Class 9 · PhysicsSet 3 · Answer key

Answer Key – Set 3 ഉത്തരസൂചിക

Step-by-step answers and the marking scheme for every question of Set 3, 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 3

Standard IX · Physics · Annual Examination

Total score: 60 · Time: 2 hours

Section AAnswer all questions · 10 × 1 = 10 marks

When light passes from glass into air, the angle of incidence for which the angle of refraction becomes 90° is called the

(a) angle of deviation(b) angle of reflection(c) critical angle(d) refractive index

[1]

Step-by-step solution

For light going from a denser medium (glass) to a rarer medium (air), the angle of refraction grows as the angle of incidence grows.
The angle of incidence for which the angle of refraction is 90° (the ray grazes along the surface) is the critical angle (ക്രിട്ടിക്കൽ കോൺ). For glass it is 42°.

✅ Answer: (c) critical angle

Marking: 1 mark for (c).

📖 Revise: Ch 1 · Critical angle · Chapter 1 – Refraction of Light

Pick the odd one out and give the reason: optical fibre, mirage, twinkling of stars, cycle reflector.

[1]

Step-by-step solution

Optical fibre, cycle reflector and mirage all involve total internal reflection.
Twinkling of stars is caused by atmospheric refraction (അന്തരീക്ഷത്തിലെ അപവർത്തനം) – irregular refraction in air layers whose optical density keeps changing. So it is the odd one out.

✅ Answer: twinkling of stars – it is due to atmospheric refraction only; the others involve total internal reflection

Marking: ½ mark for “twinkling of stars”, ½ mark for the reason.

📖 Revise: Ch 1 · Uses of total internal reflection · Ch 1 · Twinkling of stars · Chapter 1 – Refraction of Light

The velocity of a scooter decreases uniformly from 15 m/s to 5 m/s in 5 s. Its retardation is

(a) 2 m/s²(b) −2 m/s²(c) 4 m/s²(d) 10 m/s²

[1]

Step-by-step solution

a = (v − u)/t = (5 − 15)/5 = −10/5 = −2 m/s².
Retardation (മന്ദീകരണം) is the negative acceleration, so retardation = 2 m/s². (We do not write a minus sign with retardation.)

✅ Answer: (a) 2 m/s²

Marking: 1 mark for (a).

📖 Revise: Ch 2 · Retardation · Chapter 2 – Equations of Motion

When a rocket is launched, the upward force on the rocket is exerted by

(a) the air below the rocket(b) the hot gases rushing out of it(c) the launch pad(d) the Earth’s gravity

[1]

Step-by-step solution

The burning fuel pushes hot, high-pressure gases downwards out of the rocket (action).
By Newton’s third law the gases push the rocket upwards with an equal and opposite force (reaction, പ്രതിബലം).

✅ Answer: (b) the hot gases rushing out of it

Marking: 1 mark for (b).

📖 Revise: Ch 3 · Third law – rocket · Chapter 3 – Laws of Motion

Find the relation and fill in: walking : we push the ground backwards :: swimming :

[1]

Step-by-step solution

In walking we push the ground backwards (action) and the ground pushes us forwards (reaction).
In swimming the swimmer pushes the water backwards with the hands and legs; the water pushes the swimmer forwards (third law).

✅ Answer: we push the water backwards

Marking: 1 mark.

📖 Revise: Ch 3 · Action and reaction · Chapter 3 – Laws of Motion

An iron block and a copper block have the same mass. Both are fully immersed in water. (Copper is denser than iron.) Which statement is correct?

(a) the buoyant force is the same on both(b) the buoyant force on the copper block is greater(c) the buoyant force on the iron block is greater(d) there is no buoyant force on either block

[1]

Step-by-step solution

Same mass, but copper is denser, so the copper block has the smaller volume; the iron block has the larger volume.
A larger volume displaces more water, so the buoyant force (പ്ലവക്ഷമബലം) on the iron block is greater.

✅ Answer: (c) the buoyant force on the iron block is greater

Marking: 1 mark for (c).

📖 Revise: Ch 5 · Volume of the object · Chapter 5 – Buoyant Force

The readings on the scale of a hydrometer (increase / decrease) from the top to the bottom.

[1]

Step-by-step solution

A hydrometer (ഹൈഡ്രോമീറ്റർ) floats higher in a denser liquid, so the mark at the liquid surface is then a lower mark on the stem.
So the larger relative densities are marked lower down: the values increase from top to bottom (e.g. 0.50 near the top, 1.00 lower down).

✅ Answer: increase

Marking: 1 mark.

📖 Revise: Ch 5 · Hydrometer · Chapter 5 – Buoyant Force

1 watt is equal to

(a) 1 J s(b) 1 J/s(c) 1 N/m(d) 1 N s

[1]

Step-by-step solution

Power (പവർ) P = W/t = work ÷ time.
So its unit is joule per second: 1 W = 1 J/s.

✅ Answer: (b) 1 J/s

Marking: 1 mark for (b).

📖 Revise: Ch 6 · Power and the watt · Chapter 6 – Work and Energy

The vibrations of the eardrum are passed on to the

(a) outer ear (pinna)(b) ear canal(c) inner ear(d) larynx

[1]

Step-by-step solution

Sound → outer ear (ചെവിക്കുട) → ear canal (കർണ്ണനാളം) → eardrum (കർണ്ണപുടം) vibrates.
The eardrum sends the vibrations to the inner ear (ആന്തരകർണ്ണം); from there signals go to the brain.

✅ Answer: (c) inner ear

Marking: 1 mark for (c).

📖 Revise: Ch 8 · Ear and hearing · Chapter 8 – Sound

10.

The instrument used to detect hearing loss is the

(a) hearing aid(b) audiometer(c) ear muff(d) stethoscope

[1]

Step-by-step solution

Hearing loss is detected with an audiometer (ഓഡിയോമീറ്റർ).
A hearing aid helps a person hear; ear muffs protect the ears from loud noise; a stethoscope is used by doctors to listen to the heart.

✅ Answer: (b) audiometer

Marking: 1 mark for (b).

📖 Revise: Ch 8 · Audiometer and hearing aid · Chapter 8 – Sound

Section BAnswer all questions · 8 × 2 = 16 marks

11.

A lamp O at the bottom of a pond sends two rays OA and OB to the water surface. Ray OA strikes the surface at an angle of incidence of 40° and ray OB at 55° (figure). The critical angle of water is 48.6°. (a) Copy the figure and complete the paths of both rays. (b) Name the phenomenon that happens to ray OB.

[2]

Step-by-step solution

Light is going from water (denser) to air (rarer).
Ray OA: 40° is less than the critical angle 48.6°, so the ray is refracted into the air. Going from denser to rarer, it bends away from the normal (angle of refraction r greater than 40°).
Ray OB: 55° is greater than the critical angle 48.6°, so it is not refracted. It is reflected completely back into the water, making 55° with the normal on the other side.
Completed figure:
(b) This is total internal reflection (പൂർണ്ണാന്തര പ്രതിപതനം).

✅ Answer: (a) OA comes out into air, bending away from the normal; OB is reflected back into the water at 55° (b) total internal reflection

Marking: (a) ray OA drawn bending away from the normal: ½; ray OB drawn reflected back into the water: ½; angle of reflection 55° marked: ½. (b) ½.

📖 Revise: Ch 1 · Total internal reflection · Ch 1 · Denser to rarer · Chapter 1 – Refraction of Light

12.

A school bus is moving at 36 km/h. The driver sees children at a zebra crossing 25 m ahead and applies the brakes at once. The bus stops in 4 s with uniform retardation. (a) Find the retardation of the bus. (b) Find the distance the bus travels before stopping. Does it stop before the zebra crossing?

[2]

Step-by-step solution

Given: u = 36 km/h = 36 × 5/18 = 10 m/s, v = 0, t = 4 s
Formula: a = (v − u)/t ; v² = u² + 2as
Substitution: (a) a = (0 − 10)/4 = −2.5 m/s². (b) 0 = 10² + 2 × (−2.5) × s ⇒ 5s = 100
Answer: (a) retardation = 2.5 m/s² (b) s = 20 m
The crossing is 25 m ahead and the bus needs only 20 m, so it stops 5 m before the zebra crossing. (Check: average velocity (10 + 0)/2 = 5 m/s, 5 × 4 = 20 m ✓.)

✅ Answer: (a) 2.5 m/s² (b) 20 m – yes, it stops 5 m before the crossing

Marking: (a) conversion to 10 m/s ½, retardation with unit ½. (b) distance 20 m ½, conclusion ½.

📖 Revise: Ch 2 · Retardation · Ch 2 · Equations of motion · Chapter 2 – Equations of Motion

13.

Pushing a car from inside, while sitting in it, does not move the car. But when a person standing on the road pushes it, the car moves. Explain why.

[2]

Step-by-step solution

From inside, we push the car forward with our hands, but our body pushes the car’s floor/seat backwards with an equal force (action and reaction both act on the same system – the car with us in it).
These are internal forces; they are balanced (resultant zero), so the car does not move.
Standing on the road, we push the ground backwards and the ground pushes us forward; our push on the car is now an unbalanced external force, so the car moves.

✅ Answer: from inside the force is internal and balanced; from the road it is an unbalanced external force

Marking: Inside – internal / balanced forces: 1. Outside – unbalanced external force: 1.

📖 Revise: Ch 3 · Pushing the car from inside · Ch 3 · Balanced and unbalanced forces · Chapter 3 – Laws of Motion

14.

A KSRTC bus going round a hairpin bend on the Thamarassery ghat road moves along part of a circular path. (a) Which force provides the centripetal force the bus needs? (b) Why must the driver slow down at such bends?

[2]

Step-by-step solution

(a) A body moving on a circular path needs a centripetal force (അഭികേന്ദ്രബലം) towards the centre. For a vehicle on a road it comes from the friction between the tyres and the road.
(b) The centripetal force needed depends on the mass of the vehicle, its speed and the sharpness of the bend. At high speed on a sharp bend more force is needed than friction can give.
Then the bus tends to slide outwards off the curve (it tries to go along the tangent). Slowing down reduces the force needed, so the bus can turn safely.

✅ Answer: (a) friction between the tyres and the road (b) at high speed this force is not enough and the bus slides outwards

Marking: (a) 1. (b) more force needed at higher speed ½, bus slides outwards ½.

📖 Revise: Ch 4 · Circular motion and centripetal force · Chapter 4 – Gravitation

15.

A lactometer is placed first in whole milk and then in skimmed milk (milk from which the fat has been removed). (a) In which does it sink more? (b) Give the reason.

[2]

Step-by-step solution

A lactometer (ലാക്ടോമീറ്റർ) is a hydrometer for milk. It works on the principle of floatation and floats higher in a denser liquid.
Fat is lighter than water (relative density about 0.9). Removing the fat makes the milk denser, so skimmed milk has a higher relative density.
(a) So the lactometer floats higher in skimmed milk and sinks more in whole milk.

✅ Answer: (a) in whole milk (b) fat is lighter than water, so skimmed milk is denser and the lactometer floats higher in it

Marking: (a) 1. (b) skimmed milk is denser (fat is lighter) ½, floats higher in a denser liquid ½.

📖 Revise: Ch 5 · Lactometer · Chapter 5 – Buoyant Force

16.

A lift carries 8 passengers of total mass 560 kg up through a height of 15 m in 20 s. (g = 10 m/s²) (a) Find the work done in lifting the passengers against gravity. (b) Find the power used for this, in watt and in kilowatt.

[2]

Step-by-step solution

Given: m = 560 kg, h = 15 m, t = 20 s, g = 10 m/s²
Formula: W = mgh ; P = W/t
Substitution: W = 560 × 10 × 15 = 84 000 J ; P = 84 000 / 20
Answer: (a) W = 84 000 J (b) P = 4200 W = 4200/1000 = 4.2 kW

✅ Answer: (a) 84 000 J (b) 4200 W = 4.2 kW

Marking: (a) formula and substitution ½, answer ½. (b) 4200 W ½, 4.2 kW ½.

📖 Revise: Ch 6 · Power · Ch 6 · Work against gravity · Chapter 6 – Work and Energy

17.

A uniform wire has a resistance of 4 Ω. It is cut into two equal halves. (a) What is the resistance of each half? (b) The two halves are now joined side by side (in parallel). Find their effective resistance.

[2]

Step-by-step solution

(a) Resistance is proportional to length. Half the length → half the resistance: 4/2 = 2 Ω each.
Given: R₁ = R₂ = 2 Ω, in parallel
Formula: R = R₁R₂/(R₁ + R₂)
Substitution: R = (2 × 2)/(2 + 2) = 4/4
Answer: (b) R = 1 Ω. (Joining side by side is like a wire of half the length and double the thickness, so the resistance becomes one quarter.)

✅ Answer: (a) 2 Ω (b) 1 Ω

Marking: (a) 1. (b) formula ½, answer with unit ½.

📖 Revise: Ch 7 · Length and resistance · Ch 7 · Resistors in parallel · Chapter 7 – Current Electricity

18.

Both ends of a plastic bottle are cut off. A piece of balloon is stretched tightly over one end and a few mustard seeds are placed on it (figure). A student speaks loudly into the open end. (a) What happens to the mustard seeds? Why? (b) Which part of our ear works like the balloon diaphragm?

[2]

Step-by-step solution

(a) The voice makes pressure variations in the air inside the bottle. They reach the balloon and make the diaphragm vibrate, so the mustard seeds jump up and down.
(b) In our ear the eardrum (കർണ്ണപുടം) vibrates in the same way when sound reaches it, and passes the vibrations to the inner ear.

✅ Answer: (a) the seeds jump up and down because the diaphragm vibrates (b) the eardrum

Marking: (a) seeds jump ½, diaphragm vibrates ½. (b) eardrum 1.

📖 Revise: Ch 8 · Diaphragm activity and the eardrum · Chapter 8 – Sound

Section CAnswer all questions · 6 × 3 = 18 marks

19.

The figure shows an optical fibre with its layers cut back. (a) Name the parts marked 1, 2 and 3. (b) Explain how light reaches the far end of the fibre even when the fibre is bent. (c) Write one use of optical fibres in communication and one in medicine.

[3]

Step-by-step solution

(a) 1 – core (ആന്തരകോർ), 2 – cladding (ക്ലാഡിങ്), 3 – outer jacket (ബാഹ്യകവചം).
(b) Light entering one end travels in the core and strikes its wall at an angle of incidence greater than the critical angle. It is totally reflected back into the core. This total internal reflection happens again and again, even at the bends, so the light comes out only at the far end.
(c) Communication: optical fibre cables carry telephone, TV and internet signals as light pulses (e.g. KFON cables in Kerala). Medicine: endoscopy – doctors see inside the stomach.

✅ Answer: (a) 1 core, 2 cladding, 3 outer jacket (b) repeated total internal reflection (c) internet/telephone cables; endoscopy

Marking: (a) ½ for two correct names, 1 for all three. (b) 1 (angle greater than the critical angle ½, repeated TIR ½). (c) ½ + ½.

📖 Revise: Ch 1 · Optical fibre and endoscopy · Chapter 1 – Refraction of Light

20.

A gun of mass 4 kg, free to move, fires a bullet of mass 20 g. The bullet leaves the gun with a velocity of 400 m/s. Both were at rest before firing. (a) Find the momentum of the bullet. (b) Using Newton’s third law, write the momentum the gun gets, with its direction. (c) Find the recoil velocity of the gun.

[3]

Step-by-step solution

Given: m(bullet) = 20 g = 0.02 kg, v = 400 m/s; M(gun) = 4 kg; both start from rest
Formula: p = mv
Substitution: (a) p = 0.02 × 400 = 8 kg m/s (forward)
(b) While firing, the gun pushes the bullet forward (action) and the bullet pushes the gun backward with an equal and opposite force (reaction), for the same time. So the impulses F × t are equal and opposite, and the gun gets a momentum of 8 kg m/s backwards (−8 kg m/s).
(c) Recoil velocity V = p / M = 8 / 4
Answer: V = 2 m/s, backwards. The gun moves much slower than the bullet because its mass is much larger.

✅ Answer: (a) 8 kg m/s (b) 8 kg m/s, backwards (c) 2 m/s backwards

Marking: (a) ½ conversion, ½ answer. (b) ½ magnitude, ½ backward direction with third-law reason. (c) 1.

📖 Revise: Ch 3 · Third law · Ch 3 · Impulse = change in momentum · Chapter 3 – Laws of Motion

21.

A stone is dropped from a high bridge into the river 44.1 m below. Ignore air resistance. (g on the Earth = 9.8 m/s², g on the Moon = 1.62 m/s²) (a) How long does it take to reach the water? (b) Find its speed just before it hits the water. (c) How long would the stone take to fall the same height on the Moon?

[3]

Step-by-step solution

Given: u = 0, s = h = 44.1 m, g = 9.8 m/s²
Formula: s = ut + ½gt² ⇒ h = ½gt² ⇒ t = √(2h/g) ; v = u + gt
Substitution: (a) t = √(2 × 44.1 / 9.8) = √(88.2 / 9.8) = √9 = 3 s. (b) v = 0 + 9.8 × 3
Answer: (a) t = 3 s (b) v = 29.4 m/s
(c) On the Moon: t = √(2 × 44.1 / 1.62) = √(88.2 / 1.62) = √54.44 ≈ 7.38 s. It takes longer because g on the Moon is much smaller.

✅ Answer: (a) 3 s (b) 29.4 m/s (c) about 7.38 s

Marking: (a) formula ½, answer ½. (b) 1. (c) 1.

📖 Revise: Ch 4 · Free fall · Ch 4 · g on the Moon · Chapter 4 – Gravitation

22.

A metal block of volume 200 cm³ hangs from a spring balance. It is fully immersed in turn in water, kerosene and glycerine (figure). (Density: water 1000 kg/m³, kerosene 810 kg/m³, glycerine 1260 kg/m³; g = 9.8 m/s²; 1 cm³ = 10⁻⁶ m³) (a) Find the buoyant force on the block in water. (b) Find the buoyant force in kerosene and in glycerine. (c) An egg sinks in fresh water but floats in saturated salt solution. Explain using your answers.

[3]

Step-by-step solution

Given: V = 200 cm³ = 200 × 10⁻⁶ m³ = 2 × 10⁻⁴ m³ (fully immersed, so it displaces 2 × 10⁻⁴ m³ of liquid)
Formula: buoyant force = weight of liquid displaced = (density × V) × g (Archimedes’ principle)
Substitution: water: 1000 × 2 × 10⁻⁴ = 0.2 kg → 0.2 × 9.8 ; kerosene: 810 × 2 × 10⁻⁴ = 0.162 kg → 0.162 × 9.8 ; glycerine: 1260 × 2 × 10⁻⁴ = 0.252 kg → 0.252 × 9.8
Answer: (a) water 1.96 N (b) kerosene 1.5876 ≈ 1.59 N, glycerine 2.4696 ≈ 2.47 N
(c) For the same volume, the denser the liquid, the greater the buoyant force. Saturated salt solution is denser than fresh water, so the buoyant force on the egg becomes large enough to equal its weight and it floats; in fresh water the buoyant force is less than its weight, so it sinks.

✅ Answer: (a) 1.96 N (b) kerosene ≈ 1.59 N, glycerine ≈ 2.47 N (c) a denser liquid gives a bigger buoyant force

Marking: (a) 1. (b) ½ + ½. (c) denser liquid → bigger buoyant force ½, equals the egg’s weight in salt solution ½.

📖 Revise: Ch 5 · Density of the liquid · Ch 5 · Archimedes’ principle · Chapter 5 – Buoyant Force

23.

A bulb of resistance 6 Ω is connected in series with a rheostat, an ammeter and a 9 V battery (figure). The rheostat wire has 60 turns, each of resistance 0.2 Ω. (a) What is the largest resistance the rheostat can put into the circuit? (b) Find the ammeter reading when all 60 turns are in the circuit. (c) The sliding contact is moved so that only 15 turns are in the circuit. Find the ammeter reading. What happens to the brightness of the bulb?

[3]

Step-by-step solution

(a) Largest resistance = 60 × 0.2 = 12 Ω.
Given: V = 9 V; bulb 6 Ω in series with the rheostat
Formula: series: R = R₁ + R₂ ; Ohm’s law I = V/R
Substitution: (b) R = 6 + 12 = 18 Ω, I = 9/18 (c) rheostat = 15 × 0.2 = 3 Ω, R = 6 + 3 = 9 Ω, I = 9/9
Answer: (b) I = 0.5 A (c) I = 1 A
In (c) less length of resistance wire is in the circuit, so the resistance is less and the current is more – the bulb glows brighter. A rheostat works because resistance increases with length.

✅ Answer: (a) 12 Ω (b) 0.5 A (c) 1 A; the bulb becomes brighter

Marking: (a) 1. (b) total resistance ½, current ½. (c) current ½, brighter ½.

📖 Revise: Ch 7 · Rheostat · Ch 7 · Resistors in series · Chapter 7 – Current Electricity

24.

(a) Classify these as music or noise: a veena recital, a horn blaring at a traffic junction, the drilling at a construction site, chenda melam at a temple festival. (b) Can music become noise? Explain. (c) Loudspeakers are not allowed in public places before 6 am and after 10 pm, and air horns are banned in vehicles. Why are such rules made?

[3]

Step-by-step solution

(a) Music (സംഗീതം) – regular vibrations, pleasant: veena recital, chenda melam. Noise (ഒച്ച) – irregular vibrations, unpleasant: traffic horn, drilling.
(b) Yes. When music is played too loud (beyond a limit) it becomes unpleasant and harmful to the ears – it is then noise.
(c) These rules reduce noise pollution (ശബ്ദമലിനീകരണം). Loud sound at night spoils sleep and causes headache, stress, high blood pressure and hearing loss; patients, babies, the elderly and students are affected the most.

✅ Answer: (a) music: veena, chenda melam; noise: horn, drilling (b) yes – when it is too loud (c) to reduce noise pollution

Marking: (a) ½ for each correct pair (music / noise): 1. (b) 1. (c) 1.

📖 Revise: Ch 8 · Music, noise and noise pollution · Chapter 8 – Sound

Section DAnswer all questions · 4 × 4 = 16 marks

25.

A KSRTC bus leaves a bus stop and stops at the next bus stop on a straight road. Its velocity–time graph is shown. (a) Find its acceleration from O to A. (b) Find its retardation from B to C. (c) Using the area under the graph, find the distance between the two bus stops.

[4]

Step-by-step solution

Given (from the graph): O (0 s, 0) → A (25 s, 20 m/s) → B (65 s, 20 m/s) → C (85 s, 0)
Formula: a = (v − u)/t ; displacement = area under the v–t graph
Substitution: (a) a = (20 − 0)/25 (b) a = (0 − 20)/(85 − 65) = −20/20
Answer: (a) 0.8 m/s² (b) a = −1 m/s², so retardation = 1 m/s²
(c) Area OA (triangle) = ½ × 25 × 20 = 250 m; area AB (rectangle) = (65 − 25) × 20 = 40 × 20 = 800 m; area BC (triangle) = ½ × 20 × 20 = 200 m.
Answer: distance = 250 + 800 + 200 = 1250 m (1.25 km).

✅ Answer: (a) 0.8 m/s² (b) 1 m/s² (c) 1250 m

Marking: (a) 1. (b) 1. (c) ½ for each of the three areas, ½ for the total 1250 m.

📖 Revise: Ch 2 · Area under the v–t graph · Ch 2 · Acceleration and retardation · Chapter 2 – Equations of Motion

26.

Scientists find a planet X whose mass is 3 times the mass of the Earth and whose radius is 2 times the radius of the Earth. A spacecraft first goes round planet X in a circular orbit and then lands on it. (g on the Earth = 9.8 m/s²) (a) What provides the centripetal force that keeps the spacecraft moving in its circular orbit? (b) Using g = GM/R², find the acceleration due to gravity on planet X. (c) An astronaut of mass 60 kg steps out on planet X. What are her mass and her weight there? (d) She drops a hammer from a height of 14.7 m on planet X. How long does it take to reach the ground?

[4]

Step-by-step solution

(a) The gravitational force of planet X on the spacecraft acts towards the centre of the planet and works as the centripetal force (അഭികേന്ദ്രബലം).
Given: mass of X = 3M, radius of X = 2R, where g = GM/R² = 9.8 m/s² on the Earth
Formula: g_X = G(mass of X)/(radius of X)²
Substitution: g_X = G(3M)/(2R)² = (3/4) × GM/R² = (3/4) × 9.8
Answer: (b) g_X = 7.35 m/s²
(c) Mass is the quantity of matter – it does not change: 60 kg. Weight W = mg_X = 60 × 7.35 = 441 N (on the Earth it would be 60 × 9.8 = 588 N).
(d) Free fall from rest: h = ½ g_X t² ⇒ t = √(2h/g_X) = √(2 × 14.7 / 7.35) = √(29.4 / 7.35) = √4 = 2 s.

✅ Answer: (a) the gravitational force of planet X (b) 7.35 m/s² (c) 60 kg, 441 N (d) 2 s

Marking: (a) 1. (b) substitution ½, answer ½. (c) mass ½, weight ½. (d) formula ½, answer ½.

📖 Revise: Ch 4 · Satellites and free fall · Ch 4 · g = GM/R² · Chapter 4 – Gravitation

27.

The figure shows a small hydroelectric power station. Every second, 4000 kg of water flows from the reservoir P down the penstock Q and falls through a height of 50 m to the turbine R, which turns the generator S. (g = 10 m/s²) (a) How much potential energy does the water lose every second? (b) Find the power available from the falling water, in watt and in megawatt. (c) Write the energy change that takes place in the turbine and in the generator. (d) From where did the water in the reservoir get its potential energy?

[4]

Step-by-step solution

Given: m = 4000 kg every second, h = 50 m, g = 10 m/s², t = 1 s
Formula: E_p = mgh ; P = W/t
Substitution: E_p = 4000 × 10 × 50 = 2 000 000 J ; P = 2 000 000 / 1
Answer: (a) 2 000 000 J (2 × 10⁶ J) every second (b) P = 2 000 000 W = 2 × 10⁶ W = 2 MW (1 MW = 10⁶ W)
(c) Turbine (ടർബൈൻ): kinetic energy of the flowing water → mechanical energy (rotation). Generator: mechanical energy → electrical energy.
(d) From the Sun: the Sun’s heat evaporates water; the vapour rises, falls as rain on the hills and is stored high up behind the dam, so it has potential energy.

✅ Answer: (a) 2 000 000 J (b) 2 000 000 W = 2 MW (c) turbine: KE of water → mechanical; generator: mechanical → electrical (d) from the Sun

Marking: (a) 1. (b) watt ½, megawatt ½. (c) ½ + ½. (d) 1 (Sun ½, how – evaporation and rain ½).

📖 Revise: Ch 6 · Hydroelectric power and power · Ch 6 · Potential energy mgh · Chapter 6 – Work and Energy

28.

In the circuit shown, R₁ = 4 Ω is in series with a parallel combination of R₂ = 6 Ω and R₃ = 12 Ω. The battery gives 12 V. (a) Find the effective resistance of the parallel part. (b) Find the total resistance of the circuit and the ammeter reading. (c) Find the current through R₂ and through R₃. (d) R₃ burns out and its wire breaks. What does the ammeter read now? Does the current through R₂ increase or decrease?

[4]

Step-by-step solution

Given: R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 12 Ω, V = 12 V
Formula: parallel R = R₂R₃/(R₂ + R₃); series R = R₁ + R_p; I = V/R
Substitution: (a) R_p = (6 × 12)/(6 + 12) = 72/18 (b) R = 4 + 4 = 8 Ω, I = 12/8
Answer: (a) R_p = 4 Ω (b) R = 8 Ω, ammeter reads 1.5 A
(c) p.d. across the parallel part = I × R_p = 1.5 × 4 = 6 V. I(R₂) = 6/6 = 1 A; I(R₃) = 6/12 = 0.5 A (1 + 0.5 = 1.5 A ✓).
(d) Now only R₁ and R₂ are in series: R = 4 + 6 = 10 Ω, I = 12/10 = 1.2 A. All of it flows through R₂, so the current through R₂ increases from 1 A to 1.2 A.

✅ Answer: (a) 4 Ω (b) 8 Ω, 1.5 A (c) 1 A and 0.5 A (d) 1.2 A; it increases (from 1 A to 1.2 A)

Marking: (a) 1. (b) total resistance ½, current ½. (c) ½ + ½. (d) 1.2 A ½, increases ½.

📖 Revise: Ch 7 · Series and parallel combinations · Ch 7 · Ohm’s law · Chapter 7 – Current Electricity

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	12	–	25	7
3. Laws of Motion	4, 5	13	20	–	7
4. Gravitation	–	14	21	26	9
5. Buoyant Force	6, 7	15	22	–	7
6. Work and Energy	8	16	–	27	7
7. Current Electricity	–	17	23	28	9
8. Sound	9, 10	18	24	–	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.