CBSE Sample Papers for Class 10 Science Set 3 with Solutions - #NCSOLVE 📚

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 3 with Solutions

Time : 3 Hrs.
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.
Select the group in which all processes are directly involved in the transport of water in plants. [1]
(a) Transpiration, osmosis, diffusion, root pressure
(b) Photosynthesis, transpiration, osmosis, respiration
(c) Translocation, stomatal opening, transpiration, phototropism,
(d) Root hair absorption, translocation, transpiration, seed germination
Answer:
(a) Transpiration, osmosis, diffusion, root pressure

Explanation:
The group in which all processes are directly involved in the transport of water in plants is Transpiration, osmosis, diffusion, root pressure. These processes help in the absorption and upward movement of water through the xylem from roots to leaves.

Question 2.
Adrenaline hormone is secreted by the adrenal glands in emergency situations. Which of the following is an effect of adrenaline on the body? [1]
(a) Increases heart beat rate and breathing rate
(b) Decreases heart beat rate
(c) Reduces blood supply to muscles
(d) Slows down breathing rate
Answer:
(a) Increases heart beat rate and breathing rate

Explanation:
Adrenaline is released during emergency situations and prepares the body for fight or flight by increasing the heartbeat rate, breathing rate and blood supply to muscles so that the body can respond quickly.

Question 3.
Name the type of movement that helps the food to move through the digestive tract. [1]
(a) Amoeboid movement
(b) Peristaltic movement
(c) Cilliated movement
(d) Pseudopodial movement
Answer:
(b) Peristaltic movement

Explanation:
Peristaltsis is defined as a series of wave-like muscle contractions that moves food through the digestive tract.

Question 4.
What is the role of microvilli present in the small intestine? [1]
(a) They help to move the food along
(b) They increase the surface area for absorption
(c) They protect against bacteria
(d) They move mucus over the surface
Answer:
(b) They increase the surface area for absorption

Explanation:
Microvilli increases the surface area for the absorption of products of digestion, i.e. amino acid, glucose, etc.

Question 5.
Which of the following is a correct combination of an endocrine gland and its function? [1]
(a) Sleep-wake cycle regulation—Pituitary gland
(b) Stress response (fight or flight)—Adrenal gland
(c) Sugar metabolism—Thyroid gland
(d) Growth regulation—Pancreas
Answer:
(b) Stress response (fight or flight)—Adrenal gland

Explanation:
The correct combination of endocrine gland and its function is stress response (fight or flight): Adrenal gland, as
The adrenal gland secretes adrenaline which helps the body to respond to stress by increasing heartbeat, breathing rate and energy availability.

Question 6.
The decomposers are not included in the grazing food chain. The correct reason of the same is because decomposers [1]
(a) act at every trophic level of the food chain
(b) do not breakdown organic compounds
(c) convert organic material to inorganic forms
(d) release enzymes outside their body to convert organic material to inorganic forms
Answer:
(a) act at every trophic level of the food chain

Explanation:
The decomposers are not included in the grazing food chain, because decomposers act at every trophic level of the food chain as they are saprophytes who use dead bodies or organic materials of each trophic level as food source.

Question 7.
Which of the following statements related to food chains are correct? ‘[1]
(i) Food chains always start with producers.
(ii) Decomposers are at the beginning of a food chain..
(iii) Herbivores are primary consumers.
(iv) Carnivores directly consume producers.
(v) Energy flows from producers to consumers in a food chain.
Options
(a) (i), (ii), (iii)
(b) (ii), (iii), (iv)
(c) (i), (iii), (v)
(d) (i), (iv), (v)
Answer:
(c) (i), (iii), (v)

Explanation:
Food chains always begin with producers that make their own food using sunlight. Herbivores are primary consumers that feed on producers. Energy flow’s in one direction from producers to consumers in a food chain.

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, and 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) Urethra in human male act as urinogenital canal.
Reason (R) Urethra carries only urine, while sperms are carried by vas deferens only.
Answer:
(c) A is true but R is false.

Explanation:
R can be corrected as
The urethra is a tube like structure that runs through the penis and is the common passage for both urine and sperm.

Question 9.
Assertion (A) Mendel self-crossed F₁ – progeny to obtain F₂-generation. [1]
Reason (R) F, -progeny of a tall plant with round seeds and a dwarf plant with wrinkled seeds are all dwarf plants having wrinkled seeds.
Answer:
(c) A is true but R is false.

Explanation:
R can be corrected as The F₁ -progeny of a tall plant with round seeds and a dwarf plant with wrinkled seeds are all tall plants with round seeds.

Question 10.
Males are responsible for determining the sex of a child in humans. Comment on the statement with proper justification. [2]
Answer:
Sex determination in humans depends on the type of sperm that fertilises the egg. Human males have XY sex chromosomes, while females have XX chromosomes. During fertilisation, the female contributes only an X-chromosome through her egg, but the male can contribute either an X or a Y chromosome through his sperm. If the sperm carrying an X-chromosome fertilises the egg, the child will be a girl (XX). If the sperm carries a Y- chromosome, the child will be a boy (XY), Hence, it is the male who determines the sex of the child.

Mistake Alert
Students should be careful in the case of sex determination in humans, as males are heterogametic with two different sex chromosome (XY) and female are homogametic with two X-chromosomes.

Question 11.
Attempt either A or B.
A. In plants, certain movements take place in response to light and gravity. What type of movement is shown by [2]
(i) Stems when they bend towards light
(ii) Roots when they grow downwards.
Answer:
(i) The movement shown by the stems when they bend towards light is known as positive phototropism. It occurs because the plant hormone auxin accumulates on the shaded side of the stem, causing the cells on that side to elongate more, which bends the stem towards the light.

(ii) The movement shown by the roots in response to gravity is known as positive geotropism.
This happens because more auxin collects on the lower side of the root, which slows down the growth there. The upper side grows faster, so the root bends and grows in the direction of gravity.

Or

B. Name and define two plant hormones that control growth movements in plants.
Answer:
Plant hormones are chemical substances produced in plants that help to regulate growth and responses to stimuli.

Two plant hormones that control growth movements in plants are
Auxins Help in cell elongation and bending of shoots towards light.
Gibberellins Promote stem elongation and are involved in seed germination.

Question 12.
Explain how plants ensure sufficient exchange of gases for photosynthesis and respiration. [2]
Answer:
Plants ensure sufficient exchange of gases through tiny openings called stomata found mainly on the surface of leaves. These stomata open to allow carbon dioxide to enter for photosynthesis and oxygen to leave as a by product. Similarly, during respiration, oxygen enters and carbon dioxide exits through these stomata. In addition to stomata, small pores called lenticels on stems and roots also help in the exchange of gases, ensuring that all parts of the plant receive the gases needed for vital processes.

Question 13.
Draw and explain the flow of energy in an ecosystem. [3]
Answer:
In an ecosystem, the flow of energy is always unidirectional. Tire primary source of energy is the Sun. Green plants (producers) capture solar energy through the process of photosynthesis and convert it into chemical energy in the form of food. Herbivores (primary consumers) obtain this energy by feeding on plants. Carnivores (secondary and tertiary consumers) in turn derive their energy by feeding on herbivores and other carnivores. At each trophic level, a major part of the energy is lost to the surroundings as heat and only about 10% of the energy is transferred to the next trophic level. This is known as the 10% law of energy transfer. As energy decreases at each successive level, the number of trophic levels in a food chain is limited.

Value Point:
To get maximum marks, one should mention the concepts of 10% law of energy transfer

Question 14.
In the process of digestion of food in human beings, proteins are broken down with the help of enzymes. [3]
(i) Name the two protein-digesting enzymes involved in this process.
Answer:
The two protein-digesting enzymes involved in human digestion are pepsin and trypsin. Pepsin acts in the stomach, while trypsin acts in the small intestine and breaks proteins into simpler forms.

(ii) Name the glands which secrete these enzyme.
Answer:
Pepsin is secreted by the gastric glands present in the wall of the stomach and trypsin is secreted by the pancreas,which acts in the small intestine. These enzymes help in converting complex proteins into simpler forms that can be absorbed by the body.

Question 15.
Ravi was walking in the park when he suddenly lost his balance after tripping over a stone. He quickly regained control without falling. Later, he heard a loud barking sound and noticed a dog running towards him. He ran away from the dog, his heartbeat increased and he began sweating. Help Ravi understand some concepts related to control and coordination by answering the following questions. [4]
Attempt either subpart A or B.
A. Which part of Ravi’s brain helped him maintain balance after tripping? Name this part and explain its function.
Answer:
The part of Ravi’s brain that helped him maintain balance after tripping is the cerebellum.The cerebellum is a part of the hindbrain. It is responsible for maintaining posture, balance, and coordination of voluntary muscular movements.
When Ravi tripped, the cerebellum helped his body adjust quickly to prevent a fall by coordinating the actions of his muscles and ensuring smooth movement.

Or

B. When Ravi saw a dog running towards him, his heartbeat increased.
Which hormone is responsible for this response? Name the gland that secretes it and explain how it prepares the body to deal with such situations.
Answer:
The hormone responsible for the increase in Ravi’s heartbeat upon seeing the dog is adrenaline.
Adrenaline is secreted by the adrenal glands, which are located on top of the kidneys.
This hormone prepares the body to respond to an emergency situation by increasing the heartbeat, raising blood pressure, and supplying more oxygen to muscles. It is also known as the fight-or-flight hormone, as it enables the body to act quickly in stressful situations.

C. Ravi started sweating, while running in the park. Why is this process important for the body? Which part of the brain regulates this response?
Answer:
Ravi began sweating, while, running in the park, and this process is important to maintain the body temperature.
Sweating helps cool down the body through the process of evaporation, thus preventing overheating.
This action is regulated by the hypothalamus, which is a part of the forebrain. The hypothalamus acts like the body’s thermostat and helps maintain homeostasis by controlling thirst, temperature and other involuntary activities.

D. The figure given below represents the structure of the human brain. Which of these parts is responsible for controlling the involuntary activities of the body?

Answer:
The part of the brain responsible for controlling involuntary activities of the body, such as heartbeat, breathing, and digestion, is D which is the medulla oblongata.
The medulla is a part of the hindbrain and functions without our conscious control. It ensures that vital functions like respiration and circulation continue to occur smoothly even when we are not aware of them.

Question 16.
Attempt either A or B
A. Ramesh was reading about sexually transmitted diseases in his science textbook. [5]
(i) Name any two sexually transmitted diseases caused bv bacteria and any two caused by viruses.
(ii) Mention any three preventive measures for sexually transmitted diseases given in your textbook.
Answer:
(i) Sexually transmitted diseases caused by bacteria are gonorrhoea and syphilis and that caused by viruses are warts and HIV-AIDS.

(ii) Three preventive measures for sexually transmitted diseases are as follows:

• Avoid sexual contact with infected persons.
• Avoid sharing injection needles with infected persons.
• Use of condoms during sex can prevent these infections.

Or
B. During a health awareness programme, students were informed about sexually transmitted diseases.
(i) List any three ways by which sexually transmitted diseases can spread.
(ii) Explain how an infected mother can transmit HIV to her child.
Answer:
(i) Three ways by which sexually transmitted diseases can spread are as follows

• Through sexual contact with an infected person.
• By sharing injection needles with infected person.
• From an infected mother to her baby during pregnancy or through breast feeding.

(ii) An infected mother can transmit HIV to her child when the virus passes from the mother’s blood into the baby’s body during pregnancy or through breast feeding after birth. This is why necessary precautions are advised for HIV-infected mothers.

Section – B

Question 17.
Identify the colour of gas evolved in the following experiment. [1]

(a) Colourless
(b) Yellow
(c) Whitish grey
(d) Black
Answer:
(a) Colourless

Explanation:
When potassium nitrate is heated, then colourless gas is evolved, i.e., oxygen.

Question 18.
Which of the following is an example of decomposition reaction? [1]
(a) Melting of glaciers.
(b) Rusting of old bridges.
(c) Rotting of fruits and vegetables.
(d) Absorption of carbon dioxide by oceans.
Answer:
(c) Rotting of fruits and vegetables.

Explanation:
Among the given options, rotting of fruits and vegetables is a decomposition reaction. As in this, one reactant breaks into two or more products.

Question 19.
If 10 mL of H₂SO₄ is mixed with 10 mL of Mg(OH)₂ of the same concentration, the resultant solution will give the following colour with a universal indicator. [1]
(a) Red
(b) Yellow
(c) Green
(d) Blue
Answer:
(c) Green

Explanation:
Mixing of H₂SO₄ and Mg(OH)₂ is a neutralisation reaction resulting in salt and water as its products. In a neutral solution, the universal indicator turns green.

Question 20.
Four statements about the properties of non- metals are listed below. [1]
I. They are neither malleable nor ductile.
II. They are brittle.
III. They are sonorous.
IV. They are good conductor of heat and electricity (except graphite).

Which statements are correct?
(a) I and II
(b) II and IV
(c) II and III
(d) I and IV
Answer:
(a) I and II

Explanation:
Non-metals can not be made into sheets or drawn into wires. They get easily broken and do not produce soun’d when struck with each other. They are generally poor conductors of heat and electricity except graphite.

Question 21.
Friedrich Wohler accidentally prepared an organic compound which later refuted the vital force theory and came to be known as [1]
(a) alcohol
(b) urea
(c) vinegar
(d) benzene
Answer:
(b) urea

Explanation:
Friedrich Wohler accidentally prepared urea from ammonium cyanate and the synthesis of urea discarded the vital force theory.

Question 22.
A metal ribbon X bums in oxygen with a dazzling white flame forming a white ash Y. The correct description of X, Y and the type of reaction are
X(s) + O₂(g) → Y(s)
(a) X = Ca, Y = CaO, Type of reaction = Decomposition
(b) X = Mg, Y = MgO, Type of reaction = Combination
(c) X = Al, Y = A1203, Type of reaction = Thermal decomposition
(d) X = Zn, Y = ZnO, Type of reaction = Endothermic
Answer:
(b) X = Mg, Y = MgO, Type of reaction = Combination

Explanation:
When a magnesium ribbon is burned with oxygen present in air, it forms a white powdered compound called magnesium oxide.

This is an example of combination reaction

Question 23.
The bleaching powder gives a pungent smell when exposed to air because of [1]
(a) formation of chlorine on exposure to atmosphere.
(b) unstability of bleaching powder.
(c) formation of excess of calcium carbonate.
(d) displacement reaction.
Tie following question consists of two statements—Assertion (A) and Reason (R). Answer the question by selecting the ppropriate 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.
Answer:
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).

Explanation:
Bleaching powder (CaOCl₂) reacts with carbon dioxide present in atmosphere and produce calcium carbonate and chlorine gas, that’s why it gives smell of chlorine.

Bleaching Carbon Calcium Chlorine powder dioxide carbonate gas

Question 24.
Assertion (A) When HCl is added to zinc granules, a chemical reaction occurs. [1]
Reason (R) Evolution of a gas and change in colour indicate that a chemical reaction is taking place.
Answer:
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).

Explanation:
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
The reaction takes place as follow
Zn + 2HCl → ZnCll₂ + H₂ ↑

Question 25.
(a) Write the chemical name and chemical formula of the salt used to remove permanent hardness of water. [2]
Answer:
Washing soda salt is used to remove permanent hardness of water. Its chemical name is sodium carbonate decahydrate with chemical formula Na₂CO₃ • 10H₂O.

Mistake Alert
Generally, one can make mistake in writing the correct formula of washing soda, as they may get confused between washing soda and washing powder.

(b) How is gypsum prepared? Explain with the help of chemical reaction. Give the number of water of crystallisation in gypsum.
Answer:
Plaster of Paris is a white powder and on mixing with water it changes to gypsum.

Gypsum has two molecules of water of crystallisation.

Question 26.
Attempt either A or B
A. An organic compound A has the molecular formula C₄H₈O₂ on hydrolysis gives product B and C having molecular formula C₂H₄O₂ and C₂H₆O, respectively. B is used as vinegar in cooking. [3]
(i) Identify A, B and C.
(ii) Write the chemical equation involved in the formation of B and C.
(iii) Give two uses of compund C.
Answer:
(i) Compound A is ethylacetate, compound B is ethanoic acid and compound C is ethanol.
(ii) Esters on hydrolysis in the presence of dilute sulphuric acid give carboxylic acid and alcohol.
A balanced chemical reaction illustrating the process is given as

(iii) Two uses of ethanol are
1. It is used as solvent in a variety of chemical reactions.
2. It is used in medicine as tincture of iodine.

Or

B. Name the compound formed when ethanol is heated in excess of concentrated sulphuric acid at 443 K. Also, write the chemical equation of the reaction stating the role of concentrated sulphuric acid in it. What would happen, if hydrogen is added to the product of this reaction in the presence of catalysts such as palladium or nickel?
Answer:
When ethanol is heated in excess of concentrated sulphuric acid at 443 K, ethene is formed.
The reaction involved is

Cone. H₂SO₄ acts as dehydrating agent, it removes water molecules from a compound.
If hydrogen is added to the product, then ethane is formed.

Value Point
To get maximum marks, one should write the correct IUPAC name of the reactants and products involved.

Question 27.
The diagram shown below is of electrolytic refining of copper.

In this process, a thick block of impure metal is used as anode and a thin strip of pure metal is used as cathode. A solution of metal salt is used as an electrolyte.
(i) What happens when electric current is passed in the given system?
(ii) Draw your own conclusion from the given figure.
(iii) Illustrate the reaction of the process with the help of a chemical equation.
Answer:
(i) When electric current is passed, copper ions from the electrolyte are reduced as copper which get deposited on the cathode.
(ii) An equivalent amount of impure copper from the anode gets oxidised to copper ion and goes into the solution and from there, it goes to cathode and gets deposited.
(iii) The reactions taking place during refining of copper is as follows
At cathode,
Cu^2- (aq) + 2e^– → Cu(s)
At anode, Cu(s) → Cu²⁺(aq) + 2e^–
This cycle is repeated until whole of the copper ion from impure block is dissolved and deposited on cathode.

Question 28.
Aarav took a clean test tube and added 5 mL of copper(II) sulphate solution (CuSO₄) to it. He observed the blue colour of the solution. Then, he added a small piece of iron nail into the test tube and left it undisturbed for 20 minutes. After some time, he observed that the blue colour of the solution had faded and a reddish-brown layer was deposited on the iron nail. Aarav repeated the activity using different volumes of CuSO₄ solution and recorded his observations as shown in the table below: [4]

Volume of CuS04 solution (mL)	Time taken for visible change (min)
5	20
10	35
15	50

Answer the following questions based on the above information:
A. If Aarav uses a concentrated CuSO₄ solution, what will happen to the time required for the change to be observed?
(a) Increase
(b) Decrease
(c) Remain the same Justify your answer.
Answer:
(b) A concentrated CuSO₄ solution has more
Cu²⁺ ions available, so the displacement reaction occurs faster. Thus, the time taken to observe the change will decrease.

B. Aarav added 5 mL of ZnSO₄ solution instead of CuSO₄ and repeated the experiment using an iron nail. What change would he observe and why?
Or
Aarav placed a copper wire in FeSO₄ solution. Will any reaction take place? Why or why not?
Answer:
When Aarav adds ZnSO₄ solution and dips an iron nail into it, no reaction will be observed.
Iron is less reactive than zinc. A displacement reaction only occurs when a more reactive metal displaces a less reactive one from its salt solution. Hence, iron cannot displace zinc from zinc sulphate.
Or
When copper wire is placed in FeSO₄ solution, no reaction will take place because copper is less reactive than iron, so it cannot displace iron from FeSO₄ solution. Therefore, there will be no change observed.

C. Write the balanced chemical equation for the reaction observed between iron and copper sulphate. Which of the following reaction types is correct and why?
(a) Combination and displacement reaction
(b) Decomposition and redox reaction
(c) Displacement and redox reaction
(d) Displacement and precipitation reaction
Answer:
(c) Balanced chemical equation
Fe(s) + CuSO₄(aq) → FeSO₄(ag) + Cu(s)
Iron displaces copper from copper sulphate solution. Iron becomes Fe²⁺ (oxidation), and Cu²⁺ becomes Cu (reduction).
The correct option is (c) Displacement and redox reaction.

• It is a displacement reaction because a more reactive metal (iron) displaces a less reactive one (copper).
• It is also a redox reaction because iron is oxidised (loses electrons), and copper is reduced (gains electrons).

Question 29.
Attempt either A or B
A. (a) Complete the following reactions and name the main product formed in each case. [5]

(b) Write the names of the following compounds.

State the functional group present in each compound.
Answer:

(b) (i) Butan-2-one, functional group—Ketone
(ii) Ethyl ethanoate, functional group—Ester

Or

B. (a) Explain how is vinegar made?
(b) What is glacial acetic acid?
(c) Give reason as why is butanoic acid a weak acid?
(d) What is a saponification reaction?
(e) Write the name and the formula of the two compounds formed when the ester, CH₃COOC₂H₅ undergoes saponification.
Answer:
(a) 5-8% solution of acetic acid in water is called vinegar. It is obtained by dissolving 5 g acetic acid in 100 mL water.
(b) Pure ethanoic acid is called glacial acetic acid.
(c) Butanoic acid is a weak acid because it does not ionise completely in solution.
(d) Saponification is the chemical reaction in which a fat or oil (ester) reacts with a base (usually sodium hydroxide, NaOH) to form glycerol (a type of alcohol) and soap (a sodium salt of a fatty acid).
Fat/Oil (ester) + NaOH → Soap + Glycerol
(e) Ethanol (C₂H₅OH) and sodium ethanoate (CH₃COONa).

Section – C

Question 30.
Identify the correct statement that explains, why the path of a beam of sunlight becomes visible in a smoke-filled room? [1]
(a) Light gets scattered by tiny particles in mist or smoke, so we can see the path of the beam.
(b) Mist and smoke produce their own light, making the beam glow.
(c) Light reflects off the ground and appears as a beam in the air.
(d) Light travels in a straight line and appears as a visible beam on its own.
Answer:
(a) Light gets scattered by tiny particles in mist or smoke, so we can see the path of the beam.

Explanation:
In a smoke-filled room, the air contains tiny particles like dust, smoke, or water droplets. When a beam of sunlight enters such a space, these particles scatter the light in different directions. This scattering makes the path of the beam visible to our eyes.

Question 31.
A light ray enters from medium A to medium B as shown in the figure. The refractive index of medium B relative to A will be [1]

(a) greater than unity
(b) less than unity
(c) equal to unity
(d) zero
Answer:
(a) greater than unity

Explanation:
Since, the light rays in the medium B goes towards normal. So, it has greater refractive index. Hence, refractive index of medium B w.r.t. medium A is greater than unity.

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 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 32.
Assertion (A) The given diagram is of correction of a myopic eye. [1]

Reason (R) Myopia arises due to excessive curvature of the eye lens or elongation of the eyeball.
Answer:
(d) A is false, but R is true.

Explanation:
The given diagram shows the correction of a hypermetropic eye. Hence, A is false because a concave lens is used for correcting a myopic eye, while R is true because myopia arises due to excessive curvature of the eye lens or elongation of the eyeball.

Question 33.
In the figure, a narrow beam of white light passes through a triangular glass prism. After emerging through the prism, it produces a spectrum XY on a screen. [2]

(a) Which colours appear at point X and Y.
(b) Explain why the various colours in white light deviate at different angles when passing through the prism?
Answer:
(a) The colour seen at X is violet, and the colour seen at Y is red.
(b) Different colours bend through different angles because the speed of light of each colour is different as they move through glass. Hence, the refractive index of glass is different for each colour.

Question 34.
Attempt either A or B
A. An electric lamp, whose resistance is 20 Ω and a conductor of 4 Ω resistance are connected to a 6 V battery as shown in figure. Calculate [2]
(i) the total resistance of the circuit and
(ii) the current through the circuit.

Answer:
(i) Lamp resistance R₁ = 20 Ω and conductor
resistance R₂ = 4 Ω are in series.
Total resistance, R_eq = 20 + 4 = 24 Ω

(ii) Current in the circuit,
I = \(\frac{V}{R_{\text {eq }}}=\frac{6}{24}\) = 0.25 Ω

Or

B. If a 12 V battery is connected to the arrangement of resistances given below, calculate
(i) the total effective resistance of the arrangement and
(ii) the total current flowing in the circuit.

Answer:
(i) In circuit, 10 Ω and 20 Ω are in series.
R_s = 10 + 20 = 30 Ω

5 Ω and 25 Ω are in series,
R’_s = 5 + 25 = 30 Ω

Now, R_s and R’_s are in parallel,
\(\frac{1}{R_p}=\frac{1}{R_s}+\frac{1}{R_s^{\prime}}=\frac{1}{30}+\frac{1}{30}\)
\(\frac{1}{R_p}=\frac{2}{30}\)
⇒ R_p = 15 Ω

(ii) Apply Ohm’s law,
I = \(\frac{V}{R_p}=\frac{12}{15}\) = 0.8 A

Question 35.

Use the diagram to answer the following [3]
(i) Calculate the refractive index of slab with respect to air. (Take, sin 50° =0.77)
(ii) Using result (i), find the speed of light in slab.
Answer:
(i) In figure, angle of incidence, i = 90° – 40° = 50° r = 30°, n_air = 1
Using Snell’s law,
n_airsin i = n_slab sin r
1 × sin 50° = n_slab sin 30°
n_slab = \(\frac{\sin 50^{\circ}}{\sin 30^{\circ}}=\frac{0.77}{0.5}\)
n_slab = 1.54

(ii) We Know,
n = \(\frac{c}{v}\)
⇒ v = \(\frac{c}{n}\)
v = \(\frac{3 \times 10^8}{1.54}\) = 1.94 × 10⁸ m/s

Mistake Alert
Students often confuse the angle of incidence, as it is measured between the incident ray and the normal, not the surface.

Question 36.
In the circuit given below.

(i) Will any of the bulb light up when the switch is in the open position?
(ii) When the key is closed, how would you rank the brightness of the bulbs? Give reason.
Answer:
(i) No bulb will glow when plug key is in open position, as no current would flow through the circuit.

(ii) Power of bulb, P = I²R
For the same current, P ∝ R
But for the same voltage, P ∝ \(\frac{1}{R}\) or R ∝ \(\frac{1}{P}\)
So, resistance order of all bulb is R₂₅ > R₄₀ > R₆₀
According to Joule’s law of heating, H ∝ R
(for the same current and time)
Hence, order of heating produced is
H₂₅ > H₄₀ > H₆₀
Which is order of brightness of the bulb when key is closed.

Question 37.
(i) A student is unable to see clearly the words written on the blackboard placed at a distance of approximately 5m from him. Name and define the causes of the defect of vision the boy is suffering from. [3]
(ii) Look at the below diagram and state the focal length of the lens required to corred this defect of vision.

Answer:
(i) The student is suffering from myopia/near or short sightedness.
Near sightedness is caused due to
(a) too high converging power of the eye lens.
(b) eyeball being too long.

(ii) Given, object distance, u = – ∞
Image distance, v = – 5 m = – 500 cm
Focal length, f = ?
Using lens formula,
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
⇒ \(\frac{1}{-500}-\frac{1}{-\infty}=\frac{1}{f}\)
\(\frac{1}{f}=\frac{1}{-500}\)
f = – 500 cm or – 5 m

Question 38.
Take a concave mirror and hold it in such a way that its reflecting surface faces the sun. Place a white sheet of paper or cardboard in front of the mirror. SlowIy move the sheet forward or backward until you notice a very bright and sharp spot of light formed on it. This bright spot represents the image of the sun produced by the mirror. The point at which this sharp image is obtained is known as the principal focus of the concave mirror. [4]

A. List two applications of concave mirror.
Answer:
Uses of Concave Mirrors
. Concave mirrors are commonly used in torches, search lights and headlights of vehicles, to get powerful parallel beams of light.
» Concave mirrors are used as shaving mirrors to see larger image of the face.

B. If the distance between the mirror and the principal focus is 15 cm, find the radius of curvature of the mirror. Attempt either subpart C or D
Answer:
Given, focal length, f = 15 cm
Radius of curvature, R = 2f = 2 × 15 = 30 cm

C. Draw a ray diagram to show the type of image formed when an object is placed between pole and focus of a concave mirror.
Answer:

Or

D. An object 10 cm in size is placed at 100 cm in front of a concave mirror. If its image is formed at the same point, where the object is located, find
(a) focal length of the mirror, and
(b) magnification of the image formed with sign as per New Cartesian sign convention.
Answer:
Given, size of object, he = 10 cm
Distance of object, u = -100 cm
Distance of image, v = -100 cm

(a) Using mirror formula,
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}=\frac{1}{-100}+\frac{1}{-100}=\frac{-2}{100}\)
f = -50 cm

(b) Magnification of the image,
m = \(\frac{-v}{u}=\frac{-(-100)}{-100}\) = -1

Question 39.
Attempt either A or B.

A. The diagram above shows an electric circuit in which resistances R₁, R₂, R₃, R₄ and R₅ are connected. There is some air gap at the joint of wires.
(i) What is the resistance of an air gap?
(ii) Derive an expression for equivalent resistance for the circuit.
(iii) If the wire of resistance R₁ and resistivity p is stretched to double of its length, how does its resistance and resistivity will be affected?
Answer:
(i) The resistance of an air gap is very large almost infinite. Since, there is no contact between the conducting wires.

(ii) R₂ and R₃ are in series. Thus, for this combination, R’ =R₂ + R₃.
Similarly, R₄ and R₅ are in series, so R” = R₄ + R₅.
R’ and R” are in parallel, then
R”’ = \(\frac{R^{\prime} R^{\prime \prime}}{R^{\prime}+R^{\prime \prime}}=\frac{\left(R_2+R_3\right)\left(R_4+R_5\right)}{R_2+R_3+R_4+R_5}\)
Rj and R”’ are in series, then
R_eq = R₁ + \(\frac{\left(R_2+R_3\right)\left(R_4+R_5\right)}{R_2+R_3+R_4+R_5}\)

(iii) Let the initial length of wire is l and area of cross-section is A.
When the wire is stretched to double its length, the final length is l₂ =2l₁.
But volume of wire remains same.
∴ l₁A₁ = l₂A₂
⇒ l₁A₁ = 2l₁A₂
∴ A₂ = \(\frac{A_1}{2}\)
Resistance of the wire, R = ρ\(\frac{l}{A}\)
Initially, R₁ = ρ\(\frac{l_1}{A_1}\)
After stretching, R₂ = ρ\(\frac{l_2}{A_2}\) = ρ\(\frac{2 l_1}{\frac{A_1}{2}}\) = 4ρ\(\frac{l_1}{A_1}\)
= 4R₁ [from Eq.(i)]
∴ Resistance is increased by 4 times.
Resistivity remains constant, as it is the property of the material and does not depend on the dimensions of the material.

Value Point
To get maximum marks, correctly combine R₂ and R₃ (series), R₄ and R₅5 (series), then find their parallel combination and finally add R₁ to get R_eq.

Or

B. There identical bulbs are connected in parallel to a 4.5 V battery as shown in the figure. Ammeters A₁, A₂ and A₃ are connected at suitable positions to measure the currents. The total current recorded by ammeter is 2 A when all the three bulbs glow.

Answer the following:
(i) What happens to the brightness of the remaining two bulbs, if bulb B, fuses?
(ii) How do the readings of, ammeter A₁, A₂ and A₃ change when bulb B2 fuses?
(iii) Calculate the total power dissipated in the circuit when all three bulbs glow together.
Answer:
Resistance of combination of three bulbs in parallel,
R_eq = \(\frac{V}{I}=\frac{4.5}{3}\) = 1.5 Ω
If R is the resistance of each wire, then
\(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\)
or \(\frac{1}{R_{\mathrm{eq}}}=\frac{3}{R}\)
or R = 3R_eq = 3 × 1.5 = 4.5 Ω
Current in each bulb
I = \(\frac{V}{R}=\frac{4.5 \mathrm{~V}}{4.5 \Omega}\) = 1A

(i) When bulb B₁ gets fused, then the currents in B₂ and B₃ remain same I₂ = I₃ = 1A, so their glow remains unaffected.
(ii) When bulb B₂ gets fused, then the current in B₂ becomes zero and currents in B₁ and B₃ remain 1 A.
∴ Total current, I = I₁ + I₂ + I₃ = 1 + 0 + 1 = 2A
Current in ammeter A₁, I₁ = 1A
Current in ammeter A₂, I₂ = 0
Current in ammeter A₃, I₃ = 1A
Current in ammeter A, I = 2A

(iii) When all the three bulbs are connected, then power dissipated,
P = \(\frac{V^2}{R_{\mathrm{eq}}}=\frac{(4.5)^2}{1.5}\) = 13.5 W

The post CBSE Sample Papers for Class 10 Science Set 3 with Solutions appeared first on Learn CBSE.

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