CBSE Sample Papers for Class 9 Science Set 6 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 9 Science with Solutions Set 6 are designed as per the revised syllabus.

CBSE Sample Papers for Class 9 Science Set 6 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.
What is the composition of the hard matrix in which bone cells are embedded? [1]
(a) Magnesium and phosphate
(b) Calcium and phosphate
(c) Calcium and phosphorus
(d) Manganese and phosphorus
Answer:
(c) Calcium and phosphorus

Explanation:
Bone cells are embedded in a hard matrix that is composed of calcium and phosphorus compound.

Question 2.
Identify which of the following are dead cells of phloem? [1]
(a) Sieve cells
(b) Sieve tubes
(c) Companion cells
(d) Phloem fibres
Answer:
(d) Phloem fibres

Explanation:
Among the given phloem elements, only phloem fibres are dead at maturity and they provide mechanical support to the plant.

Question 3.
Which of the following is a correct combination of function and connective tissue component? [1]
(a) Joining bones to muscles: Ligaments
(b) Providing strength and flexibility: Collagen fibres
(c) Storing fat: Cartilage
(d) Transporting oxygen: Tendons
Answer:
(b) Providing strength and flexibility: Collagen fibres

Explanation:
Providing strength and flexibility: Collagen fibres is the correct combination as it strengthen connective tissues and gives them flexibility.

Question 4.
The plant cell wall is composed of [1]
(a) cellulose
(b) lipid
(c) protein
(d) chitin
Answer:
(a) cellulose

Explanation:
The plant cell wall is mainly made of cellulose, which provides rigidity and structural support.

Question 5.
Which are the correct statements related to the structure and function of a cell? [1]
(i) Lysosomes are responsible for photosynthesis in plant cells.
(ii) The nucleus is absent in all eukaryotic cells
(iii) Mitochondria are known as the powerhouse of the cell
(iv) Ribosomes are found only in the nucleus
(v) Cell wall is present in all animal cells
(a) (ii) and (iv)
(b) (iii) only
(c) (iii) and (iv)
(d) (ii) and (iii)
Answer:
(b) (iii) only

Explanation:
The statement (iii) is correct as mitochondria are called the powerhouse of the cell because they produce energy.

Question 6.
Select the group in which all the crops belong to the same cropping season.
(a) Wheat, barley, gram, mustard, pea
(b) Rice, maize, sugarcane, cotton, bajra
(c) Wheat, rice, mustard, gram, barley
(d) Gram, rice, cotton, wheat, mustard
Answer:
(a) Wheat, barley, gram, mustard, pea

Explanation:
Wheat, barley, gram, mustard and pea belong to the same group as all of them are rabi crops.

Don’t get confused while choosing option a, as rice and cotton are karif crops that are sown at the onset of the monsoon.

Question 7.
Which of the following is an example of animal husbandry? [1]
(a) Growing wheat and rice
(b) Rearing cows and buffaloes
(c) Planting trees in a forest
(d) Using chemical fertilisers
Answer:
(b) Rearing cows and buffaloes

Explanation:
Animal husbandry is the practice of breeding and rearing animals like cows, buffaloes, sheep and goats for food, wool and other products.

The following two 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.

Question 8.
Assertion (A) Mitochondria are able to make some of their own proteins [1]
Reason (R) Mitochondria is known as the powerhouse of the cells.
Answer:
(b) Both A and R are true but R is not the correct explanation of A.

Explanation:
R can be corrected as Mitochondria have their own DNA and ribosomes.

Question 9.
Assertion (A) In desert the entire surface of a plant has an outer covering called epidermis. [1]
Reason (R) Epidermis protects plants from water loss.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Question 10.
Lysosomes are the scavengers of the cell. Comment upon the statement with justification. [2]
Answer:
Lysosomes are called the scavengers of the cell because they contain powerful digestive enzymes capable of breaking down all organic material, harmful substances, and any foreign particles entering the cell. This helps to keep the cell clean by removing waste materials. If a cell is damaged beyond repair, lysosomes can burst and digest the entire cell, ensuring proper disposal of cell debris and maintaining the health of the organism.

Question 11.
Attempt either A or B. [2]
A. Name one example each of the following practices in animal husbandry. Also, state one main advantage of each.
(i) Poultry farming
(ii) Cattle rearing
Answer:
(i) In poultry farming, birds such as hens are reared to obtain eggs and meat, which provide nutritious food and serve as a source of income for farmers.
(ii) In cattle rearing, animals such as cows or buffaloes are reared to obtain milk and milk products, which are important for the human diet and help in supporting the livelihood of farmers.

Or

B. Name any two types of cropping patterns. Explain how each pattern helps in increasing crop yield.
Answer:
The two main types of cropping patterns are as follows
Mixed cropping – Growing two or more crops together in the same field which reduces the risk of total crop failure due to adverse weather or pests.
Intercropping – Growing two or more crops in alternate rows which allows better utilisation of nutrients and sunlight.
Crop rotation – Growing different crops in the same field in a planned sequence which helps maintain soil fertility and reduces pest and disease build up. [any two]

Question 12.
The following cell structures are found in living organisms. [2]
Cell wall, cell membrane, cytoplasm, nucleus, chloroplast, mitochondria, ribosome, vacuole.
Create two separate examples of a simple cell, one for a plant cell and one for an animal cell, by using the above cell structures present in a plant cell and an animal cell.
Answer:
The structures present in a plant cell and an animal cell are as follows
Plant cell A plant cell has a cell wall, cell membrane, cytoplasm, nucleus, chloroplast, mitochondria, ribosomes and a large vacuole.
Animal cell An animal cell has a cell membrane, cytoplasm, nucleus, mitochondria and ribosomes. It does not have a cell wall or chloroplasts and its vacuoles are small.

Question 13.
Draw the structure of a smooth muscle and explain its function. [3]
Answer:
Smooth muscle is an involuntary muscle found in the walls of internal organs such as the stomach, intestine and blood vessels. Its cells are long, spindle-shaped, and non-striated. These muscles contract slowly and work without fatigue to help in functions like movement of food in the alimentary canal and the flow of blood in vessels, which are involuntarity controlled.

Question 14.
In an experiment, a raisin was kept in plain water for 2 hours and another raisin was kept in a concentrated sugar solution for the same time. [3]
(i) Mention the changes observed in each raisin after the experiment.
(ii) Explain why the size of the raisin changes in both cases.
Answer:
(i) The raisin kept in plain water swells up, while the raisin kept in concentrated sugar solution gets shrinked .
(ii) In plain water, the concentration of water outside the raisin is higher than inside, so water enters the raisin through its semi-permeable membrane, causing it to swell. In concentrated sugar solution, the concentration of water inside the raisin is higher than outside, so water moves out of the raisin, causing it to shrink.

To secure maximum marks, students should mention the concept of osmosis.

Question 15.
Ravi observed onion peel cells and human cheek cells under a microscope in his school laboratory. Help him to understand some points about these cells by answering the questions given below. [4]
Attempt either subpart A or B.
A. Which of these cells has a cell wall? State its function.
Answer:
The onion peel cell has a cell wall. It provides shape, rigidity and protection to the cell.

Or

B. Which of these plant cells lack chloroplasts, and why are they absent in them?
Answer:
The human cheek cell lacks chloroplasts because it is an animal cell and does not perform photosynthesis.

C. Which of these cells has a large vacuole? What is its role in the cell.
Answer:
The onion peel cell has a large vacuole. It helps maintain the shape of the cell and stores cell sap, which contains water, sugar and other substances.

D. Given below is a diagram of a plant cell. Which of these following parts is called as the powerhouse of the cell?

Answer:
The given figure shows plant cell with parts labelled as A,B,C and D. Out of these C represents mitochondria which is known as the powerhouse of the cell as it is responsible for production of energy.

Question 16.
Attempt either A or B. [5]
A. Ramesh wanted to improve the yield of his wheat crop.
(i) Should he rely only on chemical fertilisers or use a combination of manures and fertilisers? Justify your answer.
(ii) Explain how the use of manures improves soil structure and fertility.
Answer:
(i) Ramesh should use a combination of manures and chemical fertilisers. Manures improve the soil structure , while fertilisers provide specific nutrients quickly. Using both together ensures better crop yield and maintains soil health.
(ii) Manures add organic matter to the soil, which makes it porous, improves water retention and increases the activity of soil microorganisms. This helps in maintaining soil fertility for a longer period.

Or

B. Sunita grew two sets of potted plants of the same species and labelled them as Set A and Set B. She supplied nitrogen, phosphorus and potassium in required amounts to Set A, but supplied only water to Set B.
(i) What difference will she observe in the growth of plants in the two sets? Give the reason for this difference.
(ii) State any three disadvantages of excessive use of chemical fertilisers.
Answer:
(i) The plants in Set A will grow faster, healthier, and produce more yield, while the plants in Set B will be weak and show poor growth. This is because nitrogen, phosphorus, and potassium are essential nutrients needed for proper plant growth and only water cannot supply all the essential nutrients.

(ii) Three disadvantages of excessive use of chemical fertilisers are as follows.

• It can reduce soil fertility
• Excessive use can cause water pollution and
• It harms beneficial soil microorganisms.

Section – B

Question 17.
In which of the following processes do particles move closer together but remain in motion? [1]
(a) Condensation
(b) Diffusion
(c) Evaporation
(d) Freezing
Answer:
(a) Condensation

Explanation:
The process where particles move close together but remain in motion is condensation. This is because condensation occurs when gas particles convert into liquid at a specific temperature and pressure.

Question 18.
Four samples are given in the table. Which of the following is not an example of suspension? [1]

Sample	Component A	Component B
(a)	Chalk powder	Water
(b)	Sand	Water
(c)	Dust particles	Air
(d)	Ink	Water

Answer:
(d) A suspension is a heterogeneous mixture in which the solute particles do not dissolve, but remain suspended throughout the bulk of the medium. So, among the given samples, the mixture of ink and water is not an example of suspension because they f^rm a homogeneous mixture.

Question 19.
What is the formula unit mass of potassium carbonate? [1]
(a) 111 u
(b) 128 u
(c) 142 u
(d) 138 u
Answer:
(d) 138 u

Explanation:
The formula unit mass of potassium carbonate,
i.e., (K₂CO₃) = 2 × (Atomic mass of K) + (Atomic mass of C) + 3 × (Atomic mass ofO)
= 2 × 39 u + 12 u + 3 × 16 u
= 78 u + 12 u + 48 u
= 138 u

Don’t make mistake in calculating the correct formula unit mass by taking wrong atomic masses of different atoms.

Question 20.
When a solid is heated to form a liquid and then a gas, the force of attraction between particles decreases because [1]
(a) particles gain energy, move further apart and intermolecular forces weaken.
(b) particles lose energy, come closer and intermolecular forces strengthen.
(c) particles remain stationary and intermolecular forces stay constant.
(d) particles break into atoms and eliminating all forces.
Answer:
(a) particles gain energy, move further apart and intermolecular forces weaken.

Explanation:
When a solid is heated to form a liquid and then a gas, the force of attraction between particles decreases because particles gain energy, move further apart and intermolecular forces weaken.

Question 21.
Which laws of chemical combination apply to these reactions ? [1]
Reaction 1 : C(s) + O₂(g) → CO₂(g)
(Carbon bums in oxygen to form carbon dioxide.)
Reaction 2 : 2H₂(g) + O₂(g) → 2H₂O(l)
(Hydrogen and oxygen combine to form water.)

(a) Only reaction 1 follows the law of conservation of mass.
(b) Both reactions follow the law of constant proportions.
(c) Only reaction 2 follows the law of conservation of mass.
(d) Both reactions follow the law of conservation of mass.
Answer:
(d) Both reactions follow the law of conservation of mass;

Since total mass of reactants is equal to total mass of products so they follow law of chemical combination.

Question 22.
Four statements about atomic and molecular masses are listed. [1]
I. A molecule of nitrogen (N₂) has a mass equal to twice the atomic mass of nitrogen.
II. A molecule of oxygen (O₂) has a mass equal to twice the atomic mass of oxygen.
III. Neon is a monoatomic gas, so its atomic mass is equal to its molecular mass.
IV. A molecule of chlorine (Cl₂) has a mass equal to twice the atomic mass of chlorine.
Which of the above statements correctly describes an element whose atomic mass and molecular mass are equal?
(a) I and II
(b) II and III
(c) III only
(d) III and IV
Answer:
(c) III only

Explanation:
Neon is a noble gas and exists as single atoms (monoatomic). Hence, its atomic mass and molecular mass are the same. All other elements mentioned (N2,02,C12) exist as diatomic molecules, so their molecular mass is twice their atomic mass.

Question 23.
The number of electrons in the atom of an element X is 15 and the number of neutrons is 16. The correct representation of an atom of this element will be [1]
(a) ¹⁵₁₆X
(b) ¹⁶₁₅X
(c) ³¹₁₅X
(d) ³¹₁₆X
Answer:
(c) ³¹₁₅X

Explanation:
Number of electrons = Number of protons = Atomic number
i.e., Number of protons = Atomic number = 15.
Moreover, Number of neutrons = 16
Mass number = Number of protons + Number of neutrons = 15 + 16 = 31
Thus, the symbol of the element is ³¹₁₅X.

The following question consists of two statements – Assertion (A) and Reason (R). Answer the 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 24.
Assertion (A) Pure water obtained from different sources such as river, well spring, sea, etc. always contains hydrogen and oxygen in the ratio of 1 : 8 by mass. [1]
Reason (R) A chemical compound always contains elements combined in a fixed proportions by mass.
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Question 25.
The given diagram shows energy levels in an atom. These energy levels are called shells or orbits. [2]

(a) Identify the correct order of electron shells based on their distance from the nucleus in the given diagram.
(b) What is the maximum capacity of electrons in the K and L shell?
Answer:
(a) The correct order of electron shells based on their distance from the nucleus (from closest to farthest) is give below
K shell (n = 1) → L shell (n = 2) → M shell (n = 3) → N shell (n = 4).

(b) The maximum capacity of electrons in the K and L shell is 2 and 8.

Question 26.
Attempt either A or B. [3]
A. A substance ‘X’ exists as a solid that converts directly into gas under normal pressure and is used for refrigeration.
(i) Can ‘X’ be stored in an open container at room temperature? Give reason for your answer.
(ii) Identify ‘X’ and write the chemical equation for its formation.
Answer:
(i) No, ‘X’ cannot be stored in an open container at room temperature because it sublimes (directly changes
(ii) ‘X’ is dry ice (solid CO₂). Chemical equation for the formation of ‘X’ is given below
CO₂(g) → CO₂(s)

Or

B. During the burning of a candle, three changes occur the wax melts into liquid, the liquid wax rises up the wick, the wax burns producing flame, carbon dioxide, and water vapour.
(i) Which of these changes is a physical change and why?
(ii) Why is the burning of wax considered a chemical change?
(iii) How can you prove that both physical and chemical changes occur simultaneously during candle burning?
Answer:
(i) The melting of wax is a physical change because only the state changes from solid to liquid, no new substance is formed, and the process is reversible.
(ii) The burning of wax is a chemical change because it produces new substances (CO₂ and H₂O), is irreversible, and involves a change in chemical composition.
(iii) Melting (physical) and burning (chemical) occur simultaneously during candle burning.

Question 27.
Arpita placed torch through two solutions (a) and (b), but path of light became visible in solution (b) only as shown below in the diagram. [3]
Cal Solution of copper sulphate, (b) mixture of water and milk
Answer the following questions based on the information given above.
(a) In which container, (a) or (b) does the Tyndall effect occur? Give reason.
(b) Identify which is a true solution and which is a colloid among the given mixtures.
(c) Why is the path of light visible only in the mixture of water and milk?
Answer:
(a) The Tyndall effect occurs in container (b) because the milk particles scatter the light. This scattering makes the path of light visible to the eye. It happens due to the presence of larger dispersed particles in the colloid.
(b) The copper sulphate solution (a) is a true solution as particles are too small to scatter light. The milk and water mixture (b) is a colloid due to larger dispersed particles.
(c) The path is visible because the colloidal particles in milk scatter the light. True solutions like copper sulphate do not scatter light, so no path is seen.

Question 28.
Isotopes are atoms of the same element with the same atomic number but different mass numbers. They have the same number of protons but differ in the number of neutrons. Since they possess the same number of electrons and the same electronic configuration, their chemical properties are identical. However, due to the difference in mass, they show variation in physical properties. [4]

(a) Solution of copper sulphate.
(b) mixture of water and milk

Answer the following questions based on the information given above.
A. Why do isotopes show identical chemical properties even though their mass numbers differ?
(a) Because they have different neutrons
(b) Because they have the same number of protons and neutrons
(c) Because they have the same electronic configuration Justify your answer.
Answer:
(c) Isotopes show identical chemical properties even though their mass numbers differ because they have the same electronic configuration.

B. Give an example of any one pair of isotopes other than that of hydrogen.
Answer:
The two isotopes of carbon are ¹²₆C and ¹⁴₆C.

Or

Calculate mass number in all the three isotopes of hydrogen.
Answer:
Mass number can be calculated by using the the following relation
Mass number = Number of protons + Number of neutrons
For protium, Number of protons = 1, Number of neutrons = 0
So, Mass number = 1+0=1 ,
For deuterium, Number of protons = 1, Number of neutrons = 1
So, Mass number = 1 + 1 = 2
For tritium, Number of protons = 1, Number of neutrons = 2
So, Mass number =1+2 = 3

To get maximum marks, one should write the formula of mass number and correctly calculate it for all the three isotopes of hydrogen.

C. Which isotope is used in the treatment of cancer? Isotope of which of the following element is used in the treatment of goitre?
(a) Cobalt – 60
(b) Iodine -131
(c) Uranium – 235
(d) Carbon -14
Answer:
(b) An isotope of cobalt – 60 is used in the treatment of cancer.
Among the given isotopes, an isotope of iodine (1-131) is used in the treatment of goitre.

Question 29.
Attempt either A or B [5]
A. J.J. Thomson was the first to propose a model of the atom in 1898 after the discovery of electrons. He described the atom as a sphere of positive charge with electrons embedded in it, like seeds in a watermelon.
(a) What analogy did Thomson use to describe his model of an atom?
(b) How are electrons arranged in Thomson’s model of the atom?
(c) According to Thomson, why is an atom electrically neutral?
(d) What part of the atom in Thomson’s model is considered positively charged?
(e) Why was Thomson’s model eventually rejected?
Answer:
(a) Thomson compared the atom to a watermelon or plum pudding, where the positive charge is spread like the red part of the watermelon and electrons are embedded like seeds.
(b) Electrons are uniformly embedded in a sphere of positive charge, like plums in a pudding or seeds in a watermelon.
(c) An atom is electrically neutral because the negative charge of electrons is balanced by the uniform positive charge of the rest of the atom.
(d) The entire sphere of the atom (excluding electrons) is positively charged in Thomson’s model, holding the electrons inside it.
(e) Thomson’s model was rejected because it could not explain the results of Rutherford’s gold foil experiment, which showed that the positive charge is concentrated in a nucleus.

Or

B. The electronic structures of atoms P and Q are shown below.

Based on the information given above, answer the following questions.
(a) Write the electronic configuration of atoms P and Q.
(b) State one difference between valence electrons and valency?
(c) What is the valency of atom P?
(d) Which of the above atoms has valency of four similar to that of carbon?
(e) Write the name and symbols of both the atoms P and Q.
Answer:
(a) Electronic configuration of atom P is as follows.
(atomic number = 13)
K = 2, L = 8, M = 3
Electronic configuration of Q is as follows
(atomic number = 14)
K = 2, L = 8, M = 4

(b) Valence electrons are the electrons present in the outermost shell of an atom. They determine the chemical ireactivity of the element and valency is the combining capacity of an element. It is usually equal to the number of electrons lost, gained, or shared to complete the outermost shell.

(c) The valency of atom P is three as it has three electrons in its outermost shell and can easily lose these electrons to attain noble gas configuration.

(d) Atom Q has valency of four similar to that of carbon as it can lose or gain those four electrons to complete its octet and become stable.

(e) The name and symbols of both the atoms P and Q are aluminium(Al) and silicon(Si) respectively.

Section – C

Question 30.
Khushi was making notes and she wrote down the following statements from her understanding of motion of an object. [1]
I. Velocity is a vector quantity and has both magnitude and direction.
II. Acceleration can be negative if the object is slowing down.
III. An object with zero velocity must have zero acceleration.
IV. A body moving in a straight line with increasing speed has positive acceleration.

Choose all the correct statements related to velocity’ and acceleration.
(a) I, II, IV
(b) I, II
(c) II, IV
(d) II, III, IV
Answer:
(a) I, II, IV

Explanation:
Velocity is a vector quantity because it has both magnitude and direction. Acceleration can be negative when an object slows down, which is known as retardation. A body moving in a straight line with increasing speed experiences positive acceleration. However, statement III is incorrect, because an object can have zero velocity but still be accelerating, such as at the highest point in vertical motion.

Question 31.
Choose the correct statement from the following related to the propagation of sound. [1]
(a) Sound can travel through vacuum
(b) Sound travels fastest in gases and slowest in solids
(c) Sound needs a medium to travel
(d) Sound travels in the form of transverse waves in air
Answer:
(c) Sound needs a medium to travel

Explanation:
Sound needs a medium like solid, liquid, or gas to travel because it propagates through vibrations of particles, and cannot travel through a Vacuum.

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 correct explanation of A.
(b) both A and R are true but R is not 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 velocity of an object can be zero while the acceleration is non-zero. [1]
Reason (R) An object thrown vertically upward has zero velocity at its highest point but the acceleration due to gravity is non-zero.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
When an object is thrown upwards, at its highest point, its velocity is zero but it is still under the influence of gravitational acceleration.

Question 33.
A worker pushes a box weighing 50 kg by applying a constant horizontal force of 120 N. The box moves 6 meters along the floor. [2]

Answer the following. .
(i) Calculate the work done by the worker on the box.
(ii) If the box starts from rest and all the work done is converted into kinetic energy, calculate the velocity of the box after moving 6 meters.
(iii) If the worker applies the same force but moves the box 8 meters instead of 6 meters, calculate the new work done.
Answer:
Given, Mass of box, m = 50 kg
Force applied, F = 120 N
Displacement, d₁ = 6 m, d₂ = 8 m
Acceleration due to gravity, g = 9.8 m/s²

(i) Using W = F x d₁
W = 120 × 6 = 720 J

(ii) Using W = \(\frac{1}{2}\)mv²
v² = 2 × \(\frac{W}{m}\) = 2 × \(\frac{720}{50}\) = 28.8
v = 5.37 m/s
W = F × d₂ = 120 × 8 = 960 J

Question 34.
Attempt either A or B.
A. Observe the diagram given below and answer the following questions.

(i) Name the medium required for the propagation of sound and explain what would happen if this medium were absent?
(ii) If the frequency of the tuning fork is increased, what change will you observe in the wave pattern?
Answer:
(i) Sound requires a material medium such as solid, liquid, or gas to propagate, as it travels by vibrations of particles in the medium. If the medium is absent (for example, in a vacuum), sound cannot travel because there are no particles to carry the vibrations.
(ii) If the frequency of the tuning fork is increased, the sound waves will become closer together, meaning the wavelength decreases, and more compressions and rarefactions will appear in the same space. This results in a wave pattern showing more waves per unit length and a sound with a higher pitch.

Or

B. A sound wave travels a distance of 850 meters in 2.5 seconds through a certain medium.
(i) Calculate the speed of sound in the medium.
(ii) What happens to the wavelength if the frequency is halved?
Answer:
Given
Distance travelled by sound, d = 850 m
Time taken, t = 25 s
Frequency of sound, f = 425 Hz

(i) Using v = \(\frac{\text { Distance }}{\text { Time }}\)
v = \(\frac{850}{25}\) = 340 m/s

(ii) If frequency is halved, f’ = \(\frac{425}{2}\) = 212.5 Hz
New wavelength λ’ = \(\frac{v}{f^{\prime}}=\frac{340}{212.5}\) = 1.6 m
Hence, when frequency decreases, wavelength increases.

Question 35.
A wooden block of volume 500 cm³ floats on water. The density of wood is 0.6g/cm³, and the density of water is 1 g/cm³.

(i) What volume of the wooden block remains submerged in water?
(ii) What is the weight of the wooden block?
Answer:
Given, Volume of wooden block, V_block = 500 cm³
Density of wood, ρ_wood = 0.6 g/cm³
Density of water, ρ_water = 1 g / cm³

(i) From Archimedes’ Principle, a floating body displaces a volume of liquid equal to its own weight.
Let V_sub be the submerged volume of wood.
Then, Weight of block = Weight of displaced water
ρ_wood × V_block = ρ_water × V_sub
Substituting the values 0.6 × 500 = 1 × V_sub
Vsub= 300 cm³

(ii) Mass of block = ρ_wood × V_block
= 0.6 × 500 = 300 g or 0.3 kg
Weight = m × g = 0.3 kg × 9.8 m / s²
= 2.94 N

Question 36.
A10 kg rock is lifted to the top of a cliff 20 meters high.
(i) Calculate the potential energy of the rock at the top of the cliff.
(ii) If the rock is dropped from the cliff, what will be its kinetic energy when it is 5 meters above the ground?
Answer:
Given
Mass of rock, m = 10 kg ;
Height of cliff, h = 20 m Gravitational acceleration, g = 9.8 m/s²
(i) Using potential energy, PE = mgh = 10 × 9.8 × 20 = 1960 J
(ii) At a height of 5 m, some potential energy has been converted into kinetic energy.
Remaining potential energy at 5 m, PE = 10 × 9.8 × 5 = 490 J
By conservation of mechanical energy, KE = initial PE – final PE = 1960 – 490 = 1470 J

Question 37.
A person is standing between two parallel cliffs. He produces a sound and hears the first echo after 2 sec and the second echo 3 sec after first echo. [3]
(i) What is the time taken to hear the second echo after producing the sound?
(ii) If the speed of sound in air is 340 m/s, what is the distance of the first cliff from the person?
Answer:
(i) Time taken for second echo = Time for first echo + 3 sec = 2 + 3 = 5
(ii) Time for sound to go and return = 2 s
So, time for one way = \(\frac{1}{2}\) = 1 s
Distance = speed × time = 340 × 1 = 340 m

Students usually forget that echo time is for the round trip and often forget to divide by 2 for one way distance.

Question 38.
In a school playground, a player strikes a stationary 0.5 kg hockey ball with a 25 N force for 0.2 seconds. The ball accelerates rapidly from rest due to the applied force. After die contact ends, with negligible friction on the ground, the ball continues moving forward at constant velocity. [4]
A. State what happens
(i) During application of force?
(ii) After the ball starts moving?
Answer:
(i) When the hockey stick applies force, the stationary ball experiences an unbalanced force, causing it to accelerate rapidly from rest.
(ii) After losing contact with stick, ball moves with constant velocity, as there is negligible friction to oppose its uniform motion.

B. Find acceleration of ball.
Answer:
We know F = m × a
F = 25N
m = 0.5 kg
a = \(\frac{F}{m}=\frac{25}{0.5}\) = 50m/s²

Attempt either C or D
C. Calculate
(i) Velocity of ball after the hit
(ii) Momentum after the hit
Answer:
(i) Velocity of ball after the hit.
Using first equation of motion v = u + a × t
u = 0 m/s
a = 50 m /s²
t = 0.2 sec
v = 0 + 50 × 0.2 = 10 m/s
Velocity after the hit = 10 m/s

(ii) Momentum after the hit
Momentum p = m × v
p = 0.5 × 10 = 5 kg m/s

Or

D. Calculate the impulse imparted to the ball.
Answer:
We know
Given F = 25 N
t = 0.2 s
Impulse = force × time
Impulse = 25 × 0.2 = 5 Ns
Impulse imparted to the ball = 5 Ns

Question 39.
Attempts either A or B
A. In the figure below, the card is flicked with a push.

(i) What do you observe in this case and explain your observation?
(ii) State the law involved in this case.
(iii) What will be your observation if the above coin is replaced by a heavy five-rupee coin? Justify your answer.
Answer:
(i) When the card is given a sharp horizontal flick with finger, the card underneath the coin shoot away. The coin due to its inertia of rest, tends to remain in the state of rest. Thus, the coin falls vertically into the glass.
(ii) The law involved is Newton’s first law of motion, which state that a body contiues in its state of rest or uniform motion in a straight line unless compelled to change that state by an external unbalanced force.
(iii) If the coin is replaced by a heavier five-rupee coin, the same result will be observed. The heavier coin possesses greater inertia of rest, so it resists any horizontal motion even more strongly. Hence, when the card is flicked away, the heavy coin remains at rest momentarily and falls vertically into the glass due to gravity.

B. A 2 kg box is placed on a smooth and frictionless surface. Initially, the box is at rest. A constant horizontal force of 10 N is applied to the box for 0.5 seconds. After this time, the force is suddenly removed, and the box continues to move on the frictionless surface.

(i) Calculate the acceleration of the box while the force is applied.
(ii) Explain how the momentum of the box changes during the force application?
(iii) If the force had acted for twice the time, how would the final velocity of the box change? Calculate the new velocity.
Answer:
Given, Mass of the box, m = 2 kg
Force applied, F = 10 N
Time of force application, t = 0.5 s
Initial velocity, u = 0 m/s
Surface is smooth and frictionless

(i) Using Newton’s second law,
a = \(\frac{F}{m}=\frac{10}{2}\) = 5 m/s²

(ii) Initially, the box is at rest, so its momentum is zero.
While the force is applied, the box accelerates, gaining velocity. As a result, its momentum increases from zero to a final value.
Using the impulse momentum theorem
Change in momentum = F × f
= 10 × 0.5
= 5 kg m/s
So, the momentum increases by 5 kg m/s in the direction of the force.

(iii) New time, t’ = 2 × 0.5 = 1 s
Acceleration, a = 5 m/s²
Using the first equation of motion v = u + at
= 0 + 5 × 1
= 5 m/s j
So, the new final velocity would be 5 m/s.

For full marks, mention using v = u + at with new time to calculate the updated final velocity.