ p = ± √48
⇒ p = 4√3 or -4√3

Ch 4 Maths Class 10 Extra Questions Question 12.
Solve for x: √13x? – 2√3x – 2√3 = 0
Solution:
√3x² – 2√3x – 2√3 = 0
⇒ √3x² – 3√2x + √2x – 2√3 = 0
⇒ √3x(x – √6) + √2(x – √6) = 0
⇒ (√3x + 2)(x – √6) = 0
⇒ √3x + √2 = 0 or x – √6 = 0
⇒ x = \(\frac{-√2}{√3}\) or x = √6

Chapter 4 Maths Class 10 Extra Questions Question 13.
If x = \(\frac{2}{3}\) and x = -3 are roots of the quadratic equation ax² + 7x + b = 0, find the values of a and b.
Solution:
Let us assume the quadratic equation be Ax² + Bx + C = 0.
Sum of the roots = –\(\frac{B}{A}\)

Class 10 Maths Ch 4 Extra Questions Question 14.
A two-digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.
Solution:
Let the ten’s digit be x and unit’s digit = y
Number 10x + y

Extra Questions For Quadratic Equation Class 10 Question 15.
Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other.
Solution:
Let the roots of the given equation be a and 6α.
Thus the quadratic equation is (x – a) (x – 6α) = 0
⇒ x² – 7αx + 6α² = 0 …(i)
Given equation can be written as

Question 16.
If ad ≠ bc, then prove that the equation
(a² + b²) x² + 2(ac + bd)x + (c²+ d²) = 0 has no real roots.
Solution:
The given quadratic equation is (a² + b²⁾x² + 2(ac + bd)x +(c²+ d²) = 0
D = b² – 4ac
= 4(ac + bd)² – 4(a² + b²) (c²+ d²)
= -4(a²d² + b²c²– 2abcd) = – 4(ad – bc)²
Since ad ≠ bc
Therefore D < 0
Hence, the equation has no real roots.

Question 17.
Solve for x: √13x² – 2x – 8√3 = 0
Solution:
√3x² – 2x – 8√3 = 0
By mid term splitting
⇒ √3x² – 6x + 4x – 8√3 = 0
⇒ √3x(x – 2/3) + 4 (x – 2/3) = 0
⇒ (x – 2√3)(√3x + 4) = 0
⇒ Either (x – 2√3) = 0 or (√3x + 4) = 0
⇒ x = \(\frac{-4}{\sqrt{3}}\), 2√3

Quadratic Equations Class 10 Extra Questions Short Answer Type 2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) √2x² + 7x + 5√2 = 0 (ii) 2x² – x + \(\frac{1}{8}\) = 0
Solution:
(i) We have, √2x² + 7x + 5√2 = 0
= √2x² + 5x + 2x + 5√2 = 0
x(√2x + 5) + √2 (√2x + 5) = 0
= (√2x + 5)(x + √2) = 0
∴ Either √2x + 5 = 0 or x + √2 = 0
∴ x = – \(\frac{5}{\sqrt{2}}\) or x = -√2
Hence, the roots are – \(\frac{5}{\sqrt{2}}\) and -√2.
(ii) We have, 2x² – x + \(\frac{1}{8}\) = 0

Question 2.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² + x – 4 = 0
(ii) 4x² + 4√3x + 3 = 0
Solution:
(i) We have, 2x² + x – 4 = 0
On dividing both sides by 2, we have

(ii) We have, 4x² + 4√3x + 3 = 0

Question 3.
Find the roots of the following quadratic equations by applying the quadratic formula.
(i) 2x² – 7x + 3 = 0
(ii) 4x² + 4√3x + 3 = 0
Solution:
(i) We have, 2x² – 7x + 3 = 0
Here, a = 2, b = -7 and c = 3
Therefore, D = b² – 4ac
⇒ D = (-7)² – 4 × 2 × 3 = 49 – 24 = 25
∵ D > 0, ∴ roots exist.


So, the roots of given equation are 3 and \(\frac{1}{2}\)

(ii) We have, 4x² + 4√3x + 3 = 0
Here, a = 4, b = 4√3 and c = 3
Therefore, D = b² – 4ac = (4√3)² – 4 × 4 × 3 = 48 – 48 = 0
∴ D = 0, roots exist and are equal.

Question 4.
Using quadratic formula solve the following quadratic equation:
p²x² + (p² – q²) x – q² = 0
Solution:
We have, p²x² + (p² – q²) x – q² = 0
Comparing this equation with ax² + bx + c = 0, we have
a = p², b = p² – q² and c = – q²
∴ D = b² – 4ac
⇒ (p² – q)² – 4 × p² × (-q²)
⇒ (p² – q²)² + 4p²q²
⇒ (p² + q³)² > 0
So, the given equation has real roots given by

Question 5.
Find the roots of the following equation:

Solution:

⇒ (x + 3) (x – 6)
⇒ -20 or x² – 3x + 2 = 0
⇒ x² – 2x -x + 2 = 0
⇒ x(x – 2) -1(x – 2) = 0)
⇒ (x – 1) (x – 2) = 0
⇒ x = 1 or x = 2
Both x = 1 and x = 2 are satisfying the given equation. Hence, x = 1, 2 are the solutions of the equation.

Question 6.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 3x² – 4√3x + 4 = 0) (ii) 2x² – 6x + 3 = 0
Solution:
(i) We have, 3x² – 4√3x + 4 = 1
Here, a = 3, b = – 4√3 and c = 4
Therefore,
D = b² – 4ac
⇒ (- 4√3)² – 4 × 3 × 4
⇒ 48 – 48 = 0
Hence, the given quadratic equation has real and equal roots.

(ii) Wehave, 2x² – 6x + 3 = 0
Here, a = 2, b = -6, c = 3
Therefore, D = b² – 4ac
= (-6)² 4 × 2 × 3 = 36 – 24 = 12 > 0
Hence, given quadratic equation has real and distinct roots.

Question 7.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
Solution:
(i) We have, 2x² + kx + 3 = 0
Here, a = 2, b = k, c = 3
D = b² – 4ac = k² – 4 × 2 × 3 = k² – 24 For equal roots
D = 0
i.e., k² – 24 = 0
⇒ ķ² = 24
⇒ k = ± √24
⇒ k = + 2√6

(ii) We have, kx(x – 2) + 6 = 0
⇒ kx² – 2kx + 6 = 0
Here, a = k, b = – 2k, c = 6
For equal roots, we have
D = 0
i.e., b² – 4ac = 0
⇒ (-2k)² – 4 × k × 6 = 0
⇒ 4k² – 24k = 0
⇒ 4k (k – 6) = 0
Either 4k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
But k = 0 6)ecause if k = 0 then given equation will not be a quadratic equation).
So, k = 6.

Question 8.
If the roots of the quadratic equation (a – b) x² + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c.
Solution:
Since the equation (a – b)x² + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant
D = (b – c)² – 4(a – b) (c – a) = 0
⇒ b² + c² – 2bc – 4(ac – a² – bc + ab)
⇒ b² + c² – 2bc – 4ac + 4a² + 4bc – 4ab = 0
⇒ 4a² + b² + c² – 4ab + 2bc – 4ac = 0
⇒ (2a)² + (- b)² + (-c)² + 2(2a) (-b) + 2(-b) (-c) + 2(-c) 2a = 0
⇒ (2a – b – c)² = 0
⇒ 2a – b – c = 0
⇒ 2a = b + c. Hence Proved

Question 9.
If the equation (1 + m²)x² + 2mcx + c²– a² = 0 has equal roots, show that c² = a² (1 + m²).
Solution:
The given equation is (1 + m²) x² + (2mc) x + (c²– a²) = 0
Here, A = 1 + m², B = 2mc and C = c² – a²
Since the given equation has equal roots, therefore D = 0 = B² – 4AC = 0.
⇒ (2mc)² – 4(1 + m²) (c² – a²) = 0
⇒ 4m²c² – 4(c² – a² + m²c² – m²a²) = 0
⇒ m²c² – c² + a² – m²c² + m²a² = 0. [Dividing throughout by 4]
⇒ – c² + a² (1 + m²) = 0
⇒ c² = a(1 + m²) Hence Proved

Question 10.
If sin θ and cos θ are roots of the equation ax² + bx + c = 0, prove that a² – b² + 2ac = 0.
Solution:

Question 11.
Determine the condition for one root of the quadratic equation ax² + bx + c = 0 to be thrice the other.
Solution:
Let the roots of the given equation be a and 3α.

Question 12.
Salve for

Solution:

⇒ (4x – 2)(2x – 1) – (3x + 9)(x + 3) = 5(x + 3)(2x – 1)
⇒ (8x² – 4x – 4x + 2) – (3x² + 9x + 9x + 27) = 5(2x² – x + 6x – 3)
⇒ 8x² – 8x + 2 – 3x² – 18x – 27 = 10x² + 25x – 15
⇒ 5x² – 26x – 25 = 10x² + 25x – 15
⇒ 5x² + 51x + 10 = 0
⇒ 5x² + 50x + x + 10 = 0
⇒ 5x (x + 10) + 1 (x + 10) = 0
⇒ (5x + 1) (x + 10) = 0
⇒ 5x + 1 = 0 or x + 10 = 0
⇒ x = \(\frac{-1}{5}\) or x = -10

Question 13.
Solve the equation

Solution:

⇒ (4 – 3x) (2x + 3) = 5x ⇒ 8x – 6x² + 12 – 9x = 5x
⇒ 6x² + 6x – 12 = 0
⇒ x² + x – 2 = 0
⇒ x² + 2x + x – 2 = 0
⇒ x(x + 2)-1(x + 2) = 0
⇒ (x – 1)(x + 2) = 0
⇒ x – 1 = 0 or x + 2 = 0
⇒ x = 1 or x = -2

Question 14.
Solve for

Solution:

⇒ (16 – x) (x + 1) = 15x
⇒ 16x – x² + 16 – x = 15x
⇒ x² + 15x – 15x – 16 = 0
⇒ x² = 16
⇒ x = ± 4

Question 15.
Solve for x: + 5x – (a² + a – 6) = 0
Solution:
⇒ x² + 5x – (a² + a – 6) = 0
⇒ x² + 5x – (a? + 3a – 2a – 6) = 0
⇒ x² + 5x – [a(a + 3) -2 (a + 3)] = 0
⇒ x² + 5x – (a – 2) (a + 3) = 0
∴ x² + (a + 3)x – (a – 2) x – (a – 2) (a + 3) = 0
⇒ x[x + (a + 3)]-(a – 2) [X + (a + 3)] = 0
⇒ [{x + (a + 3)} {x – (a – 2)}] = 0
∴ x = -(a + 3) or x = (a -2)
⇒ -(a + 3), (a – 2)
Alternative method
x² + 5x -(a² + a – 6) = 0

Question 16.
Solve for

Solution:

⇒ 2x(2x + 3) + (x – 3) + (3x + 9) = 0
⇒ 4x² + 10x + 6 = 0
⇒ 2x² + 5x + 3 = 0
⇒ (x + 1) (2x + 3) = 0
⇒ x = -1, x = – \(\frac{3}{2}\)
But x ≠ – \(\frac{3}{2}\)
∴ x = -1

Question 17.
Solve for

Solution:

⇒ 3(x – 3 + x – 1) = 2(x – 1)(x – 2)(x – 3)
⇒ 3(2x – 4) = 2(x – 1)(x – 2)(x – 3)
⇒ 3 × 2(x – 2) = 2(x – 1)(x − 2)(x – 3)
⇒ 3 = (x – 1)(x – 3) i.e., x² – 4x = 0
⇒ x(x – 4) = 0 ∴ x = 0, x = 4

Question 18.
If the roots of the equation (c- ab)x² – 2(a²– bc)x + b² – ac = 0 in x are equal, then show that either a = 0 or a³ + b³ + c³ = 3abc.
Solution:
For equal roots D = 0
Therefore 4(a² – bc)² – 4(c² – ab) (b² – ac) = 0
⇒ 4[a⁴ + b²c² – 2a²bc – b²c² + ac³ + ab³ – a²bc] = 0.
⇒ a(a³ + b³ + c³ – 3abc) = 0
Either a = 0 or a³ + b³ + c³ = 3abc

Question 19.
If the roots of the quadratic equation
(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
are equal, then show that a = b = c.
Solution:
Given (x – a) (x – b) + (x – b) (x – c) + (x – 6) (x – a) = 0
⇒ x² – ax – bx + ab + x² – bx – cx + bc + x² – cx – ax + ac = 0
⇒ 3x² – 2(a + b + c)x + ab + bc + ca = 0
Now, for equal roots D = 0
⇒ B² – 4AC = 0
⇒ 4(a + b + c)² – 12(ab + bc + ca) = 0
4a² + 4b² + 4c² + 8ab + 8bc + 8ca – 12ab – 12bc – 12ca = 0
⇒ 2[2a² + 2b² + 2co – 2ab – 2bc – 2ca] = 0
⇒ 2[(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca)] = 0
⇒ [(a – b)² + (b – c)² + (c – a)²] = 0
⇒ a – b = 0, b – c = 0, c – a = 0
⇒ a = b, b = c, c = a
⇒ a = b = c

Quadratic Equations Class 10 Extra Questions Long Answers

Question 1.
Using quadratic formula, solve the following equation for x:
abx² + (b² – ac) x – bc = 0
Solution:
We have, abx² + (b² – ac) x – bc = 0
Here, A = ab, B = b² – ac, C = – bc

Question 2.
Find the value of p for which the quadratic equation
(2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.
Solution:
Since the quadratic equation has equal roots, D = 0
i.e., b² – 4ac = 0
In (2p + 1 )x² – (7p + 2)x + (7p – 3) = 0
Here, a = (2p + 1), b = -(7p + 2), c = (7p – 3)

Question 3.
Solve for

Solution:

Question 4.
The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is Find his present age.
Solution:
Let the present age of Rehman be x years.
So, 3 years ago, Rehman’s age = (x – 3) years
And 5 years from now, Rehman’s age = (x + 5) years
Now, according to question, we have

But x ≠ -3 (age cannot be negative)
Therefore, present age of Rehman = 7 years.

Question 5.
The difference of two natural numbers is 5 and the difference of their reciprocals is \(\frac{1}{10}\). Find the numbers.
Solution:
Let the two natural numbers be x and y such that x > y.
According to the question,
Difference of numbers, x – y = 5 ⇒ x = 5 + y …..(i)
Difference of the reciprocals,

∴ y is a natural number.
∵ y = 5
Putting the value of y in (i), we have
⇒ x = 5 + 5
⇒ x = 10
The required numbers are 10 and 5.

Question 6.
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Solution:
Let the two consecutive odd numbers be x and x + 2.

Hence, the numbers are 13 and 15 or -15 and -13.

Question 7.
The sum of two numbers is 15 and the sum of their reciprocals is 3. Find the numbers.
Solution:
Let the numbers be x and 15 – x.
According to given condition,

⇒ 150 = 3x(15 – x)
⇒ 50 = 15x – x²
⇒ x² – 15x + 50 = 0
⇒ x² – 5x – 10x + 50 = 0
⇒ x(x – 5) -10(x – 5) = 0
⇒ (x – 5)(x – 10) = 0
⇒ x = 5 or 10.
When x = 5, then 15 – x = 15 – 5 = 10
When x = 10, then 15 – x = 15 – 10 = 5
Hence, the two numbers are 5 and 10.

Question 8.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics be x.
Therefore, Shefali’s marks in English is (30 – x).
Now, according to question,
⇒ (x + 2) (30 – x – 3) = 210
⇒ (x + 2) (27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ 25x – x² + 54 – 210 = 0
⇒ 25x – x² – 156 = 0
⇒ -(x² – 25x + 156) = 0
⇒ x² – 25x + 156 = 0
= x² – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0
⇒ (x – 13) (x – 12) = 0
Either x – 13 or x – 12 = 0
∴ x = 13 or x = 12
Therefore, Shefali’s marks in Mathematics = 13
Marks in English = 30 – 13 = 17
or Shefali’s marks in Mathematics = 12
marks in English = 30 – 12 = 18.

Question 9.
A train travels 360 km at a uniform speed. If the speed has been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let the uniform speed of the train be x km/h.

But x cannot be negative, so x ≠ – 45
therefore, x = 40
Hence, the uniform speed of train is 40 km/h.

Question 10.
The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let x be the length of the side of first square and y be the length of side of the second square.
Then, x² + y² = 468 …(i)
Let x be the length of the side of the bigger square.
4x – 4y = 24
⇒ x – y = 6 or x = y + 6 …(ii)
Putting the value of x in terms of y from equation (ii), in equation (i), we get
(y + 6)² + y² = 468
⇒ y² + 12y + 36 + y² = 468 or 232 + 12y – 432 = 0
⇒ y² + 6y – 216 = 0
⇒ y² + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
Either y + 18 = 0 or y – 12 = 0
⇒ y = -18 or y = 12
But, sides cannot be negative, so y = 12
Therefore, x = 12 + 6 = 18
Hence, sides of two squares are 18 m and 12 m.

Question 11.
Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence, Swati’s age will be two-fifth of Varun’s age. Find their present ages.
Solution:
Seven years ago, let Swati’s age be x years. Then, seven years ago Varun’s age was 5x² years.
∴ Swati’s present age = (x + 7) years
Varun’s present age = (5x² + 7) years
Three years hence,
Swati’s age = (x + 7 + 3) years = (x + 10) years
Varun’s age (5x² + 7 + 3) years = (5x² + 10) years
According to the question,

Hence, Swati’s present age = (2 + 7) years = 9 years
and Varun’s present age = (5 × 2² + 7) years = 27 years

Question 12.
A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.
Solution:
Let the original speed of the train = x km/h.


Therefore, the usual speed of the train = 25 km/h.

Question 13.
A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Solution:
Let the digit at tens place be x.

Question 14.
If twice the area of a smaller square is subtracted from the area of a larger square; the result is 14 cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of the two squares.
Solution:
Let the sides of the larger and smaller squares be x and y respectively. Then
x² – 2y² = 14 …..(i)
and 2x² + 3y² = 203 ……(ii)
Operating (ii) -2 × (i), we get
⇒ 2x² + 3y² – (2x² – 4y²) = 203 – 2 × 14
⇒ 2x² + 3y² – 2x² + 4y² = 203 – 28
⇒ 7y² = 175
⇒ y² = 25
⇒ y ± 15
⇒ y = 5 [∵ Side cannot be negative]
By putting the value of y in equation (i), we get
x² – 2 × 5 = 14
⇒ x² – 50 = 14 or x² = 64
∴ x = 8 or x = 8
∴ Sides of the two squares are 8 cm and 5 cm.

Question 15.
If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?
Solution:
Let the present age of Zeba be x years
Age before 5 years = (x – 5) years According to given condition,
⇒ (x – 5)² = 5x + 11
⇒ x² + 25 – 10x = 5x + 11
⇒ x² – 10x – 5x + 25 – 11 = 0
⇒ x² – 15x + 14 = 0
⇒ x² – 14x – x + 14 = 0
⇒ x(x – 14) -1(x – 14) = 0
⇒ (x – 1)(x – 14) = 0
⇒ x – 1 = 0 or x – 14 = 0
x = 1 or x = 14
But present age cannot be 1 year.
∴ Present age of Zeba is 14 years.

Question 16.
A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Solution:
Let the speed of the stream be x km/h.
Speed of motorboat in still water = 18 km/h
Speed of motorboat in upstream = (18 – x) km/h
Speed of motorboat in downstream = (18 + x) km/h
Distance travelled = 24 km.

Speed of motorboat = 6 km/h. (∵Speed cannot be negative)

Question 17.
A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Solution:
Let the natural number be x
According to the question,
⇒  x + 12 = \(\frac{60}{x}\)
= x² + 12x – 160 = 0
= x² + 20x – 8x – 160 = 0
= x(x + 20) -8(x + 20) = 0
= (x + 20) (x – 8) = 0
x = -20 (Not possible) or x = 8
Hence, the required natural number is 8.

Question 18.
The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
Solution:
Let the two consecutive multiples of 7 be x and x +7.
x² + (x + 7)² = 637
= x² + x² + 49 + 14x = 637
= 2x² + 14x – 588 = 0
= x² + 7x – 294 = 0
= x² + 21x – 14x – 294 = 0
= x(x + 21) – 14(x + 21) = 0
= x(x + 14) (x + 21) = 0
= x = 14 or x = -21
The multiples are 14 and 21.

Question 19.
Solve for

Solution:

Question 20.
Find the positive value(s) of k for which both quadratic equations x² + kx + 64 = 0 and x² – 8x + k = 0 will have real roots.
Solution:
(i) For x² + kx + 64 = 0 to have real roots
⇒ k² – 4(1)(64) ≥ 0 i.e., k² – 256 ≥ 0
⇒ k ≥ ± 16

(ii) For x² – 8x + k = 0 to have real roots
⇒ (-8)² – 4(k) ≥ 0 i.e., 64 – 4k ≥ 0
⇒ k ≤ ± 16
For (i) and (ii) to hold simultaneously k = 16

Question 21.
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Solution:
Let the speed of stream be x km/h.
∴ Speed of boat upstream = (15 – x) km/h.
Speed of boat downstream = (15 + x) km/h.
According to question,

⇒ 30 × 2 × 30 = 9(225 – x²)
⇒ 100 × 2 = 225 – x²
⇒ 200 = 225 – x²
⇒ x² = 25
⇒ x = ±5
⇒ x = 5 (Rejecting – 5)
∴ Speed of stream = 5 km/h

Question 22.
Ram takes 6 days less than Bhagat to finish a piece of work. If both of them together can finish the work in 4 days, in how many days Bhagat alone can finish the work.
Solution:
Let Bhagat alone can do the work in x number of days
∴ Ram takes (x – 6) number of days
Work done by Bhagat in 1 day = \(\frac{1}{x}\)
Work done by Ram in 1 day = \(\frac{1}{x-6}\)
According to the question,

Quadratic Equations Class 10 Extra Questions HOTS

Question 1.
One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.
Solution:
Let x be the total number of camels.
Then, number of camels in the forest = \(\frac{x}{4}\)
Number of camels on mountains = 2√x
and number of camels on the bank of river = 15

But, the number of camels cannot be a fraction.
∴ y = 6
⇒ x = x² = 36
Hence, the number of camels = 36

Question 2.
Solve the following quadratic equation:
9x² – 9 (a + b) x + [2a² + 5ab + 2b²] = 0.
Solution:
Consider the equation 9x² – 9 (a + b) x + [2a² + 5ab + 2b²] = 0)
Now comparing with Ax² + Bx + C = 0, we get
A = 9, B = -9 (a + b) and C = [2a² + 5ab + 2b²]
Now discriminant, .
D = B² – 4AC

Question 3.
Two taps running together can fill a tank in 3 hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?
Solution:
Let, time taken by faster tap to fill the tank be x hours
Therefore, time taken by slower tap to fill the tank = (x + 3) hours
Since the faster tap takes x hours to fill the tank.

Hence, time taken by faster tap to fill the tank = x = 5 hours
and time taken by slower tap = x + 3 = 5 + 3 = 8 hours.

Question 4.
In a rectangular park of dimensions 50 m × 40 m, a rectangular pond is constructed so that the area of grass strip of uniform width surrounding the pond would be 1184 m². Find the length and breadth of the pond.
Solution:

Let ABCD be rectangular lawn and EFGH be rectangular pond. Let x m be the width of grass area, which is same around the pond.
Given, Length of lawn = 50 m
Width of lawn = 40 m
Length of pond = (50 – 2x)m
Breadth of pond = (40 – 2x)m
Also given,
Area of grass surrounding the pond = 1184 m²
⇒ Area of rectangular lawn – Area of pond = 1184 m²
⇒ 50 × 40 – {(50 – 2x) × (40 – 2x)} = 1184
⇒ 2000 – (2000 – 80x – 100x + 4x²) = 1184
⇒ 2000 – 2000 + 180x – 4x² = 1184
⇒ 4x² – 180x + 1184 = 0
⇒ x² – 45x + 296 = 0
⇒ x² – 37x – 8x + 296 = 0
⇒ x(x – 37) – 8(x – 37) = 0
⇒ (x – 37) (x – 8) = 0
⇒ x- 37 = 0 or x – 8 = 0)
⇒ x = 37 or x = 8
x = 37 is not possible as in this case length of pond becomes 50 – 2 × 37 = -24 (not possible) Hence, x = 8 is acceptable
∴ Length of pond = 50 – 2 × 8 = 34 m
Breadth of pond = 40 – 2 × 8 = 24 m

Question 5.
A car covers a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.
Solution:
Let speed of the car be x km/h
According to question
Time taken = \(\frac{x}{2}\) h.

⇒ x = 72 km/h [Taking square root both sides]
∴ Time taken = \(\frac{x}{2}\) = 36 hours.

Here we are providing Probability Class 10 Extra Questions Maths Chapter 15 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-maths/

Extra Questions for Class 10 Maths Probability with Answers Solutions

Extra Questions for Class 10 Maths Chapter 15 Probability with Solutions Answers

Probability Class 10 Extra Questions Very Short Answer Type

Probability Class 10 Extra Questions Question 1.
State true or false and give the reason. If I toss a coin 3 times and get head each tir ne, then I should expect a tail to have a higher chance in the 4th toss.
Solution:
False, because the outcomes ‘head’ and ‘tail are equally likely. So, every time the probability of getting head or tail is \(\frac{1}{2}\)

Class 10 Probability Extra Questions Question 2.
A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since, this situation has only two possible outcomes, so the probability of each is \(\frac{1}{2}\) Justify.
Solution:
True, because the outcomes odd number’ and `even number’ are equally likely here.

Probability Class 10 Extra Questions With Answers Pdf Question 3.
In a family, having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is \(\frac{1}{4}\). Is this correct? Justify your answer.
Solution:
False, because the outcomes are not equally, likely. For no girl, outcome is bbb, for one girl, it is
bgb, gbb, bbg, for two girls, it is bgg, ggb, gbg and for all girls, it is ggg.

Probability Extra Questions Class 10 Question 4.
A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig. 15.1). Are the outcomes 1, 2 and 3 equally likely to occur? Give reason.
Solution:

False, because the outcome 3 is more likely than the other numbers.

Extra Questions Of Probability Class 10 Question 5.
Two coins are tossed simultaneously. Find the probability of getting exactly one head.
Solution:
Possible outcomes are {HH, HT, TH, TT}.
(exactly one head) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

Extra Questions On Probability Class 10 Question 6.
From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen.
Solution:
Number of black queens in a pack of cards = 2
∴ P (black queen) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Probability Extra Questions Question 7.
If P (E) = 0.05, what is the probability of ‘not E’ ?
Solution:
As we know that,
P (E) + P (not E) = 1
P (not E) = 1 – P (E) = 1 – 0.05 = 0.95

Probability Class 10 Extra Questions With Solutions Question 8.
What is the probability of getting no head when two coins are tossed simultaneously?
Solution:
Favourable outcome is TT;
∴ P (no head) = \(\frac{1}{4}\)

Class 10 Maths Probability Extra Questions Question 9.
In a single throw of a pair of dice, what is the probability of getting the sum a perfect square?
Solution:
Total outcomes = 36
Favourable outcomes are {(1,3), (3, 1), (2, 2), (3, 6), (6,3), (4, 5), (5, 4)}
∴ Required probability = \(\frac{7}{36}\)

Probability Class 10 Important Questions Question 10.
Someone is asked to choose a number from 1 to 100. What is the probability of it being a prime number?
Solution:
Total prime numbers between 1 to 100 = 25
∴ P (Prime number) = \(\frac{25}{100}\) = \(\frac{1}{4}\)

Extra Sums Of Probability Class 10 Question 11.
Cards marked with number 3, 4, 5, …., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
Solution:
Possible outcomes are 4, 9, 16, 25, 36, 49, i.e., 6.
∴ P (perfect square number) = \(\frac{6}{48}\) or \(\frac{1}{8}\)

Probability Questions Class 10 Question 12.
A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.
Solution:
Number of possible outcomes = 52
Number of red cards and queens = 28
Number of favourable outcomes = 52 – 28 = 24
P (getting neither a red card nor a queen) = \(\frac{24}{52}\) = \(\frac{6}{13}\)

Questions On Probability Class 10 Question 13.
20 tickets, on which numbers 1 to 20 are written, are mixed throughly and then a ticket is drawn at random out of them. Find the probability that the number on the drawn ticket is a multiple of 3 or 7.
Solution:
n(s) = 20, Multiples of 3 or 7, A: {3, 6, 9, 12, 15, 18, 7, 14), n(A) = 8
∴ Required probability = \(\frac{8}{20}\) or \(\frac{2}{5}\)

Probability Important Questions Class 10 Question 14.
A number is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What will be the probability that square of this number is less then or equal to 1?
Solution:
Favourable outcomes are -1, 0, 1 = 3
Total outcomes = 7
∴ Required probability = \(\frac{3}{7}\)

Probability Class 10 Extra Questions Short Answer Type 1

Class 10 Maths Ch 15 Extra Questions Question 1.
Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is a prime number.
Solution:
Product of the number on the dice is prime number, i.e., 2, 3, 5.
The possible ways are, (1, 2), (2, 1), (1, 3), (3, 1), (5, 1), (1, 5)
So, number of possible ways = 6
∴ Required probability = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Class 10th Probability Extra Questions Question 2.
Find the probability that a number selected from the numbers 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.
Solution:
Total prime numbers from 1 to 25 = 9.
∴ Non-prime numbers from 1 to 25 = 25 – 9 = 16.
⇒ P (non-prime number) = \(\frac{16}{25}\)

Probability Class 10 Important Questions 2020 Question 3.
One card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is an ace and black.
Solution:
Number of black aces in a pack of cards = 2
∴ P (an ace and black card) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Ch 15 Maths Class 10 Extra Questions Question 4.
A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither an ace nor a king.
Solution:
Let E be the event card drawn is neither an ace nor a king.
Then, the number of outcomes favourable to the event E = 44 (4 kings and 4 aces are not there)
∴ P(E) = \(\frac{44}{52}\) = \(\frac{11}{13}\)

Probability Class 10 Questions Question 5.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange flavoured candy? (ii) a lemon flavoured candy?
Solution:
(i) As the bag contains only lemon flavoured candies. So, the event related to the experiment of taking out an orange flavoured candy is an impossible event. So, its probability is 0.

(ii) As the bag contains only lemon flavoured candies. So, the event related to the experiment of taking out lemon flavoured candies is certain event. So, its probability is 1.

Question 6.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Here, total number of pens = 132 + 12 = 144
∴ Total number of elementary outcomes = 144
Now, favourable number of elementary events = 132
∴ Probability that a pen taken out is good one = \(\frac{132}{144}\) = \(\frac{11}{12}\)

Question 7.
Two players, Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeeta’s winning the match is 0.62. What is the probability of Reshma’s winning the match?
Solution:
Let S and R denote the events that Sangeeta and Reshma wins the match, respectively.
The probability of Sangeeta’s winning = P(S) = 0.62
As the events R and S are complementary
∴ The probability of Reshma’s winning = P(R) = 1 – P(S)
= 1 – 0.62 = 0.38.

Question 8.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
The total number of elementary events associated with random experiment of throwing a die is 6.
(i) Let E be the event of getting a letter A.
∴ Favourable number of elementary events = 2
∴ P(E) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
(ii) Let E be the event of getting a letter D.
∴ Favourable number of elementary events = 1
∴ P(E) = \(\frac{1}{6}\)

Question 9.
A card is drawn at random from a pack of 52 playing cards. Find the probability that the card drawn is neither a red card nor a black king.
Solution:
Let E be the event card drawn is neither a red card nor a black king’
The number of outcomes favourable to the event E = 24 (26 red cards and 2 black kings are not there, so 52 – 28 = 24)
∴ P(E) = \(\frac{24}{52}\) = \(\frac{16}{13}\)

Question 10.
Out of 400 bulbs in a box, 15 bulbs are defective. One bulb is taken out at random from the box. Find the probability that the drawn bulb is not defective.
Solution:
Total number of bulbs in the box = 400
Total number of defective bulbs in the box = 15
Total number of non-defective bulbs in the box = 400 – 15 = 385

Question 11.
Rahim tosses two different coins simultaneously. Find the probability of getting at least one
tail.
Solution:
The sample space is {HH, HT, TH, TT}
Total number of outcomes = 4
Outcomes for getting at least one tail is {HT, TH, TT}
Number of favourable outcomes = 3

= \(\frac{3}{4}\)

Probability Class 10 Extra Questions Short Answer Type 2

Question 1.
Harpreet tosses two different coins simultaneously (say, one is of 1 and other of 2). What is the probability that she gets at least one head?
Solution:
When two coins are tossed simultaneously, the possible outcomes are (H, H), (H, T), (T, H), (T, T) which are all equally likely. Here (H, H) means head up on the first coin (say on ₹ 1) and head up on the second coin (₹ 2). Similarly (H, T) means head up on the first coin and tail up on the second coin and so on.
The outcomes favourable to the event E, ‘at least one head’ are (H, H), (H, T) and (T, H). So, the number of outcomes favourable to E is 3.
Therefore, P(E) = \(\frac{3}{4}\)
i.e., the probability that Harpreet gets at least one head is \(\frac{3}{4}\).

Question 2.
A game consists of tossing a one-rupee coin 3 times and noting the outcome each time. Ramesh wins the game if all the tosses give the same result (i.e. three heads or three tails) and loses otherwise. Find the probability of Ramesh losing the game.
Solution:
The outcomes associated with this experiment are given by
HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
∴ Total number of possible outcomes = 8
Now, Ramesh will lose the game if he gets
HHT, HTH, THH, TTH, THT, HTT
∴ Favourable number of events = 6
∴ Probability that he lose the game = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Question 3.
Three unbiased coins are tossed together. Find the probability of getting:
(i) all heads.
(ii) exactly two heads.
(iii) exactly one head.
(iv) at least two heads.
(v) at least two tails
Solution:
Elementary events associated to random experiment of tossing three coins are
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
∴ Total number of elementary events = 8

(i) The event “getting all heads” is said to occur, if the elementary event HHH occurs, i.e., HHH is an outcome.
∴ Favourable number of elementary events = 1
Hence, required probability = \(\frac{1}{8}\)

(ii) The event “getting two heads” will occur, if one of the elementary events HHT, THH, HTH occurs.
∴ Favourable number of elementary events = 3
Hence, required probability = \(\frac{3}{8}\)

(iii) The event of “getting one head”, when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH, occurs.
Favourable number of elementary events = 3
Hence, required probability = \(\frac{3}{8}\)

(iv) If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that
the event “getting at least two heads” occurs.
∴ Favourable number of elementary events = 4
honom 4 Hence, required probability = \(\frac{4}{8}\) = \(\frac{1}{2}\)

(v) Similar as (iv) P (getting at least two tails) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Question 4.
A die is thrown once. Find the probability of getting:
(i) a prime number.
(ii) a number lying between 2 and 6.
(iii) an odd number.
Solution:
We have, the total number of possible outcomes associated with the random experiment of throwing a die is 6 (i.e., 1, 2, 3, 4, 5, 6).
(i) Let E denotes the event of getting a prime number.
So, favourable number of outcomes = 3 (i.e., 2, 3, 5)
∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
(ii) Let E be the event of getting a number lying between 2 and 6.
∴ Favourable number of elementary events (outcomes) = 3 (i.e., 3, 4, 5)
∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
(iii) Let E be the event of getting an odd number.
∴ Favourable number of elementary events = 3 (i.e., 1, 3, 5)
∴ P(E) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 5.
Suppose we throw a die once.
(i) What is the probability of getting a number greater than 4?
(ii) What is the probability of getting a number less than or equal to 4?
Solution:
(i) Here, let E be the event getting a number greater than 4′. The number of possible outcomes are six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. Therefore, the number of outcomes favourable to E is 2. So,
P(E) = P (number greater than 4) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
(ii) Let F be the event ‘getting a number less than or equal to 4’.
Number of possible outcomes = 6
Outcomes favourable to the event F are 1, 2, 3, 4.
So, the number of outcomes favourable to F is 4.
Therefore, P(F) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Question 6.
One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will:
(i) be an ace.
(ii) not be an ace.
Solution:
Well-shuffling ensures equally likely outcomes.
(i) There are 4 aces in a deck. Let E be the event ‘the card is an ace’.
The number of outcomes favourable to E = 4.
The number of possible outcomes = 52
Therefore, P(E) = \(\frac{4}{52}\) = \(\frac{1}{3}\).
(ii) Let Ē be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event Ē = 52 – 4 = 48.
The number of possible outcomes = 52.
Therefore, P(Ē) = \(\frac{48}{52}\) = \(\frac{12}{13}\)

Question 7.
Five cards – the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
Here, the total number of possible outcomes = 5.
(i) Since, there is only one queen
∴ Favourable number of elementary events = 1
∴ Probability of getting the card of queen = \(\frac{1}{5}\)

(ii) Now, the total number of possible outcomes = 4.
(a) Since, there is only one ace
∴ Favourable number of elementary events = 1
∴ Probability of getting an ace card = \(\frac{1}{4}\)

(b) Since, there is no queen (as queen is put aside)
∴ Favourable number of elementary events = 0
∴ Probability of getting a queen = \(\frac{0}{4}\)

Question 8.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) white? (iii) not green?
Solution:
Here, total number of marbles = 17
∴ Total number of possible outcomes = 17.
(i) Since, there are 5 red marbles in the box.
∴ Favourable number of elementary events = 5
∴ Probability of getting red marble = \(\frac{5}{17}\)

(ii) Since, there are 8 white marbles in the box.
∴ Favourable number of elementary events = 8
∴ Probability of getting white marble = \(\frac{8}{17}\)

(iii) Since, there are 5 + 8 = 13 marbles which are not green in the box.
∴ Favourable number of elementary events = 13
∴ Probability of not getting a green marble = \(\frac{13}{17}\)

Question 9.
A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:
(i) red.
(ii) black or white.
(iii) not black.
Solution:
Total number of balls = 5 + 7 + 3 = 15
Number of red balls = 7, Number of black or white = 5 + 3 = 8 balls
Number of not black = 7 + 3 = 10 balls

(1) P (red ball) = \(\frac{7}{15}\)
(ii) P (black or white ball) = \(\frac{8}{15}\)
(iii) P (not black ball) = \(\frac{10}{15}\) = \(\frac{2}{2}\)

Question 10.
A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is: (i) red or white. (ii) not black. (iii) neither white nor black.
Solution:
Total number of balls = 5 + 8 + 7 = 20
(i) P (red or white) = \(\frac{5+8}{20}\) = \(\frac{13}{20}\)
(ii) P (not black) = 1 – P (black) = 1 – \(\frac{7}{20}\) = \(\frac{13}{20}\)
(iii) P (neither white nor black) = P (Red balls) = \(\frac{5}{20}\) = \(\frac{1}{4}\)

Question 11.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let E be the event of having the same birthday.
Therefore, Ē is the event of not having the same birthday.
i.e., P (Ē) = 0.992 (Given)
Now, we have
P(E) + P(Ē) = 1 ⇒ P(E) = 1 – P(E) = 1 -0.992 = 0.008.

Question 12.
1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Saket has purchased one lottery ticket, what is the probability of winning a prize ?
Solution:
Out of 1000 lottery tickets, one ticket can be chosen in 1000 ways.
∴ Total number of elementary events = 1000
It is given that there are 5 prizes on these 1000 tickets.
Therefore, number of ways of selecting a prize ticket = 5 .
∴ Favourable number of elementary events
Hence, P (Winning a prize) = \(\frac{5}{1000}\) = \(\frac{1}{200}\)

Question 13.
In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice ? (ii) a total of 9 or 11 ?
Solution:
Total outcomes = 36
(i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5) (5, 2) (5, 3) (5, 5) i.e., 9 outcomes.
P (a prime number on each die) = \(\frac{9}{36}\) or \(\frac{1}{4}\)
(ii) Favourable outcomes are (3,6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) i.e., 6 outcomes
P (a total of 9 or 11) = \(\frac{6}{36}\) or \(\frac{1}{6}\)

Question 14.
Two different dice are thrown together. Find the probability that the numbers obtained
(i) have a sum less than 7
(ii) have a product less than 16
(iii) is a doublet of odd numbers.
Solution:
Total number of outcomes = 36
(i) Favourable outcomes are
(1, 1,) (1, 2) (1, 3) (1, 4) (1,5) (2, 1) (2, 2) (2, 3)
(2, 4) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (5, 1) i.e., 15
∴ P (sum less than 7) = \(\frac{15}{36}\) or \(\frac{5}{12}\)
(ii) Favourable outcomes are
(1, 1) (1, 2) (1,3) (1, 4) (1,5) (1,6) (2, 1) (2, 2) (2,3)
(2, 4) (2,5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3,5) (4, 1)
(4, 2) (4, 3) (5, 1) (5, 2) (5, 3) (6, 1) (6, 2) i.e., 25
∴ P (product less than 16) = \(\frac{25}{36}\)
(iii) Favourable outcomes are
(1, 1) (3, 3) (5, 5) 1.e, 3
∴ (doublet of odd number) = \(\frac{3}{36}\) or \(\frac{1}{12}\)
3 or 1

Probability Class 10 Extra Questions Long Answer Type

Question 1.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
(i) a king of red colour.
(ii) a face card.
(iii) a red face card.
(iv) the jack of hearts.
(v) a spade.
(vi) the queen of diamonds.
Solution:
Here, total number of possible outcomes = 52
(i) As we know that there are two suits of red card, i.e., diamond and heart and each suit contains one king.
∴ Favourable number of outcomes = 2
∴ Probability of getting a king of red colour = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(ii) As we know that kings, queens and jacks are called face cards. Therefore, there are 12 face cards.
∴ Favourable number of elementary events = 12
∴ Probability of getting a face card = \(\frac{12}{52}\) = \(\frac{3}{13}\)

(iii) As we know there are two suits of red cards, i.e., diamond and heart and each suit contains 3 face cards.
∴ Favourable number of elementary events = 2 × 3 = 6
∴ Probability of getting red face card = \(\frac{6}{52}\) = \(\frac{3}{26}\)

(iv) Since, there is only one jack of hearts.
∴ Favourable number of elementary events = 1
∴ Probability of getting the jack of heart = \(\frac{1}{52}\)

(v) Since, there are 13 cards of spade.
∴ Favourable number of elementary events = 13
∴ Probability of getting a spade = \(\frac{13}{52}\) = \(\frac{1}{4}\)

(vi) Since, there is only one queen of diamonds.
∴ Favourable number of outcomes (elementary events) = 1
∴ Probability of getting a queen of diamond = \(\frac{1}{52}\)

Question 2.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is:
(i) an ace.
(ii) red.
(iii) either red or king.
(iv) red and a king.
(v) a face card.
(vi) a red face card.
(vii) “2′ of spades.
(viii) ’10’ of a black suit.
Solution:
Out of 52 cards, one card can be drawn in 52 ways.
So, total number of elementary events = 52
(i) There are four ace cards in a pack of 52 cards. So, one ace can be chosen in 4 ways.
∴ Favourable number of elementary events = 4
Hence, required probability = \(\frac{4}{52}\) = \(\frac{1}{13}\)

(ii) There are 26 red cards in a pack of 52 cards. Out of 26 red cards, one card can be chosen in 26 ways.
∴ Favourable number of elementary events = 26
Hence, required probability = \(\frac{26}{52}\) = \(\frac{1}{2}\)

(iii) There are 26 red cards, including two red kings, in a pack of 52 playing cards. Also, there are 4 kings, two red and two black. Therefore, card drawn will be a red card or a king if it is any one of 28 cards (26 red cards and 2 black kings).
∴ Favourable number of elementary events = 28
Hence, required probability = \(\frac{28}{52}\) = \(\frac{7}{13}\)

(iv) A card drawn will be red as well as king, if it is a red king. There are 2 red kings in a pack of 52 playing cards.
∴ Favourable number of elementary events = 2
Hence, required probability = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(v) In a deck of 52 cards: kings, queens, and jacks are called face cards. Thus, there are 12 face cards. So, one face card can be chosen in 12 ways.
Favourable number of elementary events = 12
Hence, required probability = \(\frac{12}{52}\) = \(\frac{3}{13}\)

(vi) There are 6 red face cards 3 each from diamonds and hearts. Out of these 6 red face cards, one card can be chosen in 6 ways.
∴ Favourable number of elementary events = 6
Hence, required probability = \(\frac{6}{52}\) = \(\frac{3}{26}\)

(vii) There is only one ‘2’ of spades.
∴ Favourable number of elementary events = 1 Hence, required probability = 2

(viii) There are two suits of black cards viz. spades and clubs. Each suit contains one card bearing number 10.
∴ Favourable number of elementary events = 2
Hence, required probability = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 3.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,8 see Fig, and these are equally likely outcomes. What is the probability that it will point at: (i) 8? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?
Solution:

Here, total number of elementary events (possible outcomes) = 8
(i) We have only one ‘P’ on the spining plant.
∴ Favourable number of outcomes = 1
Hence, the probability that arrow points at 8 = \(\frac{1}{26}\).

(ii) We have four odd points (i.e., 1, 3, 5 and 7)
∴ Favourable number of outcomes = 4
∴ Probability that arrow points at an odd number = \(\frac{4}{8}\) = \(\frac{1}{2}\)

(iii) We have 6 numbers greater than 2, i.e., 3, 4, 5, 6, 7 and 8.
Therefore, favourable number of outcomes = 6
∴ Probability that arrow points at a number greater than 2 = \(\frac{6}{8}\) = \(\frac{3}{4}\)

(iv) We have 8 numbers less than 9, i.e, 1, 2, 3, … 8.
∴ Favourable number of outcomes = 8
∴ Probability that arrow points at a number less than 9 = \(\frac{8}{8}\) = 1.

Question 4.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is: (i) 8? (ii) 13? (iii) less than or equal to 12?
Solution:
When the blue die shows ‘l’, the grey die could show any one of the numbers 1, 2, 3, 4, 5, 6.
The same is true when the blue die shows ‘2’, ‘3’, ‘4’, ‘5’ or ‘6’. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die. So, the number of possible outcomes = 6 × 6 = 36.

(i) The outcomes favourable to the event the sum of the two numbers is 8′ denoted by E, are :
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (see figure)
i.e., the number of outcomes favourable to E = 5.
Hence, P(E) = \(\frac{5}{36}\)

(ii) As you can see from figure, there is no outcome favourable to the event F, ‘the sum of two numbers is 13’.
So, P(F) = \(\frac{0}{36}\) = 0

(iii) As you can see from figure, all the outcomes are favourable to the event G, ‘sum of two numbers ≤ 12.
So, P(G) = \(\frac{36}{36}\) = 1.

Question 5.
A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards throughly. Find the probability that the number on the drawn card is:
(i) an odd number.
(ii) a multiple of 5.
(iii) a perfect square.
(iv) an even prime number.
Solution:
Total number of cards = 49
Total number of outcomes = 49
(i) Odd number
Favourable outcomes : 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49
Number of favourable outcomes = 25

Question 6.
All the black face cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a:
(i) face card.
(ii) red card.
(iii) black card.
(iv) king.
Solution:
Cards remaining after removing black face cards = red cards + black cards excluding face cards
= 26 + 20 = 46
Total number of possible outcomes = 46
(i) Face Card
Favourable outcomes: 6 red face cards (king, queen and jack of diamond and heart suits)

Question 7.
Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is:
(i) an odd number.
(ii) a perfect square number.
(iii) divisible by 5.
(iv) a prime number less than 20.
Solution:
No. of possible outcomes = 60 – 11 + 1 = 50.
(i) An odd number
Favourable outcomes : 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59
No. of favourable outcomes = 25
Number of favourable outcomes

= \(\frac{25}{50}\) = \(\frac{1}{2}\)

(ii) A perfect square number
Favourable outcomes : 16, 25, 36, 49
No. of favourable outcomes = 04

= \(\frac{4}{50}\) = \(\frac{2}{25}\)

(iii) Divisible by 5
Favourable outcomes : 15, 20, 25, 30, 35, 40, 45, 50, 55, 60
No. of favourable outcomes = 10

= \(\frac{10}{50}\) = \(\frac{1}{5}\)

(iv) A prime number less than 20
Favourable outcomes : 11, 13, 17, 19
No. of favourable outcomes = 4

= \(\frac{4}{50}\) = \(\frac{2}{25}\)

Question 8.
A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.
Solution:
x can be any one of 1, 2, 3 or 4.
y can be any one of 1, 4, 9 or 16
Total number of cases of product of x and y = 16
Product less than 16 = (1 × 1, 1 × 4, 1 × 9, 2 × 1, 2 × 4, 3 × 1, 3 × 4, 4 × 1)
Number of cases, where product is less than 16 = 8
∴ Required probability = \(\frac{8}{16}\) or \(\frac{1}{2}\)

Question 9.
In Fig, shown a disc on which a player spins an arrow twice. The function \(\frac{a}{b}\) is formed, where ‘a’ is the number of sector on which arrow stops on the first spin and ‘b’ is the number of the sector in which the arrow stops on second spin. On each spin, each sector has equal chance of selection by the arrow. Find the probability that the fraction \(\frac{a}{b}\) > 1.
Solution:

For alb > 1, when a = 1, b can not take any value,
a = 2, b can take 1 value,
a = 3, b can take 2 values,
a = 4, b can take 3 values,
a = 5, b can take 4 values,
a = 6, b can take 5 values.
Total possible outcomes = 36

Question 10.
Two different dice are thrown together. Find the probability that the numbers obtained have
(i) even sum, and (ii) even product.
Solution:
Total number of outcomes = 36[(1, 1), (1, 2) … (6,6)]
Number of outcomes when sum is even = 18 [(1, 1), (1, 3) …(6, 6)]
Number of outcomes when product is even = 27 [(1, 2), (1, 4) … (6,6)]

Probability Class 10 Extra Questions HOTS

Question 1.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
Solution:
(i) Favourable outcomes are
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3,6), (4, 1), (4,2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)]
Total outcomes = 36
∴ Required probability = \(\frac{25}{36}\)
(ii) Probability that 5 will come atleast once = 1 – P (5 will not come up either time)
= 1 – \(\frac{25}{36}\) = \(\frac{11}{36}\)

Question 2.
Find the probability that in a leap year there will be 53 Tuesdays.
Solution:
Leap year = 366 days = (52 × 7 + 2) days = 52 weeks and 2 days.
Thus, a leap year always has 52 Tuesdays. The remaining 2 days can be:

1. Sunday and Monday
2. Monday and Tuesday
3. Tuesday and Wednesday
4. Wednesday and Thursday
5. Thursday and Friday
6. Friday and Saturday
7. Saturday and Sunday

Out of these 7 cases, we have Tuesdays in two cases
∴ P(53 Tuesdays) = \(\frac{2}{7}\)

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.
Solution:
Let there be x blue balls in the bag.
∴ Total number of balls in the bag = (5 + x)

Hence, there are 15 blue balls in the bag.

Question 4.
Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why?
Solution:
Apoorv throws two dice once.
So, total number of outcomes, n(S) = 36.
Number of outcomes for getting product 36,
n(E₂) = 1 [(6 × 6)]
∴ Probability for Apoorv getting the number 36

Also, Peehu throw one die.
So, total number of outcomes n(S) = 6
Number of outcomes for getting square of a number as 36.
n(E₂) = 1 (∵ 6² = 36)
∴ Probability for Peehu getting the number 36

Hence, Peehu has better chance of getting the number 36.

In this page, we are providing Chemical Reactions and Equations Class 10 Extra Questions and Answers Science Chapter 1 pdf download. NCERT Extra Questions for Class 10 Science Chapter 1 Chemical Reactions and Equations with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-10-science/

Class 10 Science Chapter 1 Extra Questions and Answers Chemical Reactions and Equations

Extra Questions for Class 10 Science Chapter 1 Chemical Reactions and Equations with Answers Solutions

Extra Questions for Class 10 Science Chapter 1 Very Short Answer Type

Class 10 Science Chapter 1 Extra Questions Question 1.
How does the food become rancid?
Answer:
Food becomes rancid when fat and oils present in the food are oxidised.

Chemical Reaction And Equation Class 10 Extra Questions Question 2.
A student burnt a metal A found in the form of ribbon. The ribbon burnt with a dazzling flame and a white powder B was formed which was basic in nature. Identify A and B. Write the balanced chemical equation.
Answer:
X = Mg,  Y = MgO,  Mg + O₂ → 2MgO

Chemical Reactions And Equations Extra Questions Question 3.
What is a balanced chemical equation?
Answer:
An equation that has equal number of atoms of each element on both the sides of the equation is called a balanced chemical equation, i.e., mass of the reactants is equal to mass of the products.
For example, \(2 \mathrm{Mg}+\mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{MgO}\)

Chemical Reactions And Equations Class 10 Extra Questions Answers Question 4.
Write a balanced equation for a chemical reaction that can be characterised as precipitation.
Answer:
BaCl₂(aq) + Na₂SO₄ (aq) → BaSO₄(s) + 2NaCl(aq)

Chemical Reactions And Equations Class 10 Extra Questions Pdf Question 5.
What is rust?
Answer:
It is a brown mass known as hydrated ferric oxide. Its formula is Fe₂O₃. xH₂O.

Extra Questions Of Chemical Reactions And Equations Question 6.
A zinc rod is left for nearly 20 minutes in a copper sulphate solution. What change would you observe in the zinc rod?
Answer:
The zinc rod will change into zinc sulphate.

Chemical Reaction And Equation Extra Questions Question 7.
Name two salts that are used in black and white photography.
Answer:
Both silver chloride and silver bromide are used in black and white photography.

Extra Questions For Class 10 Science Chapter 1 Question 8.
Which chemical process is used for obtaining a metal from its oxide?
Answer:
The process is known as the reduction of metal oxide.

Class 10 Chemical Reactions And Equations Extra Questions Question 9.
If you collect silver coins and copper coins you may have seen that after some days a black coating forms on silver coins and a green coating on copper coins. Which chemical phenomenon is responsible for these coatings? Write the chemical name of the black and green coatings.
Answer:
Corrosion is responsible for the formation of this coating. Black coating is due to formation of Ag₂S and green coating is due to formation of CuCO₃.Cu(OH)₂.

Class 10 Chemistry Chapter 1 Extra Questions Question 10.
When carbon dioxide is passed through lime water, it turns milky, why?
Answer:
Lime water (calcium hydroxide) combines with carbon dioxide to form a suspension of calcium carbonate which makes lime water milky.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Chemical Reactions And Equations Class 10 Extra Questions Question 11.
Identify the most reactive and least reactive metal: Al, K, Ca, Au.
Answer:
Most reactive metal: K(Potassium); least reactive metal: Au(gold).

Class 10 Science Ch 1 Extra Questions Question 12.
X + Y SO₄ → X SO₄ + Y
Y + X SO₄ → No reaction
Of the two elements T and Y which is more reactive and why?
Answer:
‘X’ is more reactive than ‘Y since it has displaced ‘Y’ in the displacement reaction.

Chapter 1 Science Class 10 Extra Questions Question 13.
Why is it necessary to balance a chemical equation?
Answer:
An equation is balanced in order to satisfy the law of conservation of mass according to which total mass of the reactants is equal to the total mass of the products, i.e., mass can neither be created nor be destroyed during any chemical change.

Class 10 Science Chapter 1 Extra Questions With Solutions Question 14.
During electrolysis of water, the gas collected in one test tube is double than the other, why?
Answer:
On electrolysis, water decomposes into hydrogen and oxygen in the ratio 2 : 1 by volume so, H₂ gas collected in one test tube is double than O₂.

Extra Questions For Class 10 Science Chapter 1 With Answers Question 15.
Represent decomposition of ferrous sulphate with the help of balanced chemical equation.
Answer:
2FeSO₄ (s) → Fe₂O₃ (s) + SO₂ (g) + SO₃ (g)

Question 16.
What is a chemical equation?
Answer:
A chemical equation is a symbolic notation that uses formulae instead of words to represent a chemical equation.

Question 17.
A teacher took a few crystals of sugar in a dry test tube and heated the test tube over a flame. The colour of sugar turned black. Explain why?
Answer:
Sugar is a complex compound which on heating undergoes decomposition. Water gets evaporated thereby leaving behind black carbon in the test tube.

Question 18.
Name two metals which do not get corroded.
Answer:
Gold (Au) and platinum (Pt) do not get corroded.

Question 19.
Identify the compound oxidised in the following reaction:
H₂S (g) + Cl₂ → S(s) + 2HCl (g)
Answer:
H₂S is oxidised.

Question 20.
Why is a magnesium ribbon cleaned before burning?
Answer:
Magnesium reacts with moist air and forms a layer of oxide, MgO (white), on its surface. So, a magnesium ribbon is cleaned to remove the oxide layer before burning.

Question 21.
State the chemical change that takes place when limestone is heated.
Answer:
Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide.

Question 22.
On which chemical law, balancing of chemical equation is based?
Answer:
Balancing of a chemical equation is based on the law of conservation of mass.

Question 23.
Name the term used for the solution of the reactants or products when dissolved in water.
Answer:
Aqueous

Question 24.
What happens when magnesium ribbon burns in air?
Ans. When magnesium ribbon burns in air, it combines with the oxygen to form magnesium oxide.
2Mg(s) + O₂(g) → 2MgO(s)

Question 25.
Give an example of an exothermic reaction.
Answer:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) + Heat (evolved)

Question 26.
What type of reaction is this: Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl
Answer:
It is a double displacement reaction.

Question 27.
Give an example of exothermic reaction.
Answer:
N₂ (g) + O₂ (g) → 2NO (g) – Heat (absorbed)

Question 28.
A substance X used for coating iron articles is added to a blue solution of a reddish brown metal Y. The colour of the solution gets discharged. Identify X and Y and also the type of reaction.
Answer:
X = Zn, Y = Cu, Displacement reaction.

Question 29.
Name the gas evolved when zinc reacts with dil. HCl.
Answer:
Hydrogen gas is evolved.

Extra Questions for Class 10 Science Chapter 1 Short Answer Type I

Question 1.
You are given the following materials
(i) Marble chips (ii) dilute hydrochloric acid (iii) Zinc granules
Identify the type of reaction when marble chips and zinc granules are added separately to acid taken in two test tubes.
Answer:
(i) Marble chips react with dilute hydrochloric acid to form calcium chloride and carbon dioxide. It is a double displacement reaction.
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

(ii) Zinc granules react with dilute hydrochloric acid to give hydrogen gas. It is a displacement reaction.
Zn (s) + 2HCl → ZnCl₂ (aq) + H₂ (g)

Question 2.
What do you understand by precipitation reaction? Explain with suitable examples.
Answer:
The reaction in which two compounds in their aqueous state react to form an insoluble compound. When two reactants react and product formed remains insoluble and settles as a solid it is substance (precipitate) is called a precipitation reaction.

For example,
(i) When aqueous solution of sodium sulphate is mixed with an aqueous solution or barium chloride, barium sulphate is obtained as a white precipitate.
Na₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) + 2NaCl (ag)

(ii) When aqueous solution of sodium chloride is mixed with an aqueous solution of silver nitrate, silver chloride is obtained as a white precipitate.

Question 3.
What happens when aqueous solutions of sodium sulphate and barium chloride are mixed? What type of reaction is it?
Answer:
On mixing the solutions of sodium sulphate and barium chloride, a white precipitate of barium sulphate is obtained.

It is a double displacement reaction.

Question 4.
Explain the following terms with suitable examples.
(a) Oxidation
(b) Reduction
Answer:
(a) Oxidation is a process of addition of oxygen to a substance or removal of hydrogen from a substance, for example,

Copper is oxidised to CuO, as oxygen is added to copper.

(b) It is the process of removal of oxygen from a substance or addition of hydrogen to a substance, for example,

Copper oxide is reduced to copper as it involves removal of oxygen.

Question 5.
Complete the missing components/variables given as x and y in the following reactions.  [NCERT Exemplar]
(a) Pb(NO₃)₂ (aq) + 2Kl (aq) → PbI₂ (x) + 2KNO₃ (y)
(b) Cu (s) + 2AgNO₃ (aq) → Cu(NO₃)₂ (aq) + x (s)
(c) Zn (s) + H₂SO₄ (aq) → ZnSO₄ (x) + H₂ (y)

Answer:
(a) x = (s), y = (aq)
(b) x = 2Ag
(c) x = (aq); y = (g)
(d) x = heat

Question 6.
An iron knife kept dipped in a blue copper sulphate solution turns the blue solution light green. Why?
Answer:
We know that iron is more reactive than copper, so it displaces copper from copper sulphate solution and forms ferrous sulphate which is of light green colour.

Question 7.
A, B and C are three elements which undergo chemical reactions in the following way.
A₂O₃ + 2B → B₂O₃ + 2A
3CSO₄ + 2B → B₂(SO₄)₃ + 3C
3CO + 2A → A₂O₃ + 3C
Answer the following:
(a) Which element is most reactive?
(b) Which element is least reactive?
Answer:
(a) The most reactive element is ‘B’. It has displaced both ‘A’ and ‘C’ from their compounds.
(b) The least reactive element is ‘C’ as it has been displaced by both ‘A’ and ‘B’.

Question 8.
Write the balanced chemical equations for the following reactions and identify the type of reaction in each case.
(a) Nitrogen gas is treated with hydrogen gas in the presence of a catalyst at 773 K to form ammonia gas.
(b) Sodium hydroxide solution is treated with acetic acid to form sodium acetate and water.
(c) Ethanol is warmed with ethanoic acid to form ethyl acetate in the presence of concentrated H₂SO₄.
(d) Ethene is burnt in the presence of oxygen to form carbon dioxide, water and releases heat and light.  [NCERT Exemplar]
Answer:

Question 9.
What is lime water test for the detection of carbon dioxide?
Answer:
When carbon dioxide gas is passed through lime water, it turns milky due to the formation of a milky suspension (precipitate) of calcium carbonate.
Ca(OH)₂ (aq) + CO₂ (g) → CaCO₃ (s) + CO₂ (g)

Question 10.
During the reaction of some metals with dilute hydrochloric acid, following observations were made,
(a) Silver metal does not show any change.
(b) The temperature of the reaction mixture rises when aluminium (Al) is added.
(c) The reaction of sodium metal is found to be highly explosive.
(d) Some bubbles of a gas are seen when lead (Pb) is reacted with the acid.
Explain these observations giving suitable reasons.   [NCERT Exemplar]
Answer:
(a) No change takes place because silver metal does not react with hydrochloric acid in normal situations.
(b) The reaction between hydrochloric acid and aluminium is exothermic, thus the temperature of the reaction mixture rises when aluminium is added.
(c) Since, sodium is a highly reactive metal, thus it reacts with hydrochloric acid vigorously and produces a large amount of heat. Thus, the reaction is exothermic.
(d) Bubbles of hydrogen gas are formed when lead react with dilute hydrochloric acid.
Pb + 2HCl → PbCl₂ + H₂

Question 11.
A copper coin is kept in a solution of silver nitrate for some time. What will happen to the coin and the colour of the solution?
Answer:
We know that copper is more reactive than silver, so it will displace silver from its salt solution.
Cu (s) + 2AgNO₃ (aq) → Cu(NO₃)₂ (aq) + 2Ag (s)
So, the solution will turn blue due to the formation of copper nitrate.

Question 12.
An aqueous solution of metal nitrate P reacts with sodium bromide solution to form yellow ppt. of compound Q which is used in photography. Q on exposure to sunlight undergoes decomposition reaction to form metal present in P along with reddish brown gas. Identify P and Q. Write the chemical reaction and type of chemical reaction.
Answer:
P = Ag NO₃, Q = AgBr
AgNO₃ (aq) + NaBr (aq) → NaNO₃ (aq) + AgBr (s); Double decomposition reaction
2AgBr (s) → 2Ag (s) + Br₂ (g); Photochemical decomposition reaction

Question 13.
What happens when iron nails are immersed in copper sulphate solution? What type of reaction is it?
Answer:
When iron nails are immersed in copper sulphate solution, iron ions displace copper ions and a new compound ferrous sulphate is formed.

It is a diplacement reaction.

Question 14.
Which of the following reaction is possible. Explain giving suitable reason.
(i) Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)
(ii) Fe (s) + ZnSO₄ (aq) → FeSO₄ (aq) + Zn (s)
(iii) Zn (s) + FeSO₄ (s) → ZnSO₄ (aq) + Fe (s)
Answer:
Reaction (i) and (iii) are possible.
(i)

Zinc is more reactive than copper, therefore, it can displace copper from copper sulphate solution.

(iii) Zn (s) + FeSO₄ (aq) → ZnSO₄ (aq) + Fe (s)
Zn is more reactive than Fe, therefore, it can displace iron from ferrous sulphate solution.

Reaction (ii) is not possible as iron is less reactive than zinc, hence, it cannot displace Zn.

Question 15.
Which among the following changes are exothermic or endothermic in nature?
(a) Decomposition of ferrous sulphate
(b) Dilution of sulphuric acid
(c) Dissolution of sodium hydroxide in water
(d) Dissolution of ammonium chloride in water  [NCERT Exemplar]
Answer:
(a) endothermic
(b) exothermic
(c) exothermic
(d) endothermic.

Question 16.
State one advantage and one disadvantage of corrosion.
Answer:
Advantage. In some metals a protective layer is formed on its surface due to corrosion which prevent it from further corrosion.

Example, Aluminium (Al) forms a layer of aluminium oxide (Al₂O₃) by corrosion. This layer prevents further corrosion of aluminium.
Disadvantage: Loss of the metal.

Extra Questions for Class 10 Science Chapter 1 Short Answer Type II

Question 1.
What is corrosion? State condiyions necessary for rusting of iron. How is rusting harmful?  [NCERT Exemplar]
Answer:
Corrosion. The process of eating away of the metal surface by the action of atmospheric reagents like water, oxygen and acids changing the metal into its compound is called corrosion.

Rusting of iron. When iron objects are exposed to atmosphere, they are attacked by air and moisture (water) of the atmosphere and a brown and orange coloured layer is formed on the surface. It is called rust which is mainly hydrated iron (III) oxide Fe₂O₃. xH₂O.

Harmful effect of rusting. Hydrated iron (III) oxide is a brittle substance and falls off from the surface of iron and thus the object is damaged. Holes, cavities and roughness of surfaces are the result of rusting of an iron object.

Conditions necessary for rusting:

• Open surface of the metal
• Presence of air (oxygen)
• Presence of moisture (water).

Question 2.
The gases hydrogen and chlorine do not react with each other even if kept together for a long time. However, in the presence of sunlight, they readily combine. What actually happens?   [NCERT Exemplar]
Answer:
In chemical reactions, energy is needed to break the bonds present in the reacting molecules so that they may combine to form the products. In this reaction, sunlight is the source of energy in the form of photons. The energy made available by sunlight helps in breaking the bonds and this leads to chemical reaction between hydrogen and chlorine.

Question 3.
Write the balanced chemical equations for the following reactions and identify the type of reaction in each case.
(a) Thermite reaction, iron (III) oxide reacts with aluminium and gives molten iron and aluminium oxide.
(b) Magnesium ribbon is burnt in an atmosphere of nitrogen gas to form solid magnesium nitride.
(c) Chlorine gas is passed in an aqueous potassium iodide solution to form potassium chloride solution and solid iodine.
(d) Ethanol is burnt in air to form carbon dioxide, water and releases heat.   [NCERT Exemplar]
Answer:
The balanced equations are as under:
(a) Fe₂O₃ (s) + 2Al (s) → 2Fe (l) + Al₂O₃ (s) + Heat
It is a redox reaction / displacement reaction.

(b) 3Mg (s) + N₂ (g) → Mg₃N₂ (s)
It is a combination reaction as well as redox reaction.

(c) Cl₂ (g) + 2KI (aq) → 2KCl (aq) + I₂ (s)
It is a displacement reaction as well as redox reaction.

(d) C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (l) + Heat
Redox reaction/combustion reaction

Question 4.
What is rancidity? Write the common method to prevent it.
Answer:
When food item are kept unprotected for some time, their smell and taste changes. This process is called rancidity. Actually, the microorganisms oxidise the fat and oils present in them. So oxidation of food items need to be prevented to protect them.

Common methods to prevent rancidity of food item:

• Keeping the food at low temperature.
• Keeping food item in air tight containers.
• By filling nitrogen in the food storage bags.

Question 5.
(a) What happens chemically when quick lime is added to water?
(b) Write the chemical equation in balanced form.
MnO₂ + HCl → MnCl₂ + Cl₂ + H₂O
(c) What is decomposition reaction? Explain it with a suitable example.
Answer:
(a) When quick lime (CaO) is added to water, slaked lime Ca(OH)₂ is formed. The reaction is highly exothermic in nature.

(b) The balanced chemical equation is:
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

(c) Decomposition reaction is a chemical reaction in which a single substance splits or breaks into two or more substances under suitable conditions. For example,
2FeSO₄(s) → Fe₂O₃(s) + SO₂(g) + SO₃ (g)

Question 6.
Identify the oxidising agent (oxidant) in the following reactions:
(a) Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O
(b) 2Mg + O₂ → 2MgO
(c) CuSO₄ + Zn → Cu + ZnSO₄
(d) V₂O₅ + 5Ca → 2V + 5CaO
(e) 3Fe + 4H₂O → Fe₃O₄ + 4H₂
(f) CuO + H₂ → Cu + H₂O [NCERT Exemplar]
Answer:
(a) Pb₃O₄
(b) O₂
(c) CuSO₄
(d) V₂O₅
(e) 4H₂O
(f) CuO

Question 7.
Write the balanced chemical equations for the following reactions:  [NCERT Exemplar]
(a) Sodium carbonate on reaction with hydrochloric acid in equal molar concentrations gives sodium chloride and sodium hydrogencarbonate.
(b) Sodium hydrogencarbonate on reaction with hydrochloric acid gives sodium chloride, water and liberates carbon dioxide.
(c) Copper sulphate on treatment with potassium iodide precipitates cuprous iodide (Cu₂I₂), liberates iodine gas and also forms potassium sulphate.
Answer:
(a) Na₂CO₃ + HCl → NaCl + NaHCO₃
(b) NaHCO₃ + HCl → NaCl + H₂O + CO₂
(c) 2CUSO₄ + 4KI → 2K₂SO₄ + Cu₂I₂ + I₂

Question 8.
Write chemical equations for the reactions taking place when:
(i) Iron reacts with steam
(ii) Magnesium reacts with dilute HCl
(iii) Copper is heated in air
Answer:

Question 9.
fialance the following chemical equations and identify the type of chemical reaction.
(a) Mg (s) + Cl₂ (g) → MgCl₂ (s)

(d) TiCl₄ (l) + Mg (s) → Ti (s) + MgCl₂ (s)
(e) CaO (s) + SiO₂ (s) → CaSiO₃ (s)

[NCERT Exemplar]
Answer:
The chemical equation in their balanced form may be written as follows:
(a) Mg (s) + Cl₂ (g) → MgCl₂ (s), Combination reaction

(d) TiCl₄ (l) + 2Mg (s) → Ti (s) + 2MgCl₂ (s); Displacement reaction
(e) CaO (s) + SiO₂ (s) → CaSiO₃ (s); Combination reaction

Question 10.
A silver article generally turns black when kept in the open for a few days. The article when rubbed with toothpaste again starts shining.
(a) Why do silver articles turn black when kept in the open for a few days? Name the phenomenon involved.
(b) Name the black substance formed and give its chemical formula.   [NCERT Exemplar]
Answer:
(a) Silver articles turn black when kept in the air for a few days because H₂S gas present in the air attacks silver forming a coating of black silver sulphide. The phenomenon is called corrosion.

(b) Black substance formed is silver sulphide (Ag₂S)
2Ag (s) + H₂S (g) → Ag₂S (s) + H₂ (g).

Question 11.
Which among the following are physical or chemical changes?
(a) Evaporation of petrol
(b) Burning of Liquefied Petroleum Gas (LPG)
(c) Heating of an iron rod to red hot
(d) Curdling of milk
(e) Sublimation of solid ammonium chloride  [NCERT Exemplar]
Answer:
(a) Physical change
(b) Chemical change
(c) Physical change
(d) Chemical change
(e) Physical change

Extra Questions for Class 10 Science Chapter 1 Long Answer Type

Question 1.
Balance the following equations:
(a) Bacl₂ + H₂SO₄ → BaSO₄ + HCl
(b) CH₄ + O₂ → CO₂ + H₂O



Answer:
(a) Bacl₂ + H₂SO₄ → BaSO₄ + 2HCl
(b) CH₄ + 2O₂ → CO₂ + 2H₂O

Question 2.
On heating blue coloured powder of copper (II) nitrate in a boiling tube, copper oxide (black), oxygen gas and a brown gas X is formed.
(а) Write a balanced chemical equation of the reaction.
(b) Identify the brown gas X evolved.
(c) Identify the type of reaction.
(d) What could be the pH range of aqueous solution of the gas X?  [NCERT Exemplar]
Answer:

(b) Brown gas X is nitrogen dioxide (NO₂).
(c) It is a thermal decomposition reaction.
(d) The gas (NO₂) is an oxide of a non-metal. Hence, its aqueous solution will be acidic, i.e., pH range would be between 0 and 7.

Question 3.
(A) Name the type of chemical reaction represented by the following equation:

(c) Zn(s) + H₂SO₄(ag) → ZnSO₄(aq) + H₂(g)
(B) “A solution of potassium chloride when mixed with silver nitrate solution, and an insoluble white substance is formed”.   [CBSE 2010, 2012]
(i) Translate the above statement into a chemical equation.
(ii) State two types for the classification of this reaction.
Answer:
(A) (a) Decomposition reaction
(b) Combination reaction
(c) Displacement reaction.

(B) (i) KCl (aq) + AgNO₃ (aq) → AgCl (s) + KNO₃ (aq)
(ii) It is a double displacement reaction also called precipitation reaction.

Question 4.
What happens when zinc granules are treated with dilute solution of H₂SO₄, HCl, HNO₃, NaCl and NaOH, also write the chemical equations if reaction occurs.  [NCERT Exemplar]
Answer:

Question 5.
(A) Write the balanced chemical equations for the following chemical reactions:
(i) Hydrogen + Chlorine → Hydrogen chloride
(ii) Lead + Copper chloride → Lead chloride + Copper
(iii) Zinc oxide + Carbon → Zinc + Carbon monoxide
(B) Write balanced chemical equations for the following reactions:
(a) Silver bromide on exposure to sunlight decomposes into silver and bromine.
(b) Sodium metal reacts with water to form sodium hydroxide and hydrogen gas.
Answer:
(A) (i) H₂ + Cl₂ → 2HCl
(ii) Pb + CuCl₂ → PbCl₂ + Cu
(iii) ZnO + C → Zn + CO
(B)

(b) 2Na + 2H₂O→ 2NaOH + H₂

Question 6.
(a) Why cannot a chemical change be normally reversed?
(b) Why is it always essential to balance a chemical equation?
(c) What happens when CO₂ gas is passed through lime water and why does it disappear on passing excess CO₂?
(d) Can rusting of iron takes place in distilled water?
Answer:
(a) In a chemical change some bonds are broken and some bonds are formed. The products are quite different from the reactants. Therefore, it normally can’t be reversed.
(b) A chemical equation has to be balanced to satisfy the law of conservation of mass.
(c) On passing CO₂ gas through lime water, it turns milky due to formation of insoluble calcium carbonate which dissolves on passing excess CO₂ due to formation of soluble calcium bicarbonate.
Ca(OH)₂ + CO₂(g) → CaCO₃(s) + H₂O(l)
CaCO₃(s) + H₂O(l) + CO₂(g) → Ca(HCO₃)₂ (soluble)
(d) No

Question 7.
What happens when a piece of
(a) zinc metal is added to copper sulphate solution?
(b) aluminium metal is added to dilute hydrochloric acid?
(c) silver metal is added to copper sulphate solution?
Also, write the balanced chemical equation if the reaction occurs.  [NCERT Exemplar]
Answer:
(a) Zinc is more reactive than copper. It displaces Cu from CuSO₄ solution forming colourless zinc sulphate. Thus, blue colour of CuSO₄ solution starts fading and ultimately blue colour disappears.

(b) Aluminium reacts with dilute HCl acid forming AlCl₃ along with evolution of bubbles of H₂ gas.

(c) Silver is less reactive than copper. Hence, Ag cannot displace Cu from CuSO₄ solution. Thus, no reaction occurs.

Question 8.
(A) A brown substance ‘X’ on heating in air forms a substance ‘Y’. When hydrogen gas is passed over heated ‘Y’, it again changes back into ‘X’.
(i) Name the substances X and Y.
(ii) Name the chemical processes occurring during both the changes.
(iii) Write the chemical equations.   [CBSE 2011]
(B) A metal is treated with dil. H₂SO₄. The gas evolved is collected by the method shown in the figure. Answer the following:

(i) Name the gas.
(ii) Name the method of collection of the gas.
(iii) Is the gas soluble or insoluble in water?
(iv) Is the gas lighter or heavier than air?
Answer:
(A) (i) The substance X is copper and Y is copper (II) oxide or CuO.
(ii) The process of change of X into Y is oxidation. The process of change of Y into X is reduction.
(iii) The chemical equations are:

(B) (i) The gas evolved is hydrogen.
(ii) The method of collection of the gas is the downward displacement of water.
(iii) The gas is insoluble in water. That is why, it can be collected over water.
(iv) The gas is lighter than air.

Chemical Reactions and Equations HOTS Questions With Answers

Question 1.
The marble statues often slowly get corroded when kept in open for a long time. Assign a suitable explanation.
Answer:
SO₂, NO₂ gases are released into the atmosphere from various sources. These dissolve in rain water to give acid which corrodes marble statues.
2SO₂ + O₂ → 2SO₃
H₂O + SO₃ → H₂SO₄
2NO₂ + H₂O → 2HNO₃
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂

Question 2.
Observe the following activity and identify the following:

(a) Type of chemical reaction.
(b) Write a chemical equation to represent the above change.
(c) Name the gas evolved in the above reaction.
(d) Do you observe any colour change in the above reaction? If yes, mention the colour change.
(e) Name the coloured substance formed.
Answer:
(a) Decomposition reaction.
(b) 2Pb(NO₃)₂ (s) → 2PbO (s) + 4NO₂ (g) + O₂ (g)
(c) Reddish brown fumes of nitrogen dioxide along with oxygen gas is evolved.
(d) The colour of the substance changes from white to yellow due to formation of lead oxide.
(e) Lead oxide which is yellow coloured.

Question 3.
A substance X, which is an oxide of a group 2 element, is used intensively in the cement industry. This element is present in bones also. On treatment with water it forms a solution which turns red litmus blue. Identify X and also write the chemical reactions involved.   [NCERT Exemplar]
Answer:
The above information suggests that the substance ‘A’ is oxide of the element calcium (Ca) which is present in group 2 of the periodic table. Calcium is also a constituent of our bones in the form of calcium phosphate. Calcium oxide (CaO) reacts with water to form calcium hydroxide (basic in nature). It forms a basic solution which turns red litmus blue.
CaO (s) + H₂O (aq) → Ca(OH)₂ (aq)

Question 4.
On adding a drop of barium chloride solution to an aqueous solution of sodium sulphite, a white precipitate is obtained.
(a) Write a balanced chemical equation of the reaction involved.
(b) What other name can be given to this precipitation reaction?
(c) On adding dilute hydrochloric acid to the reaction mixture, white precipitate disappears. Why? [NCERT Exemplar]
Answer:
(a) A white precipitate of barium sulphite is formed when barium chloride is added to the solution of sodium sulphite.
BaCl₂ (aq) + Na₂SO₃ (aq) → BaSO₃ (s) + 2NaCl (aq)

(b) This precipitation reaction is also a double displacement reaction.

(c) Barium chloride, sulphur dioxide and water are formed when dilute hydrochloric acid is added to this solution of barium sulphate and sodium chloride. Since barium chloride is a soluble substance, thus white precipitate of barium sulphite disappears.
BaSO₃ (s) + HCl (ag) → BaCl₂ (ag) + SO₂ (g) + H₂O

Question 5.
A water insoluble substance ‘X’ on reacting with dilute H₂SO₄ released a colourless and odourless gas accompanied by brisk effervescence. When the gas was passed through water, the solution obtained turned blue litmus red. On bubbling the gas through lime water, it initially became milky and the milkiness disappeared when the gas was passed in excess. Identify the substance ‘X’ and write the chemical equations of the reaction involved.
Answer:
The water insoluble substance ‘X’ is most probably the metal carbonate (CaCO₃). The chemical reaction that were involved are given below.
CaCO₃ (s) + H₂SO₄ (aq) → CaSO₄ (ag) + H₂O (aq) + CO₂ (g)
Ca(OH)₂ (aq) + CO₂ (g) → CaCO₃ (s) (milky) + H₂O (l)
CaCO₃ (s) + CO₂ (g) + H₂O (aq) → Ca (HCO₃)₂ (milkiness)

Question 6.
The given set up shows the electrolysis of water taking place.

(a) Identify the gases evolved at cathode and anode.
(b) Why is the amount of gas collected in one of the test tubes double the amount in the other? Name this gas.
(c) How will you test the evolved gases?
Answer:
(a) The gases evolved at anode and cathode are oxygen and hydrogen.

(b) On electrolysis, water decomposes to form hydrogen and oxygen gas in the ratio of 2 : 1 by volume according to the equation given below:

The gas collected in double volume is hydrogen.

(c) Test for oxygen. If we bring a burning splinter near the mouth of test tube containing oxygen gas it burns more brightly.
Test for hydrogen. On bringing a burning splinter near the mouth of the test tube containing hydrogen, the gas burns with a pop sound.

Question 7.
A magnesium ribbon is burnt in oxygen to give a white compound X accompanied by emission of light. If the burning ribbon is now placed in an atmosphere of nitrogen, it continues to burn and forms a compound Y.
(a) Write the chemical formulae of X and Y.
(b) Write a balanced chemical equation, when X is dissolved in water.   [NCERT Exemplar]
Answer:

Question 8.
You are provided with two containers made up of copper and aluminium. You are also provided with solutions of dilute HCl, dilute HNO₃, ZnCl₂ and H₂O. In which of the above containers these solutions can be kept? [NCERT Exemplar]
Answer:
(A) When solutions are kept in copper container
(a) Dilute HCl
Copper does not react with dilute HCl. Therefore, it can be kept.

(b) Dilute HNO₃
Nitric acid acts as a strong oxidising agent and reacts with copper vessel, therefore cannot be kept.

(c) ZnCl₂
Zinc is more reactive than copper (Cu) therefore, no displacement reaction occurs and hence can be kept.

(d) H₂O
Copper does not react with water. Therefore, can be kept.

(B) When solutions are kept in aluminium containers
(a) Dilute HCl
Aluminium reacts with dilute HCl to form its salt and hydrogen is evolved. Therefore, cannot be kept.
2 Al + 6HCl → 2AlCl₃ + 3H₂

(b) Dilute HNO₃
Aluminium gets oxidised by dilute HNO₃ to form a layer of Al₂O₃ and can be kept.

(c) ZnCl₂
Aluminium being more reactive than zinc can displace zinc ion from the solution. Therefore, the solution cannot be kept.
2Al + 3ZnCl₂ → 2AlCl₃ + 3Zn

(d) H₂O
Aluminium does not react with cold or hot water. Therefore, water can be kept.
Aluminium is attacked by steam to form aluminium oxide and hydrogen.
2Al(s) + 3H₂O(g) → Al₂O₃(S) + 3H₂ (g)

Extra Questions for Class 10 Science Chapter 1 Value Based

Question 1.
Rakesh wanted to give a coating of white wash on the walls of his house. He purchased quick lime (CaO) and dropped it in container of water and immediately applied the same on the walls. In this process, he spoiled his hands and even suffered minor bums. His friend Kapil advised him to keep the container overnight before applying a coating on the wall.
(i) What mistake was committed by the Rakesh?
(iii) Why did he suffer from minor bums?
(iii) How was Kapil’s advice useful?
(iv) What values are exhibited by Kapil?
Answer:
(i) Rakesh should have waited for a few hours because when quick lime dissolves in water, slaked lime is formed and the process is highly exothermic (a lot of energy is released).

(ii) Because the reaction is exothermic and a lot of energy is released and container becomes hot so he suffered from minor burns.

(iii) Quick lime (CaO) reacts with water to form Ca(OH)₂ which is known as slaked lime. The dissolution process is highly exothermic. So a large amount of heat is evolved. By keeping container overnight, the chemical reaction subsides and the heat dissipates. So the coating of slaked lime can be applied safely on the walls.

(iv) Knowledge of Chemistry, helpful and caring nature.

Question 2.
Palak is an eleven year old girl. She purchased a packet of potato chips and had a few from it more than a month back. She had kept away the open packet containing the remaining potato chips then. She wanted to eat the remaining potato chips now. Her elder sister Anjali found that the potato chips were giving out an unpleasant odour. When she put one of them in her mouth, she found that it had an unpleasant taste too. Anjali threw away the packet and did not allow Palak to eat the potato chips.
(a) What name is given to the condition in which potato chips kept in the open for a long time give out unpleasant smell and taste?
(b) Which chemical reaction is responsible for the spoilage of potato chips?
(c) Explain the reason for the unpleasant smell as well as unpleasant taste of the potato chips.
(d) Mention the values exhibited by Anjali.
Answer:
(a) Rancidity.
(b) Oxidation reaction.
(c) Potato chips contain oil. The oil present in potato chips (which have been kept exposed to air for a considerable time) gets oxidised by the oxygen of air. The oxidation products have unpleasant smell and taste. The potato chips are then said to have turned rancid. They become unfit to eat.
(d) (i) Awareness (or knowledge of Chemistry)
(ii) Concern for the health of her sister.