CBSE Sample Papers for Class 10 Science Set 2 with Solutions - #NCSOLVE 📚

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 2 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 organisms act as primary consumers in a food web. [1]
(a) Grass, rabbit, deer, mouse
(b) Rabbit, grasshopper, deer, caterpillar
(c) Rabbit, fox, mouse, lion
(d) Tiger, rabbit, deer, owl
Answer:
(b) Rabbit, grasshopper, deer, caterpillar

Explanation:
Rabbit, grasshopper, deer, caterpillar act as primary consumers because these organisms are herbivores and directly feed on plants.

Question 2.
Which of the following is a correct combination of process and reason for higher carbon dioxide uptake during daytime in plants? [1]
(a) Transpiration – Due to root pressure
(b) Photosynthesis – Due to sunlight availability
(c) Respiration – Due to stomatal closure
(d) Water absorption – Due to leaf surface area
Answer:
(b) Photosynthesis – Due to sunlight availability

Explanation:
The correct combination is Photosynthesis – Due to sunlight availability. During daytime, the presence of sunlight enables the process of photosynthesis in plants, which leads to higher uptake of carbon dioxide.

Question 3.
Which of the following are correct statements related to the result of a dihybrid cross? [1]
(i) The ratio of phenotypes is 9 :3 : 3 :1
(ii) Round yellow seeds appear most frequently
(iii) Wrinkled green seeds appear in the smallest number
(iv) All offspring are of one type only
(v) This cross involves one contrasting trait
(a) (i), (ii), (iii)
(b) (ii), (iii), (iv)
(c) (i), (iii), (v)
(d) (i), (ii), (iv)
Answer:
(a) (i), (ii), (iii)

Explanation:
The statements (i), (ii), (iii) are correct as in a dihybrid cross, the phenotypic ratio obtained is 9:3:3:1. Among the progeny, round yellow seeds are the most common, while wrinkled green seeds are the least common.

Question 4.
The region of the brain responsible for memory power will be [1]
(a) cerebellum
(b) medulla
(c) cerebrum
(d) pons
Answer:
(c) cerebrum

Explanation:
The region of brain responsible for memory power is cerebrum as it is the largest part of the brain which is responsible for thinking, intelligence, memory and voluntary actions.

Question 5.
A student learned that variation in organisms plays a major role in evolution. Which of the following statements best explains this concept? [1]
(a) All variations help a species to survive better.
(b) Variations arise due to environmental changes only.
(c) Variations selected by the environment form the basis of evolution.
(d) Variations are not linked to genetic changes.
Answer:
(c) Variations selected by the environment form the basis of evolution.

Explanation:
Variations selected by the environment form the basis of evolution because variations that are better suited to the environment are selected and passed on, to the next generation.

Question 6.
Food web is the [1]
(a) food that a spider collects using its web
(b) network of interlinked trophic levels
(c) network of interlinked food chains
(d) display of food items on a website
Answer:
(c) network of interlinked food chains

Explanation:
A food web is a network of interlinked food chains operating at various trophic levels.

Question 7.
The manufacturing of chlorofluorocarbons free refrigerators is mandatory throughout the world. How does this help prevent ozone depletion? [1]
(a) This will help convert oxygen molecules into ozone
(b) This will help convert the CFCs into ozone molecules
(c) This will reduce the production of CFCs from oxygen molecules
(d) This will reduce the release of CFCs that reacts with ozone molecules
Answer:
(d) This will reduce the release of CFCs that reacts with ozone molecules

Explanation:
The manufacturing of chlorofluorocarbons free refrigerators is mandatory throughout the world as this will reduce the release of CFCs that reacts with ozone molecules of ozone layer to cause its depletion.

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) Root tips grow downward into the soil even when the direction of light is from above in plant. [1]
Reason (R) Auxin inhibits cell elongation in the lower side of the root, causing the root to bend downward.
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Question 9.
Assertion (A) In a dihybrid cross between round yellow and wrinkled green seeds, new combinations like round green and wrinkled yellow appear in the F,-generation. [1]
Reason (R) Characters are inherited together and never show independent assortment.
Answer:
(c) (A) is true but (R) is false.

Explanation:
(A) is true but (R) is false. (R) can be corrected as Characters governed by different genes are inherited independently of each other.

Question 10.
Adrenaline is secreted in response to emergency situations. Comment upon the statement with justification. [2]
Answer:
Adrenaline is secreted in response to emergency situations because adrenaline is released by the adrenal glands in situations of stress, fear or excitement. It prepares the body for “fight or flight” by increasing heart rate, blood flow to muscles and breathing rate, helping the body to respond quickly in emergency conditions.

Question 11.
Attempt either A or B. [2]
A. Pollination is an important process in the reproduction of flowering plants.
(i) Name some common agents of pollination.
(ii) Write any one advantage of cross-pollination over self-pollination.
Answer:
(i) Some common agents of pollination are wind, water, insects and animals.
(ii) One advantage of cross-pollination over self-pollination is that it promotes genetic variation which helps in the development of better-adapted plant varieties.

Or
B. What is pollination? State its importance in plants.
Answer:
Pollination is the transfer of pollen grains from the anther to the stigma of a flower. It is important because it enables fertilisation, which leads to the formation of seeds and fruits in flowering plants.

Question 12.
During a visit to an environmental awareness exhibition, students displayed a model depicting the ozone layer and its depletion due to pollutants. [2]
Using this context, explain the importance of the ozone layer in the atmosphere.
Answer:
The ozone layer is important because it protects life on Earth by absorbing harmful ultraviolet (UV) rays from the Sun. Its depletion allows more UV radiation to reach the Earth’s surface, which can cause health problems like skin cancer or cataract and can also affect immunity. It also affects plant growth and marine life.

Value Point
Understanding the concept of ozone depletion is important, as it can have serious effects on life on Earth.

Question 13.
Draw a labelled diagram of a neuron and explain how receptors help in transmission of information in the human body. [3]
Answer:
Receptors are specialised cells located in our sense organs such as skin, eyes, nose, tongue and ears.
They detect specific stimuli like touch, light, smell, taste, or sound, and convert them into electrical impulses. These impulses are transmitted by sensory neurons to the brain or spinal cord, where the information is processed. The brain then sends suitable instructions to effectors through motor neurons, enabling the body to respond.

Question 14.
In a genetic experiment, pure tall pea plants with white flowers (TTww) were crossed with pure dwarf pea plants having violet flowers (ttWW). [3]
(i) Mention the different combinations of gametes formed by the F₁ plants during self-pollination.
Answer:
The parent plants are

• TTww (Tall with white flowers)
• ttWW (Dwarf with violet flowers)

All plants in the F₁ generation will be TtWw (Tall with violet flowers).
The gametes formed from F₁ plants will be TW, Tw, tW, tw.

(ii) A total of 256 plants were obtained in the F₂ -generation. Show the ratio in which these traits are independently inherited in these 256 plants.
Answer:
On self-pollinating F₁ plants, the F₂ generation will follow a 9:3:3:1 phenotypic ratio, i.e.,

• 9 Tall Violet
• 3 Tall White
• 3 Dwarf Violet
• 1 Dwarf White

Out of 256 plants

• 144 will be Tall Violet
• 48 will be Tall White
• 48 will be Dwarf Violet
• 16 will be Dwarf White

Mistake Alert
Don’t get confused between the ratios from a monohybrid cross and a dihybrid cross. A monohybrid cross forms a 3 :1 ratio, whereas a dihybrid cross forms a 9:3 :3 :1 ratio.

Question 15.
Ravi visited a botanical garden and observed the following plants. [4]
A plant kept in a dark shed with yellow leaves and stunted growth.
A cactus plant with thick and waxy leaves.
A large-leaved plant growing under the shade of a tree.
Help Ravi understand these observations by answering the questions given below.

Attempt either subpart A or B.
A. Why does a plant kept in the dark for many days show yellow leaves and poor growth? Name the process that is affected.
Answer:
A plant kept in the dark for many days shows yellow leaves and poor growth because the process of photosynthesis is affected. In the absence of sunlight, chlorophyll is not formed properly, leading to yellow leaves, and the plant is unable to prepare sufficient food, resulting in poor growth.

Or
B. Why do leaves of some desert plants have a thick waxy coating? How does this adaptation help in survival?
Answer:
Leaves of some desert plants have a thick waxy coating to reduce water loss. This adaptation helps the plants to survive in hot and dry conditions by minimising transpiration.

C. Why do plants growing in shady areas often have large green leaves? Explain.
Answer:
Plants growing in shady areas often have large green leaves to capture more sunlight for photosynthesis. Tire large surface area of the leaves allows them to absorb maximum light available in low-light conditions.

D. The figure below shows an experimental setup to demonstrate a process in plants. Identify the aim of this experiment and what will be the expected observation in both setups after a few hours.

Answer:
The aim of this experiment is to demonstrate that carbon dioxide is necessary for photosynthesis in plants.
In setup (a), the watch-glass contains potassium hydroxide, which absorbs carbon dioxide from the air inside the bell jar. Since CO₂ is absent, the plant will not be able to perform photosynthesis, and no starch will be formed in its leaves.
In setup (b), carbon dioxide is available, so the plant will carry out photosynthesis and starch will be formed in the leaves.

Question 16.
Attempt either A or B. [5]
A. A married woman, Reema, chose a surgical method of birth control called ‘X’, which is a permanent method and was performed in a government hospital. Despite its effectiveness, her family had concerns about the procedure.
(i) Explain about the surgical method ‘X’ used by Reema
(ii) Mention one advantage of this method.
Answer:
(i) The surgical method of birth control ‘X’ chosen by Reema is called tubectomy. It is a permanent method used for female sterilisation. In this method, a small portion of the fallopian tubes is surgically removed or tied off to block the path of the egg from the ovary to the uterus. As a result, the egg is unable to meet the sperm and fertilisation does not take place.

(ii) One major advantage of tubectomy is that it is a highly effective and permanent method of birth control. It does not affect the hormonal balance of the body, so the menstrual cycle remain normal.

Or

B. A married woman, Radha, opted for a birth control method ‘X’ that need to be taken orally and prevents ovulation by altering the hormonal levels in the body.
(i) How does method ‘X’ prevent pregnancy?
(ii) Mention one disadvantage of using this method.
Answer:
(i) The birth control method ‘X’ used by Radha is oral hormonal pills. These pills contain synthetic female hormones like oestrogen and progesterone. They work mainly by preventing ovulation, which means the ovary does not release an egg. Without an egg, there is no chance for fertilisation by sperm. These pills also make the uterine lining thinner and thicken the mucus in the cervix, making it harder for sperm to enter the uterus.

(ii) A disadvantage of using oral pills is that they need to be taken regularly, and missing a dose can reduce their effectiveness. In some women, they may also cause side effects such as nausea, headache, weight gain, or hormonal imbalance.

Section – B

Question 17.
Caustic soda is termed an alkali while ferric hydroxide is not because [1]
(a) caustic soda is a strong base, while ferric hydroxide is a weak base
(b) caustic soda is a base which is soluble in water while ferric hydroxide is also a base but it is not soluble in water
(c) caustic soda is a strong base while ferric hydroxide is a strong acid
(d) caustic soda and ferric hydroxide both are strong base but the solubility of sodium hydroxide in water is comparatively higher than that of ferric hydroxide
Answer:
(b) Alkali are the bases that dissolve in water. Thus, caustic soda is an alkali, while ferric hydroxide is not.

Question 18.
Butanone is a four carbon containing compound with functional group. [1]
(a) Carboxylic acid
(b) Aldehyde
(c) Ketone
(d) Alcohol
Answer:
(c) Ketone

Explanation:
In butanone, functional group is ketone (-one) [C=0],

Question 19.
Which of the following can be used as an acid-base indicator by a visually impaired student? [1]
(a) Litmus
(b) Vanilla essence
(c) Turmeric
(d) Petunia leaves
Answer:
(b) Vanilla essence

Explanation:
Vanilla essence is an olfactory indicator. So, its smell is different in acidic and basic media which can be detected easily by a visually impaired student. It has a characteristics pleasant smell. If a basic solution like sodium hydroxide solution is added to it, the smell disappears or weakens and in acidic solution like hydrochloric acid, however does not destroy the smell of vanilla extract.

Question 20.
A student performed the following four experiments in four different test tubes as shown below. [1]

Which of the following observation is correct ?
(a) The displacement reaction occurs in all experiments except III.
(b) The solid deposition is observed in experiments I, II and III.
(c) The solid deposition is observed in experiments I and II.
(d) The displacement reaction is observed in experiment TV.
Answer:
In I, II and III experiments, there is the formation of a solid deposition because Al, Zn and Fe are more reactive than Cu. They will displace Cu from its solution and Cu will deposit as solid.

Question 21.
Zinc granules are added to dilute sulphuric acid and the mixture is observed. What is the type of chemical reaction and the evidence that a chemical change has occurred? [1]

Sample Type of Reaction	Observable Evidence
(a) Decomposition	Solution turns green
(b) Double displacement	A white precipitate forms
(c) Displacement	Gas bubbles are released
(d) Combination	A solid black residue forms

Answer:
(c) Zinc displaces hydrogen from dilute sulphuric acid forming zinc sulphate and hydrogen gas and thus the reaction is called a displacement reaction.
The release of gas bubbles indicates a chemical change.

Mistake Alert:
Don’t misidentify the reaction as a combination reaction just because two substances (zinc and acid) are reacting together and without analysing the products formed.

Question 22.
The IUPAC name of acetaldehyde is
(a) ethanol
(b) ethanal
(c) methanal
(d) propanal
Answer:
(b) ethanal

Explanation:
The IUPAC name of acetaldehyde is ethanal.

Question 23.
On adding some copper turnings to silver nitrate solution what will you observe?
(a) Tire solution turned blue
(b) Yellow precipitate was formed
(c) White precipitate was formed
(d) The solution turned red

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 tme, but (R) is false.
(d) (A) is false, but (R) is true.
Answer:
When copper turnings are added to silver nitrate solution, a blue coloured solution is formed after sometime because copper is oxidised to Cu2+ ions and forms copper nitrate. It displaces silver from its solution and forms blue coloured solution of CU(NO₃)₂.

Question 24.
Assertion (A) Diamond does not conduct electricity. [1]
Reason (R) Diamond has a high refractive index.
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Due to the absence of free electrons, diamond does not conduct electricity.

Question 25.
The open chain structure of one hydrocarbon is given below. [2]

(a) Write the IUPAC name and valency of each carbon atom in the above structure.
(b) Name the products formed when ethane bums in air. Write the balanced chemical equation for the reaction showing the types of energies liberated.
Answer:
(a) The IUPAC name of the above structure is ethane and valency of each carbon atom is four.
(b) Carbon dioxide and water are formed when ethane bums in air. The balanced chemical equation is

Question 26.
Attempt either A or B. [3]
A. An element ‘X’ reacts vigorously with cold water, while element ‘Y reacts only with steam. Element ‘7! shows no reaction with any form of water.
(a) Write balanced chemical equations for the reactions of elements ‘X’ and ‘Y with water.
Answer:
With cold water

(b) Identify two metals from the reactivity series that behave like element ‘Z’ and explain why they don’t react with water.
Answer:
Gold and silver do not react with any form of water. These metals are placed at the bottom of the reactivity series. Their low reactivity is due to their inability to lose electrons easily.

Or

B. The extraction method of metals depends on their reactivity. Highly reactive metals require special processes as they form stable compounds, while moderately reactive metals can be extracted using simpler methods.
(a) How is the method of extraction of metals high up in the reactivity series different from that for metals in the middle?
Answer:
Highly reactive metals (Na, Ca, Al) are extracted through electrolysis of their molten chlorides or oxides, while moderately reactive metals (Fe, Zn, Pb) are extracted by reducing their oxides with carbon/carbon monoxide.

(b) Explain why the reduction process using carbon cannot be applied for extracting metals high in the reactivity series.
Answer:
Carbon cannot reduce oxides of highly reactive metals because these metals have greater affinity for oxygen than carbon. Their metal-oxygen bonds are too strong to be broken by carbon.

(c) Name the specific processes used for extracting highly reactive and moderately reactive metal.
Answer:
Highly reactive metals are extracted by electrolytic reduction and moderately reactive metals by smelting.

Value Point
Understanding the reactivity of metals helps in selecting an appropriate extraction method, ensuring efficient use of resources and energy during metal extraction.

Question 27.
Amrita took two test-tubes, test-tube (A) with red coloured solution and test-tube (B) with pink coloured solution as shown in the diagram below. [3]

Answer the following questions based on the information given above.
(a) What is meant by pH?
(b) Two solutions A and Shave pH values of 3.0 and 9.5 respectively. Which of these will turn litmus solution from blue to red and which will turn phenolphthalein from colourless to pink?
(c) Water is a neutral substance. What colour will you get when you add a few drops of universal indicator to a test tube containing distilled water?
Answer:
(a) The p in pH stands for potenz which means power in German. pH is a number which indicates the acidic or basic nature of a solution.
(b) As the pH value of solution A is 3.0, i.e., acidic in nature. Hence, it turns litmus solution from blue to red and phenolphthalein indicator will change tire colour of solution B with pH value 9.5,
i.e., basic medium from colourless to pink.
(c) Distilled water’ turns the universal indicator solution green as its pH value is 7.

Question 28.
Regular cooking leads to the formation of a black layer over cooking utensils and causes the burner flame to become yellowish. The observations are recorded in the table below [4]

Observation	Reason
1. Black layer on utensils	Incomplete combustion
2. Yellowish flame colour	Presence of unbumt carbon particles

Answer the following questions based on the above information:

A. What causes the burner flame to turn yellowish during this process?
(a) Complete combustion of fuel
(b) Incomplete combustion of fuel
(c) Presence of unbumt carbon particles Justify your answer.
Answer:
(c) Presence of unbumt carbon particles causes the burner flame to turn yellowish during this process.

B. What are the three parts of flame?
Or
What precaution should be taken to avoid this process?
Answer:
The three parts of flame are following.
(a) Inner part
(b) Middle part
(c) Outer part
Or
To prevent this situation, burners of gas or stoves should be cleaned timely so that inlets do not block.

C. Incomplete combustion leads to the formation of which of the following major air pollutants?
(a) Oxygen and nitrogen
(b) Carbon dioxide and water vapour
(c) Methane and hydrogen gas .
(d) Carbon monoxide and carbon dioxide
Answer:
(d) Incomplete combustion is very harmful to our environment. This leads to the formation of carbon monoxides and carbon dioxide which are major pollutants.

Question 29.
Attempt either A or B. [5]
A. A hydrocarbon with the formula CH₄ (methane) reacts with chlorine which is shown in the following equation:
CH₄ + Cl₂ → CH₃Cl + HCI
(a) Mention the name of the given reaction and explain it.
(b) What is the essential condition required for the given reaction of methane with chlorine?
(c) Why are higher homologues of alkanes not preferred for substitution reactions?
(d) In the above reaction, what is the the product CH₃Cl called and also name the byproduct formed?
(e) What is the final product of this reaction? Write all the steps involved.
Answer:
(a) The given reaction of methane with chlorine is called substitution reaction. It is a reaction in which a reagent substitutes (replace) atoms or a group of atoms from the reactant (substrate) are called substitution reactions.
(b) The essential condition required for the given reaction of methane with chlorine is the presence of sunlight.
(c) The higher homologues of alkanes are not preferred for substitution reactions cause they form a mixture of multiple products.
(d) The product CH₃Cl is called chloromethane and the the byproduct formed is hydrochloric acid (HCl).
(e) When chlorine is added to hydrocarbons in the presence of sunlight, Cl replaces H-atoms one by one to give carbon tetrachloride as the final product.

Or

B. The structures of two hydrocarbons are shown below

(a) What is the process of addition of hydrogen to an unsaturated hydrocarbon called?
(b) What is the catalyst and which catalyst is used in the above reaction?
(c) Mention the condition where the above reaction is commonly used.
(d) State one difference between vegetable oils and animal fats.
(e) Why are vegetable oils with unsaturated fatty acids considered healthier?
Answer:
(a) The process of addition of hydrogen to an unsaturated hydrocarbon is called hydrogenation.
(b) Catalysts are substances that cause a reaction to occur or proceed at a different rate without the reaction, itself being affected. This reaction is commonly carried out using a nickel catalyst.
(c) The above reaction is commonly used in the hydrogenation of vegetable oils.
(d) Vegetable oils generally have long unsaturated carbon chains while animal fats have saturated carbon chains.
(e) Vegetable oils with unsaturated fatty acids are healthier than animal fats because they do not raise cholesterol levels, unlike saturated fats which are harmful to health.

Section – C

Question 30.
A scientist is studying the dispersion of light through a prism to understand how different colors of light bend at different angles. He observes that violet light bends the most and red light bends the least. Based on his observations, he writes down the following statements: [1]
Statement I Violet light has the shortest wavelength and bends the most when passing through a prism.
Statement II Red light has the longest wavelength and bends the least due to its lower refractive index compared to violet light.
Statement III The angle of deviation increases as the wavelength of light increases.

Choose from the following the correct option that lists the true statements about light dispersion in a prism,
(a) Statement I and II are correct.
(b) Statement I and III are correct.
(c) Statement II and III are correct.
(d) Statement I, II and III are correct.
Answer:
(a) Tire correct answer is (a). Statement I and II are correct.

Violet light has the shortest wavelength and bends the most when passing through a prism due to its higher refractive index. On the other hand, red light has the longest wavelength and bends the least because it has a lower refractive index. The statement about the angle of deviation increasing with wavelength is incorrect, as shorter wavelengths (like violet) bend more and thus have a larger angle of deviation, while longer wavelengths (like red) bend less, resulting in a smaller angle.

Question 31.
Which of the following statements explains the reason why a concave mirror form different types of images depending on the position of the object? [1]
(a) A concave mirror always forms real images regardless of the object’s position.
(b) The position of the object relative to the focal point determines whether the image is real or virtual.
(c) Concave mirrors form only virtual images, and their size remains the same at all positions.
(d) The size of the image formed by a concave mirror is always smaller than the object, regardless of the object’s position.

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.
Answer:
(b) The correct answer is (b). The position of the object relative to the focal point determines whether a concave mirror forms a real or virtual image. If the object is beyond the focal point, a real, inverted image is formed. If the object is between the focal point and the pole of the mirror, a virtual, upright image is formed. The image’s size and nature depend on this position.

Question 32.
Assertion (A) A concave mirror can form both real and virtual images depending on the object’s position relative to the focal point. [1]
Reason (R) The image formed by a concave mirror is always real when the object is placed beyond the focal point and virtual when the object is placed between the focal point and the mirror.
Answer:
(a) The correct answer is (a). Both Assertion (A) and Reason (R) are true, and Reason is the correct explanation of Assertion.

A concave mirror can form both real and virtual images depending on the position of the object.
When the object is beyond tire focal point, the mirror forms a real, inverted image. When the object is between the focal point and the pole of the mirror, the image formed is virtual and upright. The reason correctly explains the assertion as it describes how the image type changes based on the object’s position.

Question 33.
Study the diagram below and answer the following: [2]

(i) Identify the type of lens represented byXX’ in the diagram.
(ii) Determine the position of the point on the principal axis OO’ where the image of the object AB is formed.
(iii) Explain the nature of the lens and describe where the image of the object AB will be located on the principal axis OO’. Discuss the reason for the image’s position in the context of the lens’s type.
Answer:
(i) The lens represented by XX’ is a concave lens.
(ii) The image of the object AB will be formed on the principal axis OO’ between O’ and Y, which is closer to the focal point of the concave lens.
(iii) A concave lens is a diverging lens, and it forms an erect, virtual, and diminished image of the object. Since concave lenses diverge light rays, the image formed is always virtual and located on the same side as the object. In this case, the image of the object AB will appear to be formed between O’ and Y along the principal axis OO’.

Question 34.
Attempt either A or B. [2]
A. Justify the following statements.
(i) The relative strength of the magnetic field is shown by the degree of closeness of the field lines.
(ii) Magnetic field lines are closed curves.
Answer:
(i) Closer the lines more will be the strength and farther the lines, less will be the magnetic field strength.
(ii) The direction of field lines outside a magnet is from north pole to south pole while it is from south to north pole inside the magnet and thus forms closed curves.

Or

B. (i) Draw the labelled diagram to show magnetic field lines due to current in a straight wire.
Answer:

(ii) How to increase the magnetic field around the wire?
Answer:
Increasing current, I passing through the wire automatically increase magnetic field around it.

Question 35.
A person needs a lens of power -5.5 D for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 D [3]
(i) Define the defect of vision.
Answer:
The defect of vision is presbyopia. In this defect a person suffers from both myopia as well as hypermetropia due to less power of accommodation of eye.

(ii) Find the focal length of the lens for distant and near vision.
Answer:
For distant vision we use concave lens,
Focal length = \(\frac{1}{\text { power }}=\frac{1}{-5.5}\) [given, power P = – 5.5 D]
f= – 18 cm
For near vision vve use convex lens.
Focal length = \(\frac{1}{\text { Power }}=+\frac{1}{1.5}\) = 0.67 m
[given, power, P = + 1.5 D]
f = 67cm [convex lens]

Mistake Alert
Assuming image distance equals object distance leads to wrong focal length and distance calculation.

Question 36.
A circuit diagram is given as shown below. [3]

Calculate
(i) The total effective resistance of the circuit.
(ii) The total current in the circuit.
Answer:
Given, R₁ = 2Ω, R₂ = 5Ω, R₃ 3= 10Ω, V = 10V
(i) Total effective resistance as the combination is in parallel.

(ii) Total current, I = \(\frac{V}{R_{\mathrm{eff}}}=\frac{10}{125}\) = 8A

Value Point
Always mention in parallel combination current divides across branches and voltage same across all the resistors in parallel.

Question 37.
Study the diagram given below and answer the questions that follow. [3]

(i) Why do the iron filings arrange in such a pattern and what does this pattern demonstrate?
(ii) Why do the iron filings near the bar magnet seem to align in the shape of closed curves?
Answer:
(i) The iron fillings arrange in such a pattern because they get magnetised by the bar magnet and align themselves along the magnetic field lines produced by the magnet. This pattern demonstrate the shape of magnetic field lines.
(ii) Iron filings near a bar magnet align in the shape of closed curves because they trace out the magnetic lines which are the visual representation of the magnetic field around the magnet.

Question 38.
Rohit, a class 10 student is performing a physics experiment on the magnetic effects of current. He placed a straight, horizontal conductor on the table and keeps a magnetic compass near it. Initially, the compass needle points north. When he doses the switch to allow current to flow through the conductor he noticed that the compass needle deflects to one side. Curious about the phenomenon he reverses the direction of the current. He also wonders what would happen if he increases the current. He recorded all his observation to analyse the effect of current on magnetic needle.
A. Why does the compass needle deflect near a current carrying conductor?
B. What does the increased deflection of the needle indicate?

Attempt either subpart C or D.
C. If the student reverse the direction of current in the wire, how will the compass needle behave?
Or
D. What will happen to the deflection if the student moves the compass further away from the wire?
Answer:
A. The compass needle deflects near a current-carrying conductor because the electric current in the wire creates a magnetic field around it. This magnetic field affects the compass needle, which is a tiny magnet, causing it to move and point in a different direction.
B. The increased deflection of the needle means the magnetic field around the wire is stronger. This happens because a higher current produces a stronger magnetic field.
C. If the student reverses the direction of the current in the wire, the magnetic field’s direction also reverses. This will make the compass needle
D. If the student moves the compass farther away from the wire, the deflection of the needle will become smaller. This is because the magnetic field gets weaker as you move away from the wire.

Question 39.
Attempt either A or B [5]
A. (i) How is electric current related to the potential difference across the terminals of a conductor? Draw a labelled circuit diagram to verify this relationship.
(ii) Why should an ammeter have low resistance?
(iii) Two V-I graphs A and B for series and parallel combinations of two resistors are as shown. Giving reason state which graph shows (a) series, (b) parallel combination of the resistors.

Answer:
(i) V = IR, where R is constant and it is called resistance. Hence, current I is directly proportional to V.

(ii) An ammeter should have low resistance because it should not significantly affect the current flowing through the circuit being measured. If an ammeter has high resistance, It will draw a large amount of current from the circuit, causing the measured value to be lower than the actual value.
Hence, the resistance of an ammeter should be very low compared to the resistance of the circuit being measured.

(iii) (a) Graph B represents parallel combination because B has lower resistance.
(b) Graph A shows series combination A has higher resistance.

Or

B. Study the following circuit and find

Here, R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 4 Ω, R₄ = 6 Ω, R₅ = 10 Ω
(i) effective resistance of the circuit
(ii) current drawn from the battery
(iii) potential difference across the 5Ω resistor.
Answer:
(i) Resistance R₃ and R₄ are in series,
R’= R₃ + R₄₁
=4 + 6
= 10Ω

Again R’ and R, are in parallel
\(\frac{1}{R^{\prime \prime}}=\frac{1}{R^{\prime}}+\frac{1}{R_2}\)
= \(\frac{1}{10}+\frac{1}{10}=\frac{2}{10}\)
R” = 5Ω

Now resistance Rv R”and R5 are in series
R = R₁ + R” + R₅
= 5 + 5 + 10
= 20Ω

(ii) Current drawn from the battery
I = \(\frac{V}{R}=\frac{20}{20}\)
= 1A

(iii) Potential difference across 5Ω resistor
V = IR
V = 1 × 5 = 5 V

The post CBSE Sample Papers for Class 10 Science Set 2 with Solutions appeared first on Learn CBSE.

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