CBSE Sample Papers for Class 10 Science Set 4 with Solutions - #NCSOLVE 📚

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 4 with Solutions

Time : 3Hrs
Max. Marks: 80

General Instructions

• This question paper consists of 39 questions in 3 sections. Section A is Biology, Section B is Chemistry and Section C is Physics.
• All questions are compulsory. However, an internal choice is provided in some questions.
• A student is expected to attempt only one of these questions.

Section – A

Question 1.
The characteristic feature observed in anaerobic respiration is [1]
(a) release of oxygen
(b) lack of carbon dioxide
(c) release of high amount of energy
(d) release of lactic acid or ethyl alcohol
Answer:
(d) release of lactic acid or ethyl alcohol

Explanation:
A characteristic feature of anaerobic respiration is the release of lactic acid or ethyl alcohol formed by partial breakdown of glucose.

Question 2.
Which of the following correctly shows the role of organisms at different trophic levels in a food chain? [1]
(a) Primary consumer – Eats plants
(b) Producer – Feeds on herbivores
(c) Decomposer – Makes food using sunlight
(d) Secondary consumer – Converts solar energy to food
Answer:
(a) Primary consumer – Eats plants

Explanation:
Primary consumer- Eats plants shows the correct combination as primary consumers are herbivores that feed directly on producers like green plants.

Question 3.
Which of the following statements are correct about the process of urine formation in human beings? [1]
(i) Filtration of blood takes place in bowman’s capsule.
(ii) Useful substances are reabsorbed in the tubules.
(iii) Urine is formed directly in the glomerulus.
(iv) Additional wastes are secreted into the tubules.
(v) Urine is directly collected in the urinary bladder from the tubules.
(a) (i), (ii), (iv)
(b) (ii), (iii), (v)
(c) (i), (iii), (iv)
(d) (ii), (iv), (v)
Answer:
(a) (i), (ii), (iv)

Explanation:
The statements (i), (ii) and (iv) are correct as urine formation involves filtration in Bowman’s capsule, reabsorption of useful substances in the tubules, and secretion of additional wastes into the tubules.

Question 4.
Which of the following represents reflex action? [1]
(a) Beating of heart
(b) Withdrawing your hand immediately on touching hot object
(c) Riding a bicycle
(d) Kicking a football while playing
Answer:
(b) Withdrawing your hand immediately on touching hot object

Explanation:
Withdrawing your hand immediately on touching a hot object is a reflex action as reflex actions are rapid response to a stimulus.

Question 5.
Select the group in which all parts belong to the human heart. [1]
(a) Aorta, alveoli, left atrium, pulmonary vein
(b) Right ventricle, vena cava, bronchi, pulmonary artery
(c) Left ventricle, right atrium, pulmonary artery, vena cava
(d) Pulmonary vein, trachea, aorta, right atrium
Answer:
(c) Left ventricle, right atrium, pulmonary artery, vena cava

Explanation:
Left ventricle, right atrium, pulmonary artery’ and vena cava belong to the same group as all of them are involved in blood circulation in the heart.

Question 6.
During a Mendelian experiment, a cross-breeding is done between tall pea plants bearing violet flower and dwarf pea plant with white flowers. The F, -generation produced have all violet flowers, but half of them were short. What can be the genotype of the tall parent? [1]
(a) TTww
(b) TTWW
(c) TtWW
(d) TtWw
Answer:
(c) TtWW

Explanation:
The genotype of the tall parent is TtWW, as all offspring have violet flowers, indicating that the parent is homozygous dominant (WW) for flower colour. Since only half the offspring are tall, the parent must be heterozygous (Tt) for height.

Question 7.
Which one is not a reason for the success of Mendel? [1]
(a) Made statistical analysis of the offspring
(b) Kept accurate records
(c) Selection of pea plant
(d) Performed only self-pollination in plants.
Answer:
(d) Performed only self-pollination in plants.

Explanation:
Performing self-pollination in plants is not the reason for the success of Mendel because he performed both self-pollination and cross-pollination in pea plants.

The following two questions consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.

Question 8.
Assertion (A) Damage to the cerebellum can lead to difficulty in maintaining posture and body balance.
Reason (R) The cerebellum controls voluntary actions like writing and walking. [1]
Answer:
(c) (A) is true but (R) is false.

Explanation:
A is true but R is false. R can be corrected as The cerebrum controls voluntary’ activities like writing,walking and balance and coordination are controlled by cerebellum.

Question 9.
Assertion (A) Ureter forms the common passage for both the sperms and urine.
Reason (R) Ureter never carries sperms. [1]
Answer:
(d) (A) is false but (R) is true.

Explanation:
A is false, but R is true. A can be corrected as In males, urethra forms the common passage for both sperms and urine.

Question 10.
Recessive traits may not appear in the first generation, but can reappear in the second generation. Comment on the statement with justification. [2]
Answer:
Recessive traits remain hidden in the F₁ generation because dominant traits mask their expression. However, when F₁ plants are self-pollinated, the recessive traits reappear in the F₂-generation. This is because the recessive alleles get a chance to pair together. Mendel showed this by crossing pure tall and dwarf pea plants in which all the F₁, were tall, but dwarf plants reappeared in F₂.

Question 11.
Attempt either A or B
A. In the human male reproductive system, certain organs are responsible for the production of male gametes and for maintaining the conditions needed for their proper development. With reference to the human male reproductive system, answer the following questions. [2]
(i) Name the organ that releases sperms and state its function.
(ii) Why are testes located outside the abdominal cavity in the scrotum?
Answer:
(i) The testes are the organs that release sperms.
Their function is to produce sperms and the hormone testosterone.

(ii) Testes are located outside the abdominal cavity in the scrotum because sperm formation requires a slightly lower temperature than the normal body temperature.

Or

B. Describe the menstrual cycle in human females. What is its significance?
Answer:
The menstrual cycle is a 28-day cycle in human females during which the uterus prepares to receive a fertilised egg. If fertilisation does not occur, the thick lining of the uterus breaks down and is released as blood and mucus. It is significant as it prepares the female body every month for pregnancy.

Question 12.
About 100 acres of agricultural land were planted with pea plants for experimental purposes. It was observed that the tendrils of the pea plants coiled around nearby supports. [2]
How do the tendrils of a pea plant coil around a support?
Answer:
When a tendril touches a support, the part of the tendril that is in contact grows slowly’, while the other side grows faster. This causes the tendril to bend and wrap around the support. This type of movement happens due to growth.

Value Point
To get maximum marks, students should mention that tendrils are sensitive to touch and their growth is directional.

Question 13.
Draw the diagram of human heart and explain the flow of blood through the heart. [3]
Answer:
The human heart is a muscular organ that pumps blood throughout the body. It receives deoxygenated blood from the body into the right atrium, which then flows into the right ventricle. This blood is pumped to the lungs for oxygenation. The oxygen-rich blood returns to the left atrium, moves into the left ventricle, and is finally pumped to the rest of the body. This process maintains continuous circulation of blood.

Question 14.
Mendel studied the inheritance of traits in pea plants and explained how characters are passed from parents to offspring. His experiments helped in understanding the basic principles of heredity. [3]
(i) What is heredity?
(ii) What decides whether a trait is dominant or recessive in Mendelian inheritance?
Answer:
(i) Heredity is the transmission of characters or traits from the parents to their offspring. It is the reason that offspring resemble their parents.
(ii) In Mendel’s experiments, a trait is said to be dominant if it is expressed in the F₁, generation, and a trait which is not expressed in the F₁, generation but reappears in the F₂ generation is called recessive.

Question 15.
Ravi and Ananya are studying the reproductive system in humans. They discussed the different organs and processes involved in human reproduction. Help them understand the concepts better by answering the questions given below. [4]
Attempt either subpart A or B
A. Which organ in the human male reproductive system produces sperms? Name the hormone secreted by this organ.
Answer:
The organ in the male reproductive system that produces sperms is the testis. It also produces a hormone called testosterone.

Or

B. What is the role of the uterus in the female reproductive system? Name the part where fertilisation takes place.
Answer:
The uterus is a muscular organ in the female reproductive system. Its main role is to hold and nourish the developing embryo. Fertilisation takes place in the fallopian tube (oviduct).

C. What is puberty? Mention two changes observed in girls during puberty.
Answer:
Puberty is the stage when the body becomes sexually mature and capable of reproduction. In girls, it begins around the age of 10 to 14 and brings both physical and hormonal changes. Two important changes observed are the development of breasts and the beginning of the menstrual cycle.

D. The diagram given below shows the female reproductive system. Which part releases the egg and which part is the site of implantation?

Answer:
The given diagram shows the female reproductive system with parts labelled as A, B, C, D and E. Out of these the part labelled as B is the ovary which releases the egg and the part labelled as A is the uterus which is the site for implantation.

Question 16.
Attempt either A or B
A. A forest officer named Riya observed a decline in the number of snakes and owls in a protected forest area. She also noticed an increase in the population of rats and insects. [5]
(i) Based on your understanding of food webs, explain why the population of rats and insects increased.
(ii) How is the balance in a food web maintained?
Answer:
(i) In a food web, snakes and owls are predators that feed on rats and insects. When the number of snakes and owls decreases, their prey rats and insects are not controlled. As a result, the population of rats and insects increases because there are fewer predators to feed on them.

(ii) The balance in a food web is maintained through interdependence among organisms. Each organism plays a role, the producers make food, herbivores consume plants, and carnivores control the population of herbivores.
If one species is removed or reduced, it affects all the others. Thus, a balanced food web keeps the ecosystem stable.

Or

B. During a field study, Aarav prepared a food web using the following organisms: grass, grasshopper, frog, snake, eagle and mushroom.
(i) Identify one food chain from this web and name the producer, primary consumer and top carnivore.
(ii) Aarav noticed that some organisms in his food web were a part of more than one food chain. What does this indicate about the structure of a food web as compared to a food chain? Explain with an example from his food web.
Answer:
(i) One food chain from the given food web is:
Grass → Grasshopper → Frog → Snake → Eagle.
In this food chain, grass acts as the producer as it prepares food through photosynthesis. The grasshopper is the primary consumer because it feeds directly on the producer. At the top of this food chain is the eagle, which is the top carnivore as it feeds on the snake.

(ii) Aarav’s observation shows that a food web is a combination of several interconnected food chains. Unlike a food chain, which shows a single path of energy flow, a food web gives a complete picture of feeding relationships in an ecosystem.
For example, in his food web, the frog eats grasshoppers and is eaten by both snakes and eagles. This shows that organisms can occupy different positions in different chains, making the food web more stable.

Mistake Alert
Don’t confuse a food chain with a food web. A food web is the interlinking of many food chains, whereas a food chain is a single straight pathway of energy flow in an ecosystem.

Section – B

Question 17.
Sarita heated the limestone and observed that a product is formed along with a gas. The product formed is [1]
(a) slaked lime
(b) quick lime
(c) lime water
(d) caustic soda
Answer:
(b) quick lime

Explanation:
When lime stone (also called calcium carbonate) is heated, then it decomposes into calcium oxide and carbon dioxide. Calcium oxide is also known as quick lime.

Question 18.
Methyl orange is added to two test tubes ‘P’ and ‘Q’. Test tube ‘P’ contains dilute H₂SO₄, and test tube ‘Q’ contains dilute KOH. Which of the following observations is correct? [1]
(a) In test tube the solution turns red, and in test tube ‘Q’ it turns yellow.
(b) In test tube ‘P’ the solution turns yellow, and in test tube ‘Q’ it turns red.
(c) There is no colour change in either of the test tubes.
(d) Both test tubes show orange colour.
Answer:
(a) In test tube the solution turns red, and in test tube ‘Q’ it turns yellow.

Explanation:
Methyl orange is an”acid-base indicator that shows red colour in acidic solutions and yellow colour in basic solutions. Since test tube ‘P’ contains dilute H2S04 (acid), it turns red, and test tube ‘Q’ contains dilute KOH (base), it turns yellow.

Mistake Alert
Don’t confuse in the action of methyl orange with that of litmus. A common mistake students make is assuming methyl orange turns blue in bases and red in acids, which is incorrect.

Question 19.
Four statements about metal oxides and non-metal oxides are listed below. [1]
I. Aluminium oxide is amphoteric in nature.
II. Sodium oxide is acidic in nature.
III. Carbon dioxide is an acidic oxide.
IV. Zinc oxide reacts with both acids and bases.
Which statements are correct?
(a) I and II
(b) I and III
(c) I, III, and IV
(d) II and III
Answer:
(c) I, III, and IV

Explanation:
Al₂O₃ is amphoteric, CO₂ is acidic, and ZnO reacts with both acids and bases.

Value Point
Acidic oxides react with bases, basic oxides react with acids, while amphoteric oxides react with both acids and bases to form salt and water.

Question 20.
A hydrocarbon with molecular formula C₃H₈ is commonly used as an LPG fuel. Which homologous series does it belong to? [1]

Sample	Homologous series
(a) A	Alkene
(b) B	Alkyne
(c) C	Alkane
(d) D	Alcohol

Answer:
(c) C – Alkane

Explanation:
Alkane, C₃H₈ is propane, a saturated hydrocarbon with single bonds, and it is the main component of LPG fuel.

Question 21.
Select a metal which will displace hydrogen from dilute acids. [1]
(a) Copper
(b) Gold
(c) Zinc
(d) Silver
Answer:
(c) Zinc

Explanation:
Zinc will displace hydrogen from dilute acids as it has high reactivity than hydrogen.

Question 22.
When a soil solution turns bluish-black with pH paper, its colour changes to greenish-blue on adding a certain substance because [1]
(a) baking soda neutralises the base present in the soil.
(b) vinegar partially neutralises the base present in the soil.
(c) common salt reacts to form an acidic solution.
(d) an antacid increases the basicity of the solution.
Answer:
(b) vinegar partially neutralises the base present in the soil.

Explanation:
The soil solution is basic. When vinegar, which is a weak acid, is added, it partially neutralises the base and lowers the pH. This changes the colour of the pH paper to greenish blue.

Question 23.
Which of the following reactions involved the combination of two elements? [1]
(a) CaO(s)+ CO₂(g) → CaCO₃(s)
(b) 4Na(s) + O₁(s) → 2Na₂O(s)
(c) SO₂(S) + \(\frac{1}{2}\)O₂(£) → SO₃(g)
(d) NH₃(g) + HCl(g) → NH₄Cl(S)
Answer:
(b) 4Na(s) + O₁(s) → 2Na₂O(s)

Explanation:
In this reaction, two elements sodium and oxygen participate and form sodium oxide. However, in other reactions, the compounds are combined together and form a single product.

The following question consists of two statements – Assertion (A) and Reason (R). Answer the question by selecting the appropriate option given below.
(a) Both (A) and (R) are true arid (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Question 24.
Assertion (A) Baking soda is used in soda-acid fire extinguishers.
Reason (R) On heating, it releases carbon dioxide which helps extinguish fire.
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Explanation:
Baking soda is used in soda acid fire extinguishers because it releases carbon dioxide gas upon reaction with H2S04. C02 helps put out fire by cutting off the supply of oxygen, which is essential for burning.

Question 25.
The following activity is set-up in the science lab by the teacher.

He took a thick block of impure copper as anode and a thin strip of pure copper as cathode in an acidified copper sulphate solution and connected the circuit for electrolysis. After some time, he observed that the cathode became thicker, and a layer of impurities was seen at the bottom of the tank.
(a) If the teacher replaces impure copper with a pure copper block as the anode, will the students’ observation change? Justify your answer.
Answer:
If the teacher feplaces the impure copper with a pure copper block as the anode, the observation will change because no impurities will be left to settle at the bottom, and there will be no anode mud formation.

(b) What is the name of the layer collected at the bottom of the tank and why does it form?
Answer:
The layer collected at the bottom of the tank is called anode mud, and it forms due to the insoluble impurities from the impure copper anode that do not dissolve in the electrolyte.

Question 26.
Attempt either A or B
A. Name the functional groups present in the following compounds. [3]
(a) CH₃COCH₂CH₂CH₂CH₃
(b) CH₃CH₂CH₂COOH
(c) CH₃CH₂CH₂CH₂CHO
Answer:
(a) In CH₃COCH₂CH₂CH₂CH₃, the functional group present is ketone (—CO—) because the carbonyl group is attached to two carbon atoms.
(b) In CH₃CH₂CH₂COOH, the functional group present is carboxylic acid (—COOH) because it contains a carboxyl group.
(c) In CH₃CH₂CH₂CH₂CHO, the functional group present is aldehyde (—CHO) because the carbonyl group is attached to at least one hydrogen atom.

Or

B. Draw the electron dot structures for
(a) ethanoic acid
(b) propanone
(c) H₂S
Answer:
(a) Electron dot structure of ethanoic acid (CH₃COOH) is

(b) Electron dot structure of propanone (CH₃COCH₃) is

(c) Electron dot structure of H₂S is

Question 27.
Rohit wanted to check whether all compounds containing hydrogen act as acids. He set up the circuit as shown in the figure below to test the conductivity of different solutions. [3]

He first poured dilute HCl solution into the beaker, and the bulb glowed. Then he replaced HCl with alcohol and glucose solutions, but the bulb did not glow.
Answer the following questions based on the activity described above:
(a) Why does the bulb glow when dilute HCl is used in the beaker?
(b) Why does the bulb not glow when alcohol or glucose is used instead of HCl?
(c) What does this activity prove about compounds like alcohol and glucose even though they contain hydrogen?
Answer:
(a) The bulb glows in dilute HCl solution because HCl ionises in water to produce H⁺ (aq) ions, which conduct electricity by completing the circuit.
(b) The bulb does not glow with alcohol or glucose because these compounds do not ionise in water to produce free ions and hence cannot conduct electricity.
(c) This activity proves that not all compounds containing hydrogen are acids, as only those which produce H⁺ (aq) ions in solution show acidic behaviour.

Question 28.
Rohan placed small pieces of iron (Fe), zinc (Zn), and copper (Cu) in separate test tubes containing aqueous copper(II) sulphate solution. His observations are recorded in the table below [4]

Metal sample	Observation in CuSO4 solution
Iron (Fe)	Brown deposit forms on iron
Zinc (Zn)	Brown deposit forms on zinc
Copper (Cu)	No visible change

Answer the following questions based on the above information:
A. Which of the following metals are more reactive than copper based on Rohan’s observations with CuS04 solution?
(a) Only Fe
(b) Fe and Zn
(c) Cu and Fe
Justify your answer
Answer:
(b) Fe and Zn

Explanation:
Iron and zinc are more reactive than copper because they displace copper from copper sulphate solution, forming a brown deposit of copper on their surfaces. Copper does not react as it is less reactive than copper itself.

B. Rohan now immerses an iron nail in zinc sulphate solution will he observe any change? Why or why not?
Answer:
No change will be observed when an iron nail is dipped in zinc sulphate solution because iron is less reactive than zinc and cannot displace zinc from its salt solution.

Or

If an iron nail is left in the open for a few days, it gets covered with a reddish-brown deposit. Name this process and the compound responsible for the colour change.
Answer:
When an iron nail is left in the open for a few days, it undergoes rusting, forming hydrated ferric oxide (Fe₂O₃. xH₂0), which appears as a reddish-brown coating. This is due to the reaction of iron with oxygen and moisture in the air.

C. Which of the following statements is true about the reaction between iron and copper sulphate solution, and why?
(a) It is a displacement and redox reaction.
(b) It is a precipitation and double displacement reaction.
(c) It is neutralisation and double displacement reaction.
(d) It is neutralisation and precipitation reaction.
Answer:
(a) It is a displacement and redox reaction.

Explanation:
It is a displacement and redox reaction, because iron displaces copper from copper sulphate solution. Iron is oxidised to Fe²⁺ ions, and Cu²⁺ ions are reduced to copper metal.

Question 29.
Attempt either A or B. [5]
A. Alcohols and carboxylic acids are important classes of carbon compounds. Alcohols contain the functional group —OH (hydroxyl group), while carboxylic acids contain the —COOH(carboxyl group).
(a) Distinguish between alcohols and carboxylic acids with the help of two reactions and examples.
(b) Riya added ethanol in a test tube containing sodium metal and observed vigorous bubbles. Later, she heated ethanol with cone. H₂SO₄ and observed a gas.
(i) Write the reaction of ethanol with sodium metal.
(ii) Which gas is formed when ethanol reacts with sodium?
(iii) Name the gas formed on heating ethanol with cone. H₂SO₄.
Answer:
(a) Alcohol and carboxylic acid can be distinguished experimentally by two reactions:
Reaction with base Alcohols do not react with base like NaOH, KOH while carboxylic acids react with base to form salt and water.
E.g. CH₃COOH + NaOH → CH₃COO^–Na⁺ + H₂O
CH₃CH₂OH + NaOH → No reaction [1]

(b) (i) Reaction of ethanol with sodium metal:
E.g. CH₃COOH + Na → 2CH₃CH₂ONa + H₂ ↑
(ii) The gas formed when ethanol reacts with sodium is hydrogen (H₂).
(iii) The gas formed on heating ethanol with cone.
H₂SO₄ is ethene (CH₂ = CH₂).

B. Most dirt is oily in nature and as you know, oil does not dissolve in water. The molecules of soap are sodium or potassium salts of long chain carboxylic acids. The ionic end of soap interacts with water while the carbon chain interacts with oil. The soap molecules, thus form structures called micelles, where one end of the molecules is towards the oil droplet while the ionic end faces outside. This forms an emulsion in water.

Based on the information given above, answer the following questions:
(a) What Answer:is used with hard water detergent or soap?
Answer:
Detergents are preferred in hard water because they contain sodium salts of sulphonic acids or ammonium salts with chlorides or bromides, which do not form insoluble scum with calcium and magnesium ions present in hard water.

(b) Draw the structure of a soap molecule showing its hydrophilic and hydrophobic parts.
Answer:

(c) People use a variety of methods to wash clothes. Why is agitation necessary to get clean clothes?
Answer:
Soap molecules form micelles that trap dirt and grease particles from clothes. When the clothes are agitated or scrubbed, the micelles detach from the cloth surface and are removed with water, leaving the clothes clean.

(d) Why are soaps not suitable for washing clothes when water is hard?
Answer:
The formation of lather is necessary for removing dirt from clothes during the washing of clothes. Soap does not give lather with hard water as it reacts with calcium and magnesium ions present in the hard water to form insoluble precipitates of Ca and Mg salts of fatty acid. The precipitate sticks to clothes being washed and it interferes with the cleaning ability of soap.

(e) What does the hydrophilic end of a soap molecule interacts with?
Answer:
The hydrophilic end of a soap molecule interacts with water. This end is attracted to water molecules and helps to dissolve the soap in water, allowing the soap to form micelles and trap oily dirt for clearing.

Section – C

Question 30.
Amav studied the magnetic effects of current and wrote the following statements: [1]
I. A current-carrying conductor experiences a force in a magnetic field.
II. The force on the conductor is independent of the direction of the current, m. The force is maximum when the conductor is parallel to the magnetic field.
Choose the correct option.
(a) I and II
(b) I and III
(c) II and III
(d) Only I
Answer:
(d) Only I

Explanation:
Statement is correct, as a current-carrying conductor experiences a force in a magnetic field. Statement II and III are incorrect because the force depends on the direction of the current and is maximum when the conductor is perpendicular to the magnetic field not parallel.

Question 31.
Choose the correct option that explains the reason for the deflection of the needle of a compass when placed near a current-carrying conductor. [1]
(a) The needle aligns with the magnetic field produced by the current in the conductor.
(b) The current induces an electric field in the conductor that moves the needle.
(c) The needle deflects due to the gravitational force acting on it.
(d) The deflection is caused by the force between the conductor and the magnetic field of the Earth.
Answer:
(a) The needle aligns with the magnetic field produced by the current in the conductor.

Explanation:
The magnetic field produced by the current in the conductor interacts with the magnetic needle of the compass, causing it to align in the direction of the field.

Question 32.
The following question consists of two statements – Assertion (A) and Reason (R). Answer the question by selecting the appropriate option given below
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are hue, but (R) is not the correct explanation of (A).
(c) (A) is hue, but (R) is false.
(d) (A) is false, but (R) is true.

Assertion (A) Myopia is due to the increased converging power of the eye lens. [1]
Reason (R) Myopia can be corrected by using spectacles made from concave lenses.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
In myopia, due to the increased converging power of eye lens, the image of a far object is formed in front of the retina. Myopia defect can be corrected by using a concave lens of suitable power.

Question 33.

The above image shows the formation of an image with an optical instrument.
Answer the following questions
(a) Identify the optical instrument shown in the diagram.
(b) What type of image is formed when the object is placed beyond the centre of curvature (2F) of the concave mirror?
(c) Based on the measurements given in the diagram, if the object distance (u) is 60 cm, calculate the image distance of the concave mirror
Answer:
(a) The optical instrument shown is a concave mirror.
(b) When the object is placed beyond the centre of curvature (2F) of a concave mirror, the image formed is real, inverted and diminished.
(c) The mirror formula is
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

In this case, the object is placed beyond 2 F, so we know the image will form between F and 2 F and the image will be real, inverted and diminished.
Now, let’s assume the focal length (f) of the concave mirror is 30 cm (since, it typically follows a similar pattern in such diagrams).
Substituting the given values,
\(\frac{1}{30}=\frac{1}{v}+\frac{1}{-60}\)
Simplifying the equation,
\(\frac{1}{30}=\frac{1}{v}-\frac{1}{60}\)
\(\frac{1}{v}=\frac{1}{30}+\frac{1}{60}=\frac{2}{60}+\frac{1}{60}=\frac{3}{60}\)
v = \(\frac{60}{3}\) = 20 cm
The image distance (v) is 20 cm.

Question 34.
Attempt either A or B [2]

A. Find out the following in the electric circuit given in the figure :
(a) Calculate the effective resistance of the two resistors (R₁ =8 Ω and R₂ = 8 Ω) in the given parallel circuit.
(b) If the circuit is connected to a 12 V power supply, calculate the total current flowing through the circuit using Ohm’s law.
Answer:
(a) The formula for effective resistance in a parallel circuit is,
\(\frac{1}{R_{\mathrm{eff}}}=\frac{1}{R_1}+\frac{1}{R_2}\)
Given, R₁ = 8 Ω and R₂ = 8 Ω
\(\frac{1}{R_{\mathrm{eff}}}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}\) Ω
R_eff = 4 Ω

(b) Using Ohm’s law,
V = I × R_eff
Given, V = 12 V and R_eff = 4 Ω
I = \(\frac{V}{R_{\text {eff }}}=\frac{12}{4}\) = 3A

Or

B. Study the circuit and find out

The diagram below shows a magnetic compass needle placed in the plane of the paper near point A.
A. In which plane should a straight current-carrying conductor be placed, so that it passes through point A and there is no change in the deflection of the compass needle?
B. Under what condition will the deflection of the compass needle be maximum? Explain, why?
Answer:
(a) Given,
R₁ = 10 Ω, R₂ = 20 Ω, R₃ = 30 Ω
To calculate the total resistance R_eff of the parallel combination of resistors: [1]

(b) Using Ohm’s law, calculate the total current supplied by the 12 V battery: [1]
I_total = \(\frac{V}{R_{\text {eff }}}\)
I_total = \(\frac{12}{5.45}\) = 2.2 A

Question 35.

The diagram below shows a magnetic compass needle placed in the plane of the paper near point A.
A. In which plane should a straight current-carrying conductor be placed, so that it passes through point A and there is no change in the deflection of the compass needle?
B. Under what condition will the deflection of the compass needle be maximum? Explain, why?
Answer:
A. The current-carrying wire should be placed perpendicular (at a right angle) to the paper. This way, the magnetic field created by the wire will pass through point A without changing the compass needle’s direction.

B. The deflection of the compass needle will be maximum when the wire is placed parallel to the compass needle. In this position, the magnetic field produced by the current will have the greatest effect on the needle, making it move the most.

Question 36.

A circuit diagram is given below, answer the following questions based on the diagram.
(i) Name the device which is connected in series in the circuit and the component which controls the amount of current in the circuit.
(ii) State and explain Ohm’s law.
Answer:
(a) (i) The name of the device which is connected in series in the circuit is ammeter. It measures the current, so it is connected in series in the circuit. Rheostat is the component of circuit which controls the amount of current in the circuit.

(ii) According to Ohm’s law, the electric current flowing through a conductor is directly proportional to the potential difference applied across its ends, providing the physical conditions (such as temperature) remains unchanged.
If V is the potential difference applied across the ends of a conductor through which current I flows, then according to Ohm’s law,
V ∝ I (at constant temperature)
or
V = IR or I = \(\frac{V}{R}\)
where, R is the constant of proportionality called resistance of the conductor at a given temperature.

Question 37.
The diagram below shows the dispersion of light through a glass prism. [3]

A student makes the following statements :
(a) When white light passes through the prism, it splits into its component colours. Why?
(b) Red light bends the least and violet light bends the most. Explain, why this happens?
Answer:
(i) When white light passes through the prism, it splits into its component colours due to dispersion. White light consists of different colours that have different wavelengths, and each wavelength bends by a different amount when passing through the prism.

(ii) Different colours of light have different wavelengths, which causes them to refract (bend) by different amounts. The angle of deviation is greater for shorter wavelengths (violet) and smaller for longer wavelengths (red).

Red light bends the least because it has the longest wavelength, while violet light bends the most because it has the shortest wavelength. Shorter wavelengths of light are refracted more due to their higher frequency and lower speed in the prism material, causing a greater deviation.

Question 38.
The ability of a medium to refract light is expressed in terms of its optical density. Optical density has a definite connotation. It is not the same as mass density. On comparing two media, the one with the larger refractive index is optically denser medium than the other. The other medium with a lower refractive index is optically rarer. Also, the speed of light through a given medium is inversely proportional to its optical density. [4]
A. Determine the speed of light in diamond, if the refractive index of diamond with respect to vacuum is 2.42. Speed of light in vacuum is 3 × 10⁸ m/s.
Answer:
Refractive index of diamond
= \(\frac{\text { Speed of light in vacuum }}{\text { Speed of light in diamond }}\)
Speed of light in diamond
= \(\frac{3 \times 10^8}{2.42}\)
= 1.23 × 10⁸ m/s

B. Refractive indices of glass, water and carbon disulphide are 1.5,1.33 and 1.62, respectively. If a ray of light is incident in these media at the same angle (say 0), then write the increasing order of the angle of refraction in these media.
Answer:
We have, n = \(\frac{\sin i}{\sin r}\)
⇒ sin r = \(\frac{\sin i}{n}\)
Hence, ∠r in carbon disulphide < ∠r in glass < ∠r in water.

Attempt either C or D
C (i) The speed of light in glass is 2 × 10⁸ m/s and in water is 2.25 × 10⁸ m/s.
Which one of the two is optically denser and why?
(ii) A ray of light is incident normally at the water-glass interface when it enters a thick glass container filled with water. What will happen to the path of the ray after entering the glass? Give reason.
Answer:
(i) Glass. The speed of light in water is more than the speed of light in glass. So, refractive index of glass is more than the refractive index of water.
(ii) Light will enter from water to glass without bending (undeviated) because in this case ∠i = 0; ∠r = 0.

Or

D. The absolute refractive indices of water and glass are 4/3 and 3/2, respectively. If the speed of light in glass is 2 × 10⁸ m/s, find the speed of light in
(i) vacuum and
(ii) water.
Answer:
(i) Refractive index of glass = \(\frac{3}{2}\)
Refractive index of water = \(\frac{4}{3}\)
Speed of light in glass = 2 × 10⁸ m/s

Refractive index of glass = \(\frac{\text { Speed of light in vacuum }}{\text { Speed of light in glass }}\)
or c = n_glass × v_glass
= \(\frac{3}{2}\) × 2 × 10⁸
= 3 × 10⁸ m/s

(ii) V_water = \(\frac{c}{n_{\text {water }}}=\frac{3 \times 10^8}{\frac{4}{3}}\)
= \(\frac{9}{4}\) × 10⁸ m/s
= 2.25 × 10⁸ m/s

Mistake Alert
Students should be cautious about the optical densities of the media through which the light passes and in which order and use the formula for finding the refractive index accordingly.

Question 39.
Attempt either A or B [5]
A. (i) How will you infer with the help of an experiment that the same current flows through every part of the circuit containing the resistors R₁, R₂ and R₃ in series connected to a battery of V volts?
(ii) Find current in 12Ω. resistor.
(iii) Find difference in the reading of A₁ and A₂ (if any).

Answer:
(i) The experimental set-up comprise three resistors R₁, R₂ and R₃ of three different values which are connected in series.
Connect them with a battery of V, an ammeter and plug key, as shown in figure.

The key K is closed and the ammeter reading is recorded. Now, the position of ammeter is changed to anywhere in between the resistors again, the ammeter reading is recorded each time.
It’s observed that there was identical reading each time, which shows that same current flows through every part of the circuit containing three resistances in series connected to a battery.

(ii) Equivalent resistance of given circuit is R, then
R = (24 ∥ 24) + 12 = \(\frac{24 \times 24}{24+24}\) + 12
= 12 + 12
= 24 Ω

∴Current through 12 resistors
I = \(\frac{V}{R}=\frac{6}{24}\) = 0.25 A

(iii) Difference in reading of A₁ and A₂
= (0.25 – 0.25)A
= 0 A

Or

B. In the given figure, R₁ = 5 Ω,R₂ = 10 Ω, R₃ = 15 Ω, R₄ = 20 Ω, R₅ = 25Ω and a 15V battery is connected to the arrangement. Calculate
(i) the total resistance in the circuit,
(ii) the total current flowing in the circuit and
(iii) power consumed in the combination of R₁ and R₂.

Answer:
(i) Resistors R₁ and R₂ are in parallel. [2]
So, \(\frac{1}{R^{\prime}}=\frac{1}{R_1}+\frac{1}{R_2}\)
⇒ \(\frac{1}{R^{\prime}}=\frac{1}{5}+\frac{1}{10}\)
⇒ R’ = \(\frac{10}{3}\)

Similarly, R, R and R are i parallel.
\(\frac{1}{R^{\prime \prime}}=\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}=\frac{1}{15}+\frac{1}{20}+\frac{1}{25}\)
⇒ \(\frac{1}{R^{\prime \prime}}=\frac{20+15+12}{300}\)
⇒ R” = \(\frac{300}{47}\) Ω

Thus, the total resistance,
R = R’ + R”
= \(\frac{10}{3}+\frac{300}{47}\)
= 3.33 + 6.38
= 9.71 Ω

(ii) The total current, I = \(\frac{V}{R}=\frac{15}{9.71}\) = 154 A

(iii) R₁ and R₂ are in parallel,
R_eq = \(\frac{R_1 R_2}{R_1+R_2}=\frac{5 \times 10}{5+10}\) = 3.33

⇒ Power consumed in R₁ ∥ R₂
P = \(\frac{\left(v^{\prime}\right)^2}{R_{\text {eq }}}=\frac{(5.15)^2}{3.33}\) = 7.96 W

The post CBSE Sample Papers for Class 10 Science Set 4 with Solutions appeared first on Learn CBSE.

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