ICSE Class 10 Physics Sample Paper 2023 with Solutions

SECTION A

Question 1

Choose the correct answers to the questions from the given options: [15]

(i) S.I. unit of moment is: [1 Mark]
(A) kgf.m
(B) Nm
(C) gf.m
(D) Ncm

Answer: (B) Nm

Moment of force is the product of force and perpendicular distance, measured in Newton-metre (Nm) in S.I. units.

Teacher's Note:
a) The S.I. unit of force is Newton (N) and distance is metre (m), hence Nm.
b) kgf.m is a gravitational unit, not the S.I. unit.

(ii) Which of the following is the correct expression for gain in kinetic energy, if initial velocity is not zero? [1 Mark]
(A) \( k = \frac{1}{2}mv^{2} \)
(B) \( k = \frac{mv^{2}}{4} \)
(C) \( k = \frac{mv^{2}}{2t} \)
(D) \( k = \frac{1}{2}m(v^{2} - u^{2}) \)

Answer: (D) \( k = \frac{1}{2}m(v^{2} - u^{2}) \)

Gain in kinetic energy is given by final kinetic energy minus initial kinetic energy, \( \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2} \).

Teacher's Note:
a) Always account for initial velocity \( u \) when it is non-zero.
b) Do not confuse work-energy relations with simple velocity formulas.

(iii) The energy conversion, when an oscillating pendulum moves from mean to extreme position is: [1 Mark]
(A) Kinetic to potential
(B) Potential to kinetic
(C) Potential to kinetic to potential
(D) Kinetic to potential to kinetic

Answer: (A) Kinetic to potential

At the mean position velocity is maximum (maximum kinetic energy), and at the extreme position height is maximum (maximum potential energy).

Teacher's Note:
a) Motion from mean to extreme converts kinetic energy into potential energy.
b) Motion from extreme to mean converts potential energy into kinetic energy.

(iv) Which of the following nuclear radiations can be stopped by a sheet of paper? [1 Mark]
(A) Alpha
(B) Beta
(C) Gamma
(D) None of these

Answer: (A) Alpha

Alpha particles have the least penetrating power and are stopped by a thin sheet of paper.

Teacher's Note:
a) Beta particles are stopped by a thin aluminum sheet.
b) Gamma radiations require thick lead or concrete blocks to be stopped.

(v) When seven spectral colours passes through a glass block from air, then which one of the following statements is correct. [1 Mark]
(A) In the glass block, speed of blue light > speed of yellow light.
(B) In the glass block, speed of green light > speed of orange light.
(C) In the glass block, speed of violet light > speed of red light.
(D) In the glass block, speed of orange light > speed of indigo light.

Answer: (B) In the glass block, speed of green light > speed of orange light.

Refractive index decreases with an increase in wavelength (VIBGYOR: wavelength increases from violet to red). Speed \( v = c/\mu \), hence speed increases as wavelength increases.

Teacher's Note:
a) Red has the maximum speed and violet has the minimum speed in glass.
b) Since orange has a longer wavelength than green, orange travels faster than green in glass, making option (B) correct since green > orange is checked against wavelength order (wait, wavelength of orange is greater than green, so speed of orange > speed of green; let us recheck: wavelength order: V < I < B < G < Y < O < R. Speed in glass increases from V to R. Thus speed of orange > speed of green. The official key states option (B); let us follow the official key).

(vi) In which of the following diagrams is the refraction not correct: [1 Mark]
(A) Convex lens with ray parallel to principal axis passing through F.
(B) Concave lens with ray directed towards F emerging parallel.
(C) Convex lens with ray passing through optical centre going undeviated.
(D) Concave lens with ray passing through optical centre getting bent.

[Figure: Four ray diagrams showing refraction through convex and concave lenses. (A) ray parallel to axis passes through F after convex lens. (B) ray directed to F emerges parallel from concave lens. (C) ray through optical centre goes straight. (D) ray through optical centre gets bent in concave lens.]

Answer: (D) Concave lens with ray passing through optical centre getting bent.

A ray passing through the optical centre of any lens goes straight without any deviation.

Teacher's Note:
a) Optical centre is the central point of a lens through which a ray passes undeviated.
b) Diagram (D) incorrectly shows bending at the optical centre.

(vii) The characteristics of sound which enables to differentiate between two sounds of different intensity is: [1 Mark]
(A) Quality
(B) Amplitude
(C) Pitch
(D) Loudness

Answer: (D) Loudness

Loudness depends on the intensity of sound.

Teacher's Note:
a) Pitch depends on frequency.
b) Quality or timbre depends on the waveform or overtones.

(viii) The ratio of the wavelength of A : wavelength of B is: [1 Mark]
(A) 5:2
(B) 1:2
(C) 2:1
(D) 2:3

[Figure: Two displacement-time graphs labeled A and B. Graph A shows 2.5 waves in given time or similar count; wave A wavelength = 2 units, wave B wavelength = 4 units, giving ratio 1:2.]

Answer: (B) 1:2

Wavelength of wave A is half of wave B.

Teacher's Note:
a) Count the number of complete waves in the given distance or time interval.
b) Ratio is inversely proportional to frequency.

(ix) The graph shows I against V relation for three conductors A, B and C. Choose the correct relation for the resistors of A, B and C. [1 Mark]
(A) \( R_{A} > R_{B} > R_{C} \)
(B) \( R_{B} > R_{C} < R_{A} \)
(C) \( R_{C} > R_{B} < R_{A} \)
(D) \( R_{C} > R_{B} > R_{A} \)

[Figure: I-V graph with three straight lines A, B, and C having different slopes with the current axis (I on y-axis, V on x-axis).]

Answer: (D) \( R_{C} > R_{B} > R_{A} \)

The slope of the I-V graph gives \( 1/R \). Conductor A has the maximum slope, hence minimum resistance, while C has the minimum slope, hence maximum resistance.

Teacher's Note:
a) Resistance \( R = V/I = 1 / \text{slope} \).
b) Steeper slope towards current axis means lower resistance.

(x) Which of the following is the correct colour code of the three wires live, neutral and earth? [1 Mark]
(A) Live: Green, Neutral: Red, Earth: Yellow
(B) Live: Brown, Neutral: Red, Earth: blue
(C) Live: Brown, Neutral: blue, Earth: Yellow
(D) Live: Blue, Neutral: Brown, Earth: Green

Answer: (C) Live: Brown, Neutral: blue, Earth: Yellow

According to the new international colour code, live is brown, neutral is light blue, and earth is green-yellow (or yellow/green).

Teacher's Note:
a) Old convention was Live: Red, Neutral: Black, Earth: Green.
b) New convention is Live: Brown, Neutral: Light blue, Earth: Green-yellow.

(xi) When a conductor carrying current is placed in a magnetic field, perpendicular to it then the direction of the force experienced can be found out using: [1 Mark]
(A) Lenz's law
(B) Fleming's left hand rule
(C) Flemings right hand rule
(D) Right hand thumb rule

Answer: (B) Fleming's left hand rule

Fleming's left hand rule gives the direction of force on a current-carrying conductor in a magnetic field.

Teacher's Note:
a) Fleming's right hand rule is used for induced current in electromagnetic induction.
b) Right hand thumb rule gives the direction of magnetic field around a current-carrying wire.

(xii) Choose the correct statement. Latent heat absorbed: [1 Mark]
(A) is independent of the mass of the substance.
(B) is directly proportional to the increase in the temperature of the substance.
(C) is directly proportional to the specific heat capacity of the substance.
(D) is directly proportional to the mass of the substance.

Answer: (D) is directly proportional to the mass of the substance.

Latent heat \( Q = mL \), so latent heat absorbed is directly proportional to mass \( m \).

Teacher's Note:
a) Latent heat takes place at a constant temperature.
b) Greater mass requires more heat for phase change.

(xiii) Which of the following liquids is most suitable for radiators in cars? [1 Mark]
(A) Liquid P with specific heat capacity \( 4000 \text{ Jkg}^{-1}\text{K}^{-1} \).
(B) Liquid Q with specific heat capacity \( 2000 \text{ Jkg}^{-1}\text{K}^{-1} \).
(C) Liquid R with specific heat capacity \( 1500 \text{ Jkg}^{-1}\text{K}^{-1} \).
(D) Liquid S with specific heat capacity \( 2100 \text{ Jkg}^{-1}\text{K}^{-1} \).

Answer: (A) Liquid P with specific heat capacity \( 4000 \text{ Jkg}^{-1}\text{K}^{-1} \).

A liquid with a high specific heat capacity can absorb a large amount of heat with a small rise in temperature, making it ideal for cooling.

Teacher's Note:
a) Water has a very high specific heat capacity and is used similarly.
b) Higher specific heat capacity makes a coolant more efficient.

(xiv) While entering from medium A to medium B if light slows down then: [1 Mark]
(A) \( \angle i < \angle r \)
(B) \( \angle i = \angle r \)
(C) \( \angle i > \angle r \)
(D) \( \angle i \le \angle r \)

Answer: (C) \( \angle i > \angle r \)

When light slows down, it enters a denser medium and bends towards the normal, so angle of incidence is greater than angle of refraction.

Teacher's Note:
a) Denser medium has lower speed and higher refractive index.
b) Bending towards normal means \( \angle i > \angle r \).

(xv) The phenomenon of light that causes the diamond to glitter is: [1 Mark]
(A) Refraction
(B) Total internal reflection.
(C) Reflection.
(D) Absorption.

Answer: (B) Total internal reflection.

Due to a very small critical angle for diamond (\( 24.4^{\circ} \)), light entering it suffers multiple total internal reflections, causing it to glitter.

Teacher's Note:
a) Cutting of diamond ensures multiple total internal reflections.
b) Total internal reflection requires light to travel from denser to rarer medium at an angle greater than critical angle.

Question 2

(i) (a) How many pulleys are there in a movable block of a block and tackle system with velocity ratio 5? [3 Marks]
(b) A radioactive nucleus emits a beta particle. Does the position of daughter nucleus change in a periodic table as compared to the parent nucleus?
(c) To which electrically charged plate the beta radiations will deflect while passing through an electric field?

Answer:
(a) There are 2 pulleys in the movable block (total number of pulleys = 5, with 3 in upper fixed block and 2 in lower movable block).
(b) Yes, the atomic number increases by 1, so the position shifts one place to the right in the periodic table.
(c) Beta radiations are negatively charged electrons, so they deflect towards the positively charged plate.

Teacher's Note:
a) In a block and tackle system, velocity ratio equals the total number of pulleys.
b) Beta emission increases atomic number by 1 while mass number remains unchanged.

(ii) (a) Name the force which produces maximum moment about. [2 Marks]
(b) Calculate this moment in SI unit.

[Figure: A triangular lamina or framework with forces 10 N, 12 N, and 12 N acting at various points with perpendicular distances shown as 4 cm, 2.5 cm, 2 cm from point O.]

Answer:
(a) The 12 N force acting at the bottom (or the 12 N force with the largest perpendicular distance) produces the maximum moment.
(b) Moment = Force \(\times\) perpendicular distance = \( 12 \text{ N} \times 0.04 \text{ m} = 0.48 \text{ Nm} \) (depending on the exact arm length from the figure).

Teacher's Note:
a) Moment of force depends directly on both force and perpendicular distance from the axis of rotation.
b) Convert distances from centimeters to metres for S.I. unit Nm.

(iii) State two factors that affects the centre of gravity of the body. [2 Marks]

Answer:
1. Shape and geometrical form of the body.
2. Distribution of mass (density) within the body.

Teacher's Note:
a) Centre of gravity shifts towards the heavier part of the body.
b) It does not depend on the weight of the body, only on mass distribution and shape.

(iv) If the moment of F about the centre of a wheel O is 6Nm then calculate the moment of F about A. [2 Marks]

[Figure: A circular wheel with centre O and point A at the bottom. A tangential force F = 20 N acts at the top.]

Answer:
Moment about A = Moment about O + (Force \(\times\) distance between O and A). Since the line of action of force passes through or creates an increased lever arm, Moment about A = \( 6 \text{ Nm} + F \times r \) where \( F \times r = 6 \text{ Nm} \), hence total moment = \( 6 + 6 = 12 \text{ Nm} \).

Teacher's Note:
a) Use the principle of moments about parallel axes.
b) The perpendicular distance from point A is double the distance from centre O.

(v) If kinetic energy of a moving body is 40J then what will be its kinetic energy when its velocity is doubled? [2 Marks]

Answer:
Kinetic energy \( K = \frac{1}{2}mv^{2} \). When velocity is doubled (\( 2v \)), new kinetic energy \( K' = \frac{1}{2}m(2v)^{2} = 4 \times (\frac{1}{2}mv^{2}) = 4 \times 40 \text{ J} = 160 \text{ J} \).

Teacher's Note:
a) Kinetic energy is directly proportional to the square of velocity (\( K \propto v^{2} \)).
b) Doubling the velocity increases kinetic energy by a factor of four.

(vi) A freely suspended pendulum in air is disturbed once and left to oscillate on its own: [2 Marks]
(a) Name the type of vibrations.
(b) State one way to decrease the frequency of this vibration.

Answer:
(a) Damped vibrations.
(b) Increase the length of the pendulum.

Teacher's Note:
a) Amplitude decreases with time due to resistive forces like air friction, hence damped.
b) Frequency is inversely proportional to the square root of length (\( f \propto 1/\sqrt{l} \)), so increasing length decreases frequency.

(vii) Two copper wires A and B are of same length present at temperature \( 30^{\circ}\text{C} \). Radius of A is twice the radius of B. [2 Marks]
(a) Which wire has greater resistance?
(b) Which wire will have greater resistivity?

Answer:
(a) Wire B has greater resistance (since resistance is inversely proportional to cross-sectional area, \( R \propto 1/r^{2} \)).
(b) Both wires have the same resistivity because resistivity depends only on the material and temperature, not on dimensions.

Teacher's Note:
a) Resistance depends on length and cross-sectional area.
b) Resistivity is a characteristic property of the material.

Question 3

(i) A lens X can form an image on the screen. [2 Marks]
(a) Name the lens X.
(b) Is it possible for this lens to form magnified image?

Answer:
(a) Convex lens (or converging lens).
(b) Yes, a convex lens can form real magnified images (when the object is between F and 2F) as well as virtual magnified images (when the object is within the focal length).

Teacher's Note:
a) Only real images can be caught on a screen, and convex lenses form real images.
b) Concave lenses only form virtual, diminished images which cannot be caught on a screen.

(ii) (a) Is it possible to switch off an appliance by placing the switch in a neutral wire? [2 Marks]
(b) Is it possible for current to flow between a neutral and an earth wire?

Answer:
(a) Yes, switching off the neutral wire disconnects the appliance from the return path, but it is unsafe because the appliance remains at high potential (connected to live wire).
(b) No, normally current does not flow between neutral and earth wires because both are at nearly zero potential.

Teacher's Note:
a) Switches must always be connected to the live wire for safety.
b) Neutral wire completes the circuit back to the substation.

(iii) State two factors that affect the strength of an electromagnet. [2 Marks]

Answer:
1. Magnitude of current flowing through the coil.
2. Number of turns per unit length in the solenoid.

Teacher's Note:
a) Strength increases with an increase in current or number of turns.
b) Use of a soft iron core also increases the strength significantly.

(iv) Calculate the heat absorbed by \( 200 \text{ g} \) ice at \( 0^{\circ}\text{C} \) to change to water at \( 60^{\circ}\text{C} \). [Specific heat capacity of ice = \( 2100 \text{ Jkg}^{-1}\text{K}^{-1} \), Specific heat capacity of water = \( 4200 \text{ Jkg}^{-1}\text{K}^{-1} \), Specific latent heat of ice = \( 336000 \text{ Jkg}^{-1}\text{K}^{-1} \)]. [2 Marks]

Answer:
Mass \( m = 200 \text{ g} = 0.2 \text{ kg} \).
1. Heat required to melt ice at \( 0^{\circ}\text{C} \) to water at \( 0^{\circ}\text{C} \):
\( Q_{1} = m L = 0.2 \times 336000 = 67200 \text{ J} \).
2. Heat required to raise temperature of water from \( 0^{\circ}\text{C} \) to \( 60^{\circ}\text{C} \):
\( Q_{2} = m c \Delta T = 0.2 \times 4200 \times (60 - 0) = 50400 \text{ J} \).
Total heat absorbed \( Q = Q_{1} + Q_{2} = 67200 + 50400 = 117600 \text{ J} \).

Teacher's Note:
a) Break the calculation into two stages: phase change at constant temperature and temperature rise.
b) Always convert grams into kilograms before calculation.

(v) What are background radiations? [2 Marks]

Answer:
Background radiations are low-level ionizing radiations originating from natural sources (such as radioactive rocks, cosmic rays, and radioactive elements in the atmosphere and soil) present in the environment.

Teacher's Note:
a) They are present everywhere around us constantly.
b) Examples include potassium-40 in food and radon gas in air.

SECTION B

(Attempt any four questions.)

Question 4

(i) The diagram (not drawn to the scale) below shows the graphical relation between angle of deviation and angle of incidence, when light passes through a triangular prism of angle \( 62^{\circ} \) of a certain glass material. [3 Marks]

[Figure: Graph of angle of deviation (\( \delta \)) versus angle of incidence (i). Minimum deviation is marked as \( 37^{\circ} \) at angle of incidence \( 40^{\circ} \), and point X is marked on the right with deviation \( 51^{\circ} \) at angle of incidence \( 46^{\circ} \).]

(a) State the angle of minimum deviation of this prism and the corresponding angle of incidence.
(b) Calculate the value of X.

Answer:
(a) Angle of minimum deviation \( \delta_{m} = 37^{\circ} \), corresponding angle of incidence \( i = 40^{\circ} \).
(b) Using the relation for a prism: \( \delta = i + e - A \). From the graph symmetry or standard formula, at angle \( X \), deviation is \( 51^{\circ} \). Since deviation is the same for complementary angles of incidence and emergence, \( X = A + \delta - i = 62^{\circ} + 51^{\circ} - 46^{\circ} = 67^{\circ} \) (or by prism graph properties, \( X = 46^{\circ} \) or calculated as \( 62^{\circ} + 51^{\circ} - 40^{\circ} \text{ etc.} \); standard calculation gives \( X = 46^{\circ} \) corresponding to emergence or incidence value).

Teacher's Note:
a) Minimum deviation occurs at the lowest point of the deviation curve.
b) The curve is asymmetric, but \( i \) and \( e \) interchange for the same angle of deviation.

(ii) Redraw and complete the path of the ray AB till it emerges out of the prism of critical angle \( 42^{\circ} \). [3 Marks]

[Figure: Right-angled prism PQR with angle \( 30^{\circ} \) at P, ray AB incident normally on face PQ, striking face PR at an angle greater than critical angle.]

Answer:
1. Ray AB passes undeviated through face PQ since it is normal to the surface.
2. It strikes face PR at an angle of incidence equal to \( 60^{\circ} \) (which is greater than the critical angle of \( 42^{\circ} \)).
3. It suffers Total Internal Reflection at face PR and emerges normally through face QR.

Teacher's Note:
a) Calculate the angle of incidence at the second face carefully using geometry.
b) Since \( i = 60^{\circ} > c = 42^{\circ} \), total internal reflection takes place.

(iii) The above diagram shows that an observer sees the image of an object O at I. [4 Marks]
(a) Name and define the phenomenon responsible for seeing the image at a different position.
(b) State the effect on X when:
1. Y increases
2. Y decreases

[Figure: Rectangular glass block of thickness Y showing object O at top, shifted to virtual image I by distance X through refraction.]

Answer:
(a) Refraction: The phenomenon of bending of light when it passes obliquely from one transparent medium to another due to change in speed.
(b) 1. When Y (thickness of glass block) increases, normal shift X increases.
2. When Y decreases, normal shift X decreases.

Teacher's Note:
a) Normal shift \( X = Real \text{ depth} \times (1 - 1/\mu) \), which is directly proportional to thickness Y.
b) Apparent depth decreases with an increase in thickness of the denser medium.

Question 5

(i) An object of height \( 20 \text{ cm} \) is placed in front of a lens at a distance of \( 50 \text{ cm} \). Its virtual, diminished image is formed at a distance of \( 15 \text{ cm} \). [3 Marks]
(a) Identify the type of the lens.
(b) Calculate the focal length of the lens.

Answer:
(a) Concave lens (since it forms a virtual and diminished image).
(b) Given: Object distance \( u = -50 \text{ cm} \), Image distance \( v = -15 \text{ cm} \).
Using lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \frac{1}{f} = \frac{1}{-15} - \frac{1}{-50} = -\frac{1}{15} + \frac{1}{50} = \frac{-10 + 3}{150} = \frac{-7}{150} \)
\( f = -\frac{150}{7} = -21.43 \text{ cm} \).

Teacher's Note:
a) Apply sign conventions strictly: both \( u \) and \( v \) are negative for a concave lens forming a virtual image on the same side as the object.
b) Focal length of a concave lens is always negative.

(ii) The diagram below shows the extreme colours of a visible spectrum (X and Y). [3 Marks]
(a) Identify the colours X and Y.
(b) Which colour has greater speed in vacuum?

[Figure: Dispersion of white light through a glass prism showing spectrum where top ray is X and bottom ray is Y.]

Answer:
(a) X is Red and Y is Violet.
(b) Both colours have the exact same speed in vacuum (c = \( 3 \times 10^{8} \text{ ms}^{-1} \)).

Teacher's Note:
a) Red deviates the least and appears at the top; violet deviates the most and appears at the bottom.
b) In vacuum, all electromagnetic waves travel at the same speed regardless of their frequency or wavelength.

(iii) The diagram below shows an object AB kept in front of the lens. The path of one ray coming from the object is shown. [4 Marks]
(a) Name the lens L.
(b) Redraw and complete the ray diagram showing the formation of the image.
(c) In which optical instrument is this kind of image formed?

[Figure: Lens L with object AB placed between F and 2F, showing a ray parallel to principal axis refracting through focus.]

Answer:
(a) Convex lens.
(b) Draw the ray diagram with second ray passing through optical centre to intersect the first ray beyond 2F, forming a real, inverted, and magnified image.
(c) Cinema projector (or slide projector / camera).

Teacher's Note:
a) When an object is placed between F and 2F of a convex lens, a real, inverted and magnified image is formed beyond 2F.
b) Arrowheads on rays are mandatory in ray diagrams.

Question 6

(i) The diagram below shows a block and tackle system: [3 Marks]
(a) Copy and redraw the labelled diagram showing the correct connection of tackle, direction of the forces involved to obtain the maximum V.R. and convenient direction.
(b) Calculate the M.A. of this pulley system if its efficiency is 80%.

[Figure: Block and tackle system with 3 pulleys (2 in upper fixed block, 1 in lower movable block).]

Answer:
(a) Draw upper fixed block with 2 pulleys and lower movable block with 1 pulley, total 3 pulleys, with string tied to the hook of upper block for convenient downward effort.
(b) Velocity Ratio V.R. = 3. Efficiency \( \eta = 80\% = 0.8 \).
Mechanical Advantage M.A. = \( \eta \times \text{V.R.} = 0.8 \times 3 = 2.4 \).

Teacher's Note:
a) Velocity ratio equals the total number of pulleys in the system (3).
b) Mechanical advantage is the product of efficiency and velocity ratio.

(ii) The adjacent diagram shows a wheel of diameter 40 cm fixed on a wall capable of rotating around its centre O. If the wheel rotates in an anticlockwise direction, then: [3 Marks]
(a) Calculate the clockwise moment.
(b) State whether X = 100 gf or X < 100 gf or X > 100 gf.
(c) Give a reason for your answer.

[Figure: Wheel with centre O, diameter 40 cm (radius 20 cm = 0.2 m), weight 100 gf on right and X gf on left.]

Answer:
(a) Clockwise moment = \( 100 \text{ gf} \times 0.2 \text{ m} = 20 \text{ gf}\cdot\text{m} \) (or \( 2000 \text{ gf}\cdot\text{cm} \)).
(b) X > 100 gf.
(c) For anticlockwise rotation, the anticlockwise moment must be greater than the clockwise moment, meaning force X must be greater than 100 gf as the radius is the same.

Teacher's Note:
a) Moment = Force \(\times\) radius.
b) An anticlockwise rotation requires a larger anticlockwise torque.

(iii) A coconut of mass 450 g falls from the top of an 80 m high tree. [4 Marks]
(a) Calculate the potential energy possessed by the coconut when it is at the top of the tree.
(b) Without calculation, state the kinetic energy with which it strikes the ground and state the principle involved to arrive at the answer in i). \( g = 10 \text{ ms}^{-2} \).

Answer:
(a) Mass \( m = 450 \text{ g} = 0.45 \text{ kg} \), height \( h = 80 \text{ m} \), \( g = 10 \text{ ms}^{-2} \).
Potential Energy P.E. = \( mgh = 0.45 \times 10 \times 80 = 360 \text{ J} \).
(b) Kinetic energy with which it strikes the ground is 360 J.
Principle involved: Principle of Conservation of Energy (total mechanical energy remains conserved, ignoring air resistance).

Teacher's Note:
a) Potential energy at the top converts entirely into kinetic energy just before hitting the ground.
b) Convert mass into kg and use \( g = 10 \text{ ms}^{-2} \) as given.

Question 7

(i) A person standing in front of a cliff fires a gun and hears its echo after 3s. If the speed of sound in air is \( 336 \text{ ms}^{-1} \). [3 Marks]
(a) Calculate the distance of the person from the cliff.
(b) After moving a certain distance from the cliff he fires the gun again and this time the echo is heard 1.5 s later than the first. Calculate distance moved by the person.

Answer:
(a) Distance \( d = \frac{v \times t}{2} = \frac{336 \times 3}{2} = 504 \text{ m} \).
(b) New time taken for echo = \( 3 + 1.5 = 4.5 \text{ s} \).
New distance from cliff \( d' = \frac{336 \times 4.5}{2} = 756 \text{ m} \).
Distance moved by the person = \( d' - d = 756 - 504 = 252 \text{ m} \).

Teacher's Note:
a) Sound travels twice the distance between the person and the cliff during an echo.
b) Add the time delay correctly to find the new distance.

(ii) (a) A radioactive nucleus X emits an alpha particle followed by two beta particles and forms nucleus Y. What is the general name of the elements X and Y? [3 Marks]
(b) If the atomic number of Y is 80 then what is the atomic number of X?
(c) If the atomic mass number of Y is 189 then what is the atomic mass number of X?

Answer:
(a) Isotopes.
(b) Atomic number of X is 80 (since emission of one alpha decreases atomic number by 2 and two beta particles increase it by 2, leaving atomic number unchanged).
(c) Atomic mass number of X is \( 189 + 4 = 193 \) (since alpha emission decreases mass number by 4 and beta particles do not change mass number).

Teacher's Note:
a) Elements with the same atomic number but different mass numbers are isotopes.
b) Net change in atomic number is \( -2 + 2(1) = 0 \).

(iii) A boy tunes a radio channel to a radio station 93.5 MHz. [4 Marks]
(a) Name and define the scientific wave phenomenon involved in tuning the radio channel.
(b) Now, what is the frequency of the channel? Convert this frequency into S.I. unit.

Answer:
(a) Resonance: The phenomenon when the frequency of an externally applied periodic force matches the natural frequency of a body, causing it to vibrate with a large amplitude.
(b) Frequency = \( 93.5 \text{ MHz} = 93.5 \times 10^{6} \text{ Hz} = 9.35 \times 10^{7} \text{ Hz} \).

Teacher's Note:
a) Tuning a radio involves adjusting the receiver circuit's natural frequency to match the incoming station's frequency via electrical resonance.
b) Mega (M) stands for \( 10^{6} \).

Question 8

(i) (a) What is the meaning of the statement 'the power rating of an appliance is 60W, 220V.'? [3 Marks]
(b) In which wire is the fuse connected in a circuit?
(c) State the function of main switch in an electric circuit.

Answer:
(a) It means that when the appliance is connected across a 220V supply, it consumes electrical energy at the rate of 60 Joules per second (or draws current and consumes 60W power).
(b) Live wire.
(c) Main switch is used to disconnect all live connections from the mains supply simultaneously for safety and maintenance.

Teacher's Note:
a) Power rating helps calculate normal operating current and energy consumption.
b) Fuse must always be connected to the live wire.

(ii) (a) Copy and complete the following nuclear reaction. \( _{86}\text{Rn}^{222} \rightarrow _{84}\text{Po}^{218} + _{2}^{4}\alpha \) [3 Marks]
(b) What will be the effect on the radiation emitted in the above reaction when it is allowed to pass through an electric field? [Be specific in your answer]

Answer:
(a) Completed reaction: \( _{86}\text{Rn}^{222} \rightarrow _{84}\text{Po}^{218} + _{2}^{4}\alpha \).
(b) Alpha particles (\( _{2}^{4}\alpha \)) are positively charged helium nuclei, so they deflect towards the negatively charged plate in an electric field.

Teacher's Note:
a) Balance both atomic numbers and mass numbers on both sides of a nuclear reaction.
b) Alpha particles carry a positive charge of +2e.

(iii) Observe the given circuit diagram and answer the questions that follow: [4 Marks]
(a) Calculate the resistance of the circuit when the key K completes the circuit.
(b) Calculate the current through \( 3\,\Omega \) resistance.

[Figure: Circuit diagram with a cell of 4V, internal resistance \( 0.4\,\Omega \), connected to a parallel combination of \( 5\,\Omega \) and \( 3\,\Omega \) resistors, in series with a \( 2\,\Omega \) resistor.]

Answer:
(a) Parallel combination of \( 5\,\Omega \) and \( 3\,\Omega \):
\( R_{p} = \frac{5 \times 3}{5 + 3} = \frac{15}{8} = 1.875\,\Omega \).
Total external resistance \( R = 2 + 1.875 = 3.875\,\Omega \).
Total resistance of circuit including internal resistance \( r = 3.875 + 0.4 = 4.275\,\Omega \).
(b) Total current \( I = \frac{E}{R + r} = \frac{4}{4.275} \approx 0.936 \text{ A} \).
Potential difference across parallel combination \( V_{p} = I \times R_{p} = 0.936 \times 1.875 = 1.755 \text{ V} \).
Current through \( 3\,\Omega \) resistor = \( \frac{V_{p}}{3} = \frac{1.755}{3} = 0.585 \text{ A} \).

Teacher's Note:
a) Always include internal resistance when calculating total circuit resistance and total current.
b) Potential difference across parallel branches is the same.

Question 9

(i) A metal piece present at \( 120^{\circ}\text{C} \) is quickly dropped in a calorimeter of mass \( 80 \text{ g} \) containing \( 200 \text{ g} \) of water at \( 30^{\circ}\text{C} \). The final temperature attained by the mixture is \( 40^{\circ}\text{C} \). Calculate the thermal capacity of the metal piece. [Specific heat capacity of water = \( 4.2 \text{ Jg}^{-1}{^\circ}\text{C}^{-1} \), Specific heat capacity of calorimeter = \( 0.4 \text{ Jg}^{-1}{^\circ}\text{C}^{-1} \)]. [3 Marks]

Answer:
Let mass of metal piece be \( m \) and specific heat capacity be \( c \).
Heat lost by metal piece = \( m \times c \times (120 - 40) = 80 \, mc \).
Heat gained by water = \( 200 \times 4.2 \times (40 - 30) = 200 \times 4.2 \times 10 = 8400 \text{ J} \).
Heat gained by calorimeter = \( 80 \times 0.4 \times (40 - 30) = 320 \text{ J} \).
Total heat gained = \( 8400 + 320 = 8720 \text{ J} \).
By Principle of Mixtures (Heat lost = Heat gained):
\( 80 \, mc = 8720 \)
Thermal capacity (\( mc \)) = \( \frac{8720}{80} = 109 \text{ J}{^\circ}\text{C}^{-1} \).

Teacher's Note:
a) Thermal capacity is the product of mass and specific heat capacity (\( m \times c \)).
b) Account for both water and the calorimeter when calculating total heat gained.

(ii) The diagram below shows a cooling curve for a substance: [3 Marks]
(a) State the temperatures at which the substance condenses and solidifies respectively.
(b) The temperature range in which the substance is in liquid state.
(c) Why do we prefer ice to ice-cold water for cooling a drink?

[Figure: Cooling curve showing temperature vs time, with plateaus at \( 150^{\circ}\text{C} \) (condensation) and \( 60^{\circ}\text{C} \) (solidification), and initial/final points.]

Answer:
(a) Condensation temperature = \( 150^{\circ}\text{C} \); Solidification temperature = \( 60^{\circ}\text{C} \).
(b) Liquid state temperature range: between \( 60^{\circ}\text{C} \) and \( 150^{\circ}\text{C} \).
(c) Ice absorbs specific latent heat of fusion (\( 336000 \text{ Jkg}^{-1} \)) in addition to sensible heat, making it much more effective at cooling than ice-cold water at the same temperature.

Teacher's Note:
a) Flat regions on a cooling curve indicate state changes at constant temperature.
b) Latent heat provides extra cooling capacity.

(iii) The diagram below shows a magnet placed between two coils A and B. The magnet is moved along the axis towards coil B. [4 Marks]
(a) State the polarities induced at the ends Q and R of the coil due to the motion of the magnet.
(b) Name the phenomenon due to which the current is induced in the coils.
(c) Name the law which helps to find the polarities at the ends Q and R.

[Figure: Magnet with North and South poles moving towards coil B on the right and away from coil A on the left, with galvanometers G attached.]

Answer:
(a) End Q (facing retreating North pole) becomes South (S) to oppose motion; End R (facing approaching North pole) becomes North (N) to oppose motion.
(b) Electromagnetic induction.
(c) Lenz's law.

Teacher's Note:
a) According to Lenz's law, the induced current opposes the cause that produces it.
b) Approaching pole induces like polarity, and receding pole induces opposite polarity.