CBSE Sample Papers for Class 9 Science Set 1 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 9 Science with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 9 Science Set 1 with Solutions

Time: 3Hrs.
Max.Marks: 80

General Instructions

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

Question 1.
Select the group in which all the tissues belong to the same type.
(a) Bone, cartilage, blood, adipose tissue and tendons
(b) Bone, cartilage, muscles, nerves and blood
(c) Cartilage, ligaments, tendons, muscles and skin
(d) Blood, bones, cartilage, epidermis and adipose tissue
Answer:
(a) Bone, cartilage, blood, adipose tissue and tendons

Explanation:
Bone, cartilage, blood, adipose tissue and tendons belong to the same group as all of them are connective tissues.

Question 2.
Which of the following correctly matches the process with its function in organisms?
(a) Formation of gametes: Mitosis
(b) Daughter cells having same number of chromosomes as mother cell: Meiosis
(c) Helps in growth and repair of tissues: Mitosis
(d) Cell division for asexual reproduction: Meiosis
Answer:
(c) Helps in growth and repair of tissues: Mitosis

Explanation:
Helps in growth and repair of tissues : Mitosis is the correct combination as mitosis produces geneticall y identical cells, essential for growth and tissue repair.

Question 3.
Consider the following statements related to manure and identify the option with correct statement.
(i) It contains small quantities of organic matter.
(ii) It is prepared from synthetic chemicals in laboratories.
(iii) It decreases the water holding capacity of soil.
(iv) It harms soil organisms and increases population. .
(v) It is prepared by the decomposition of animal excreta and plant waste.

(a) (v) only
(b) (iv) and (v)
(c) (i) and (ii)
(d) (i) only
Answer:
(a) (v) only

Explanation:
Statement (v) is correct as manure is prepared by the decomposition of animal excreta and plant waste. It is natural and supports soil organisms.

Question 4.
Which of the following is involved in transpiration?
(a) Cell wall
(b) Guard cells
(c) Cork cell
(d) Epidermal cell
Answer:
(b) Guard cells

Explanation:
Two kidney shaped cells called guard cells are responsible for opening and closing of stomata. Which aids in transpiration.

Question 5.
Lysosomes are formed by
(a) RER
(b) SER
(c) golgi apparatus
(d) ribosomes
Answer:
(c) golgi apparatus

Explanation:
Lysosomes are membrane- bound sacs, filled with digestive enzymes. Which are formed by golgi bodies.

Question 6.
Which of the following is a macronutrient for plants?
(a) Zinc
(b) Sulphur
(c) Chlorine
(d) Manganese
Answer:
(b) Sulphur

Explanation:
Macronuitrients are required in larger quantities by plants. N itrogen, phosphorus, sulphur and potassium are the primary macronutrients.

Question 7.
Which of the following is an example of an exotic breed of cattle?
(a) Sahiwal
(b) Red sindhi
(c) Brown swiss
(d) Gir
Answer:
(c) Brown swiss

Explanation:
Brown swiss is an example of an exotic breed of dairy cattle. Exotic breeds are those that have been introduced from other regions.

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, but (R) is the correct explanation of A.
(b) Both A and R are true, and (R) is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 8.
Assertion (A) Regular cleaning and proper shelter are essential in dairy animal rearing. [1]
Reason (R) Unclean conditions reduce milk yield and increase chances of disease in animals.
Answer:
(a) Both A and R are true, but (R) is the correct explanation of A.

Question 9.
Assertion (A) All cell organelles are surrounded by double membranes. [1]
Reason (R) Presence of membrane bound organelles is the characteristic feature of all the eukaryotic cells.
Answer:
(d) A is false but R is true.

Explanation:
A is false but R is true. A can be corrected as some cell organelles are enclosed by double membranes, while others h ave a single membrane or lack a membrane entirely.

Question 10.
Unlike muscles, connective tissues do not contract, but play a vital role in support and connection in the body. Comment upon this statement with justification. [2]
Answer:
Connective tissues do not have the ability to contract like muscles because their primary role is to provide support, corvne ct different tissues and organs, and protect the bod y.
For example, tendons connect muscles to bones and have a strong, fibrous structure that resists stretching, enabling efficient force transmission during moveme nt.
Thus, despite lacking contractile ability, connective tissues are essen tial for maintaining the structural integrity and function of the body.

Question 11.
Attempt either A or B [2]
A. What are the functions of the following cell organelles?
How do these organelles assist in the synthesis and processing of cellular molecules?
(i) Ribosome
(ii) Rough Endoplasmic Reticulum (RER)
Answer:
Ribosomes synthesise proteins and RER is involved in folding and modification of proteins.
Organelles like ribosomes synthesise proteins which are tran sported to RER where these protein undergo modification and folding.

Or

B. Why does the rough endoplasmic reticulum appear rough under a microscope, and why is the endoplasmic reticulum considered a transport system?
Answer:
The rough end oplasmic reticulum appears rough under a microscope due to the presence of ribosomes attached to its surface, which gives it a rough texture. The endoplasmic reticulum is
considered a transport system because it forms a network of membranes that facilitates the movement of substances such as proteins (in the RER) and lipids (in the SER) within the cell.

Question 12.
A team of biologists examines two cells – one from a prokaryote and another from Amoeba. Using this information, write down the differences the biologists might observe between the two cells. [2]
Answer:
The differences the biologists might observe between the two cells are as follows.

Feature	Prokaryotic Cell	Eukaryotic Cell
1. Size	Generally small (1-10 pm)	Generally large (5-100 pm)
2. Nuclear region	Not well defined and known as nucleoid; lacks nuclear membrane	Well defined and surrounded by a nuclear membrane

To obtain maximum marks, students should mention that prokaryotic cells have a nucleoid region without a nuclear membrane and eukaryotic cells have well defined nucleus with nuclear membrane.

Question 13.
Draw a neat diagram of the tissue responsible for the translocation of food from the leaves to different parts of the plant. [3]
Answer:
Phloem is the part of complex permanent plant tissue which is responsible for the translocation of food from leaves to different parts of the plant.

Question 14.
In eukaryotic cells, organelle A plays a crucial role in synthesising and transporting important molecules, organelle A is studded with organelle B, which is also found in the cytoplasm, mitochondria and plastids, and is mainly involved in producing proteins. Meanwhile, organelle A without organelle B is responsible for lipid synthesis and detoxification. [3]
(i) Name the organelle A and organelle B mentioned in the passage.
(ii) Is organelle A found in prokaryotic cells? Justify your answer.
Answer:
(i) Organelle A is the endoplasmic reticulum (ER) and organelle B is the ribosome.
(ii) Organelle A (Endoplasmic Reticulum) is not found in prokaryotic cells. This is because prokaryotes lack membrane-bound organelles and ER is a membrane-bound structure present only in eukaryotic cells. Prokaryotic cells perform their functions without such compartmentalisation.

Question 15.
Dried raisins or apricots are soaked in plain water for some time and then transferred into a concentrated sugar or salt solution. When placed in plain water, they swell up by gaining water. However, when placed in the concentrated solution, they lose water and shrink. [4]
Attempt either subpart A or B
A. Explain why raisins or apricots swell when placed in plain water.
Answer:
Raisins swell because water moves by osmosis from the surrounding dilute plain water into the fruit cells, where the concentration of dissolved substances is higher. This movement of water causes the fruit to absorb water and swell.

Or
B. Why do they shrink when placed in a concentrated sugar or salt solution?
Answer:
They shrink because water moves out of the fruit cells by osmosis into the surrounding concentrated sugar or salt solution. Since the external solution has a higher concentration of solutes, water leaves the cells, leading to shrinkage.

C. Just as water moves into cells by osmosis, how do gases like oxygen and carbon dioxide move in and out of cells? Explain the process.
Answer:
Gases like oxygen and carbon dioxide move in and out of cells by diffusion. Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration.

D. The diagram given below shows a type of cell division. Identify the type of cell division and explain why is it important?

Answer:
The given diagram represents meiosis, a type of cell division that produces four daughter cells, each with half the number of chromosomes. It is important because it helps in the formation of gametes (sperm and egg cells) and ensures that the chromosome number remains constant from one generation to the next. It also introduces variation, which is essential for evolution.

Students should not confuse between mitosis and meiosis. Mitosis produces cells with the same chromosome number, while meiosis produces cells with half the chromosome number.

Question 16.
Attempt either A or B
A. Cattle farming is a common practice in india where exotic breeds are crossed with indigenous ones to produce offspring with desired traits. [5]
(i) Why is cattle farming done on a large scale in india? Name two indigenous and two exotic breeds of cattle.
(ii) What is the purpose of crossbreeding an indigenous cattle breed with an exotic breed in cattle farming? How can the nutritional requirements of the cattle be met?
Answer:
(i) Cattle farming is done on a large scale in india because cattle provide milk, meat, hides and helps in agricultural work such as ploughing fields and transporting goods.
Indigenous breeds are Red sindhi, Sahiwal Exotic breeds are Jersey, Brown swiss
(ii) The purpose of crossbreeding an indigenous cattle breed with an exotic breed is to combine the best traits of both the breeds. The nutritional requirements of the cattle are met by providing a balanced diet containing green fodder, dry fodder, concentrates, minerals, and clean water.

Or

B. Ravi collected grains and seeds from cereals, pulses and oilseeds and noted the seasons in which they are sown and harvested. He observed that crops like rice, maize and cotton are grown in one season, while crops like wheat, gram, and mustard are grown in another, depending on their climate and water requirements.
(i) Name the crops given above that are sown in the kharif season and those sown in the rabi season. Why do these crops grow in different seasons?
(ii) How does understanding the difference between kharif and rabi crops help farmers?
Answer:
(i) Kharif crops are rice, maize, cotton and rabi crops are wheat, gram, mustard.
These crops grow in different seasons because they have different temperature, rainfall and daylight requirements. Kharif crops need warm weather and plenty of rain during the monsoon, while rabi crops require cooler temperatures and moderate irrigation during winter.
(ii) Understanding the difference between kharif and rabi crops helps farmers to choose the right crop for the right season, plan irrigation schedules and use resources efficiently. This improves crop yield and reduces the risk of crop failure.

Section – B

Question 17.
Which of the following sets of ions comprise of ions which possess the same valency? [1]

	Ion 1	Ion 2
(a)	Plumbic	Stannic
(b)	Ferrous	Aluminium
(c)	Sodium	Calcium
(d)	Cupric	Ammonium

Answer:
(a) Among the given sets of ions, plumbic and stannic has same valency, i.e. 4.

• Plumbic Pb⁴⁺ → 4 valency
• Stannic Sn⁴⁺ → 4 valency
• Ferrous Fe²⁺ → 2 valency
• Aluminium Al³⁺ → 3 valency
• Sodium Na⁺ → 1 valency
• Calcium Ca²⁺ → 2 valency
• Cupric Cu²⁺ → 2 valency
• Ammonium NH₄⁺ → 1 valency

Question 18.
The melting points of three substances X, Y, Z are 120°C, 75°C and 100°C respectively. Arrange each of them in increasing order of intermolecular force of attraction. [1]
(a) X < Y < Z
(b) Y < Z < X
(c) Z < Y < X
(d) Y < X < Z
Answer:
(b) Y < Z < X

Explanation:
The correct increasing order of intermolecular force of attraction is Y < Z < X. As the melting point increases, the intermolecular force of attraction also increases.

Question 19.
In which conversion do water molecules lose speed? [1]

(a) Ice → water
(b) Ice → steam
(c) Steam → ice
(d) Water → steam
Answer:
(c) Steam → ice

Explanation:
When steam (gas) converts into ice (solid), the particles move closer to one another and they lose speed due to the decrease in energy. In solid state, particles do not move. They only vibrate about their mean position.

Question 20.
Heterogeneous mixture in which the solute particles do not dissolve and remain suspended throughout the solvent and the solute particles can be seen with the naked eye is known as [1]
(a) colloidal solution
(b) super saturated solution
(c) sublimation
(d) suspension
Answer:
(d) suspension

Explanation:
Suspension is a heterogeneous mixture in which the solute particles do not dissolve and remain suspended throughout the solvent and the solute particles can be seen with the naked eye.

Question 21.
Four statements about atomic and molecular masses are listed below. [1]
I. The atomic mass of oxygen is 16 u.
II. The molecular mass of water H2Ois 18 u.
III. The molecular mass of carbon dioxide C02 is 44 u.
IV. The atomic mass of hydrogen is 2 u.

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

Explanation:
Statement I: Correct-Atomic mass of oxygen = 16u
Statement II : Correct-Molecular mas of
H₂O = (1 × 2) + 16 = 18u.
Statement III: Correct-Molecular mass
CO₂ = 12 + (16 × 2) = 44 u.
Statement IV : Incorrect-Atomic mass of hydrogen is 1 u, not 2 u.

Question 22.
Which of the following is the correct chemical formula for aluminium sulphate? [1]
(a) AlSO₄
(b) Al₂(SO₄)₃
(C) Al₂SO₃
(d) Al(SO₄)₂
Answer:
(b) Al₂(SO₄)₃

Explanation:
Al₂(SO₄)₃. Aluminium has valency +3, sulphate ion has valency 2. Cross-multiplying the valencies gives the formula Al₂(SO₄)₃.

Question 23.
Take 10 mL of water in 6 test tubes each and add different samples of substances to each test tube as shown in the given figure. Shake the test tubes vigorously for a couple of seconds and leave them undisturbed. In which test tube substances will remain insoluble in water? [1]

(a) 1, 2 and 3
(b) 2, 3 and 4
(c) 3, 4 and 5
(d) 4, 5 and 6
Answer:
(c) 3, 4 and 5

Explanation:
Among the given test tubes, only 1, 2 and 6 are soluble in water while others 3, 4 and 5 are insoluble in water because sand, chalk powder and saw dust are insoluble in water.

The following question consists of two statements-Assertion (A) and Reason (R). Answer the question by selecting 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) In Rutherford’s gold foil experiment, very few a-particles are deflected back.
Reason (R) Nucleus present inside the atom is heavy.
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Explanation:
In Rutherford’s experiment, very few a-particles are deflected back because the size of the nucleus is very small in the atom.

Question 25.
(a) Identify the heterogeneous mixture from the following.
Air, soda water, soap solution, brass
Answer:
Soap solution is a heterogeneous mixture while the rest are homogeneous mixture.

(b) Write two components of a colloidal solution.
Answer:
The two components of a colloidal solution are dispersed phase and dispersion medium.

Question 26.
Attempt either A or B
A. In the given graph, the physical state of water is given with respect to temperature.

(a) Which region contains only solid?
(b) Which region contains liquid form of water?
(c) Which region shows latent heat of vaporisation?
Answer:
From the given graph between temperature and time, we get
(a) AB is the region that contains only solid in the form of ice.
(b) CD is the region that contains liquid form of water.
(c) DE is the region that shows latent heat of vaporisation.

Students often assume that temperature increases during phase change, instead of remaining constant.

Or

B. State reasons for the following.
(a) Solids have a fixed geometrical shape.
(b) Ice at 0°C appears colder in the mouth than water at 0°C.
(c) Steam at 100°C is better for heating than water at 100°C.
Answer:
(a) In solids, the inter-particles forces are very strong and particles do not move. They only vibrate about their mean position. Therefore, solids have a fixed geometrical shape. [1]
(b) Ice at 0°C absorbs heat energy as well as the latent heat of fusion while water at 0°C absorbs only heat energy. Therefore, ice absorbs more amount of energy from the mouth and therefore, feels colder.
(c) Steam is better for heating than water because it possesses the additional heat, i.e., latent heat of vaporisation.

Question 27.
In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer. [3]
Answer:
The given statement is incorrect.
According to this statement p> n> e~. But actually, the number of protons can never be greater than the number of neutrons (except protium). The number of neutrons can be equal to or greater than the number of protons because the mass number is equal to double the atomic number or greater than double the atomic number. The number of neutrons can be greater than the number of electrons because the number of electrons are equal to the number of protons in the neutral atom.

Question 28.
Riya was studying different models of the atom. She read about Dalton’s atomic theory, J.J. Thomson’s plum pudding model, Rutherford’s nuclear model, and Bohr’s model of the atom. She performed a small activity to understand the experimental basis of these models. [4]
She learnt that Dalton proposed atoms as indivisible particles that combine in whole-number ratios.
J.J. Thomson suggested that an atom consists of a positively charged sphere with electrons embedded in it like “dry fruits in a pudding”.
Rutherford, through his alpha-particle scattering experiment, found that most of the space inside an atom is empty and that the positive charge and most of the mass are concentrated in a small nucleus.
Bohr improved Rutherford’s model by suggesting that electrons revolve around the nucleus in fixed orbits without radiating energy.
Riya recorded her observations in the following table.

Model of atom	Main postulate
Dalton’s model	Atoms are indivisible, identical for a given element.
Thomson’s model	Positively charged sphere with electrons embedded in it.
Rutherford’s model	Positive charge in a dense nucleus, most space is empty.
Bohr’s model	Electrons move in fixed orbits without losing energy.

Answer the following questions based on the above information.
A. Which model of the atom first introduced the concept of a central nucleus?
(a) Dalton’s model
(b) Thomson’s model
(c) Rutherford’s model
Answer:
(c) Rutherford’s model

Explanation:
Rutherford’s alpha-particle scattering experiment introduced the concept of a small, dense, positively charged nucleus at the center of the atom.

B. If most of the alpha particles passed straight through the gold foil in Rutherford’s experiment, what does it indicate about the structure of the atom?
Answer:
Most of the space inside the atom is empty. In Rutherford’s experiment, most alpha particles passed straight through the gold foil without deflection, indicating that electrons and the nucleus occupy only a small portion of the atom, while the rest is empty space.

Or
Which postulate of Bohr’s model explains why electrons do not spiral into the nucleus?
Answer:
Bohr’s postulate states that electrons revolve in fixed energy levels (orbits) without radiating energy. This explains why electrons do not spiral into the nucleus despite electrostatic attraction.

C. Which of the following statements is true about the Thomson’s model and why?
(a) It explained the existence of a nucleus.
(b) It described electrons as embedded in a positively charged sphere.
(c) It suggested electrons revolve in fixed orbits.
(d) It proved atoms are indivisible.
Answer:
(b) It described electrons as embedded in a positively charged sphere.

Explanation:
In Thomson’s “plum pudding” model, electrons were thought to be embedded in a sphere of positive charge, like plums in a pudding.

Question 29.
Attempt either A or B [5]
A. (a) Which law states that, the sum of the masses of the reactants and products remains unchanged during a chemical reaction?
(b) What are polyatomic ions?
(c) What is the formula unit mass of Ca₃(PO₄)₂?
(d) What is the formula of sodium carbonate and ammonium sulphate?
(e) What is the ratio by mass in H₂S compound?
Answer:
(a) Law of conservation of mass states that during a chemical reaction the sum of the masses of the reactants and products remains unchanged. [1]
(b) A group of atoms carrying a fixed charge on
(c) Ca₃(PO₄)₂ = 3 × 40 + 2 × 31 + 8 × 16 = 310u
(d) Na₂CO₃; (NH₄)₂SO₄
(e) Mass of H = 1
Mass of sulphur = 32
In H₂S molecule, 2H : S
2 : 32
1 : 16
Therefore in H₂S molecule, combining elements are present in 1:16 ratio by mass.

Or

B. SO₂ is an air pollutant released during burning of fossil fuels and from automobile exhaust’.
(a) Write the names of elements present in this gas.
(b) What are the valencies of sulphur in SO₂ and SO₃?
(c) Define the term molecular mass.
(d) Determine the molecular mass of ZnSO₄ [Atomic mass of Zn = 65 y, S = 32 u and O = 16 u].
(e) Identify the error in this chemical formula and then rectify it.

Answer:
(a) The elements present in SO₂ gas are sulphur and oxygen. [1]
(b) Valency is the combining capacity of an atom.
Valency of sulphur in SO₂ = 4
Valency of sulphur in SO₃ =6 [1]
(c) Molecular mass It is the sum of the atomic masses of all the atoms present in a molecule of the substance.
(d) Molecular mass of ZnSO₄ is = 65 u + 32 u + 4(x) × 16u = 161 u
(e) The error in the chemical formula is that the valencies of the respective elements have been exchanged. The valency of Mg is +2 and that of Cl is -1
The correct representation will be

In a chemical formula of ionic compounds, the cation symbol is always written first and then anion symbol. The cation ¡s formed by metals and anion is formed by non-metals.

Section – C

Question 30.
Which of the following will increase the upthrust on an object submerged in water? [1]
(i) Increasing the depth of the object in the water.
(ii) Using a liquid with higher density instead of water.
(a) Only (i)
(b) Only (ii)
(c) (i) and (ii)
(d) None of these
Answer:
(b) Only (ii)

Explanation:
Upthrust depends on the density of the liquid and the volume displaced, but not on the depth of the object (assuming the liquid is incompressible).

Question 31.
Choose the correct option which explains, why astronauts feel weightless in space? [1]
(a) There is no gravity in space.
(b) They are far away from earth’s gravity, so no force acts on them.
(c) They are in continuous free fall along with their spacecraft.
(d) They wear special suits that cancel gravity.
Answer:
(c) They are in continuous free fall along with their spacecraft.

Explanation:
Astronauts and spacecraft fall freely under gravity, creating a sensation of weightlessness.

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

Question 32.
Assertion (A) Work is said to be done only when a force is applied on a body and the body is displaced in the direction of force. [1]
Reason (R) If there is no displacement, even if the force is applied, no work is done.
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
Work = Force × displacement × cos θ.
If displacement is zero then work done is zero regardless of force applied.

Question 33.
The velocity-time graph of an ascending passenger lift is given below. What is the acceleration of the lift?

(a) Acting between O and A
(b) Acting between A and B
(c) Acting between B and C [2]
Answer:
(a) Between O and A From the graph,
Change in speed = 46 – 0 = 46 m/s
Change in time = 2 – 0 = 2 s
Acceleration (a) = Change in speed / Time taken
= \(\frac{4.6}{2}\) = 2.3 m/s²

(b) Between A and B
From the graph, change in speed = 0
Since, A to B, the graph is parallel to x-axis showing no change in speed. Acceleration is zero.

(c) From B to C
From the graph, change in speed
= 0 – 46 = – 46 m/s
Time taken = 12 – 10 = 2 s
Acceleration (a) = \(-\frac{4.6}{2}\) = -2.3 m/s²
Negative sign shows that the body is retarding.

To ensure full marks, always interpret the meaning of negative sign in acceleration.

Question 34.
Attempt either A or B
A. Observe the diagram I and II carefully. An object of mass mis lifted from A to Bto height h along path 1 and path 2. [2]
(a) What would be the work done on the object in both the cases?
(b) Give reason.

Answer:
(a) Work done on the object in both cases as W = mgh
(b) Work done by gravity depends on the difference in vertical height between the initial and final positions of the object, not on the path along which the object is moved.

Or

B. Given below is the velocity-time graph for the motion of the car.
(a) What does the nature of the graph shown?
(b) Find acceleration.
Answer:
(a) The nature of the graph shows that velocity changes by equal amounts in equal interval of time. For a uniformly accelerated motion, velocity-time graph is always a straight line.

(b) a = \(\frac{B C}{A C}\) or a = \(\frac{\left(v_2-v_1\right)}{\left(t_2-t_1\right)}\)
a =(10.0 – 7.5)ms^-1/(20 – 15)s
or a = 25 ms^-1/5 s
a = 0.5 ms²

Question 35.
Study the graph and answer the question. [3]
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
Answer:
(a) The slope of distance-time graph gives us speed.
Hence, the graph with the highest slope will have the highest speed.
Since, Bhas the highest slope (inclination). It travels the fastest.

(b) Any point on the curve will give the distance of the object from O. Since, there is no intersection point of all the three graphs and they never meet at the same point on the road. (Although any two of them meet at some point on the road).

Question 36.
A boy is standing 200 m away from a wall and claps his hands. He hears the echo after 1.2 s. [3]
(a) Calculate the speed of sound in air from the observations.
(b) If the temperature of air increases by 10°C and the speed of sound increases by 0.6 m/s for each 1°C rise in temperature, find the new time after which the boy near the echo.
Answer:
Given, distance from boy to wall = 200 m echo heard after 1.2 s.
For an echo, the sound travels to the wall and back, so total = 2 × 200 = 400 m
(a) Speed of Sound v = \(\frac{\text { Total distance }}{\text { Time }}\)
\(\frac{400}{12}\) = 333.33 m/s

(b) New time after temperature rise 10°C.
Speed increases by 0.6 m/s per 1°C, so increase = 0.6 × 10 = 6.0 m/s
New speed, v’ = 333.33 + 6.0 = 339.33 m/s
New time for the 400 m round trip
t’ = \(\frac{400}{v^{\prime}}=\frac{400}{339.33}\) ≈ 1179 s

Question 37.
The speed-time graph for a car is shown in figure. [3]
(a) Find how far does the car travel in the first 4 s? Shade the area on the graph that represents the distance travelled by the car during this period.
(b) Which part of the graph represents uniform motion of the car ?
Answer:
(a) The area under the curve will give the distance travelled by the car. In time t = 4 s the distance travelled by the car is equal to the area under the curve from x = 0 to x = 4
Considering this part of the graph as a quarter of a circle whose radius = 4 unit.
Therefore, required area
= \(\left(\frac{1}{4}\right)\)πr² = \(\left(\frac{1}{4}\right)\)π4²
= 12.56 m

(b) In uniform motion, the speed of car become constant which is represented by line parallel to the time axis. In the given figure the straight line graph from t = 6stof =10s represents the uniform motion of the car.

Graph might be in m/s versus time in minutes must convert before area calculation.

Question 38.
While catching a fast moving cricket ball, a fielder in the ground gradually pulls his hands backwards with the moving ball. In doing so, the fielder increases the time during which the high velocity of the moving ball decreases to zero. Thus, the acceleration of the ball is decreased and therefore the impact of a catching the fast moving ball (see figure) is also reduced. If the ball is stopped suddenly, then its high velocity decreases to zero in a very short interval of time. Thus, the rate of change of momentum of the ball will be large. Therefore, a large force would have to be applied for holding the catch that may hurt the palm of the fielder. [4]
A. Given reason
(i) Why a fast moving cricket ball can cause more injuries to a cricketer than a moving tennis ball?
(ii) Why does a fielder pull his hands backwards, while catching a fast moving cricket ball?
Answer:
(i) A fast moving cricket ball causes more injuery because its momentum (p = mv) is greater than that of a tennis ball due to its larger mass. Therefore, it exerts a greater impact force than the tennis ball.

(ii) A fielder pulls his hands backward, while catching a fast-moving cricket ball to increase the time over which the ball’s velocity decreases to zero. This reduces the acceleration of the ball, thereby decreasing the force of impact on the fielder’s hands, minimizing the risk of injury.

B. A student claims that catching a cricket ball with stiff hands is effective. Do you agree? Justify your answer.
Answer:
No, the student is wrong, catching with stiff hands reduces the time to stop the ball which increases the force due to the impulse momentum relation. Pulling the hand back increases the time of impact, reducing the force and making the catch safer.

Attempt either C or D
C. State what happen when the mass of a body and the force acting on it are both doubled, what happens to the acceleration?
Answer:
We know that, acceleration is related to mass by the expression, F = ma. Now, if both force and mass are doubled, acceleration F = ma = 1 × 2 = 2, i.e. acceleration will remain the same.

Or
D. What are the benefits of pulling hands back, while catching in terms of momentum and force?
Answer:
Pulling the hands back, while catching benefits the fielder by increasing the time over which the momentum of the ball is reduced to zero. This action reduces the rate of change of momentum exerted on the hands. Thus, the force experienced by the hands is smaller, reducing the likelihood of injuery.

Question 39.
Attempt either A or B [5]
A. Study the given velocity-time graph of a moving object and answer the following
(a) Calculate the acceleration during the motion from A to B and from B to C.
(b) Find the total distance covered in the region ABE.
(c) Determine the average velocity during motion from C to D.
Answer:
(a) Acceleration from A → B and B → C From A → B,
a_AR = \(\frac{v_B-v_A}{t_B-t_A}=\frac{25-0}{3-0}=\frac{25}{3}\) = 8.33m/s²
From B → C,
a_BC = \(\frac{v_C-v_B}{t_C-t_B}=\frac{15-25}{4-3}=\frac{-10}{1}\)
= -10 m/s²

(b) Total distance covered in region ABE = Area of triangle ABE
Distance = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 3 × 25
= 37.5 m

(c) Average velocity from C → D,
Displacement fromf = 4 s to f = 6s is equal to the area under the v — t curve between C and D (a triangle of base 2 s and heights V_c = 17 and V_D = 0):

Displacement = \(\frac{1}{2}\) × 2 × (17 + 0) = 17 m
Average velocity over the interval, Δt = 2 s
v = \(\frac{\text { Displacement }}{\Delta t}=\frac{17}{2}\) = 8.5 m/s

Or
B. The speed-time graph of a vehicle moving along a straight road is shown. The vehicle starts from rest, accelerates uniformly to a speed of 25 m/s in 20 s, continues at this speed until 50 s, and then comes to rest at 60 s. Using the graph, answer the following:
(a) Calculate the acceleration during the first 20 s.
(b) Find the total distance travelled by the vehicle from 0 to 60 s.
(c) Determine the total time taken by the vehicle to cover the first 500 m of its journey and state whether it occurs before or after point B on the graph.
Answer:
(a) Acceleration (0 → 20s),
a = \(\frac{\Delta v}{\Delta t}=\frac{25-0}{20-0}=\frac{25}{20}\) = 1.25 m/s²

(b) Total distance from 0 to 60 s
Distance = Area under speed – Time graph = Area of first triangle + Rectangle + Last triangle.
First triangle (0 – 20), \(\frac{1}{2}\) × 20 × 25 = 10 × 25
= 250 m
Rectangle (20-50 s),
30s × 25 m/s = 750 m
Last triangle (50 – 60s),
\(\frac{1}{2}\) × 10 × 25 = 5 × 25 = 125 m
Total = 250 + 750 + 125 = 1125 m

(c) Cumulative distance after first 20 s = 250 m 500 m > 250 m,
So, the 500 m mark is reached during the constant-speed segment (20 – 50 s).
Remaining distance to reach 500 m after 20 s = 500 – 250 = 250 m
At 25 m/s time needed = \(\frac{250}{25}\) = 10 s
So, the 500 m point is reached at t = 20 + 10 = 30s
Point B is at f =50 s, so 30 s is before B.