Class 9 · PhysicsSet 2 · Answer key

Answer Key – Set 2 ഉത്തരസൂചിക

Step-by-step answers and the marking scheme for every question of Set 2, for teachers and parents to value the paper. Accept any correct answer in the student’s own words; give method marks even when the final number is wrong.

ANSWER KEY & MARKING SCHEME · SET 2

Standard IX · Physics · Annual Examination

Total score: 60 · Time: 2 hours

Section AAnswer all questions · 10 × 1 = 10 marks

A ray of light passes from glass into water. It falls on the boundary obliquely, at an angle of incidence of 30°. In the water, the ray

(a) bends towards the normal(b) bends away from the normal(c) goes straight on without bending(d) is reflected back into the glass

[1]

Step-by-step solution

Glass (refractive index 1.5) is optically denser than water (1.33).
The ray goes from a denser to a rarer medium obliquely, so it bends away from the normal (ലംബത്തിൽ നിന്ന് അകലുന്നു).

✅ Answer: (b) bends away from the normal

Marking: 1 mark for (b).

📖 Revise: Ch 1 · Rules of refraction · Chapter 1 – Refraction of Light

A ray of light passes obliquely from medium P into medium Q and bends towards the normal. Which statement is correct?

(a) Light travels faster in Q than in P.(b) P is optically denser than Q.(c) Q has a higher refractive index than P.(d) The angle of refraction is greater than the angle of incidence.

[1]

Step-by-step solution

A ray bends towards the normal when it goes from a rarer to a denser medium. So Q is optically denser than P.
Light is slower in the denser medium Q, and since n = c/v, Q has the higher refractive index.
Also, r is less than i here, so (d) is wrong; (a) and (b) are the opposite of the truth.

✅ Answer: (c) Q has a higher refractive index than P

Marking: 1 mark for (c).

📖 Revise: Ch 1 · Rules of refraction · Ch 1 · Refractive index · Chapter 1 – Refraction of Light

Which of the following moves with uniform velocity?

(a) a stone dropped from a height(b) a bus leaving a bus stop(c) a child on a swing(d) light travelling through vacuum

[1]

Step-by-step solution

Uniform velocity: equal displacements in equal intervals of time, in one direction – neither the speed nor the direction changes.
A falling stone speeds up, a bus leaving a stop speeds up, and a swing keeps changing its speed and direction – all non-uniform.
Light in vacuum goes in a straight line at a constant speed, so its velocity is uniform.

✅ Answer: (d) light travelling through vacuum

Marking: 1 mark for (d).

📖 Revise: Ch 2 · Uniform and non-uniform velocity · Chapter 2 – Equations of Motion

Pick the odd one out and give the reason: distance, speed, displacement, time.

[1]

Step-by-step solution

Displacement (സ്ഥാനാന്തരം) has both magnitude and direction, so it is a vector quantity.
Distance, speed and time have only magnitude – they are scalar quantities (അദിശ അളവുകൾ).

✅ Answer: Displacement – it is a vector quantity; the others are scalar quantities.

Marking: ½ mark for “displacement”, ½ mark for the reason.

📖 Revise: Ch 2 · Distance and displacement · Chapter 2 – Equations of Motion

The mass of an object is measured by comparing it with standard masses. The instrument used for this is the

(a) spring balance(b) common balance(c) platform balance(d) measuring cylinder

[1]

Step-by-step solution

Mass (മാസ്) is measured with a common balance by comparing the object with standard masses.
Spring balances and platform balances measure weight, which is a force. A measuring cylinder measures volume.

✅ Answer: (b) common balance

Marking: 1 mark for (b).

📖 Revise: Ch 4 · Mass and weight · Chapter 4 – Gravitation

More cargo is loaded on to a ship that is floating in a harbour. The ship

(a) rises a little out of the water(b) sinks a little deeper but keeps floating(c) stays at exactly the same level(d) sinks to the bottom at once

[1]

Step-by-step solution

A floating ship displaces water whose weight equals its own weight (principle of floatation).
When cargo is added, its weight increases, so it must displace more water to stay afloat. It goes a little deeper into the water.

✅ Answer: (b) sinks a little deeper but keeps floating

Marking: 1 mark for (b).

📖 Revise: Ch 5 · Principle of floatation · Chapter 5 – Buoyant Force

Fill in the blanks: Relative density of a substance = density of the substance ÷ . Relative density has unit.

[1]

Step-by-step solution

Relative density (ആപേക്ഷികസാന്ദ്രത) = density of the substance ÷ density of water.
It is a ratio of two densities with the same unit (kg/m³), so the units cancel: it has no unit.

✅ Answer: density of water; no unit

Marking: ½ mark for each blank.

📖 Revise: Ch 5 · Relative density · Chapter 5 – Buoyant Force

One ampere is equal to

(a) 1 joule per coulomb(b) 1 coulomb per second(c) 1 volt per ampere(d) 1 joule per second

[1]

Step-by-step solution

Current I = Q/t, so its unit is coulomb ÷ second = C/s, called the ampere (A).
J/C is the volt, V/A is the ohm and J/s is the watt.

✅ Answer: (b) 1 coulomb per second

Marking: 1 mark for (b).

📖 Revise: Ch 7 · Electric current · Chapter 7 – Current Electricity

Which of these is a secondary (rechargeable) cell?

(a) a torch (dry) cell(b) a button cell in a watch(c) the battery of a mobile phone(d) a voltaic cell made with zinc and copper

[1]

Step-by-step solution

Primary cells cannot be used again once they are used up – dry cells, button cells and the voltaic cell.
Secondary (storage) cells can be recharged and reused – for example the battery of a mobile phone.

✅ Answer: (c) the battery of a mobile phone

Marking: 1 mark for (c).

📖 Revise: Ch 7 · Primary and secondary cells · Chapter 7 – Current Electricity

10.

Which of these shows that sound can travel through liquids?

(a) A snake senses the vibrations of an approaching animal through the ground.(b) Dolphins and whales communicate with each other under the sea.(c) We hear the school bell when we are on the playground.(d) Astronauts in space talk to each other by radio.

[1]

Step-by-step solution

Dolphins and whales send sounds to each other through sea water – a liquid.
The snake example shows sound in a solid (the ground); the school bell shows sound in air; radio is used in space because sound cannot travel through vacuum.

✅ Answer: (b) Dolphins and whales communicate with each other under the sea.

Marking: 1 mark for (b).

📖 Revise: Ch 8 · Propagation of sound · Chapter 8 – Sound

Section BAnswer all questions · 8 × 2 = 16 marks

11.

A ray of light AO travels inside a glass block and meets the glass–air surface obliquely at O, as shown. (a) Copy and complete the diagram: draw the normal at O and the refracted ray in air, and mark the angle of incidence i and the angle of refraction r. (b) Which is greater, i or r? Why?

[2]

Step-by-step solution

Draw the normal NN′ at O, perpendicular to the glass–air surface (dashed line).
The ray goes from glass (optically denser) into air (optically rarer), so the refracted ray OB bends away from the normal.
Completed diagram:
Angle of incidence i = ∠AON′, between the incident ray and the normal in glass. Angle of refraction r = ∠BON, between the refracted ray and the normal in air.
(b) r is greater than i, because a ray going obliquely from a denser to a rarer medium bends away from the normal.

✅ Answer: The refracted ray bends away from the normal; r is greater than i.

Marking: Normal ½, refracted ray bending away from the normal ½, i and r marked correctly ½; (b) r > i with reason ½.

📖 Revise: Ch 1 · Rules of refraction · Ch 1 · Refraction in a glass slab · Chapter 1 – Refraction of Light

12.

Ammu walks from her house H 300 m east to a shop S, and then 400 m north to her school C (figure). The straight-line distance HC is 500 m. The whole walk takes 500 s. Find (a) the distance she travels and her displacement (b) her speed and her velocity.

[2]

Step-by-step solution

Given: HS = 300 m, SC = 400 m, straight line HC = 500 m, time t = 500 s
(a) Distance = length of the path = 300 + 400 = 700 m.
Displacement (സ്ഥാനാന്തരം) = straight-line distance from start to end, with direction = 500 m, from H towards C.
Formula: speed = distance ÷ time; velocity = displacement ÷ time
Substitution: speed = 700 ÷ 500; velocity = 500 ÷ 500
Answer: (b) speed = 1.4 m/s; velocity = 1 m/s, in the direction H → C.

✅ Answer: (a) distance 700 m, displacement 500 m from H towards C (b) speed 1.4 m/s, velocity 1 m/s from H towards C

Marking: (a) distance ½, displacement with direction ½; (b) speed ½, velocity with direction ½.

📖 Revise: Ch 2 · Distance and displacement · Ch 2 · Speed and velocity · Chapter 2 – Equations of Motion

13.

A bullet of mass 10 g moves with a velocity of 300 m/s. A ball of mass 5 kg rolls with a velocity of 0.5 m/s. (a) Find the momentum of each. (b) Which has more momentum?

[2]

Step-by-step solution

Given: bullet m = 10 g = 0.01 kg, v = 300 m/s; ball m = 5 kg, v = 0.5 m/s
Formula: momentum (മൊമെന്റം) p = mv
Substitution: bullet: p = 0.01 × 300; ball: p = 5 × 0.5
Answer: bullet p = 3 kg m/s; ball p = 2.5 kg m/s
(b) The bullet has more momentum. Its mass is very small, but its velocity is very large – momentum depends on both mass and velocity.

✅ Answer: (a) bullet 3 kg m/s, ball 2.5 kg m/s (b) the bullet

Marking: Formula ½, bullet ½ (with 10 g changed to 0.01 kg), ball ½, comparison ½.

📖 Revise: Ch 3 · Momentum · Chapter 3 – Laws of Motion

14.

A diver feels weightless for a moment after she leaves the diving board, until she enters the water. (a) What is meant by free fall? (b) Why does she feel weightless during this time, even though the Earth is still pulling her?

[2]

Step-by-step solution

(a) When a body moves only because of the Earth’s gravitational force, with no other force acting, its motion is called free fall (നിർബാധ പതനം).
(b) After leaving the board, the diver moves only under gravity. She and everything with her fall with the same acceleration g.
Nothing supports her and nothing presses on her body. The whole of the gravitational force is used to give her acceleration, so she feels no weight – weightlessness (ഭാരമില്ലായ്മ).
Her real weight mg still acts on her – that is why she accelerates downwards.

✅ Answer: (a) Falling only because of the Earth’s gravitational force. (b) She and everything with her fall with the same acceleration g, so nothing supports or presses on her.

Marking: (a) 1 mark; (b) same acceleration / nothing supporting her ½, whole of gravity used for acceleration ½.

📖 Revise: Ch 4 · Free fall and weightlessness · Chapter 4 – Gravitation

15.

The relative densities of candle wax, aluminium and kerosene are 0.9, 2.7 and 0.81. (a) Which of the two solids, wax or aluminium, will float in water? Why? (b) Will a piece of wax float in kerosene? Give a reason.

[2]

Step-by-step solution

Relative density less than 1 means the substance is less dense than water; more than 1 means denser than water.
When an object is fully under a liquid, the buoyant force equals the weight of the liquid it displaces (Archimedes’ principle). If the object is less dense than the liquid, this buoyant force is more than its weight, so it rises and floats. If it is denser, its weight is more, so it sinks.
(a) Wax floats in water (0.9 < 1). Aluminium is 2.7 times as dense as water, so it sinks.
(b) No. Wax (0.9) is denser than kerosene (0.81), so its weight is more than the buoyant force of kerosene and it sinks in kerosene.

✅ Answer: (a) Wax floats (relative density less than 1); aluminium sinks (2.7 > 1). (b) No – wax (0.9) is denser than kerosene (0.81), so it sinks.

Marking: (a) wax floats with reason ½, aluminium sinks ½; (b) sinks ½, reason ½.

📖 Revise: Ch 5 · Relative density · Ch 5 · Float or sink · Chapter 5 – Buoyant Force

16.

A girl riding a bicycle has a kinetic energy of 625 J. The mass of the girl and the bicycle together is 50 kg. (a) Find her speed. (b) What will the kinetic energy be if she doubles her speed?

[2]

Step-by-step solution

Given: E_k = 625 J, m = 50 kg
Formula: kinetic energy (ഗതികോർജം) E_k = ½mv²
Substitution: 625 = ½ × 50 × v² = 25 v² ⇒ v² = 625 ÷ 25 = 25
Answer: (a) v = √25 = 5 m/s
(b) New speed = 2 × 5 = 10 m/s. E_k = ½ × 50 × 10² = 25 × 100 = 2500 J – four times as much, because E_k depends on v².

✅ Answer: (a) 5 m/s (b) 2500 J

Marking: (a) formula and substitution ½, answer with unit ½; (b) 1 mark.

📖 Revise: Ch 6 · Kinetic energy · Chapter 6 – Work and Energy

17.

(a) Using the circuit symbol of a cell, draw a battery of three 1.5 V cells connected in series. Write its emf. (b) By mistake, the middle cell is put in the wrong way round. What will the emf of the battery be now?

[2]

Step-by-step solution

In series, the + terminal (long thin line) of one cell is joined to the − terminal (short thick line) of the next.
Battery of three cells in series:
(a) In series the emfs add: 1.5 + 1.5 + 1.5 = 4.5 V (ശ്രേണീരീതി).
(b) The reversed cell pushes against the other two, so its emf is subtracted: 1.5 − 1.5 + 1.5 = 1.5 V.

✅ Answer: (a) 4.5 V (b) 1.5 V

Marking: (a) correct drawing ½, emf ½; (b) 1 mark.

📖 Revise: Ch 7 · Combination of cells · Chapter 7 – Current Electricity

18.

The speed of sound in some media at 25 °C is given below (Table 8.3).

Medium	steel	air	sea water	helium
Speed (m/s)	5944	347	1525	968

(a) Arrange the media in increasing order of the speed of sound. (b) What do you conclude about the speed of sound in solids, liquids and gases?

[2]

Step-by-step solution

(a) 347 < 968 < 1525 < 5944, so the order is air < helium < sea water < steel.
(b) The speed of sound (ശബ്ദവേഗം) is different in different media. It is highest in solids (steel), less in liquids (sea water) and lowest in gases (air, helium).

✅ Answer: (a) air < helium < sea water < steel (b) sound is fastest in solids, slower in liquids and slowest in gases

Marking: (a) 1 mark (any one place wrong: 0); (b) 1 mark.

📖 Revise: Ch 8 · Speed of sound · Chapter 8 – Sound

Section CAnswer all questions · 6 × 3 = 18 marks

19.

(a) Why do stars twinkle? (b) In the figure, just before sunrise, the Sun S is still below the horizon, yet the observer sees it at S′, above the horizon. Explain why. (c) Name the phenomenon behind both (a) and (b).

[3]

Step-by-step solution

(a) Starlight passes through many layers of the atmosphere. The temperature and pressure of these layers keep changing, so their optical density keeps changing.
So the starlight is refracted irregularly and its path keeps changing. The apparent position and brightness of the star keep changing – the star twinkles.
(b) Air near the Earth is denser; higher up it is rarer. Light from the Sun below the horizon enters the atmosphere and passes into denser and denser layers. At each layer it bends towards the normal, so its path curves down and reaches the observer.
Our eye sees the light as coming straight from the direction in which it enters the eye. So we see the Sun at a higher, apparent position S′, above the horizon, a little before it actually rises.
(c) Atmospheric refraction (അന്തരീക്ഷത്തിലെ അപവർത്തനം).

✅ Answer: (a) and (b) are both due to atmospheric refraction: starlight is refracted irregularly by changing layers of air; sunlight from below the horizon bends down towards us, so the Sun appears higher.

Marking: (a) changing optical density ½, irregular refraction ½, apparent position/brightness change ½; (b) bending towards the normal in denser layers ½, Sun seen at a higher apparent position ½; (c) ½.

📖 Revise: Ch 1 · Atmospheric refraction · Chapter 1 – Refraction of Light

20.

The position–time graphs of two cyclists A and B riding along a straight road in the same direction are shown. At t = 0, B is 30 m ahead of A. (a) What kind of velocity does each cyclist have? Give a reason. (b) Find the velocity of A and of B. (c) At what time, and at what position, does A catch up with B?

[3]

Step-by-step solution

(a) Both graphs are sloping straight lines: each cyclist has equal displacements in equal intervals of time. So both have uniform velocity (സമപ്രവേഗം).
Given: A: 0 m at 0 s, 50 m at 10 s; B: 30 m at 0 s, 50 m at 10 s
Formula: velocity v = (s₂ − s₁) ÷ (t₂ − t₁)
Substitution: A: (50 − 0) ÷ (10 − 0); B: (50 − 30) ÷ (10 − 0)
Answer: (b) velocity of A = 5 m/s; velocity of B = 2 m/s
(c) A catches B where the two graphs meet: at t = 10 s, at the 50 m position.
Check: A gains 5 − 2 = 3 m on B every second, and B was 30 m ahead, so A needs 30 ÷ 3 = 10 s. Then A is at 5 × 10 = 50 m and B at 30 + 2 × 10 = 50 m ✓.

✅ Answer: (a) uniform velocity (straight-line graphs) (b) A: 5 m/s, B: 2 m/s (c) at t = 10 s, at the 50 m position

Marking: (a) uniform ½, reason ½; (b) ½ each; (c) time ½, position ½.

📖 Revise: Ch 2 · Position–time graph · Ch 2 · Uniform velocity · Chapter 2 – Equations of Motion

21.

A carpenter’s hammer of mass 0.8 kg hits a nail at 5 m/s and comes to rest in 0.02 s. (Take the direction of the hammer’s motion as positive.) (a) Find the change in momentum of the hammer. (b) Find the average force exerted on the hammer by the nail. What does the negative sign show? (c) If a thick rubber sheet were placed on the nail head, the hammer would stop in 0.1 s. Find the force now. Which way is better for driving the nail in? Why?

[3]

Step-by-step solution

Given: m = 0.8 kg, u = 5 m/s, v = 0, t = 0.02 s
Formula: change in momentum = mv − mu; F = (mv − mu) ÷ t
Substitution: mv − mu = 0.8 × 0 − 0.8 × 5 = −4; F = −4 ÷ 0.02
Answer: (a) change in momentum = −4 kg m/s (this is also the impulse, −4 N s); (b) F = −200 N
The negative sign shows that the force acts opposite to the direction of motion of the hammer – it stops the hammer. By the third law, the hammer pushes the nail forward with 200 N.
(c) Same change in momentum, longer time: F = −4 ÷ 0.1 = −40 N.
Hitting the nail directly is better: the hammer stops in a very short time, so the force (ആവേഗബലം, impulsive force) is large and drives the nail in. For the same change in momentum, the longer the time, the smaller the force.

✅ Answer: (a) −4 kg m/s (b) −200 N, opposite to the hammer’s motion (c) −40 N; hitting the nail directly is better

Marking: (a) formula ½, answer ½; (b) force ½, meaning of the sign ½; (c) force ½, conclusion with reason ½.

📖 Revise: Ch 3 · Impulsive force and impulse · Ch 3 · Second law · Chapter 3 – Laws of Motion

22.

A wooden block has a volume of 2000 cm³ and a mass of 1.2 kg. (Density of water = 1000 kg/m³; 1 m³ = 10⁶ cm³.) (a) Find the density and the relative density of the wood. (b) The block is placed in water (figure). What volume of the block is under water when it floats? (c) The block is moved to a saturated salt solution (density 1025 kg/m³). Does it float higher or lower? Find the volume under the solution now.

[3]

Step-by-step solution

Given: V = 2000 cm³ = 2000 ÷ 10⁶ = 0.002 m³, m = 1.2 kg
Formula: density = mass ÷ volume; relative density = density of wood ÷ density of water
Substitution: density = 1.2 ÷ 0.002; relative density = 600 ÷ 1000
Answer: (a) density = 600 kg/m³; relative density = 0.6 (no unit)
(b) Its relative density is less than 1, so it floats. By the principle of floatation, the weight of the water displaced = the weight of the block, so the mass of water displaced = 1.2 kg.
Volume of water displaced = mass ÷ density = 1.2 ÷ 1000 = 0.0012 m³ = 1200 cm³. This is the volume under water (the other 800 cm³ is above the water).
(c) The salt solution is denser, so a smaller volume of it has the same weight as the block. The block floats higher.
Volume under the solution = 1.2 ÷ 1025 = 0.001 171 m³ ≈ 1171 cm³.

✅ Answer: (a) 600 kg/m³, relative density 0.6 (b) 1200 cm³ (c) higher; about 1171 cm³

Marking: (a) density ½, relative density ½; (b) principle ½, volume ½; (c) floats higher ½, volume ½.

📖 Revise: Ch 5 · Principle of floatation · Ch 5 · Relative density · Chapter 5 – Buoyant Force

23.

(a) A body of mass m is at rest. A constant force F acts on it and it moves through a displacement s, reaching a velocity v. Show that the kinetic energy it gains is ½mv². (b) What kind of energy is stored in (i) a stretched catapult (ii) a coconut hanging on a coconut tree?

[3]

Step-by-step solution

(a) Work done by the force: W = F × s.
By the second law of motion, F = ma. So W = ma × s = m(as).
From v² = u² + 2as with u = 0 (it starts from rest): v² = 2as, so as = v²/2.
So W = m × v²/2 = ½mv². This work is stored in the body as kinetic energy: E_k = ½mv².
(b) (i) Stretched catapult: potential energy due to configuration (കോൺഫിഗറേഷൻ മൂലമുള്ള സ്ഥിതികോർജം) – its shape has been changed.
(ii) Coconut on the tree: potential energy due to position (സ്ഥാനം മൂലമുള്ള സ്ഥിതികോർജം) – it is at a height.

✅ Answer: (a) W = Fs = mas and as = v²/2, so E_k = ½mv² (b) (i) potential energy due to configuration (ii) potential energy due to position

Marking: (a) W = Fs and F = ma ½, v² = 2as ½, final result 1; (b) ½ + ½.

📖 Revise: Ch 6 · Kinetic energy – derivation · Ch 6 · Potential energy · Chapter 6 – Work and Energy

24.

In the circuit shown, the battery is made of two 1.5 V cells. When the switch K is closed, the ammeter reads 0.5 A and the voltmeter reads 3 V. (a) How are the two cells connected? Why? (b) How much charge flows through the bulb in 2 minutes? (c) How much work is done in moving this charge through the bulb?

[3]

Step-by-step solution

(a) The cells are in series (+ of one joined to − of the other). In series the emfs add: 1.5 + 1.5 = 3 V. (In parallel they would give only 1.5 V.)
Given: I = 0.5 A, t = 2 min = 2 × 60 = 120 s
Formula: I = Q/t ⇒ Q = I × t
Substitution: Q = 0.5 × 120
Answer: (b) Q = 60 C
(c) Given: potential difference (പൊട്ടൻഷ്യൽ വ്യത്യാസം) V = 3 V, Q = 60 C. Formula: V = W/Q ⇒ W = V × Q. Substitution: W = 3 × 60.
Answer: (c) W = 180 J

✅ Answer: (a) in series, so that their emfs add up to 3 V (b) 60 C (c) 180 J

Marking: (a) series ½, reason ½; (b) formula ½, answer with unit ½ (2 min must be changed to 120 s); (c) formula ½, answer ½.

📖 Revise: Ch 7 · Electric current · Ch 7 · Potential difference · Chapter 7 – Current Electricity

Section DAnswer all questions · 4 × 4 = 16 marks

25.

An autorickshaw with its driver has a total mass of 400 kg. Starting from rest on a straight road, it is pushed forward by a resultant (unbalanced) force of 600 N for 10 s. (a) Find its acceleration. (b) Find its velocity after 10 s. (c) Find its momentum after 10 s and the rate of change of its momentum. Compare this rate with the force. (d) Two passengers with a total mass of 200 kg get in. If the same resultant force acts, what is the new acceleration? What do you learn from this?

[4]

Step-by-step solution

Given: m = 400 kg, F = 600 N, u = 0, t = 10 s
Formula: F = ma ⇒ a = F/m
Substitution: a = 600 ÷ 400
Answer: (a) a = 1.5 m/s²
(b) v = u + at = 0 + 1.5 × 10 = 15 m/s.
(c) Momentum p = mv = 400 × 15 = 6000 kg m/s. Change in momentum = 6000 − 0 = 6000 kg m/s. Rate of change = 6000 ÷ 10 = 600 kg m/s².
This is equal to the force, 600 N – the rate of change of momentum equals the resultant force (Newton’s second law).
(d) New mass = 400 + 200 = 600 kg. a = F/m = 600 ÷ 600 = 1 m/s².
For the same force, a bigger mass gets a smaller acceleration (a = F/m) – the heavier autorickshaw speeds up more slowly.

✅ Answer: (a) 1.5 m/s² (b) 15 m/s (c) 6000 kg m/s; 600 kg m/s², equal to the force (d) 1 m/s²

Marking: (a) 1 mark; (b) 1 mark; (c) momentum ½, rate and comparison ½; (d) acceleration ½, conclusion ½.

📖 Revise: Ch 3 · Newton’s second law · Ch 3 · Rate of change of momentum · Chapter 3 – Laws of Motion

26.

Chandrayaan-3 placed a rover on the Moon. Suppose a small rover has a mass of 30 kg. (g on the Earth = 9.8 m/s²; g on the Moon = 1.62 m/s²) (a) Find its weight on the Earth. (b) What are its mass and its weight on the Moon? (c) A small tool is dropped from a height of 3.24 m above the Moon’s surface and falls freely. How long does it take to reach the surface? (d) An astronaut on the Moon drops a feather and a hammer together from the same height. Which reaches the ground first? Why?

[4]

Step-by-step solution

Given: m = 30 kg, g (Earth) = 9.8 m/s², g (Moon) = 1.62 m/s²
Formula: weight W = mg
Substitution: Earth: W = 30 × 9.8; Moon: W = 30 × 1.62
Answer: (a) weight on the Earth = 294 N (30 kgwt). (b) On the Moon the mass is still 30 kg (മാസ് does not change); weight = 48.6 N – about 1/6 of its weight on the Earth.
(c) Given: u = 0, s = 3.24 m, a = g = 1.62 m/s². Formula: s = ut + ½gt² ⇒ t² = 2s/g. Substitution: t² = 2 × 3.24 ÷ 1.62 = 4.
Answer: (c) t = √4 = 2 s
(d) They reach the ground together. The Moon has no air, so there is no air resistance to slow the feather. The acceleration g = GM/R² does not depend on the mass of the falling body, so both fall with the same acceleration.

✅ Answer: (a) 294 N (b) mass 30 kg, weight 48.6 N (c) 2 s (d) both together – no air, and g is the same for all bodies

Marking: (a) 1 mark; (b) mass ½, weight ½; (c) formula ½, answer ½; (d) together ½, reason ½.

📖 Revise: Ch 4 · Mass and weight · Ch 4 · g on the Moon · Chapter 4 – Gravitation

27.

At an Onam celebration, a child sits on a swing. The total mass of the child and the seat is 25 kg. At the highest point A the seat is 0.8 m above its lowest point B (figure). Ignore air resistance; g = 10 m/s². (a) Find the potential energy (taking B as the zero level) and the kinetic energy at A, where the swing stops for an instant. (b) Find the kinetic energy and the speed at B. (c) At what height above B will the speed of the seat be 2 m/s? (d) If nobody pushes, the swing slowly comes to rest. Does this mean that energy is destroyed? Explain.

[4]

Step-by-step solution

Given: m = 25 kg, h = 0.8 m, g = 10 m/s²
Formula: E_p = mgh; E_k = ½mv²
Substitution: E_p at A = 25 × 10 × 0.8
Answer: (a) E_p = 200 J; the swing is at rest for an instant, so E_k = 0 J. Total energy = 200 J.
(b) At B, E_p = 0. By the law of conservation of energy (ഊർജസംരക്ഷണ നിയമം), E_k = 200 J. ½ × 25 × v² = 200 ⇒ v² = 400 ÷ 25 = 16 ⇒ v = 4 m/s.
(c) At 2 m/s: E_k = ½ × 25 × 2² = 50 J. So E_p = 200 − 50 = 150 J. h = E_p ÷ mg = 150 ÷ (25 × 10) = 0.6 m.
(d) No. Air resistance and friction at the hooks slowly change the mechanical energy into heat (and a little sound). Energy is neither created nor destroyed; it only changes from one form to another.

✅ Answer: (a) PE 200 J, KE 0 (b) KE 200 J, speed 4 m/s (c) 0.6 m (d) No – it changes into heat (and sound)

Marking: (a) PE ½, KE ½; (b) KE ½, speed ½; (c) 1 mark; (d) no ½, explanation ½.

📖 Revise: Ch 6 · Conservation of energy · Ch 6 · Potential energy · Chapter 6 – Work and Energy

28.

On a monsoon evening, lightning strikes a coconut tree 1388 m away from Meena. (Speed of sound in air = 347 m/s; speed of light = 3 × 10⁸ m/s.) (a) After how many seconds does she hear the thunder? (b) She sees the flash almost at once. Why? (c) On a hotter day, would the thunder from the same distance reach her sooner or later? Why? (d) The thunder reaches her as compressions and rarefactions in the air. Do the air particles near the lightning travel all the way to her ear? Explain.

[4]

Step-by-step solution

Given: distance d = 1388 m, speed of sound v = 347 m/s
Formula: time = distance ÷ speed
Substitution: t = 1388 ÷ 347
Answer: (a) t = 4 s
(b) Light travels at 3 × 10⁸ m/s. It takes only 1388 ÷ (3 × 10⁸) ≈ 0.000 004 6 s to reach her – practically no time. Sound in air is nearly a million times slower.
(c) Sooner. The speed of sound increases when the temperature of the medium increases, so in hotter air the sound covers the same distance in less time.
(d) No. Each air particle only vibrates to and fro about its place and pushes the next particles. These pressure changes – compressions (മർദം കൂടിയ മേഖല) and rarefactions (മർദം കുറഞ്ഞ മേഖല) – pass on through the air to her ear. The energy travels; the particles do not.

✅ Answer: (a) 4 s (b) light is extremely fast (about 0.000 004 6 s for 1388 m) (c) sooner – sound is faster in hotter air (d) No – the particles only vibrate; the disturbance travels

Marking: (a) formula ½, answer ½; (b) 1 mark; (c) sooner ½, reason ½; (d) no ½, explanation ½.

📖 Revise: Ch 8 · Speed of sound · Ch 8 · Propagation of sound · Chapter 8 – Sound

Question numbers by chapter

Chapter	A (1)	B (2)	C (3)	D (4)	Total
1. Refraction of Light	1, 2	11	19	–	7
2. Equations of Motion	3, 4	12	20	–	7
3. Laws of Motion	–	13	21	25	9
4. Gravitation	5	14	–	26	7
5. Buoyant Force	6, 7	15	22	–	7
6. Work and Energy	–	16	23	27	9
7. Current Electricity	8, 9	17	24	–	7
8. Sound	10	18	–	28	7
Total	10	16	18	16	60

Score to grade

Kerala grading scale: A+ 54–60 (90 %+), A 48–53, B+ 42–47, B 36–41, C+ 30–35, C 24–29, D+ 18–23, D 12–17, E below 12. For every question the student lost marks on, the linked guide page explains the idea again.