CBSE Sample Papers for Class 10 Science Set 1 with Solutions - #NCSOLVE 📚

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 1 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 have the same mode of nutrition. [1]
(a) Cuscuta, yeast, legumes, leeches and tapeworm
(b) Cactus, ticks, lice, leeches and cow
(c) Cuscuta, ticks, lice, leeches and tapeworm
(d) Cactus, grass, lice, lion and tapeworm
Answer:
(c) Cuscuta, ticks, lice, leeches and tapeworm

Explanation:
Cuscuta, ticks, lice, leeches and tapeworms have the same mode of nutrition because all of them are parasitic in nature.

Question 2.
Which of the following options indicates the products formed after breakdown of the glucose in our muscle cells when there is lack of oxygen? [1]
(a) Ethanol + carbon dioxide + energy
(b) Lactic acid + energy
(c) Lactic acid + carbon monoxide + energy
(d) Carbon dioxide + water + energy
Answer:
(b) Lactic acid + energy

Explanation:
Lactic acid + energy indicates the products formed after the breakdown of the glucose in our muscle cells when there is a lack of oxygen. During anaerobic conditions, glucose is partially broken down to produce energy and instead of being converted to carbon dioxide and water, lactic acid is formed.

Question 3.
Which of the following is a correct combination of function and part of the brain? [1]
(a) Posture and balance: Cerebrum
(b) Salivation: Medulla in midbrain
(c) Hunger: Pons in hindbrain
(d) Blood pressure: Medulla in hindbrain
Answer:
(d) Blood pressure: Medulla in hindbrain

Explanation:
Blood pressure: Medulla in hindbrain is the correct combination, as the medulla controls involuntary functions such as heartbeat, breathing, and blood pressure.

Question 4.
The blood glucose level in a patient was very high. It may be due to inadequate secretion of [1]
(a) growth hormone from pituitary gland
(b) oestrogen from ovary
(c) insulin from pituitary gland
(d) insulin from pancreas
Answer:
(d) insulin from pancreas

Explanation:
The blood glucose level in the patient can be high due to the inadequate secretion of insulin from the pancreas. Without enough insulin, the body’s cells cannot take in glucose properly and this may lead to diabetes.

Question 5.
In a cross between black furred rabbit (B) and white furred rabbit (b), all offspring were found to have black fur. What can be inferred about the genetic makeup of the parent rabbits? [1]
(a) BB × bb
(b) Bb × Bb
(c) Bb × bb
(d) bb × bb
Answer:
(a) BB × bb

Explanation:
In a cross between black furred rabbit (B) and white furred rabbit (b), all offspring were found to have black fur. This indicates that black fur is a dominant trait and the black-furred parent is homozygous dominant (BB), while the white-furred parent is homozygous recessive (bb). Therefore, the genetic makeup of the parents is BB × bb.

Question 6.
Which are the correct statements related to ozone?
(i) Ozone layer helps in increasing the UV radiations reaching Earth.
(ii) Ozone is a deadly poison.
(iii) Ozone layer shields the Earth from UV radiations.
(iv) Ozone layer prevents UV rays which cause skin cancer.
(v) Ozone is formed with the help of chlorofluorocarbons.
(a) (i), (ii) and (iii)
(b) (ii), (iii) and (iv)
(c) (iii), (iv) and (v)
(d) (i), (iv) and (v)
Answer:
(b) (ii), (iii) and (iv)

Explanation:
The statements (ii),(iii) and (iv) are correct Ozone though poisonous at the ground level forms a protective layer in the upper atmosphere. This protective layer shields the Earth from harmful ultraviolet (UV) radiations coming from the Sun. By absorbing these UV rays, it helps prevent serious health issues like skin cancer.

Question 7.
Which of the following human activities has resulted in an increase of non-biodegrad able substances? [11
(a) Organic farming
(b) Increase in tree plantation
(c) Use of plastic as packaging material
(d) Composting of kitchen waste

The following questions consist 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:
(c) (A) is true, but (R) is false.

Explanation:
The human activity that has resulted in an increase of non-biodegradable substances is the use of plastic as packaging material. This is because plastic being a non-biodegradable substance does not decompose naturally and remains in the environment for a long time, causing pollution.

Question 8.
Assertion (A) Tallness of a pea plant is controlled by an enzyme. [1]
Reason (R) The gene for that enzyme makes proteins which help the plant to be tall.
Answer:
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).

Question 9.
Assertion (A) Vulture will always have the least amount of pesticides in a food chain. [1]
Reason (R) Vulture occupies the last trophic level and it gets only 10% of energy of the previous trophic level.
Answer:
(d) (A) is false, but (R) is true. (A) can be corrected as Vultures will always have the highest amount of pesticides in the food chain.

Question 10.
Unlike animals, plants do not have any excretory products as they do not eat food. Comment upon the statement with justification. [2]
Answer:
Plants do not eat food like animals, but they perform metabolic activities such as photosynthesis and respiration, which produce waste products. These wastes are removed in different ways. Oxygen is released during photosynthesis and carbon dioxide and water vapour from respiration are expelled through stomata. Some wastes like resins and gums are stored in old tissues or leaves, which later fall off.

Question 11.
Students to attempt either A or B. [2]
A. How many chambers are there in the heart of the following organisms? How is mixing of oxygenated and deoxygenated blood prevented in their body?
(i) Fishes
(ii) Humans
Or
B. Explain the mechanism by which the water is transported in plants?
Answer:
A. (i) Fishes have a 2-chambered heart with one auricle and one ventricle. They show single circulation, where only deoxygenated blood passes through the heart, gets oxygenated in the gills and then flows to the body.
(ii) Humans have a 4-chambered heart with double circulation. Oxygenated and deoxygenated blood are completely separated, ensuring efficient oxygen supply to the body. [1]
Or
B. Xylem transports water in plants through a continuous network of channels from roots to leaves. Water is absorbed by root hairs from the soil and moves into the xylem. The main force driving this upward movement is transpiration pull, created by the loss of water vapour through stomata. Cohesive and adhesive properties of water help maintain a continuous water column, ensuring its movement from roots to leaves.

Question 12.
About 100 acres of forest land was declared as Natural reserve park. The following organisms were predominant in the Natural reserve park. [2]
Rabbit, frog, grass, fish, fox, water insects, zebra, peacock, snake, trees, bird, owl, insects, tiger, vulture, duck.
Create a food web comprising two separate food chains with different producers by using the above data.
Answer:
In the Natural reserve park, a food web can be formed using two separate food chains with different producers.

First food chain:
Grass → Rabbit → Fox → Tiger

Second Food Chain:
Trees → Insects → Frog → Snake → Owl
These two food chains are part of a larger food web, where many organisms are interlinked through feeding relationships. This shows the interdependence of organisms in a natural ecosystem.

The following food web illustrates the interconnections between the two food chains mentioned above, highlighting the feeding relationships among various organisms in the ecosystem.

Question 13.
Draw and explain how the nerve cells help in transmission of impulses? [3]
Answer:
Nerve cells or neurons help in transmitting messages called nerve impulses throughout the body. The process starts when dendrites receive a signal from a sensory organ or another neuron.
This signal is passed to the cell body and then travels down the axon in the form of an electrical impulse.
When the impulse reaches the axon terminals. It triggers the release of neurotransmitters. These chemicals carry the message across the synapse to the dendrites of the next neuron, continuing the transmission.

(a) Structure of neuron,
(b) Neuromuscular junction

Question 14.
In a genetic experiment, plants with pure round green seeds (RRyy) were crossed with plants with wrinkled yellow seeds (rrYY). [3]
(i) Show the gametes formed when F, was self-pollinated.
(ii) A total of 144 seeds were produced which developed into saplings. Show the ratio in which these traits are independently inherited in these 144 sapling.
Answer:
(i) In a genetic experiment, plaints with pure round green seeds (RRyy) were crossed with plants having wrinkled yellow seeds (rrYY). The F₁ generation produced from this cross had the genotype RrYy, which shows round yellow seeds because round and yellow are dominant traits.
The F₁ plants (RrYy) produce four types of gametes due to independent assortment of alleles:
RY, Ry, rY, ry
When these F₁ plants were self-pollinated (RrYy × RrYy), they gave rise to different combinations of traits in the Regeneration.

(ii) A total of 144 seeds were produced which developed into saplings
According to Mendel’s law of independent assortment, the F₂-generation shows a 9 :3 :3:1 phenotypic ratio:
9 Round Yellow 3 Round Green 3 Wrinkled Yellow 1 Wrinkled Green
Out of 144 saplings, the number of each type will be:
Round Yellow = \(\left(\frac{9}{16}\right)\) × 144 = 81
Round Green = \(\left(\frac{3}{16}\right)\) × 144 = 27
Wrinkled Yellow = \(\left(\frac{3}{16}\right)\) × 144 = 27
Wrinkled Green = \(\left(\frac{1}{16}\right)\) × 144 = 9
Hence, The traits are inherited independently, and the saplings show the traits in the ratio 81: 27 : 27 : 9, which matches the 9 : 3 : 3 :1 pattern explained by Mendel.

Question 15.
Neha consumed boiled sweet potatoes and boiled eggs for breakfast. Help her to understand some steps in the process of digestion of the food taken by her by answering the questions given below. [4]
Attempt either subpart A or B.
A. Which of these food items is rich in proteins? In which part of the alimentary canal is the digestion of this component initiated? Name the enzymes, conditions required and the glands associated with the digestion here.
Or
B. Which of these food items contains fats? How is it digested?
C. Which of these food items is rich in starch? How is its digestion initiated?
D. The figure given below represents parts of the human alimentary canal. Which of these parts will have the maximum amount of digested food as soon as the process of digestion is completed?

Answer:
A. Boiled eggs are rich in proteins and the digestion of proteins begins in the stomach. The gastric glands in the stomach secrete hydrochloric acid (HCl) and an inactive enzyme called pepsinogen. HCl provides an acidic medium that converts pepsinogen into its active form, pepsin. Pepsin then starts breaking down proteins into simpler substances.
Or
B. Fats are digested mainly in the small intestine. Bile juice from the liver breaks large fat globules into smaller ones through emulsification. This helps the enzyme lipase, secreted by the pancreas, to act more efficiently. Lipase then breaks down fats into simpler forms.
C. Boiled sweet potatoes are rich in starch. The digestion of starch is initiated in the mouth. The salivary glands secrete saliva, which contains the enzyme salivary amylase. This enzyme begins the breakdown of starch into simpler sugars.
D. The given figure represents the human alimentary canal with different parts labelled as L, M, N, O and P. Out of these, the part labelled O represents the small intestine, which is the main site for the complete digestion and absorption of food. Hence, O (small intestine) will have the maximum amount of digested food just as the digestion is completed.

Question 16.
Attempt either option A or B. [5]
A. Puneet wanted to grow banana plants.
(i) Based on your knowledge on plant reproduction should he opt for seeds or any alternate method of reproduction. Justify your answer.
(ii) Offsprings of a banana plant usually show very little variation. What causes variation and are variations good or bad? Justify.
Or
B. Annie was conducting research on the number of fruits produced by watermelon under different conditions. She grew 25 watermelon plants each in both glass house A and B. She introduced pollinators in glass house A only.
(i) What difference will she observe in the number of fruits produced in the two glass houses? Explain with reason.
(ii) List 3 changes that will occur in a flower once it gets fertilised.
Answer:
A. (i) Puneet wanted to grow banana plants. Based on the mode of reproduction in banana, he should opt for an alternate method of reproduction, such as vegetative propagation, instead of using seeds. This is because banana plants generally do not produce viable seeds. They reproduce through vegetative parts which grow into new plants. This method ensures that the new plants are genetically identical to the parent plant and helps in preserving the desired characteristics.

(ii) Offsprings of a banana plant usually show very little variation because they are produced through asexual reproduction, in which the offspring are exact copies of the parent. Variation is caused by sexual reproduction, mutations, or due to environmental influence. Variations are considered beneficial for a species as they provide the ability to adapt to changing environmental conditions and enhance the chances of survival.

Or

B. (i) Annie was conducting research on the number of fruits produced by watermelon under different conditions. She will observe that more fruits are produced in glass house A as compared to glass house B. This is because pollinators were introduced only in glass house A and the process of pollination is essential for fertilisation and fruit formation. In the absence of pollinators, as in glass house B, pollination does not occur effectively, leading to a reduced number of fruits.
(a) The zygote is formed and develops into an embryo.
(b) The ovule transforms into a seed.
(c) The ovary enlarges and becomes a fruit.

Section – B

Question 17.
Which of the following equations represent redox reactions and what are the values for ‘p and ‘q’ in these equations?
Equation 1. Fe₁O₃ (s) + 2Al(s) → Al₂O₃(s)+ pFe(l) + heat
Equation 2. 2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g)+ qH₂O(g)
(a) Only equation 1 is a redox reaction, p = 1 and q = 3
(b) Both equations 1 and 2 are redox reactions, p = 2 and q = 4
(c) Only equation 2 is a redox reaction, p = 2 and q = 10
(d) Both equations 1 and 2 are redox reactions, p = 2 and q = 10
Answer:
(d) Both equations 1 and 2 are redox reactions, p = 2 and q = 10.
(1) Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(l) + heat
(2) 2C₄H₂O(g) + 13O₂(g) → 8C02(g) + 10H₂O(g)

Question 18.
Four statements about the reactions of oxides with dilute hydrochloric acid and aqueous sodium hydroxide are listed.[1]
I. Aluminium oxide reacts with both dilute hydrochloric acid and aqueous sodium hydroxide.
II. Calcium oxide reacts with dilute hydrochloric acid and aqueous sodium hydroxide.
III. Zinc oxide reacts with both dilute hydrochloric acid and aqueous sodium hydroxide.
IV. Sulphur dioxide does not react with either dilute hydrochloric acid or aqueous sodium hydroxide.
Which statements are correct?
(a) I and II
(b) I and III
(c) II and IV
(d) III and IV
Answer:
(b) Al₂O₃ and ZnO are amphoteric oxides, so they react with both dilute HCl and aqueous NaOH.

Question 19.
An iron nail is added to each of the two test tubes ‘P’ and ‘Q’ containing aqueous copper (II) sulphate, and aqueous silver nitrate respectively. Which of the following observation is correct? [1]
(a) In test tube ‘P’ iron nail is coated with a blue coating and in test tube ‘Q’ there is no reaction.
(b) Iron nail is coated with a brown coating in test tube ‘ P’ and silver coating in test tube ‘Q’.
(c) There is no reaction in either of the test tubes ‘P’or ‘Q’.
(d) There is no reaction in test tube ‘P’ but a silver coating on iron nail is seen in test tube ‘Q’.
Answer:
(b) Fe displaces Cu from CuS04 forming brown Cu, and it also displaces Ag from AgNOa, forming a silver coating on the nail.

Mistake Alert:
Don’t assume that silver is too unreactive and will not be displaced by iron which leads to an incorrect answer, because iron does displace silver from AgN03, forming a silver coating.

Question 20.
Methyl orange is added to dilute hydrochloric acid and to aqueous sodium hydroxide. What is the colour of the methyl orange in each solution? [1]

Sample	Colour in dilute hydrochloric acid	Colour in aqueous sodium hydroxide
A	Orange	Red
B	Red	Yellow
C	Red	Orange
D	Yellow	Red

Answer:
(b) Methyl orange turns red in acids and yellow in bases.

Question 21.
Which of the following substances when dissolved in equal volume of water, will have the highest pH value? [1]
(a) Sulphuric acid
(b) Acetic acid
(c) Magnesium hydroxide
(d) Sodium hydroxide
Answer:
(d) Among the given substances, sodium hydroxide is the strongest base and will have the highest pH value.

Question 22.
When excess of carbon dioxide is passed through lime water, the milkiness disappears because
(a) water soluble calcium carbonate converts to water soluble calcium bicarbonate. [1]
(b) insoluble calcium carbonate converts to water soluble calcium bicarbonate.
(c) water soluble calcium carbonate converts to insoluble calcium bicarbonate.
(d) insoluble calcium carbonate converts to insoluble calcium bicarbonate.
Answer:
(b) When CO₂ is first passed through lime water, Ca(OH)₂ reacts to form insoluble CaC0₃, which makes the solution milky. On passing excess CO₂, the CaCO₃ dissolves forming soluble Ca(HCO₃)₂, causing the milkiness to disappear.

Question 23.
In the reaction of aqueous solution of barium chloride with aqueous solution of sodium sulphate, the aqueous solution formed will be [1]
(a) BaCl₂
(b) BaSO₄
(c)Na₂SO₄
(d) NaCl
The following question consist 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:
(d) When BaCl₂(aq) reacts with Na₂SO₄(aq), an insoluble BaSO₄(s) precipitates, and the remaining solution contains NaCl(aq).

Question 24.
Assertion (A) C₄H₈, C₄H₆ and C₄H₁₀ are members of the same homologous series. [1]
Reason (R) C₄H₈, C₄H₆, C₃H₄, C₃H₆, C₂H₄, C₂H₂ are unsaturated hydrocarbons.
Answer:
(d) A is false, but R is true. C₄H₁₀ is an alkane, C₄H₈ is an alkene, and C₄H₆ is an alkyne; they belong to different homologous series, so Assertion is false.
The given compounds like C₄H₈, C₄H₆, C₃H₄, C₃H₆, C₂H₄, C₂H₂ are unsaturated hydrocarbons, so Reason is true.

Question 25.
The following activity is set-up in the science lab by the teacher.
He clamped an aluminium wire on a stand and fixed a pin to the free end of the wire using wax. Then he heated the wire with a burner from the end where the wire is clamped. Students observed the pin fall off. [2]

(i) If the teacher replaces aluminium wire by silver wire, will the students’ observation change? Justify your answer.
(ii) Will the aluminium wire melt? Give reason for your answer.
Answer:
(i) The pin will drop in less time because silver is a better conductor of heat than aluminium, so heat reaches the waxed pin faster, melting the wax quickly.
(ii) The aluminium wire will not melt because the spirit lamp flame does not reach aluminium’s high melting point (~660°C). The wire only conducts heat to the wax.

Question 26.
Attempt either option A or B. [3]
A. An element ‘X is stored in kerosene, and cannot be extracted from its ore using a reducing agent. ‘X forms an ionic compound on reaction with chlorine.
(i) Can we store ‘X’ in water? Give reason to support your answer.
(ii) Identify element ‘X’. Name the process used and write the equation for extraction of ‘X from its ore.
Or
B. The domes of many building in Europe are made of copper. These domes now appear greenish in colour.
(i) Why do the domes appear greenish though copper is orange-red in colour?
(ii) In your opinion, should the copper domes be replaced by iron domes to overcome the problem of change of colour of copper domes?
(iii) Domes used to be made from thin sheets of metals. Why did the ancient architects use copper to make domes?
Answer:
A. (i) No, we cannot store ‘X’ in water because it reacts vigorously with water, producing hydrogen gas and heat, which can lead to fire or explosion.
(ii) Element ‘X’ is sodium (Na). It is stored in kerosene as it is highly reactive. Sodium cannot be extracted using a reducing agent because it is highly electropositive; it is extracted by electrolysis of fused sodium chloride.

Or

B. (i) The domes appear greenish because copper reacts slowly with moist air and carbon dioxide and thus forming a green basic copper carbonate layer (patina).
(ii) No, iron domes should not be used because iron rusts in moist air, leading to structural damage, whereas the green patina on copper is protective and prevents further corrosion.
(iii) Copper was used for domes because it is highly malleable, can be hammered into thin sheets, and the green patina formed over time protects the metal from further corrosion.

Question 27.
Amrita electrolysed distilled water using the set-up shown in figure 1. She was expecting two gases to be evolved at the anode and cathode respectively [3]

Suddenly, she realised that the bulb in the circuit did not glow when she used distilled water (figure 2)

After this realisation, she added a substance to the distilled water for electrolysis to take place. Answer the following questions based on the information given above.
(i) Which gas was she expecting to be formed at the anode and which one at the cathode respectively?
(ii) Why did the bulb not glow when Amrita passed electricity through distilled water?
(iii) Which substance was added by Amrita to distilled water to get the expected result?
Answer:
(i) She was expecting oxygen gas to be formed at the anode and hydrogen at the cathode.
(ii) The bulb did not glow because distilled water does not conduct electricity, as it lacks free ions to carry current.
(iii) Amrita added a few drops of a strong elecrtrolyte like acid H.SO,, or HC1 to the distilled water to increase its ion content, making the water conductive and allowing electrolysis to take place.

Question 28.
Sara took 2 mL of dilute NaOH solution in a test tube and added two drops of phenolphthalein solution to it. The solution turned pink in colour. She added dilute H₂SO₄ to the above solution drop by drop until the solution in the test tube became colourless. 40 drops of dilute H₂SO₄ were used for the change in colour from pink to colourless. When Sara added a drop of NaOH to the solution, the colour changed to back to pink again. [4]

Sara now tried the activity with different volumes of NaOH and recorded her observation in the table given below.

C. No.	Volume of dil. NaOH taken (mL)	Drops of dil. H2S04 used
1	2	20
2	3	30
3	4	40

Answer the following questions based on the above information.
A. If Sara used concentrated H₂SO₄ in place of dilute H₂SO₄, how many drops will be required for the change in colour to be observed?
(a) 40
(b) < 40 (c) > 40
Justify your answer.
Answer:
(b) < 40, because concentrated H₂SO₄ gives more
H⁺ ions than dilute acid.

B. Sara measured 20 drops of dil. H₂SO₄ and found its volume to be 1 mL. If Sara observed a change in colour of NaOH solution by using 3 mL of H₂SO₄, how many mL of NaOH did she add to the test tube initially?
Or
Sara takes 10 drops of dilute H₂SO₄ in the test tube and adds two drops of phenophthalein solution to it. Then she adds NaOH dropwise. Sara observes a change in colour after adding 20 drops of NaOH. What change in colour would she observe and why?
Answer:
1 mL of H₂SO₄ = 20 drops
Now, from the table we can conclude that 1 mL of H₂SO₄ (20 drops) neutralises 2 mL NaOH. So, 3 mL of H₂SO₄ will be 60 drops, which will neutralise 6 mL of NaOH.

S. No.	Volume of dil. NaOH taken (mL)	Drops of dil. H2S04 used
1	2	20 (1 mL)
2.	3	30 (1.5 mL)
3.	4	40 (2 mL)
4	6	3 mL = 60 drops

Or
Colour will change from colourless to pink. Phenophthalein in colourless in acids and turns pink in basic solution.

C. Write a balanced chemical equation for the reaction taking place in the above experiment. Which of the following is true and why? The reaction is a
(a) neutralisation and double displacement reaction
(b) neutralisation and precipitation reaction
(c) precipitation and double displacement reaction
(d) neutralisation, double displacement as well as precipitation reaction.
Answer:
2NaOH + H₂SO₄ → Na₂SO₄+ 2H₂O

(a) Neutralisation and double displacement reaction.
Base NaOH is getting neutralised and forming salt + water. It is double displacement as Na₊ ions are being replaced by H⁺ and OH_– by SO₄^2-, It is not precipitation reaction because Na₂SO₄ is soluble in water.

Question 29.
Attempt either option A or B. [5]
A. A hydrocarbon with the formula Cx Hy undergoes complete combustion as shown in the following equation:
2C_xH_y + 9O₂ → 6CO₂ + 6H₂O.
(a) What are the values of ‘x’ and ‘y’
(b) Give the chemical (IUPAC) name of the hydrocarbon.
(c) Draw its electron dot structure.
(d) Name the alcohol which on heating with cone. H₂SO₄ will produce the above hydrocarbon C_xH_y.
(e) Write a balanced chemical equation for the reaction of C_xH_y with hydrogen gas in presence of nickel.
Answer:
(a) The values of ‘x’ and ‘y’ are x = 3, y = 6.
(b) The IUPAC name of hydrocarbon is propene.
(c) The electron dot structure is a follows

(d) The name of alcohol is propanol.
(e) The balanced chemical equation is

Or

B. The electronic structures of atoms P and Q are shown below
Based on the information given above, answer the following questions.
(a) If P and Q combine to form a compound, what type of bond is formed between them?
(b) Give the chemical formula of the compound formed.
(c) The compound so formed is dissolved in water. Is the resultant solution acidic or basic in nature? Justify your answer.
(d) Write the chemical equation for the reaction between ‘Q’ and ethanol.
(e) What will be the formula of the compound formed when ‘P’ undergoes bonding with carbon?
Answer:
(a) The bond formed after combination between P and Q is ionic bond.
(b) The chemical formula of the compound formed is Q₂P.
(c) Basic, metallic oxides are basic in nature.
(d) The chemical equation is a follows
2C₂H₅OH + 2Q → 2C₂H₅OQ + H₂
(e) The formula of the compound formed when ‘P’ undergoes bonding with carbon is CP₂.

Section – C

Question 30.
Arnav was making notes and he wrote down the following statements from his understanding of reflection from curved surfaces. [1]
I. Concave mirrors can produce both real and virtual images depending on the position of the object.
II. Convex mirrors always produce real and inverted images regardless of the object’s position.
III. In both concave and convex mirrors, the image location can be determined using the mirror formula \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) where f is the focal length, v is the image distance, and u is the object distance.
Choose from the following, the correct option that lists the correct statements about reflection from curved surfaces.
(a) I and II
(b) I, II and III
(c) II and III
(d) I and III
Answer:
(d) Statements I and III are correct because concave mirror can form both real and virtual images, and the mirror formula is applicable to both concave and convex mirrors using proper sign conventions.
Statement II is incorrect as convex mirror always form virtual, erect and diminished images.

Question 31.
Choose the correct option from the below which explains the reason for us to perceive the day sky as blue. [1]
(a) As sunlight passes through the atmosphere, shorter wavelengths, such as blue are scattered more than other colours.
(b) The sky appears blue because all colours are scattered equally, but blue light is stronger and more visible to the human eye.
(c) The blue colour of the sky is due to longer wavelengths like red and orange scattering more than shorter wavelengths, making blue stand out more.
(d) The atmosphere contains blue-coloured particles that give the sky its blue appearance.

The following questions 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:
(a) As sunlight passes through the atmosphere, scattering causes shorter wavelengths, like blue and violet, to scatter more than other colours. However, since out eyes are more sensitive to blue than violet, the sky appears blue.

Question 32.
Assertion (A) A point object is placed at a distance of 26 cm from a convex mirror of focal length 26 cm. The image will not form at infinity. [1]
Reason (R) For above given system the equation \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) gives v = ∞.
Answer:
(c) Given, f = + 26 cm, and u = – 26 cm
Using mirror formula,
⇒ \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
⇒ \(\frac{1}{26}=\frac{1}{v}-\frac{1}{26}\)
⇒ v = 13 cm
So, the image is formed at a finite distance behind the mirror, not at infinity. Hence, assertion is correct, but the reason is incorrect as the formula does not give v = ∞.

Question 33.

The above image shows the formation of an image with an optical instrument.
(a) Identify the optical instrument (shown schematically as a rectangle) in the image.
(b) What type of image is formed in this case?
(c) Based on the measurements given in the image, calculate the focal length of the instrument.
Answer:
(a) The optical instrument shown in the figure is a concave lens, as the rays diverge after passing through it and appear to come from a point on the same side as the object. [1/2]
(b) The image formed is a virtual, erect and diminished image.
(c) Given,
object distance, u = -20 cm
image distance, v = -10 cm
Using lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
⇒ \(\frac{1}{f}=\frac{1}{-10}-\frac{1}{-20}=\frac{-1}{20}\)
⇒ f = -20 cm

Question 34.
Attempt either option A or B. . [2]

A. Find out the following in the electric circuit given in the figure.
(a) Effective resistance of two 8 Ω resistors in the combination.
(b) Current flowing through the 4 Ω resistor.

Or

B. Study the circuit and find out.
(a) Current in 12 £2 resistor.
(b) Difference in the readings of ammeter A₁ and A₂ (if any).
Answer:
A. (a) Since, the two 8 fi resistors are connected in parallel, the equivalent resistance is
\(\frac{1}{R}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}\)
R = 4Ω

(b) Total resistance in circuit,
R_total = 4 + 4 = 8 Ω
Using Ohm’s Law,
I = \(\frac{V}{R}=\frac{8}{8}\)
⇒ I = 1A
Therefore, the current through the 4 Ω resistor 1A.

Or

B. (a) Effective resistance of two 24 Ω resistor in parallel.
\(\frac{1}{R}=\frac{1}{24}+\frac{1}{24}=\frac{2}{24}\) = R = 12Ω
So, total resistance
R_total = 12 + 12 = 24Ω

Current in 12 Ω resistor = \(\frac{V^{-}}{R}=\frac{6}{24}\) = 0.25 A

(b) The readings of ammeter A, and A, are the same both showing 0.25 A, so the difference is 0 A.

Question 35.

The above image shows a corrective measure for a particular defect of vision.
(a) Identify the defect of vision and state what kind of lens is used to correct this deficiency?
Answer:
The defect is hypermetropia (long-sightedness). It is corrected using a convex lens that converges light rays to form the image on the retina.

(b) Draw and label a ray diagram that shows the defect of vision in the above case before correction.
Answer:

Question 36.
A student needs to make a 0.12 Ω resistor. She has some copper wire of 0.80 mm diameter. Resistivity of copper is 18 × 10^-8 Ω m.
(a) Determine the cross-sectional area of the wire.
Answer:
Given, diameter of wire = 0.8 mm
= 0.8 × 10-3 m
The cross-sectional area, A = πr²
= π\(\left(\frac{d}{2}\right)^2\) = π\(\frac{d^2}{4}\)
A = \(\frac{\pi \times\left(0.8 \times 10^{-3}\right)^2}{4}\)
A = 5.03 × 10^-7 m²
Thus the cross-sectional area A is approximately 5 × 10^-7 m².

(b) Calculate the length of wire required for the 0.12 Ω resistor.
Answer:
To find the length of the wire, we can use the formula of resistance

l = 3.33 m

Question 37.
Magnetic field lines are shown in the given diagram. A student makes a statement that the magnetic field at X is stronger than at Y. ‘
(a) Explain with reason, if the student’s claim is correct.
Answer:
Yes, the student’s claim is correct.
The strength of a magnetic field is indicated by the degree of closeness (or density) of magnetic field lines.
At point X, the field lines are close together compared to point Y. This indicates that the magnetic field is stronger at X than at Y.

(b) Also redraw the diagram and mark the direction of magnetic field lines.

Answer:

Question 38.

The above image is that of a Digital Single Lense Reflector (DSLR) camera which are used to take high resolution photographs by professional photographers. The second image of the above two is a schematic diagram of how an image is formed on the sensor of the camera.
Based on your understanding of the lenses, answer the following questions.
(a) What type of lens is used in the DSLR camera shown in the image?
(b) What type of image is formed on the sensor?
Attempt either option (c) or (d).
(c) A photographer is using a DSLR camera with a lens of focal length f = 50 mm to take a close-up photograph of a small object. The lens projects an image onto the camera sensor that is located 60 mm behind the lens. Calculate the object distance (i.e. the distance between the object and the lens).
Or
(d) A photographer is using a DSLR camera to take a picture of a flower. The flower is positioned 150 mm away from the camera lens. The actual height of the flower is 80 mm, and the image height formed on the camera’s sensor is measured to be 20 mm. Calculate the focal length of the camera lens.
Answer:
(a) A convex lens is used in DSLR camera to converge light rays and form real images on the sensor.
(b) Convex lens forms a real and inverted image, when the object is placed beyond the focal point.
(c) Given,
focal length of lens, F = 50 mm
image distance, v = 60 mm
object distance, u = ?
Using lens formula,

⇒ u = – 300 mm
∴ The object distance is 300 mm in front of the lens (negative sign indicates object is on opposite side of image).
Or
(d) Given,
object distance, u = -150 mm
object height, h₀ = 80 mm
image height, h₁ = – 20 mm (inverted image)
m = \(\frac{h_i}{h_o}=\frac{-20}{80}\) = -0.25 mm
Now, m = \(\frac{v}{u}\)
⇒ v = m × u
= (-0.25) × (-150)
= 37.5 mm

Using magnification formula,
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
⇒ \(\frac{1}{f}=\frac{1}{37.5}-\frac{1}{(-150)}=\frac{1}{37.5}+\frac{1}{150}\)
⇒ \(\frac{1}{f}=\frac{4+1}{150}=\frac{5}{150}=\frac{1}{30}\)
⇒ f = 30 mm
Hence, the focal length of the camera lens is 30 mm.

Question 39.
A.

The arrangement of resistors shown in the above figure is connected to a battery. The power dissipation in the 100 Ω, resistor is 81 W. Calculate
(a) the current in the circuit,
(b) the reading in the voltmeter V₂ and
(c) the reading in the voltmeter V₁.
Or

Answer:
(a) Given, power across 100 Ω resistor
P = 81W , R= 100 Ω
We know
P = I²R
I² = \(\frac{P}{R}=\frac{81}{100}\)
⇒ I = \(\sqrt{\frac{81}{100}}=\frac{9}{10}\) = 0.9 A

(b) V₂ measures voltage across the two 25 Ω resistors in parallel.
For two equal resistors in parallel,
\(\frac{1}{R_{\text {eq }}}=\frac{1}{25}+\frac{1}{25}=\frac{2}{25}\)
R_eq = 12.5 Ω
Now, V₂ = I.R_eq, = 0.9 × 12.5= 11.25V

(c) V₁ measures voltage across both 100 Ω and 25 Ω branches (i.e. total circuit voltage). Voltage across 100 Ω resistor,
V₁₀₀= I. R = 0.9 × 100 = 90 V
Total voltage, V₁ = V₁₀₀ + V₂
= 90 + 11.25
V₁ = 101.25 V

Or

B. An electric heater consists of three similar heating elements A, B and C, connected as shown in the figure above. Each heating element is rated as 1.2 kW, 240 V and has constant resistance.
S₁, S₂ and S₃ are respective switches.
The circuit is connected to a 240 V supply.
(a) Calculate the resistance of one heating element.
(b) Calculate the current in each resistor when only S₁ and S₃ are closed.
(c) Calculate the power dissipated across A when S₁, S₂ and S₃ are closed.
Answer:
Given, each heating element A, B and C is rated
Power, P= 1.2 kW = 1200 W
Voltage, V = 240 V
Supply voltage = 240 V
(a) We use formula,
P = \(\frac{V^2}{R}\) R = \(\frac{V^2}{P}\)
R = \(\frac{(240)^2}{1200}=\frac{57600}{1200}\) = 48 Ω
∴ Resistance of one heating element is 48 Ω.

(b) For S₁ and S₃ are closed
Current in C (connected directly across 240 V) using Ohm’s law.
V = IR ⇒ I = \(\frac{V}{R}\)
⇒ I = \(\frac{240}{48}\) = 5A
∴ Current in C = 5 A
Current in A and B (connected in series)
Total resistance in series,
R_series = R_A + R_B =48+ 48 = 96 Ω
I = \(\frac{V}{R}=\frac{240}{96}\) = 2.5A
∴ Current through A and B = 2.5 A [2]

(c) Power across A for S₁, S₂ and S₃ are closed: Now all three element are connected directly to 240 V.
Current through A = 5 A (same as element Q using formula
P = I²R .
P= (5)² × 48 = 25 × 48 = 1200
W = 1.25 kW
∴ Power dissipated across A is 1.2 kW

The post CBSE Sample Papers for Class 10 Science Set 1 with Solutions appeared first on Learn CBSE.

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