Class 9 · PhysicsSet 1 · Answer key

Answer Key – Set 1 ഉത്തരസൂചിക

Step-by-step answers and the marking scheme for every question of Set 1, 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 1

Standard IX · Physics · Annual Examination

Total score: 60 · Time: 2 hours

Section AAnswer all questions · 10 × 1 = 10 marks

The speed of light in a transparent medium is 1.5 × 10⁸ m/s. Its refractive index is (speed of light in vacuum c = 3 × 10⁸ m/s)

(a) 0.5(b) 1.5(c) 2(d) 4.5

[1]

Step-by-step solution

Refractive index (അപവർത്തനാങ്കം) n = c/v.
n = (3 × 10⁸) ÷ (1.5 × 10⁸) = 3 ÷ 1.5 = 2. It is a ratio, so it has no unit.

✅ Answer: (c) 2

Marking: 1 mark for (c).

📖 Revise: Ch 1 · Refractive index n = c/v · Chapter 1 – Refraction of Light

A car moves along a straight, level road at a steady speed of 60 km/h. The resultant force on the car is

(a) in the direction of motion(b) opposite to the direction of motion(c) equal to the weight of the car(d) zero

[1]

Step-by-step solution

The car moves in a straight line with uniform velocity – its speed and direction do not change.
By Newton’s first law, an unbalanced external force is needed only to change the state of motion. So the forces on the car are balanced and the resultant force is zero.

✅ Answer: (d) zero

Marking: 1 mark for (d).

📖 Revise: Ch 3 · Newton’s first law · Chapter 3 – Laws of Motion

Pick the odd one out and give the reason:
(i) dust comes out when a hanging carpet is beaten (ii) passengers fall backwards when a bus starts suddenly (iii) an athlete cannot stop at once at the finishing line (iv) a coin on a card falls into the glass when the card is flicked away

[1]

Step-by-step solution

In (i), (ii) and (iv) a body at rest tends to stay at rest: inertia of rest (നിശ്ചലജഡത്വം).
In (iii) the running athlete tends to keep moving: inertia of motion (ചലനജഡത്വം). So (iii) is the odd one.

✅ Answer: (iii) – it is due to inertia of motion; the others are due to inertia of rest

Marking: ½ mark for (iii), ½ mark for the reason.

📖 Revise: Ch 3 · Inertia of rest and of motion · Chapter 3 – Laws of Motion

A girl of mass 40 kg and her father of mass 80 kg stand 2 m apart. Which statement about the gravitational force between them is correct?

(a) The father pulls the girl with a larger force(b) They pull each other with forces of equal size(c) The girl pulls the father with a larger force(d) Only the father pulls the girl

[1]

Step-by-step solution

The force F = G m₁m₂/d² contains both masses in the same way, so there is only one value of F for the pair.
The father pulls the girl with F and the girl pulls the father with the same F, in the opposite direction (as in Newton’s third law).

✅ Answer: (b) They pull each other with forces of equal size.

Marking: 1 mark for (b).

📖 Revise: Ch 4 · Universal law of gravitation · Chapter 4 – Gravitation

The value of the gravitational constant G was first found by experiment by

(a) Henry Cavendish(b) Galileo Galilei(c) Isaac Newton(d) Albert Einstein

[1]

Step-by-step solution

Newton gave the law, but the value of G (ഗുരുത്വാകർഷണസ്ഥിരാങ്കം) = 6.67 × 10⁻¹¹ N m²/kg² was first measured by Henry Cavendish in 1798.

✅ Answer: (a) Henry Cavendish

Marking: 1 mark for (a).

📖 Revise: Ch 4 · G and Cavendish · Chapter 4 – Gravitation

A wooden block and an iron block of the same volume are held fully under water. The buoyant force is

(a) greater on the iron block(b) greater on the wooden block(c) the same on both(d) zero on the iron block

[1]

Step-by-step solution

Both blocks are fully immersed and have the same volume, so they displace the same volume (and the same weight) of water.
By Archimedes’ principle (ആർക്കിമിഡീസ് തത്വം) the buoyant force equals the weight of water displaced, so it is the same on both. The material of the block does not matter.

✅ Answer: (c) the same on both

Marking: 1 mark for (c).

📖 Revise: Ch 5 · Archimedes’ principle · Chapter 5 – Buoyant Force

A box slides along a rough floor and slowly comes to rest. The work done on the box by the force of friction is . (positive / negative / zero)

[1]

Step-by-step solution

Friction acts opposite to the direction in which the box moves.
When the displacement is opposite to the force, the work done is negative (നെഗറ്റീവ് പ്രവൃത്തി).

✅ Answer: negative

Marking: 1 mark.

📖 Revise: Ch 6 · Positive and negative work · Chapter 6 – Work and Energy

A 20 kg box is raised to a platform 2 m high, once by lifting it straight up and once by pushing it up a smooth ramp 5 m long. The work done against gravity (g = 10 m/s²) is

(a) 400 J when lifted, 1000 J along the ramp(b) 400 J in both cases(c) 1000 J in both cases(d) 200 J in both cases

[1]

Step-by-step solution

Work against gravity W = mgh = 20 × 10 × 2 = 400 J.
It depends only on the height raised (2 m), not on the path, so it is 400 J both ways. (Not 5 m – the ramp length is not the height.)

✅ Answer: (b) 400 J in both cases

Marking: 1 mark for (b).

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

A tuning fork is struck on a rubber pad and its prongs are then held tightly in the hand. What happens?

(a) The sound becomes louder(b) The sound continues as before(c) The sound stops, because the prongs stop vibrating(d) The prongs start vibrating faster

[1]

Step-by-step solution

Sound is produced by vibration (കമ്പനം). Holding the prongs tightly stops their vibration.
With no vibration there is no sound.

✅ Answer: (c) The sound stops, because the prongs stop vibrating.

Marking: 1 mark for (c).

📖 Revise: Ch 8 · Tuning fork activity · Chapter 8 – Sound

10.

Pick the odd one out and give the reason: odakkuzhal (flute), bugle, conch (shankhu), violin

[1]

Step-by-step solution

In the odakkuzhal, the bugle and the conch the main part that vibrates is the air column inside.
In the violin the main vibrating part is the strings. So the violin is the odd one.

✅ Answer: violin – its main vibrating part is the strings; in the others it is the air column inside

Marking: ½ mark for violin, ½ mark for the reason.

📖 Revise: Ch 8 · Main vibrating parts of instruments · Chapter 8 – Sound

Section BAnswer all questions · 8 × 2 = 16 marks

11.

The refractive index of a type of glass is 1.6. (a) Find the speed of light in this glass. (Speed of light in vacuum c = 3 × 10⁸ m/s) (b) The refractive index of water is 1.33. In which of the two, this glass or water, does light travel faster?

[2]

Step-by-step solution

Given: n = 1.6, c = 3 × 10⁸ m/s
Formula: n = c/v, so v = c/n
Substitution: v = (3 × 10⁸) ÷ 1.6
Answer: v = 1.875 × 10⁸ m/s
(b) The smaller the refractive index, the greater the speed of light (n = c/v). Water (1.33) has a smaller refractive index than this glass (1.6), so light travels faster in water.

✅ Answer: (a) 1.875 × 10⁸ m/s (b) in water

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

📖 Revise: Ch 1 · Refractive index n = c/v · Chapter 1 – Refraction of Light

12.

The velocity–time graphs of three vehicles P, Q and R are shown. (a) Which vehicle moves with uniform velocity? (b) Which one has uniform acceleration? (c) Which one has uniform retardation? (d) What does the area under a velocity–time graph give?

[2]

Step-by-step solution

(a) P – its graph is a horizontal line, so its velocity does not change.
(b) Q – a straight line rising from the origin: the velocity increases by equal amounts in equal times.
(c) R – a straight line falling to zero: the velocity decreases by equal amounts in equal times (uniform retardation, മന്ദീകരണം).
(d) The area under a velocity–time graph (പ്രവേഗ-സമയ ഗ്രാഫ്) gives the displacement in that time interval.

✅ Answer: (a) P (b) Q (c) R (d) the displacement

Marking: ½ mark for each part.

📖 Revise: Ch 2 · Velocity–time graphs · Chapter 2 – Equations of Motion

13.

Appu pushes a heavy box towards the east with a force of 120 N. At the same time Binu and Chinnu push it towards the west with forces of 50 N and 40 N (figure). Ignore friction. (a) Find the resultant force on the box and its direction. (b) Are the forces on the box balanced or unbalanced? What happens to the box?

[2]

Step-by-step solution

Given: 120 N east; 50 N and 40 N west. Take east as positive, so the westward forces are −50 N and −40 N.
Formula: resultant force (സഫലബലം) = sum of the forces with their signs
Substitution: 120 + (−50) + (−40) = 120 − 90
Answer: resultant = +30 N, i.e. 30 N towards the east.
(b) The resultant is not zero, so the forces are unbalanced. The box starts to move (accelerate) towards the east, the direction of the resultant force.

✅ Answer: (a) 30 N towards the east (b) unbalanced; the box starts moving towards the east

Marking: (a) working ½, 30 N with direction ½; (b) unbalanced ½, box moves east ½.

📖 Revise: Ch 3 · Resultant force · Chapter 3 – Laws of Motion

14.

Two lead spheres of masses 100 kg and 200 kg are placed with their centres 0.5 m apart. Find the gravitational force between them. (G = 6.67 × 10⁻¹¹ N m²/kg²)

[2]

Step-by-step solution

Given: m₁ = 100 kg, m₂ = 200 kg, d = 0.5 m, G = 6.67 × 10⁻¹¹ N m²/kg²
Formula: F = G m₁m₂/d²
Substitution: F = 6.67 × 10⁻¹¹ × (100 × 200) ÷ 0.5² = 6.67 × 10⁻¹¹ × 20 000 ÷ 0.25 = 6.67 × 10⁻¹¹ × 80 000
= 6.67 × 10⁻¹¹ × 8 × 10⁴ = 53.36 × 10⁻⁷
Answer: F = 5.336 × 10⁻⁶ N (a very small force).

✅ Answer: F = 5.336 × 10⁻⁶ N

Marking: Formula ½, substitution ½, answer with unit 1.

📖 Revise: Ch 4 · F = Gm₁m₂/d² · Chapter 4 – Gravitation

15.

An object hangs from a spring balance. The balance reads 6 N when the object is in air and 5 N when exactly half of the object is under water. (a) Find the buoyant force on the object when it is half immersed. (b) What will the balance read when the object is fully immersed in water? Give a reason.

[2]

Step-by-step solution

Given: weight in air W₁ = 6 N; reading when half immersed W₂ = 5 N
Formula: buoyant force (പ്ലവക്ഷമബലം) = apparent loss of weight = W₁ − W₂
Substitution: 6 − 5
Answer: (a) buoyant force = 1 N
(b) When fully immersed, the object displaces twice the volume of water it displaced when half immersed. By Archimedes’ principle the buoyant force also doubles: 2 × 1 = 2 N.
Reading = 6 − 2 = 4 N.

✅ Answer: (a) 1 N (b) 4 N

Marking: (a) ½ for W₁ − W₂, ½ for 1 N; (b) reason (twice the volume displaced) ½, 4 N ½.

📖 Revise: Ch 5 · Measuring buoyant force · Ch 5 · Archimedes’ principle · Chapter 5 – Buoyant Force

16.

State whether the work done by the force named is positive, negative or zero. Give a reason in each case.
(a) the force of gravity on a coconut falling from a tree
(b) the braking force on a moving bicycle
(c) the upward force of a boy’s hand on a school bag he carries while walking along a level corridor
(d) the pull of a crane’s cable on a load it lifts

[2]

Step-by-step solution

(a) Positive – gravity acts downwards and the coconut moves downwards, in the direction of the force.
(b) Negative – the braking force acts backwards while the bicycle moves forwards, opposite to the force.
(c) Zero – the hand’s force is upwards but the bag moves horizontally; there is no displacement in the direction of the force.
(d) Positive – the cable pulls upwards and the load moves upwards.

✅ Answer: (a) positive (b) negative (c) zero (d) positive

Marking: ½ mark for each part (answer with reason).

📖 Revise: Ch 6 · Positive and negative work · Ch 6 · When is work done? · Chapter 6 – Work and Energy

17.

Four 1.5 V cells are joined in series and connected to a resistor. An ammeter in the circuit reads 0.3 A. (a) What is the emf of this battery? (b) Find the resistance of the resistor. (Take the whole emf to be across the resistor.)

[2]

Step-by-step solution

(a) In series (ശ്രേണീരീതി) the emfs add: 4 × 1.5 V = 6 V.
Given: V = 6 V, I = 0.3 A
Formula: Ohm’s law V = IR, so R = V/I
Substitution: R = 6 ÷ 0.3
Answer: R = 20 Ω

✅ Answer: (a) 6 V (b) 20 Ω

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

📖 Revise: Ch 7 · Cells in series · Ch 7 · Ohm’s law · Chapter 7 – Current Electricity

18.

Place your fingers lightly on your throat and speak loudly. (a) What do you feel? Name the part whose vibration produces our voice, and say where it is. (b) Name two other parts that also vibrate when we speak.

[2]

Step-by-step solution

(a) We feel the throat vibrating.
Our voice is produced when the vocal cords (സ്വനതന്തുക്കൾ) vibrate. They are in the larynx (voice box, സ്വനപേടകം) at the top of the windpipe.
(b) The vocal cords are the main part, but associated parts vibrate too: the air in the throat, mouth and nose, the muscles of the throat and the windpipe. (Any two.)

✅ Answer: (a) vibration; the vocal cords in the larynx (voice box) at the top of the windpipe (b) e.g. the air in the throat, mouth and nose; the muscles of the throat

Marking: (a) feeling vibration ½, vocal cords in the larynx ½; (b) ½ mark each for two associated parts.

📖 Revise: Ch 8 · Our voice · Chapter 8 – Sound

Section CAnswer all questions · 6 × 3 = 18 marks

19.

The velocity of a car on a straight stretch of the National Highway was noted every 2 seconds:

Time t (s)	0	2	4	6	8	10
Velocity v (m/s)	4	8	12	16	16	16

(a) Draw the velocity–time graph (X-axis: 1 cm = 2 s; Y-axis: 1 cm = 4 m/s). (b) Find the acceleration of the car during the first 6 s. (c) Find the total displacement of the car in the 10 s.

[3]

Step-by-step solution

(a) Take time on the X-axis and velocity on the Y-axis with the given scales. Plot the six points and join them: a sloping straight line from (0 s, 4 m/s) to (6 s, 16 m/s), then a horizontal line at 16 m/s up to 10 s.
(b) Given: u = 4 m/s, v = 16 m/s, t = 6 s
Formula: a = (v − u)/t
Substitution: a = (16 − 4) ÷ 6 = 12 ÷ 6
Answer: a = 2 m/s²
(c) Displacement = area under the graph. From 0 to 6 s (trapezium): ½ × (4 + 16) × 6 = 60 m. (Check: s = ut + ½at² = 4 × 6 + ½ × 2 × 6² = 24 + 36 = 60 m.)
From 6 s to 10 s (rectangle): 16 × 4 = 64 m.
Answer: total displacement = 60 + 64 = 124 m

✅ Answer: (b) 2 m/s² (c) 124 m

Marking: (a) axes with scale ½, correct points and lines ½; (b) formula ½, answer ½; (c) the two areas ½, total 124 m ½.

📖 Revise: Ch 2 · Velocity–time graph and area · Ch 2 · Equations of motion · Chapter 2 – Equations of Motion

20.

The figure shows Galileo’s experiment. A ball is released from A on one side of a U-shaped channel. The other side is made less and less steep (positions 1, 2 and 3). (a) Up to about what height does the ball rise on the other side? (b) What happens to the distance the ball rolls on the other side as the slope is reduced? (c) In position 3 the ball still stops after some time. Why? What would happen if there were nothing to stop it? (d) What conclusion did Galileo draw about the force needed to keep a body moving? Which law of Newton is based on it?

[3]

Step-by-step solution

(a) The ball rises to almost the same height as A – it tries to reach the height from which it was released.
(b) As the slope is reduced, the ball rolls a greater distance before stopping.
(c) It stops because of friction (and air resistance). Without friction it would keep moving in a straight line at the same speed for ever.
(d) To stop a moving body an unbalanced external force must act against its motion; no unbalanced external force is needed to keep a body moving uniformly in a straight line. Newton built his first law of motion on this.

✅ Answer: (a) almost the height of A (b) it rolls farther (c) friction; it would go on for ever at the same speed in a straight line (d) no unbalanced force is needed to keep a body in uniform motion; Newton’s first law

Marking: (a) ½; (b) ½; (c) friction ½, moves for ever ½; (d) conclusion ½, first law ½.

📖 Revise: Ch 3 · Galileo’s observations · Chapter 3 – Laws of Motion

21.

Two bodies attract each other with a gravitational force F. Complete the table, showing your working.

Case	Mass of first body	Mass of second body	Distance between them	Force
Given	m₁	m₂	d	F
(a)	2m₁	3m₂	d	?
(b)	m₁	m₂	3d	?
(c)	2m₁	2m₂	4d	?

[3]

Step-by-step solution

Use F = G m₁m₂/d²: the force is directly proportional to the product of the masses and inversely proportional to the square of the distance.
(a) Product of masses = 2m₁ × 3m₂ = 6 m₁m₂; distance unchanged → force = 6F.
(b) Distance 3d → d² becomes 9d²; masses unchanged → force = F/9.
(c) Product of masses = 4 m₁m₂ (×4); distance 4d → d² becomes 16d² (÷16) → force = 4F/16 = F/4.

✅ Answer: (a) 6F (b) F/9 (c) F/4

Marking: 1 mark for each row (½ working, ½ answer).

📖 Revise: Ch 4 · Table 4.2 – changing masses and distance · Chapter 4 – Gravitation

22.

At Ernakulam railway station, Rahul lifts his 12 kg suitcase straight up from the platform to a height of 1.5 m. Then, holding it at that height, he walks 8 m along the level platform. (g = 10 m/s²) (a) Find the work he does against gravity while lifting the suitcase. (b) Find the work done by the force of gravity on the suitcase during the lift. Is it positive or negative? (c) How much work does he do against gravity while walking the 8 m? Why?

[3]

Step-by-step solution

Given: m = 12 kg, h = 1.5 m, g = 10 m/s²
Formula: work against gravity W = mgh
Substitution: W = 12 × 10 × 1.5
Answer: (a) W = 180 J
(b) Gravity (120 N) acts downwards while the suitcase moves 1.5 m upwards – opposite to the force. Work by gravity = −120 × 1.5 = −180 J (negative work).
(c) While walking, the suitcase stays at the same height: there is no displacement in the direction of his upward force. So the work done against gravity is zero.

✅ Answer: (a) 180 J (b) −180 J (negative) (c) zero

Marking: (a) formula ½, answer with unit ½; (b) −180 J ½, negative with reason ½; (c) zero ½, reason ½.

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

23.

The graph shows how the potential difference V across two resistors A and B changes with the current I through them, at constant temperature. (a) What does the straight line through the origin show? (b) Find the resistances of A and B. (c) A and B are connected in series to a 6 V battery. Find the current in the circuit.

[3]

Step-by-step solution

(a) V is directly proportional to I – the resistors obey Ohm’s law (ഓം നിയമം); V/I is constant.
(b) Formula: R = V/I. From the graph at I = 1.00 A: A has V = 8 V and B has V = 4 V.
Answer: R_A = 8 ÷ 1 = 8 Ω; R_B = 4 ÷ 1 = 4 Ω
(c) Given: V = 6 V; in series R = R_A + R_B = 8 + 4 = 12 Ω
Formula: I = V/R
Substitution: I = 6 ÷ 12
Answer: I = 0.5 A

✅ Answer: (a) V ∝ I (Ohm’s law) (b) R_A = 8 Ω, R_B = 4 Ω (c) 0.5 A

Marking: (a) ½; (b) ½ each for R_A and R_B; (c) effective resistance ½, current with unit 1.

📖 Revise: Ch 7 · Ohm’s law and V–I graph · Ch 7 · Resistors in series · Chapter 7 – Current Electricity

24.

A rubber band is stretched tightly over a pencil box, resting on two pencils, and plucked in the middle (figure). (a) What do you hear and see? What kind of motion does the band have? (b) What conclusion about the production of sound do you get? What are such objects called? (c) A veena works in a similar way. Name its main vibrating part and two associated parts that vibrate.

[3]

Step-by-step solution

(a) We hear a sound and see the band moving rapidly to and fro. This to-and-fro motion is vibration (കമ്പനം), not straight-line motion.
(b) Sound is produced by the vibration of objects. Objects that produce sound are called sources of sound.
(c) In a veena the main vibrating part is the strings. The associated parts that vibrate are the wooden body (resonator) and the air inside it; they make the sound louder and sweeter.

✅ Answer: (a) a sound; the band moves to and fro rapidly – vibration (b) sound is produced by vibration; sources of sound (c) the strings; the wooden body and the air inside it

Marking: (a) sound and to-and-fro movement ½, vibration ½; (b) conclusion ½, sources of sound ½; (c) strings ½, two associated parts ½.

📖 Revise: Ch 8 · How sound is produced · Chapter 8 – Sound

Section DAnswer all questions · 4 × 4 = 16 marks

25.

In the physics lab, Neha places a rectangular glass slab PQRS on a sheet of paper and sends a ray of light AB from a ray box on to its face PS (figure). The speed of light in this glass is 2 × 10⁸ m/s. (a) Find the refractive index of the glass. (c = 3 × 10⁸ m/s) (b) Copy and complete the diagram: show the path of the ray inside the slab and after it comes out, with the normals. (c) At which face does the ray bend away from the normal? Why? (d) How is the ray that comes out related to the ray AB? What is meant by lateral shift? Mark it on your diagram.

[4]

Step-by-step solution

(a) Given: c = 3 × 10⁸ m/s, v = 2 × 10⁸ m/s
Formula: n = c/v Substitution: n = (3 × 10⁸) ÷ (2 × 10⁸) Answer: n = 1.5 (no unit)
(b) At B the ray goes from air into glass (rarer to denser), so it bends towards the normal: the angle r inside is less than 50° (about 31° for this glass). It meets the opposite face QR at C. There it goes from glass into air and bends away from the normal, coming out along CD at 50° to the normal.
(c) At the face QR (point C), because the ray passes from glass to air – from an optically denser to a rarer medium.
(d) The emergent ray CD is parallel to the incident ray AB but shifted sideways. The perpendicular distance between the incident ray produced (dashed) and the emergent ray is the lateral shift.

✅ Answer: (a) 1.5 (b) bends towards the normal at B, away from the normal at the second face QR (c) at QR – glass to air is denser to rarer (d) parallel to AB; the sideways (perpendicular) distance between AB produced and the emergent ray

Marking: (a) formula ½, answer ½; (b) ray bending towards the normal inside ½, emergent ray bending away and parallel to AB ½, normals drawn ½; (c) ½; (d) parallel ½, lateral shift explained and marked ½.

📖 Revise: Ch 1 · Refraction through a glass slab · Ch 1 · Towards and away from the normal · Chapter 1 – Refraction of Light

26.

An aeroplane at Cochin International Airport starts from rest and accelerates uniformly along the runway. It must reach a velocity of 288 km/h to take off, and it covers 2000 m of the runway before it lifts off. (a) Express the take-off velocity in m/s. (b) Find the acceleration of the aeroplane. (c) How long does it run on the runway before taking off? (d) Find its velocity and the distance it has covered 25 s after starting.

[4]

Step-by-step solution

(a) 1 km/h = 5/18 m/s, so 288 km/h = 288 × 5/18 = 80 m/s.
(b) Given: u = 0, v = 80 m/s, s = 2000 m. Formula: v² = u² + 2as. Substitution: 80² = 0 + 2 × a × 2000 → 6400 = 4000a. Answer: a = 6400 ÷ 4000 = 1.6 m/s²
(c) Formula: v = u + at (ചലനസമവാക്യങ്ങൾ). Substitution: 80 = 0 + 1.6 × t → t = 80 ÷ 1.6. Answer: t = 50 s
(Check: s = ut + ½at² = ½ × 1.6 × 50² = 0.8 × 2500 = 2000 m ✓)
(d) Given: u = 0, a = 1.6 m/s², t = 25 s. Velocity v = u + at = 0 + 1.6 × 25 = 40 m/s.
Distance s = ut + ½at² = 0 + ½ × 1.6 × 25² = 0.8 × 625 = 500 m. (In half the time it covers only a quarter of the 2000 m, because s grows with t².)

✅ Answer: (a) 80 m/s (b) 1.6 m/s² (c) 50 s (d) 40 m/s; 500 m

Marking: (a) 1; (b) formula ½, answer with unit ½; (c) ½ substitution, ½ answer; (d) velocity ½, distance ½.

📖 Revise: Ch 2 · Equations of motion · Ch 2 · km/h to m/s · Chapter 2 – Equations of Motion

27.

In the school lab, Fathima hangs a brass block from a spring balance. In air the balance reads 3.92 N. She lowers the block completely into an overflow jar filled with water up to its spout, and collects the water that overflows in a beaker (figure). The water collected has a mass of 50 g. (g = 9.8 m/s²; density of water = 1000 kg/m³, i.e. 1 g/cm³) (a) Find the weight of the water collected. (b) Using Archimedes’ principle, find the buoyant force on the block and the reading of the spring balance now. (c) Find the volume of the block and its density. (d) She lowers the block a little deeper, still fully under water and not touching the jar. Will the reading change? Why?

[4]

Step-by-step solution

(a) Given: mass of water m = 50 g = 0.05 kg, g = 9.8 m/s². Formula: W = mg. Substitution: W = 0.05 × 9.8. Answer: 0.49 N
(b) Archimedes’ principle (ആർക്കിമിഡീസ് തത്വം): buoyant force = weight of water displaced = 0.49 N.
Reading = weight in air − buoyant force = 3.92 − 0.49 = 3.43 N
(c) Volume of the block = volume of water displaced = 50 g ÷ 1 g/cm³ = 50 cm³ = 5 × 10⁻⁵ m³.
Mass of the block = 3.92 ÷ 9.8 = 0.4 kg. Formula: density = mass/volume. Substitution: 0.4 ÷ (5 × 10⁻⁵). Answer: density = 8000 kg/m³ (8 g/cm³).
(d) No. Once the block is fully immersed, going deeper does not change the volume of water it displaces, so the buoyant force (0.49 N) and the reading (3.43 N) stay the same.

✅ Answer: (a) 0.49 N (b) 0.49 N; 3.43 N (c) 50 cm³; 8000 kg/m³ (d) no – the volume of water displaced does not change

Marking: (a) ½ for W = mg, ½ for 0.49 N; (b) buoyant force ½, reading ½; (c) volume ½, density ½; (d) no ½, reason ½.

📖 Revise: Ch 5 · Overflow jar and Archimedes’ principle · Ch 5 · Measuring buoyant force · Chapter 5 – Buoyant Force

28.

Anu has three 1.5 V cells joined in series, a switch and two identical bulbs L₁ and L₂, each of resistance 9 Ω (take it as constant). She makes circuit (a) with the bulbs in series and circuit (b) with the bulbs in parallel (figure). (a) What is the emf of the battery? (b) For circuit (a), find the effective resistance and the current from the battery. (c) For circuit (b), find the effective resistance, the current from the battery and the current through each bulb. (d) In which circuit do the bulbs glow brighter? If L₁ fuses, what happens to L₂ in each circuit?

[4]

Step-by-step solution

(a) Cells in series: emf (വിദ്യുത്ചാലകബലം) = 3 × 1.5 = 4.5 V.
(b) Series: R = R₁ + R₂ = 9 + 9 = 18 Ω (സഫലപ്രതിരോധം). I = V/R = 4.5 ÷ 18 = 0.25 A (the same current flows through both bulbs).
(c) Parallel: R = R₁R₂/(R₁ + R₂) = (9 × 9) ÷ (9 + 9) = 81 ÷ 18 = 4.5 Ω. I = V/R = 4.5 ÷ 4.5 = 1 A.
Each bulb gets the full 4.5 V, so the current through each = 4.5 ÷ 9 = 0.5 A (0.5 + 0.5 = 1 A ✓).
(d) The bulbs glow brighter in circuit (b): each bulb gets the full 4.5 V and 0.5 A, while in series each gets only 2.25 V and 0.25 A.
If L₁ fuses: in (a) the only path is broken, so L₂ also goes off; in (b) L₂ has its own path, so it keeps glowing.

✅ Answer: (a) 4.5 V (b) 18 Ω, 0.25 A (c) 4.5 Ω, 1 A, 0.5 A each (d) circuit (b); in (a) L₂ goes off, in (b) L₂ keeps glowing

Marking: (a) ½; (b) effective resistance ½, current ½; (c) effective resistance ½, total current ½, current in each bulb ½; (d) brighter circuit ½, what happens to L₂ in both circuits ½.

📖 Revise: Ch 7 · Resistors in series and parallel · Ch 7 · Cells in series · Chapter 7 – Current Electricity

Question numbers by chapter

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