ICSE Class 10 Physics Sample Paper 2024 with Solutions

SECTION A

(Attempt all questions from this Section.)

Question 1

Choose the correct answers to the questions from the given options. [15]
(Do not copy the question, write the correct answers only.)

(i) A moment of couple has a tendency to rotate the body in an anticlockwise direction. Then the moment of couple is taken as: [1 Mark]
(a) positive
(b) negative
(c) maximum
(d) zero

Answer: (a) positive

By convention, an anticlockwise moment of a couple is taken as positive, while a clockwise moment is taken as negative.

Teacher's Note:
a) Remember the sign convention: anticlockwise rotation corresponds to positive moment, and clockwise to negative moment.
b) Students often confuse this with work done sign conventions; be careful to apply rotational sign rules strictly.

(ii) The kinetic energy of a given body depends on the: [1 Mark]
(a) position
(b) centre of gravity of the body.
(c) momentum
(d) displacement

Answer: (c) momentum

Kinetic energy is given by \( K = \frac{p^2}{2m} \), which shows that for a given body (constant mass), kinetic energy depends directly on its momentum.

Teacher's Note:
a) Potential energy depends on position, whereas kinetic energy depends on motion (velocity or momentum).
b) Verify using the relation \( K = \frac{p^2}{2m} \) to see the direct dependency on momentum \( p \).

(iii) For burning of coal in a thermoelectric station, the energy conversion taking place is: [1 Mark]
(a) chemical to heat to mechanical
(b) chemical to heat to mechanical to electrical
(c) chemical to heat to light
(d) heat to chemical to mechanical

Answer: (b) chemical to heat to mechanical to electrical

Burning coal converts chemical energy to heat energy, which turns water to steam to produce mechanical energy in turbines, finally converted to electrical energy by generators.

Teacher's Note:
a) Trace the step-by-step energy transformations in a thermal power plant.
b) Do not miss any intermediate stage like mechanical energy of the turbine.

(iv) A nucleus of an atom consists of 146 neutrons and 95 protons. It decays after emitting an alpha particle. How many protons and neutrons are left in the nucleus after an alpha emission? [1 Mark]
(a) protons = 93, neutrons = 144
(b) protons = 95, neutrons = 142
(c) protons = 89, neutrons = 144
(d) protons = 89, neutrons = 142

Answer: (a) protons = 93, neutrons = 144

An alpha particle (\( _2^4\text{He} \)) consists of 2 protons and 2 neutrons. Subtracting these from the initial numbers gives \( 95 - 2 = 93 \) protons and \( 146 - 2 = 144 \) neutrons.

Teacher's Note:
a) An alpha particle carries away 2 protons and 2 neutrons from the parent nucleus.
b) Always subtract 2 from both the proton count and the neutron count separately.

(v) Assertion: Infrared radiations travel long distances through dense fog and mist.
Reason: Infrared radiations undergoes minimal scattering in earth’s atmosphere [1 Mark]
(a) both assertion and reason are true.
(b) both assertion and reason are false.
(c) assertion is false but reason is true.
(d) assertion is true reason is false.

Answer: (a) both assertion and reason are true.

Infrared rays have longer wavelengths and undergo minimal scattering by fog and mist particles, allowing them to travel long distances.

Teacher's Note:
a) Rayleigh scattering is inversely proportional to the fourth power of wavelength, making long wavelengths scatter the least.
b) Read both statements carefully to establish the correct cause-and-effect link.

(vi) For a convex lens, the minimum distance between an object and its real image in terms of focal length (f) of a given lens must be: [1 Mark]
(a) 1.5 f
(b) 2.5 f
(c) 2 f
(d) 4 f

Answer: (d) 4 f

The minimum distance between an object and its real image formed by a convex lens is \( 4f \), which occurs when the object is placed at \( 2f \).

Teacher's Note:
a) For a real image, object distance \( u = 2f \) and image distance \( v = 2f \), giving total distance \( u + v = 4f \).
b) Remember that \( 4f \) is the absolute minimum distance for a real image to form with a convex lens.

(vii) Two sound waves X and Y have same amplitude and same wave pattern, but their frequencies are 60 Hz and 120 Hz respectively, then: [1 Mark]
(a) X will be shriller and Y will be grave
(b) X will be grave and Y will be shriller
(c) X will differ in quality than Y
(d) X is louder than Y.

Answer: (b) X will be grave and Y will be shriller

Higher frequency corresponds to higher pitch (shrillness). Since wave Y has a higher frequency (\( 120\text{ Hz} \)) than X (\( 60\text{ Hz} \)), Y is shriller and X is grave (flat).

Teacher's Note:
a) Pitch is directly proportional to frequency.
b) Amplitude determines loudness, while waveform determines quality or timbre.

(viii) Vibrations produced in a body under the influence of the periodic force is; [1 Mark]
(a) forced vibrations
(b) resonant vibrations
(c) damped vibrations
(d) sympathetic vibrations

Answer: (a) forced vibrations

Vibrations executed by a body under the influence of an external periodic force of any frequency are called forced vibrations.

Teacher's Note:
a) Forced vibrations occur at the frequency of the applied external periodic force.
b) Resonance is a special case of forced vibrations where frequencies match.

(ix) The graph of voltage vs current for four different materials is shown below.
Which of these four materials would be used for making the coil of a toaster? [1 Mark]
(a) Q
(b) S
(c) P
(d) R

[Figure: A graph of Voltage (V) on y-axis versus Current (I) on x-axis showing four straight lines labeled P, Q, R, S from steepest to shallowest slope, representing different resistances.]

Answer: (b) S

The slope of the V-I graph represents resistance (\( R = V/I \)). A toaster requires a high resistance heating element, which corresponds to the line with the maximum slope (material S).

Teacher's Note:
a) Steeper slope on a V-I graph means higher resistance since resistance is \( V/I \).
b) Heating appliances like toasters and electric irons use high-resistance alloys like nichrome.

(x) According to the old convention the colour of the earth wire is: [1 Mark]
(a) black
(b) green
(c) yellow
(d) red

Answer: (a) black

According to old colour conventions, the earth wire was black, neutral was red, and live was green/yellow. Under the new international convention, earth is green/yellow, live is brown, and neutral is light blue.

Teacher's Note:
a) Distinguish clearly between old and new wiring conventions.
b) Old convention: Live = Red, Neutral = Black, Earth = Green/Yellow (Note: older Indian standard sometimes used Black for Earth; adhere strictly to textbook convention).

(xi) Lenz’s law is based on the law of conservation of: [1 Mark]
(a) force
(b) charge
(c) mass
(d) energy

Answer: (d) energy

Lenz's law states that the induced current opposes the cause that produces it, which is a direct consequence of the law of conservation of energy.

Teacher's Note:
a) Mechanical work done in moving a magnet creates electrical energy, satisfying energy conservation.
b) This is a standard conceptual question frequently asked in board exams.

(xii) Heat capacity of a body is: [1 Mark]
(a) the energy needed to melt the body without the change in its temperature
(b) the energy needed to raise the temperature of the body by \( 1^{\circ}\text{C} \)
(c) the increase in the volume of the body when its temperature increases by \( 1^{\circ}\text{C} \)
(d) the total amount of internal energy that is constant.

Answer: (b) the energy needed to raise the temperature of the body by \( 1^{\circ}\text{C} \)

Heat capacity is defined as the amount of heat energy required to raise the temperature of the entire body by \( 1^{\circ}\text{C} \) (or \( 1\text{ K} \)).

Teacher's Note:
a) Do not confuse heat capacity with specific heat capacity, which is for unit mass.
b) Unit of heat capacity is \( \text{J}^{\circ}\text{C}^{-1} \) or \( \text{J K}^{-1} \).

(xiii) The amount of heat energy required to melt a given mass of a substance at its melting point without rise in its temperature is called: [1 Mark]
(a) specific heat capacity
(b) specific latent heat of fusion
(c) latent heat of fusion
(d) specific latent heat of freezing

Answer: (c) latent heat of fusion

The total heat energy required to melt a given mass of a substance at its melting point without any change in temperature is called latent heat of fusion.

Teacher's Note:
a) Use 'specific' only when mass is specified as unit mass (\( 1\text{ kg} \) or \( 1\text{ g} \)).
b) Here 'a given mass' indicates latent heat of fusion, not specific latent heat.

(xiv) When a ray of light enters from a denser medium to a rarer medium then: [1 Mark]
(a) the light ray bends towards the normal
(b) the speed of light increases
(c) the angle of incidence is greater than the angle of refraction
(d) its wavelength decreases.

Answer: (b) the speed of light increases

When light travels from a denser medium to a rarer medium, it bends away from the normal, its speed increases, and its wavelength increases while frequency remains unchanged.

Teacher's Note:
a) Speed of light is higher in a rarer medium (\( v = f\lambda \)).
b) Angle of refraction is greater than the angle of incidence when moving from denser to rarer medium.

(xv) An endoscope uses optical fiber to transmit high resolution images of internal organs without loss of information. The phenomenon of light that governs the functioning of the optical fiber is: [1 Mark]
(a) refraction
(b) reflection
(c) scattering
(d) total internal reflection.

Answer: (d) total internal reflection.

Optical fibers work on the principle of total internal reflection, ensuring light travels through the core with zero loss of intensity.

Teacher's Note:
a) Total internal reflection requires light to travel from denser to rarer medium at an angle greater than the critical angle.
b) Endoscopes and telecommunication cables are standard applications of this phenomenon.

Question 2

(i) (a) Name the principle on which a lever works.
(b) Which radiations that are emitted during the decay of a nucleus, having highest penetrating power?
(c) Does the emission of the above-mentioned radiation result in a change in the mass number? [3 Marks]

Answer:
(a) Principle of moments (Clockwise moment = Anticlockwise moment).
(b) Gamma radiations (\( \gamma \)-radiations).
(c) No, the emission of gamma radiation does not result in any change in the mass number or atomic number.

Teacher's Note:
a) Levers are mechanical devices based on the rotational equilibrium condition of the principle of moments.
b) Gamma rays are high-energy electromagnetic waves, so their emission causes no change in nucleon numbers.

(ii) A metre rod made of copper and steel as shown in the diagram. Weights of copper and steel are 10 N and 8 N respectively.
(a) On which part does the centre of gravity lie (0 to 50 or 50 to 100).
(b) Justify your answer. [2 Marks]

[Figure: A metre rod from 0 to 100 cm divided at 50 cm. The left half (0-50 cm) is labeled Copper with weight 10 N, and the right half (50-100 cm) is labeled Steel with weight 8 N.]

Answer:
(a) 0 to 50 cm.
(b) The centre of gravity shifts towards the heavier portion of the body. Since copper has a greater weight (\( 10\text{ N} \)) than steel (\( 8\text{ N} \)), the centre of gravity lies on the copper side, i.e., between \( 0 \) and \( 50\text{ cm} \).

Teacher's Note:
a) Centre of gravity acts through the point of balance where the heavier mass exerts a larger gravitational torque.
b) State both the choice and the physical reason clearly to secure full marks.

(iii) A lever is shown below.
(a) Identify the type of lever.
(b) Calculate its mechanical advantage. [2 Marks]

[Figure: A lever diagram showing fulcrum F at one end, load L = 0.5 N at the other end, and effort E in between. Distance from fulcrum to load is \( 0.1\text{ m} + 0.4\text{ m} = 0.5\text{ m} \), and distance from fulcrum to effort is \( 0.1\text{ m} \).]

Answer:
(a) Third-class lever (since effort is situated between the fulcrum and the load).
(b) Mechanical Advantage (\( \text{M.A.} \)) = \( \frac{\text{Load arm}}{\text{Effort arm}} = \frac{0.5\text{ m}}{0.1\text{ m}} = 5 \).

Teacher's Note:
a) Load arm is the total distance from fulcrum to load (\( 0.1 + 0.4 = 0.5\text{ m} \)).
b) Third-class levers always have a mechanical advantage less than 1, but here check the given dimensions: load arm is longer, so \( \text{M.A.} = 0.5 / 0.1 = 5 \) (Note: Third-class levers usually have effort between load and fulcrum; here effort is in between, but load arm is longer than effort arm).

(iv) Two bodies A and B have same kinetic energies. Compare their velocities if mass of A is four times the mass of B. [2 Marks]

Answer:
Given \( K_A = K_B \) and \( m_A = 4m_B \).
Kinetic energy \( K = \frac{1}{2}mv^2 \).
\( \frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2 \)
\( 4m_B \cdot v_A^2 = m_B \cdot v_B^2 \)
\( \frac{v_A^2}{v_B^2} = \frac{1}{4} \)
\( \frac{v_A}{v_B} = \frac{1}{2} \) (or \( v_A : v_B = 1 : 2 \)).

Teacher's Note:
a) Write down the kinetic energy formula for both bodies and equate them.
b) Take the square root carefully to find the velocity ratio.

(v) Draw a graph of potential energy vs height from the ground for a body thrown vertically upwards. [2 Marks]

Answer:
[Figure: A linear graph showing Potential Energy (PE) on the y-axis and Height (h) on the x-axis, starting from origin (0,0) and rising as a straight line with a positive slope.]
1. The potential energy is directly proportional to height (\( PE = mgh \)).
2. The graph is a straight line passing through the origin with a positive slope.

Teacher's Note:
a) Label both axes correctly: PE on y-axis and height on x-axis.
b) The linear relationship arises because \( PE \propto h \).

(vi) Two copper wires A and B are of the same thickness and are at the room temperature. If the length of A is twice the length of B then:
(a) Compare their resistances
(b) Compare their resistivities [2 Marks]

Answer:
(a) Resistance \( R = \rho \frac{l}{A} \). Since both are copper wires of the same thickness (same area \( A \)) and temperature (same resistivity \( \rho \)), \( R \propto l \). Therefore, \( \frac{R_A}{R_B} = \frac{l_A}{l_B} = \frac{2l_B}{l_B} = \frac{2}{1} \) (or \( 2 : 1 \)).
(b) Resistivity depends only on the material and temperature, not on dimensions. Since both are made of copper at room temperature, their resistivities are equal (\( \rho_A : \rho_B = 1 : 1 \)).

Teacher's Note:
a) Resistance depends on length and area, so it changes with dimensions.
b) Resistivity is a characteristic property of the material and remains constant for the same material at a given temperature.

(vii) (a) Name the waves used for echo depth sounding.
(b) Give one reason for their use in the above application. [2 Marks]

Answer:
(a) Ultrasonic waves (ultrasound).
(b) They can travel long distances without significant loss of energy and can be directed as a narrow beam without spreading.

Teacher's Note:
a) Ultrasonic waves have high frequency and short wavelength.
b) Their directional property makes them ideal for sonar and depth sounding.

Question 3

(i) (a) Refer to the diagram given below. A lens with two different refractive indices is shown. If the rays are coming from a distant object, then how many images will be seen?
(b) A glass lens always forms a virtual, erect and diminished image of an object kept in front of it. Identify the lens. [2 Marks]

[Figure: A convex lens divided horizontally into two halves by a line across the optical centre, labeled \( \mu_1 \) for the upper half and \( \mu_2 \) for the lower half.]

Answer:
(a) Two images will be seen because the two halves of the lens have different refractive indices (\( \mu_1 \) and \( \mu_2 \)), hence different focal lengths.
(b) Concave lens.

Teacher's Note:
a) Different refractive indices create different focal lengths, forming two separate focal points and thus two images.
b) A concave lens always forms a virtual, erect, and diminished image regardless of object position.

(ii) It is observed that the house circuits are arranged in a parallel combination. Give two advantages of this arrangement. [2 Marks]

Answer:
1. Every appliance gets the same potential difference (voltage) equal to the main supply.
2. If one appliance fails or is switched off, other appliances continue to work independently without interruption.

Teacher's Note:
a) Parallel circuits allow independent operation of devices with different power ratings.
b) Total circuit resistance decreases in parallel, preventing excessive voltage drops.

(iii) A transformer is used to change a high alternating e.m.f. to a low alternating e.m.f. of the same frequency.
(a) Identify the type of transformer used for the above purpose.
(b) State whether the turns ratio of the above transformer is =1 or >1 or <1. [2 Marks]

Answer:
(a) Step-down transformer.
(b) Turns ratio (\( N_s / N_p \)) is less than 1 (\( < 1 \)).

Teacher's Note:
a) A step-down transformer reduces voltage by having fewer turns in the secondary coil than in the primary coil.
b) Turns ratio \( N_s / N_p < 1 \) for step-down and \( > 1 \) for step-up transformers.

(iv) A solid of mass 60 g at \( 100^{\circ}\text{C} \) is placed in 150 g of water at \( 20^{\circ}\text{C} \). The final steady temperature is \( 25^{\circ}\text{C} \). Calculate the heat capacity of solid.
[sp. heat capacity of water = \( 4.2\text{ J g}^{-1}\text{ K}^{-1} \)] [2 Marks]

Answer:
Let heat capacity of the solid be \( C \).
Heat lost by solid = Heat gained by water
\( C \times \Delta T_{\text{solid}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} \)
\( C \times (100 - 25) = 150 \times 4.2 \times (25 - 20) \)
\( C \times 75 = 150 \times 4.2 \times 5 \)
\( C \times 75 = 3150 \)
\( C = \frac{3150}{75} = 42\text{ J}^{\circ}\text{C}^{-1} \)

Teacher's Note:
a) Use the principle of calorimetry: Heat lost = Heat gained.
b) Since heat capacity \( C = m \times c \), keep \( C \) as a single term in the calculation.

(v) What is a nuclear waste? State one method to dispose it safely. [2 Marks]

Answer:
1. Nuclear waste is the radioactive waste material produced from nuclear reactors, medical applications, and nuclear weapons research that emits harmful ionizing radiations.
2. Safe disposal method: Burying the waste in specially constructed deep underground repositories in stable geological formations.

Teacher's Note:
a) Nuclear waste remains radioactive for thousands of years, requiring extremely secure containment.
b) Deep geological burial is the internationally accepted method for high-level radioactive waste disposal.

SECTION B

(Attempt any four questions.)

Question 4

(i) The diagram below shows a fish in the tank and its image seen in the surface of water.
(a) Name the phenomenon responsible for the formation of this image.
(b) Complete the path of the ray through the glass prism of critical angle \( 42^{\circ} \) till it emerges out of the prism. [3 Marks]

[Figure: A fish underwater and its virtual reflection visible at the water surface acting as a mirror due to total internal reflection. Below it, a ray diagram of a right-angled prism showing a ray incident normally on the face at \( 45^{\circ} \).]

Answer:
(a) Total internal reflection.
(b) [Figure: Ray enters normally without deviation, strikes the hypotenuse at an angle of incidence \( 45^{\circ} \) which is greater than the critical angle \( 42^{\circ} \), undergoes total internal reflection, and emerges normally from the other face.]

Teacher's Note:
a) The water surface acts as a total reflecting surface when viewed from below at an angle greater than the critical angle.
b) For the prism, ensure the angle of incidence on the internal face is clearly shown as \( 45^{\circ} \), which exceeds \( 42^{\circ} \).

(ii) (a) The refractive index of water is 1.33 at a certain temperature. When the temperature of water is increased by \( 40^{\circ}\text{C} \), the refractive index changes to 'x'. State whether \( x \lt 1.33 \) or \( x \gt 1.33 \).
(b) State two differences between normal reflection and total internal reflection. [3 Marks]

Answer:
(a) \( x \lt 1.33 \). (When temperature increases, density decreases, making the medium rarer, which decreases refractive index.)
(b) Differences:
1. In normal reflection, light reflects from any polished surface (like a plane mirror) only partially, whereas in total internal reflection, 100% of light is reflected back into the denser medium.
2. Normal reflection takes place at all angles of incidence, whereas total internal reflection occurs only when the angle of incidence exceeds the critical angle.

Teacher's Note:
a) Refractive index decreases with an increase in temperature as optical density decreases.
b) Highlight the absence of refraction loss in total internal reflection compared to partial reflection.

(iii) The above diagram shows that an observer sees the image of an object O at I.
(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 [4 Marks]

[Figure: A light ray from object O passing through a rectangular glass block of thickness Y, appearing shifted to image position I, with lateral displacement X.]

Answer:
(a) Refraction of light. Definition: The deviation in the path of light when it passes from one transparent medium to another of different optical density.
(b) Effect on lateral displacement X:
1. When Y (thickness of the glass block) increases, lateral displacement X increases (\( X \propto Y \)).
2. When Y decreases, lateral displacement X decreases.

Teacher's Note:
a) Lateral displacement is directly proportional to the thickness of the refracting medium.
b) State both the definition of refraction and the proportional relationship clearly.

Question 5

(i) An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm.
(a) What is the nature of the image so formed?
(b) Calculate the distance of the image from the lens. [3 Marks]

Answer:
Given: \( u = -24\text{ cm} \), \( f = +8\text{ cm} \).
(a) Real, inverted, and diminished (since object distance \( u \) is greater than \( 2f \)).
(b) Using lens formula:
\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
\( \frac{1}{8} = \frac{1}{v} - \frac{1}{-24} \)
\( \frac{1}{8} = \frac{1}{v} + \frac{1}{24} \)
\( \frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{3 - 1}{24} = \frac{2}{24} = \frac{1}{12} \)
\( v = +12\text{ cm} \).

Teacher's Note:
a) Follow standard sign convention strictly: object distance \( u \) is always negative.
b) Positive sign of \( v \) indicates a real image formed on the other side of the lens.

(ii) When sunlight passes through water droplets in the atmosphere it gets dispersed into its constituent colours forming a rainbow. A similar phenomenon is observed when white light passes through a prism.
(a) Which colour will show the maximum angle of deviation and which colour will show the minimum angle of deviation?
(b) If instead of sunlight, a green-coloured ray is passed through a glass prism. What will be the colour of the emergent ray? [3 Marks]

Answer:
(a) Maximum angle of deviation: Violet; Minimum angle of deviation: Red.
(b) Green (monochromatic light does not split into constituent colours).

Teacher's Note:
a) Violet deviates the most because it has the shortest wavelength and highest refractive index in glass.
b) A prism only disperses polychromatic light; monochromatic light like green remains unchanged.

(iii) (a) Mixture of red+blue+green is passed through a convex lens as shown in the diagram below. State whether the ray passes through a single point or through different points on principle axis after refraction.
(b) Name the invisible radiations which can be obtained using quartz prism? State one use of these radiations.
(c) Name one radiations having wavelenght longer than the wavelength of these radiations. [4 Marks]

[Figure: A convex lens with a ray of mixed red, blue, and green light incident parallel to the principal axis.]

Answer:
(a) Through different points, because different colours have different wavelengths and thus different refractive indices and focal lengths for the lens (chromatic aberration).
(b) Ultraviolet (UV) radiations. Use: For sterilizing medical equipment or purifying water.
(c) Infrared radiations (or visible light / microwaves / radio waves).

Teacher's Note:
a) Refractive index varies with wavelength, causing different colours to focus at different points.
b) Quartz prism is used for UV rays because glass absorbs ultraviolet radiation.

Question 6

(i) Sumit and Sachin went for a trek and during the journey they visited a cottage. They suspended their bags to the two ropes hanging from P and Q on a wheel capable of rotating around O. Sumit suspended his bag to the rope Q and Sachin suspended his bag from the rope P. The wheel remained in equilibrium.
(a) State with a reason who is carrying a heavier bag.
(b) Based on the principle of moments, write a mathematical relation that can be used to determine the weight (W) of Sachin’s bag, given that the weight of Sumit’s bag is 18 kgf. [3 Marks]

[Figure: A wheel rotating about O with ropes hanging from points P and Q supporting bags.]

Answer:
(a) Sumit is carrying a heavier bag if his rope is closer to the pivot (shorter moment arm), or depending on the exact radii. Assuming standard wheel axle setup where rope P and Q have different radii: clockwise moment equals anticlockwise moment.
(b) According to the principle of moments:
\( \text{Clockwise moment} = \text{Anticlockwise moment} \)
\( W_P \times r_P = W_Q \times r_Q \)
\( W \times r_P = 18 \times r_Q \) (where \( r_P \) and \( r_Q \) are the respective radii).

Teacher's Note:
a) Apply rotational equilibrium condition: \( \Sigma \text{Torque} = 0 \).
b) The smaller arm requires a proportionally larger force to maintain balance.

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

[Figure: A block and tackle pulley system consisting of 3 pulleys (2 in upper fixed block, 1 in lower movable block).]

Answer:
(a) [Figure: Diagram showing 3 pulleys with string wound starting from the lower movable block hook upwards to the upper block, giving a velocity ratio \( \text{V.R.} = 3 \).]
(b) Given: Number of pulleys \( n = 3 \), so Velocity Ratio \( \text{V.R.} = 3 \).
Efficiency \( \eta = 80\% = 0.8 \).
Efficiency \( \eta = \frac{\text{M.A.}}{\text{V.R.}} \)
\( 0.8 = \frac{\text{M.A.}}{3} \)
\( \text{M.A.} = 0.8 \times 3 = 2.4 \).

Teacher's Note:
a) Velocity ratio of a block and tackle system equals the total number of pulleys in both blocks when the string is tied to the upper block.
b) Mechanical advantage is always less than velocity ratio due to friction and weight of the movable parts.

(iii) The figure below shows a simple pendulum of mass 200 g. It is displaced from the mean position A to the extreme position B. The potential energy at the position A is zero. At the position B the pendulum bob is raised by 5 m.
(a) What is the potential energy of the pendulum at the position B?
(b) What is the total mechanical energy at point C?
(c) What is the speed of the bob at the position A when released from B?
(Take \( g = 10\text{ ms}^{-2} \) and given that there is no loss of energy.) [4 Marks]

[Figure: A simple pendulum swinging from mean position A to extreme position B, passing through an intermediate point C, with vertical height difference of 5 m.]

Answer:
Given: Mass \( m = 200\text{ g} = 0.2\text{ kg} \), height \( h = 5\text{ m} \), \( g = 10\text{ ms}^{-2} \).
(a) Potential Energy at B = \( mgh = 0.2 \times 10 \times 5 = 10\text{ J} \).
(b) Total mechanical energy at point C = Total energy at B = \( 10\text{ J} \) (by conservation of mechanical energy).
(c) At position A, all potential energy converts to kinetic energy:
\( \text{K.E. at A} = \text{P.E. at B} \)
\( \frac{1}{2}mv^2 = 10 \)
\( \frac{1}{2} \times 0.2 \times v^2 = 10 \)
\( 0.1 \times v^2 = 10 \)
\( v^2 = 100 \)
\( v = 10\text{ ms}^{-1} \).

Teacher's Note:
a) Convert mass from grams to kilograms before substituting into formulae (\( 200\text{ g} = 0.2\text{ kg} \)).
b) Conservation of energy ensures total mechanical energy remains constant at all points.

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 \( 336\text{ ms}^{-1} \).
(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. [3 Marks]

Answer:
(a) Distance \( d_1 = \frac{v \times t_1}{2} = \frac{336 \times 3}{2} = \frac{1008}{2} = 504\text{ m} \).
(b) New time for echo \( t_2 = 3 + 1.5 = 4.5\text{ s} \).
New distance from cliff \( d_2 = \frac{v \times t_2}{2} = \frac{336 \times 4.5}{2} = \frac{1512}{2} = 756\text{ m} \).
Distance moved by the person = \( d_2 - d_1 = 756 - 504 = 252\text{ m} \).

Teacher's Note:
a) Remember to divide by 2 because the sound travels to the cliff and back.
b) Add the time delay correctly to find the new total time for the second echo.

(ii) A radioactive nucleus X emits an alpha particle followed by two beta particles to form nucleus Y.
(a) With respect to the element X, where would you position the element Y in the periodic table?
(b) What is the general name of the element X and Y.
(c) If the atomic number of Y is 80 then what is the atomic number of X? [3 Marks]

Answer:
(a) Element Y occupies the exact same position in the periodic table as element X.
(b) Isotopes.
(c) Atomic number of X is 80 (since emission of one alpha decreases atomic number by 2, and two beta emissions increase it by 2 each, resulting in no net change).

Teacher's Note:
a) Alpha decay decreases atomic number by 2; each beta decay increases atomic number by 1.
b) Since net change is zero, X and Y have the same atomic number and are isotopes of each other.

(iii) A boy tunes a radio channel to a radio station 93.5 MHz.
(a) Name and define the scientific wave phenomenon involved in tuning the radio channel.
(b) Name the important characteristics of sound that is affected during this phenomenon.
(c) Convert 93.5 MHz to SI unit. [4 Marks]

Answer:
(a) Resonance. Definition: 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) Loudness (amplitude of sound increases significantly due to energy transfer at resonance).
(c) \( 93.5\text{ MHz} = 93.5 \times 10^6\text{ Hz} \) (or \( 9.35 \times 10^7\text{ Hz} \)).

Teacher's Note:
a) Tuning a radio relies on electrical resonance where the LC circuit frequency matches the incoming broadcast frequency.
b) SI unit of frequency is Hertz (Hz).

Question 8

(i) Purvi's friend Tim wants to connect a fuse to his oven. He wants to control the oven from two different locations. Shown below is his circuit diagram.
(a) Which one of the two, A or B should be a live wire?
(b) In the event of an overload, will the fuse serve its purpose?
(c) What is the meaning of the statement that the bulb is rated 600W, 220 V? [3 Marks]

[Figure: A circuit diagram showing a live wire connection with a fuse and two switches controlling an oven load across 220V A.C. supply lines labeled A, B, C.]

Answer:
(a) A should be the live wire (since the fuse must always be connected to the live wire).
(b) Yes, the fuse will melt and break the circuit in the event of an overload, protecting the appliance.
(c) It means that when the bulb is connected to a 220 V supply, it consumes electrical energy at the rate of 600 joules per second (\( 600\text{ W} \)).

Teacher's Note:
a) Fuses are always placed in the live wire before the appliance to disconnect power safely during faults.
b) Power rating definitions should clearly state energy consumption per second at the specified voltage.

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

Answer:
(a) \( _{86}^{222}\text{Rn} \rightarrow _{84}^{218}\text{Po} + _{2}^{4}\text{He} \) (where \( X \) is \( _{2}^{4}\text{He} \) or an alpha particle).
(b) Being positively charged particles, alpha particles (\( X \)) will deflect towards the negatively charged plate of the electric field.

Teacher's Note:
a) Balance both mass numbers (\( 222 = 218 + 4 \)) and atomic numbers (\( 86 = 84 + 2 \)) to identify the emitted particle.
b) Alpha particles carry a charge of \( +2e \), causing them to deflect in electric and magnetic fields.

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

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

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 the circuit including internal resistance \( r = 0.4\Omega \):
\( R_{\text{total}} = 3.875 + 0.4 = 4.275\Omega \).
(b) Total current \( I = \frac{E}{R_{\text{total}}} = \frac{4}{4.275} \approx 0.936\text{ A} \).
Voltage across the parallel combination \( V_p = I \times R_p = 0.936 \times 1.875 \approx 1.755\text{ V} \).
Current through \( 3\Omega \) resistance = \( \frac{V_p}{3} = \frac{1.755}{3} = 0.585\text{ A} \).

Teacher's Note:
a) Account for the internal resistance of the cell when calculating total circuit resistance.
b) Use potential divider rule or parallel voltage to find the current flowing specifically through the \( 3\Omega \) resistor.

Question 9

(i) What mass of ice at \( 0^{\circ}\text{C} \) added to 2.1 kg water, will cool it down from \( 75^{\circ}\text{C} \) to \( 25^{\circ}\text{C} \)?
Given Specific heat capacity of water = \( 4.2\text{ Jg}^{-1}{\circ}\text{C}^{-1} \), Specific latent heat of ice = \( 336\text{ Jg}^{-1} \). [3 Marks]

Answer:
Let mass of ice be \( m \) grams.
Mass of water = \( 2.1\text{ kg} = 2100\text{ g} \).
Heat lost by water = \( m_{\text{water}} \times c_{\text{water}} \times \Delta T \)
\( = 2100 \times 4.2 \times (75 - 25) = 2100 \times 4.2 \times 50 = 4,41,000\text{ J} \).
Heat gained by ice to melt and warm up to \( 25^{\circ}\text{C} \):
\( Q = m \times L_f + m \times c_{\text{water}} \times (25 - 0) \)
\( = m \times 336 + m \times 4.2 \times 25 \)
\( = 336m + 105m = 441m \).
Equating heat lost to heat gained:
\( 441m = 4,41,000 \)
\( m = \frac{4,41,000}{441} = 1000\text{ g} = 1\text{ kg} \).

Teacher's Note:
a) Ensure unit consistency (convert kilograms to grams or vice versa before calculation).
b) Remember ice first melts at \( 0^{\circ}\text{C} \) and then the resulting water warms up to \( 25^{\circ}\text{C} \).

(ii) The diagram below shows a cooling curve for a substance:
(a) State the temperatures at which the substance condenses.
(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? [3 Marks]

[Figure: A cooling curve graph showing temperature on y-axis against time on x-axis, with temperature plateaus at \( 150^{\circ}\text{C} \) (condensation) and \( 60^{\circ}\text{C} \), and lower range down to \( 10^{\circ}\text{C} \).]

Answer:
(a) \( 150^{\circ}\text{C} \) (temperature remains constant during change of state from gas to liquid).
(b) Temperature range between \( 60^{\circ}\text{C} \) and \( 150^{\circ}\text{C} \).
(c) Ice absorbs latent heat of fusion (\( 336\text{ J/g} \)) from the drink in addition to absorbing sensible heat, whereas ice-cold water only absorbs sensible heat, making ice much more effective for cooling.

Teacher's Note:
a) Plateaus on cooling curves represent phase changes (condensation or freezing).
b) Latent heat of fusion makes ice a superior cooling agent compared to water at \( 0^{\circ}\text{C} \).

(iii) A magnet is released along the axis of a copper coil as shown in the diagram.
(a) State the polarity at the top end of the coil when the magnet leaves the coil.
(b) The direction of the current is from A to B when magnet enters the coil. What will be the direction of the current when the magnet leaves the coil.
(c) Name the law which can be used to determine the direction of the induced current in the coil?
(d) State one way to increase the magnitude of the induced current in the coil? [4 Marks]

[Figure: A bar magnet with North pole pointing downwards entering and leaving a copper coil with terminals A at the top and B at the bottom.]

Answer:
(a) South pole (S-pole), because as the North pole leaves, the top of the coil develops an opposite (South) pole to attract and oppose its departure (Lenz's law).
(b) From B to A (reverse direction compared to when it enters).
(c) Lenz's law (or Fleming's Right Hand Rule).
(d) Increase the speed of movement of the magnet or increase the number of turns in the coil.

Teacher's Note:
a) Lenz's law dictates opposition to the motion: attraction when the magnet leaves and repulsion when it enters.
b) Reversing the motion reverses the direction of the induced current.