CBSE Sample Papers for Class 9 Science Set 3 with Solutions - #NCSOLVE 📚

Students must start practicing the questions from CBSE Sample Papers for Class 9 Science with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 9 Science Set 3 with Solutions

Time: 3 Hrs.
Max. Marks: 80

General Instructions:

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

Section – A

Question 1.
Which is/are the correct statements related to stomatal pore? [1]
(i) Stomatal pores are present on the stem.
(ii) Stomata are surrounded by flat rectangular cells.
(iii) Transpiration takes place only through roots, not through stomata.
(iv) Guard cells play no role in stomatal movement.
(v) Stomata are enclosed by two kidney shaped cells that help in their opening and closing.

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

Explanation:
Statement (v) is correct, Stomata are enclosed by two kidney-shaped cells called gaurd cell that help in their opening and closing.

Question 2.
Select the group in which all members have the same main type of xylem element. [1]
(a) Tracheids, sieve tubes, vessels, xylem fibres
(b) Tracheids, vessels, xylem fibres, xylem parenchyma
(c) Vessels, tracheids, sieve cells, xylem fibres
(d) Xylem fibres, xylem parenchyma, sieve tubes, vessels
Answer:
(b) Tracheids, vessels, xylem fibres, xylem parenchyma

Explanation:
Tracheids, vessels, xylern fibres, xylem parenchyma are elements of xylem tissue involved in water conduction and support.

Question 3.
Which of the following options is an example of mixed cropping? [1]
(a) Wheat + Mustard
(b) Wheat + Rice
(c) Cotton + Jute
(d) Potato + Tomato
Answer:
(a) Wheat + Mustard

Explanation:
Wheat + Mustard grown together is an example of mixed cropping.

Question 4.
Which of the following is a correct combination of function and cell organelle? [1]
(a) Protein synthesis: Ribosome
(b) Photosynthesis: Mitochondria
(c) Lipid synthesis: Lysosome
(d) Packaging of proteins: Rough endoplasmic reticulum
Answer:
(a) Protein synthesis: Ribosome

Explanation:
Protein synthesis occurs in ribosome.

Question 5.
The tissue tracheid is [1]
(a) living cells that transport food in plants.
(b) thick walls and are dead cells that conduct water.
(c) found only in animals.
(d) part of phloem tissue.
Answer:
(b) thick walls and are dead cells that conduct water.

Explanation:
The tissue tracheid, have dead cells and thick wall that helps in conduct of water.

Question 6.
Which one of the following is not a kharif crop? [1]
(a) Paddy
(b) Mustard
(c) Soybean
(d) Cotton
Answer:
(b) Mustard

Explanation:
Except mustard all are kharif crop that grows in rainy season. Mustard is a rabi crop which grows in winter and harvested in spring.

Don’t get confuse between kharif and rabi crops, kharif crops are sown at the onset of the monsoon, whereas rabi crops are sown in winter, after the monsoon.

Question 7.
A farmer sows seeds at the beginning of the rainy season. The crop he might sow is [1]
(a) wheat
(b) mustard
(c) rice
(d) gram
Answer:
(c) rice

Explanation:
Rice is a kharif crop (sown with the onset of the rainy season), while wheat, mustard and gram are rabi crops (sown in winter).

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) Fungicides act against fungal pathogens and protect plants from fungal diseases. [1]
Reason (R) Fungicides are not harmful to human beings.
Answer:
(c) (A) is true but (R) is false.

Explanation:
A is true, but R is false. R can be corrected as
Fungicides are chemicals which are effective against fungal pathogens. Fungicides are divided into two major types, inorganic and organic. Most inorganic fungicides are harmful to other living beings including humans.

Question 9.
Assertion (A) Chromosomes are composed of DNA and protein. [1]
Reason (R) These are thread-like structures that found in nucleus.
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).

Explanation:
Both A and R are true, but R is not the correct explanation of A.
Chromosomes are composed of DNA and proteins. These contain information for inheritance of characters from parents to next generation in the form of DNA.

Question 10.
Unlike other epithelial types, columnar epithelium is ideal for absorption in the intestine. Comment upon the statement with justification. [2]
Answer:
Columnar epithelium is ideal for absorption in the intestine because the cells are tall and have hair-like projections on the outer surfaces of epithelial cells which greatly increase the surface area for absorption.

Question 11.
Attempt either A or B [2]
A. Flow many types of simple plant tissues are found in plants? What are the functions of the following plant tissues?
(i) Parenchyma
(ii) Collenchyma
Answer:
Three types of simple plant tissues are found in plants these are parenchyma, collenchyma, and sclerenchyma.
(i) Parenchyma consists of living cells with thin walls and is involved mainly in storage, photosynthesis and tissue repair.
(ii) Collenchyma also has living cells but with unevenly thickened walls, providing flexible mechanical support especially in young stems and leaves.

Or

B. What is stratified squamous epithelium? State its function and mention one place where it is found in the human body.
Answer:
Stratified squamous epithelium is a type of tissue made up of multiple layers of flat (squamous) cells stacked on top of each other. It provides protection against wear and tear. It is found in the lining of the mouth, skin and oesophagus.

To get maximum marks, include the concept that simple permanent tissues are made up of only one type of cell.

Question 12.
In an organism, a cell containing 8 chromosomes undergoes cell division by both meiosis and mitosis separately. [2]
How many daughter cells are produced at the end of meiosis and mitosis respectively and what will be the chromosome number in each daughter cell produced by meiosis and mitosis?
Answer:
Meiosis produces 4 daughter cells, while mitosis produces 2 daughter cells. Each daughter cell produced by meiosis has 4 chromosomes (half the original number), whereas each daughter cell produced by mitosis has 8 chromosomes (same as the original cell).

Question 13.
Draw a labelled diagram of parenchyma and explain how it helps in the functioning of the plant? [3]
Answer:
Parenchyma help in the functioning of the plant as These cells are loosely packed with intercellular spaces allowing gas exchange.
Parenchyma helps in storage of food and water and in carrying out photosynthesis in leaves.

Question 14.
A nutritionist studied the composition of two animal products, milk and egg, by analysing 100 grams of each. The recorded data showed that milk contained 3.4 grams of protein, 3.7 grams of fat, 0.1 grams of minerals, and 0.15 milligrams of vitamins. On the other hand, egg contained 13 grams of protein, 11 grams of fat, 0.05 grams of minerals, and 0.5 milligrams of vitamins. [3]
(i) Mention the vitamins present in the given animal product.
(ii) Discuss the importance of these nutritional values in human health and diet.
Answer:
(i) The vitamins present in milk are B₁, B₂, B₁₂, D and E and in egg are B₂ and D.

(ii) The importance of these nutritional values in human health and diet is

• Provide amino acids necessary for building enzymes, hormones and other vital molecules.
• Provide essential nutrients required for growth, repair and maintenance of body tissues.

Question 15.
Neha is studying different types of tissues found in plants and animals. Help her understand these tissues by answering the questions below. [4]

Attempt either subpart A or B.
A. Which plant tissues are made up of a single type of cell? Name two examples of such plant tissues and describe their functions.
Answer:
Simple permanent tissues are made up of only one type of cells. Examples of two such tissues are [2] Parenchyma functions in photosynthesis, storage and secretion.
Collenchyma provides flexibility to growing parts of the plant.

Or

B. Which plant tissues are made up of more than one type of cells? Name two examples of such tissues and describe their role in the plant.
Answer:
Complex tissues are composed of more than one type of cells. Examples are
Xylem conducts water and minerals and provides mechanical support.
Phloem transports food (sugars) throughout the plant.

C. What is connective tissue? Name two types of connective tissues found in animals and explain their functions.
Answer:
Connective tissue is a type of animal tissue that supports, binds or separates other tissues and organs. Examples are
Bone provides rigid support and protects organs.
Cartilage provides flexible support and reduces friction in joints.

D. The figure given below shows a paint tissues. Which of these parts are tubular cells with perforated walls?

Answer:
In the given figure, part B (sieve tubes) are tubular cells with perforated walls.

Question 16.
Attempt either A or B [5]
A. Ravi wanted to study the structure and function of cell organelles.
(i) Which organelle is known as the ‘powerhouse of the cell’? Justify your answer.
(ii) Some organelles contain their own DNA and ribosomes. Name any two such organelles and explain the significance of this feature.
Answer:
(i) The mitochondria is known as the powerhouse of the cell’ because it produces energy in the form of ATP (adenosine triphosphate) through the process of cellular respiration, which is essential for making new chemical compounds and for mechanical work.
(ii) Two organelles that contain their own DNA are mitochondria and chloroplasts. This feature is significant because it allows these organelles to produce some of their own proteins independently of the cell’s nuclear DNA, supporting their role in energy production in case of mitochondria and photosynthesis in case of chloroplasts.

Or

B. Simran was investigating the differences between plant and animal cells under a microscope.
(i) Name two cell structures present in plant cells, but absent in animal cells and explain their functions.
(ii) Differentiate between mitosis and meiosis in cell division.
Answer:
(i) Two cell structures present in plant cells, but absent in animal cells are
Chloroplasts They carry out photosynthesis by converting light energy into chemical energy.
Cell wall It provides structural support and protection t6 the plant cell.

(ii) The difference between mitosis and meiosis are as follow

Mitosis	Meiosis
Growth, repair, and asexual reproduction.	Formation of gametes for sexual reproduction
Number of divisions occur is 2.	Number of divisions occur is 1.
Number of daughter cells produces are 4.	Number of daughter cells produced are 2.
Daughter cells are genetically identical to parent cell.	Daughter cells have half the chromosome number and are genetically varied.
Occurs in somatic cells	Occurs in formation of gamete (sex cells).

Section – B

Question 17.
On converting 27°C, 30°C and 65°C to Kelvin scale, the correct sequence of temperature will be
(a) 300K, 335K, 345K
(b) 325K, 330K, 338K
(c) 300K, 303K, 338K
(d) 325K, 327K, 336K
Answer:
(c) 300K, 303K, 338K

Explanation:
The given temperature in kelvin scale is
(i) 27° C = 27 + 273 = 300K
(ii) 3(PC = 30 + 273 = 303K
(iii) 65° C = 65 + 273 = 338K
Therefore, the correct sequence of temperatures will be 300 K, 303 K, 338 K.

Question 18.
In which of the following, the constituents are present in any ratio?
(a) Compound
(b) Mixture
(c) Colloid
(d) Solution
Answer:
(b) Mixture

Explanation:
Mixture is constituted by more than one substance in any ratio, e.g. Sea water, soil, polluted air, etc.

Question 19.
A sample of a gaseous substance is kept in two separate containers, P and Q, each subjected to different sets of pressure and temperature conditions to study its conversion into liquid state. Which of the following conditions is most favourable for converting the gas into liquid? [1]
(a) In container P: High pressure and low temperature.
(b) In container P: Low pressure and low temperature.
(c) In container Q: Low pressure and high temperature.
(d) In container Q: High pressure and high temperature
Answer:
(a) In container P: High pressure and low temperature.

Explanation:
High pressure forces gas molecules to come closer and low temperature reduces their kinetic energy, making it easier for intermolecular forces to convert gases into liquid.

Question 20.
Four statements about sub-atomic particles and structure of atom are given below: [1]
I. Neutrons were discovered by Chadwick and are neutral particles.
II. Bohr’s model explains the fixed energy levels of electrons.
III. Electrons are present inside the nucleus.
IV. Thomson proposed the plum pudding model of the atom.

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

Explanation:
Statements I, II, and IV are correct and statement III is incorrect because electrons are present outside the nucleus, not inside.

Question 21.
A molecule of salt is found to contain sodium and chlorine combined together in the ratio of 23:35.5 by mass.
What is the formula of salt?
(a) Na₂Cl
(b) NaCl₂
(c) Na₃Cl₂
(d) NaCl
Answer:
(d) NaCl

Explanation:

Question 22.
The chemical symbol of nitrogen gas and hydrogen gas are Nitrogen gas Hydrogen gas

	Nitrogen gas	Hydrogen gas
(a)	N	H
(b)	N2	H
(c)	N2	H2
(d)	N	H2

Answer:
(c)

Explanation:
N₂ and H₂. Nitrogen atom makes a triple bond with another nitrogen atom to form a stable molecule —N₂, whereas hydrogen atom makes a single bond with another hydrogen atom to form a stable molecule —H₂ gas.

Question 23.
The number of electrons, protons and neutrons in potassium, ₁₉K³⁹ respectively are
(a) 19, 19, 19
(b) 19, 19, 20
(c) 20, 20, 19
(d) 20, 19, 20
Answer:
(b) 19, 19, 20

Explanation:
In potassium. ₁₉K³⁹
The number of protons = 19
The number of electrons = 19
∵ The mass number = 39
∴ The number of neutrons = (mass number – number of protons) = 39 – 19 = 20
Hence, the number of electrons, protons and neutrons in potassium, ₁₉K³⁹ are 19, 19 and 20 respectively.

Don’t make mistake in calcuLating the number of neutrons if does not know about the relation between mass number, number of neutrons and protons.

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) The formula of chloride of a metal M is MCl₃, then the formula of the phosphate of metal M will be M(PO₄)₃. [1]
Reason (R) The valency of metal M is +3.
Answer:
(d) (A) is false, but (R) is true.

Explanation:
(d) (A) is false, but (R) is true. Assertion can be corrected as.

The formula of chloride of a metal M is MCl₃, then the formula of the phosphate of metal M will be MPO₄.
As, valency of metal M is + 3.

Formula = MPO₄

Question 25.
Based on the figure, answer the given questions. [2]

(a) Which observation of Rutherford’s a-ray scattering experiment led to the fact that there is large empty space inside the atom?
(b) Rutherford’s model of atom was similar to that of solar system. Justify the statement.
Answer:
(a) Most of a-particles passed through the gold foil without getting deflected and very few particles were deflected from their path, which led to the fact that most of the space inside the atom is empty.
(b) Rutherford’s model was similar to that of the solar system because just like the different planets are revolving around the sun, in a similar way the electrons are revolving around the nucleus in an atom.

Question 26.
Attempt either A or B
A. By observing the diagram given below, explain the effect of change in pressure on the state of a substance. [3]

Answer:
When the particles of fluid are present under low pressure, they are in the gaseous state as shown in the [figure (I)]. On applying some high pressure on the gas, the forces of attraction between the gas particles increase and as a result of which, gas particles are bind together to form the liquid state [figure (II)]. Further, under very high pressure, the forces of attraction between the particles become so strong that the liquid changes into the solid state as shown in the [figure (III)]. Thus, on applying pressure, a gas can be first liquified and then converted into solid.

Or
B. Explain the physical state of water at the given temperatures,
(i) 250°C
(ii) 100°C
(iii) 25°C
Answer:
The physical state of water at the given temperature is
(i) At 250°C water exists as water vapour or steam. [1]
(ii) At 100°C water exists as liquid water as well as water vapour, since steam and water co-exist at 100°C. [1]
(iii) At 25°C water exists in liquid state.

To get maximum marks one should include the brief explanation of the topics.

Question 27.
Nitrogen and hydrogen atoms combine in the ratio 14:3 by mass to form ammonia molecules. [Given atomic mass of N = 14 u and H= lu] [3]
Answer the following:
(a) Calculate the number of nitrogen and hydrogen.
(b) Write the molecular formula of ammonia based on this ratio.
Answer:
(a) Number of nitrogen atom present in the molecule = \(\frac{\text { Proportion by mass }}{\text { Atomic mass }}=\frac{14 \mathrm{u}}{14 \mathrm{u}}\) = 1
Number of hydrogen atoms present m the molecule = \(\frac{\text { Proportion by mass }}{\text { Atomic mass }}=\frac{3 \mathrm{u}}{1 \mathrm{u}}\) = 3

(b) The number of nitrogen and hydrogen atoms combined in ratio of 1: 3. Thus, the formula of a molecule of ammonia is NH₃.

Question 28.
Ravi was studying the concept of isotopes in the laboratory. He took samples of two isotopes of chlorine Cl-35 and Cl-37 and measured their relative abundance using a mass spectrometer. He noted that Cl-35 has a relative abundance of 75% and Cl-37 has a relative abundance of 25%. The masses of these isotopes were measured as 35 u and 37 u, respectively. [4]

To verify the concept of average atomic mass, Ravi recorded his observations in the table given below:

S.No.	Isotope	Mass (u)	Relative Abundance (%)
1	Cl-35	35	75
2	Cl-37	37	25

Answer the following questions based on the above information:
A. If the relative abundance of Cl-37 increases to 50%, what will happen to the average atomic mass of chlorine?
(a) It will decrease
(b) It will increase
(c) It will remain the same
Justify your answer.
Answer:
(b) It will increase

Explanation:
Increasing the relative abundance of Cl-37 (heavier isotope) raises the weighted average, resulting in a higher average atomic mass for chlorine.

Or

B. Calculate the average atomic mass of chlorine from the given data.
Answer:
Calculation of average atomic mass:
Average atomic mass = \(\frac{(35 \times 75)+(37 \times 25)}{100}=\frac{2625+925}{100}\)
= \(\frac{3550}{100}\)
= 35.5 u

The average atomic mass of chlorine is 35.5 u.

Or

Ravi finds another element X having two isotopes, X-20 and X-22, in equal proportion. What will be the average atomic mass of element X?
Answer:
For X-20 and X-22 in equal proportion:
Average atomic mass = \(\frac{(20 \times 50)+(22 \times 50)}{2}=\frac{1000+1100}{100}\)
= \(\frac{2100}{100}\) = 21

The average atomic mass of element X is 21 u.

C. What are isotopes. Which of the following statements about isotopes is true and why?
(a) They have the same number of protons but different numbers of neutrons.
(b) They have different numbers of protons and electrons.
(c) They have the same mass number but different atomic numbers.
(d) They have different physical and chemical properties.
Answer:
(a) They have the same number of protons but different numbers of neutrons.

Explanation:
Isotopes are those elements which have similar atomic number but different atomic masses, e.g. C-12 and C-13. They have the same number of protons but different numbers of neutrons. Isotopes are atoms of the same element (same atomic number, same protons) but different mass numbers due to varying neutron counts; their chemical properties remain almost identical.

Question 29.
Attempt either A or B [5]
A. (a) Krish and Rohan were given the task of sorting the chemistry lab substances. They assess the names of all the elements in tabular form. Classify the following into elements, compounds and mixtures.

Krish	Rohan
Sodium	Tin
Silicon	Sugar solution
Silver	Calcium carbonate
Soil	Coal
Air	Soap
Carbon dioxide	Methane

Answer:
The given substances are classified into elements, compounds and mixtures as

• Elements → Sodium, silver, tin and silicon.
• Compounds → Calcium carbonate, methane and carbon dioxide.
• Mixtures → Soil, sugar solution, coal, air and soap.

(b) Distinguish among true solution, suspension and colloid in tabular form under the following headings.
(i) Stability
(ii) Filterability
(iii) Type of mixture
Answer:

Property	Solution	Suspension	Colloid
(i) Stability	It is stable, constituent particles do not settle down on leaving undisturbed.	It is unstable, constituent particles settle down on leaving undisturbed.	It is stable, constituent particles do not settle down on leaving undisturbed.
(ii) Filterability	Particles cannot be separated by filtration.	Particles are large, so they can be easily separated by ordinary filtration.	Particles cannot be separated by ordinary filter paper, but can be separated by ultrafiltration.
(iii) Type of mixture	Homogeneous	Heterogeneous	Heterogeneous

Or

B. (a) Identify two non-metals from the following elements:
Carbon, sodium, chlorine, neon, platinum.
(b) Identify dispersed phase and dispersion medium in foam and rubber.
(c) Define solubility.
(d) How does solubility of a solid in water change with temperature?
(e) Define a colloid. How can a colloid be distinguished from a true solution?
Answer:
(a) The two non-metals are carbon and chlorine.
(b) Dispersed phase is gas and dispersion medium is solid in foam and rubber.
(c) Solubility is the property of a solid, liquid or gaseous chemical substance called solute to dissolve in a solid, liquid or gaseous solvent to form a solution.
(d) The solubility increases with temperature. The increase in kinetic energy that comes with higher temperature allows the solvent molecules to more effectively break apart the solute particles that are then held together by force between the particles.
(e) A colloid is a heterogeneous mixture in which the particle size is intermediate between those in true solutions and suspensions. Colloids scatter light (Tyndall effect), unlike true solutions.

Section – C

Question 30.
Ravi is conducting an experiment in his school laboratory to understand the concept of buoyancy. He takes two objects – a wooden block and a metal cube and places them in a container filled with water. He observed that the wooden block floats on water and metal cube sinks to the bottom. [1]
Based on his study and understanding of buoyancy, in which of the following conditions will an object sink in a liquid?
A. When the buoyant force is zero
B. When the buoyant force is less than the weight of the object
C. When the buoyant force is greater than the weight of the object
D. When the buoyant force is equal to the weight of the object

(a) A and B
(b) A, B and C
(c) B and C
(d) A and C
Answer:
(a) A and B

Explanation:
An object sinks in a liquid when the buoyant force is zero or less than its weight, because in both cases, the upward force is not enough to balance the downward gravitational force. If the buoyant force is equal to or greater than the object’s weight, the object will float or remain suspended in the liquid.

Question 31.
A sound wave is travelling at a speed of 339 m/s, and its wavelength is 1.5 cm. Which of the following statements is correct regarding the frequency of this sound wave? [1]
(a) The frequency is 20,000 Hz, which is at the upper limit of human hearing.
(b) The frequency is 22,600 Hz, placing it in the ultrasound range (above 20,000 Hz).
(c) The frequency is 25,000 Hz, which is below the human hearing range.
(d) The frequency cannot be calculated unless the time period is given.
Answer:
(b) The frequency is 22,600 Hz, placing it in the ultrasound range (above 20,000 Hz).

Explanation:
To calculate frequency, use f = \(\frac{v}{2}\),
f = \(\frac{1}{2}\) ≈ 22,600 Hz. Since, the value is above
20,000 Hz, it’s considered ultrasound and is inaudible to most humans.

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

Question 32.
Assertion (A) The weight of an object on the moon is one-sixth of its weight on the earth. [1]
Reason (R) The acceleration due to gravity on the moon is one-sixth of that on the earth.
Answer:
(a) Both A and R is true, and R is the correct explanation of A.

Explanation:
Weight is the force due to gravity and is given by W = mg. Since, the Moon’s gravitational acceleration g is one-sixth that of Earth’s, the weight of an object on the Moon becomes one-sixth of its weight on Earth. Hence, Both the (A) and the (R) are true, and the (R) is correct explanation of (A).

Question 33.
The position of the water level in a measuring cylinder before and after immersing a solid is shown below [2]

(i) Which scientific principle is applied when the volume of an irregular solid is measured by the amount of water it displaces?
(ii) Calculate the volume of the solid immersed in the water.
(iii) If the mass of the solid is 9 g, calculate its density. Give the answer in g/cm³.
Answer:
(i) Archimedes’ principle is applied. It states that when an object is fully or partially immersed in a fluid, it displaces a volume of fluid equal to its own volume.
(ii) volume of solid = final water level – initial water level
= 4.2 cm³ – 2.4 cm³
= 1.8 cm³
(iii) Density = \(\frac{\text { mass }}{\text { volume }}=\frac{9}{18}\) = 5 g/cm

Identify the principle and explain accurately.

Question 34.
Attempt either A or B
A. The velocity-time graph of a toy car moving on a straight horizontal surface is shown below. The mass of the toy car is 200 g. [2]

(i) What does the shape of the graph tell you about the motion of the toy car?
(ii) What distance did the toy car travel during the 4 seconds?
Answer:
(i) The graph is a straight sloping line this shows that the toy car is decelerating at a uniform (constant) rate. This is called uniform retardation. [1]
(ii) To find the distance, we calculate the area under the velocity-time graph.
Since, the graph is a triangle,
Distance = area under the graph = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 4 × 20
= 40m
Therefore, the care travel 40 meter in 4 sec.

Or

B. When a car suddenly takes a sharp turn, the passengers tend to lean sideways in the opposite direction.
(i) Why do the passengers lean in the opposite direction of the turn?
(ii) Which type of inertia is involved in this case?
Answer:
(i) Due to inertia, the passengers tend to continue in their original straight-line motion while the car turns, making them appear to lean in the opposite direction. [1]
(ii) Inertia of direction is involved in this case. It is the tendency of the body to resist a change in its direction of motion.

Question 35.
(i) A sharp knife is more effective than a blunt knife why?
(ii) Why does an iron nail sinks in water but a wooden cork floats on water?
Answer:
(i) A sharp knife has a smaller surface area at its edge, so when force is applied, the pressure exerted is more. Higher pressure allows the sharp knife to cut objects more easily and effectively than a blunt knife.
(ii) An iron nail sinks because its density is greater than that of water, so it experiences less upthrust than its weight while a wooden cork floats because its density is less than water, and it experiences an upthrust greater than its weight.

Question 36.
A motorbike accelerates uniformly from 54 km/h to 72 km/h in 2 s. Calculate
(i) the acceleration
(ii) the distance covered by the motorbike in that time.
Answer:
(i) Initial velocity, u = 54 km/h
= 54 × \(\frac{5}{18}\)
= 15 m/s

Final velocity, v = 72 km/h
= 72 × \(\frac{5}{18}\)
= 20 m/s

Students often forget to convert speed from km/h to m/s, which Leads to wrong answers.

(ii) The distance covered by the motorbike
s = ut + \(\frac{1}{2}\)at²
= 15 × 2 + \(\frac{1}{2}\) × (25) × (2)²
= 30 + 5
= 35 m

Question 37.

A student was observing the motion of a stone of mass 5kg which is dropped from a height of 10 m.
(i) What is meant by law of conservation of energy? Explain with an example.
(ii) Find the kinetic energy of the stone just before it touches the ground. (Take, g = 98 m/s²)
Answer:
(i) The law of conservation of energy states that energy can neither be created nor be destroyed, it can only be transformed from one form to another, e.g. In a pendulum, potential energy changes into kinetic energy and vice -versa.

(ii) Given, mass (m) = 5 kg
height (h) = 10 m
All potential energy converts to kinetic energy
Potential energy (PE) = mgh
= 5 × 9.8 × 10 = 490 J

Question 38.
Anupana conducted an experiment using two containers. She placed a mini speaker inside each container and tightly covered the top with a thin rubber sheet. On one container, she placed rubber beads on the surface of the sheet. After turning on the speakers, she observed that the rubber beads on the first container were jumping, while no change was observed on the second container where the speaker remained off. [4]

A. Give reason why do the rubber beads jump when the speaker is turned on?
Answer:
The speaker produces sound waves which cause the rubber sheet to vibrate. These vibrations make the rubber beads on the sheet jump because the sound waves transfer energy to the beads, causing them to move up’and down.

B. If the speed of sound in air is 340 m/s and the frequency of the sound from the speaker is 170Hz, calculate the wavelength of the sound waves produced.
Answer:
Given
Speed of Sound, v = 340 m/s
Frequency, f = 170 Hz
Using, λ = \(\frac{v}{f}\)
λ = \(\frac{340}{170}\) = 2 m

Attempt either C or D
C. In another experiment, the student used a different speaker that produced sound at a higher frequency. How would this affect the behaviour of the rubber beads?
Answer:
If the frequency is increased, the wavelength becomes shorter (since X = v/f). Higher frequency sound waves cause the rubber sheet to vibrate faster, which would make the rubber beads jump more rapidly, possibly with smaller amplitude but higher frequency of jumps.

Or
D. What does the movement of rubber beads suggest about the nature of sound waves?
Answer:
If the frequency decreases, the wavelength increases (since X – v/f). Lower frequency sound waves cause the rubber sheet to vibrate more slowly, so the rubber beads will jump less frequently but may have larger jumps (greater amplitude).

Question 39.
Attempt either A or B

A. (i) Define the term inertia. Out of a cart and boy, which one has greater inertia?
(ii) A boy of mass 40 kg jumps with horizontal velocity of 5 ms^-1 on a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is his velocity when the cart starts moving? Assume that, there is no external unbalanced force working on it in horizontal direction.
(iii) To move the cart at constant speed the boy has to exert continues force. Explain why?
Answer:
(i) The tendency of a particle or an object to oppose any change in its state of rest or motion along a straight line by itself until an external unbalanced force acts on it is called inertia.
Since, inertia is directly proportional to the mass of the objects.
Now, M_cart < < M_boy Thus, the boy has greater inertia.

(ii) Let the velocity of the boy on the cart, as the cart starts moving be v Total momentuma of the boy and the cart before the interaction
= 40 kg × 5 ms^-1 + 3 kg × 0 ms^-1
(∵ Mass of boy = 40 kg, mass of cart = 3 kg, velocity of boy before the cart starts moving = 5 ms^-1)
= 200 kg ms^-1
Also, the total momentuma after the interaction = (40 + 3) kg x v ms^-1 = 43 v kg m^-1
As per the law of conservation of momentum, the total momentum is conserved during the interaction. In other words,
⇒ 43v = 200
⇒ v = \(\frac{200}{43}\) = 465 ms^-1
Thus, the boy on cart would move with a velocity of 4.65 ms’1 in the direction in which the boy jumped on to the cart. [2]

Or

B. (i) State and explain Newton’s second law of motion. How is it different from the first law of motion?
(ii) A force of 50 N is applied to a mass of 5 kg. Calculate the acceleration produced. If this force is applied for 5 s, calculate the final velocity of the object starting from rest.
(iii) While pushing a mass of 5 kg the hand of the worker starts to hurt. Explain why?
Answer:
(i) Newton’s second law of motion states that the rate of change of momentum of an object is directly proportional to the applied force and takes place in the direction in which the force acts. This law can be mathematically expressed as
F = ma
Difference from the first law
First law Describes the behaviour of objects when no net force is acting upon them. It introduces the concept of inertia.
Second law Describes the effect of a net force acting on an object providing a relationship between force, mass and acceleration.

(ii) Given, Force, F = 50 N
Mass, m = 5 kg
Using Newton second law,
F = ma
50 = 5 × a
a = \(\frac{50}{5}\)
a = 10m/s²
Now, to calculate the final velocity after 5s
Initial velocity = 0
Time, t = 5 s
Using the first equation of motion,
v = u + at
= 0 + (10 × 5)
= 50 m/s
So, the final velocity of the object after 5 s is 50 m/s.

(iii) According to Newton’s Third law of motion, for every action, there is an equal and opposite reaction. When the worker applies a force to push the 5 kg mass, the object applies an equal and opposite force back on the worker’s hand. This reaction force acts on the muscles and joints of the hand, and if the force is large or applied over time, it can cause strain or pain in the hand. Thus, the pain is due to the reaction force acting against the worker’s effort to push the object.

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