ICSE Class 10 Physics Sample Paper 2025 with Solutions

SECTION A

(Attempt all questions from this Section.)

Question 1

Choose the correct answers to the questions from the given options. [15 Marks]
(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 couple is taken as positive, while a clockwise moment is taken as negative.

Teacher's Note:
a) Remember that sign conventions in rotational motion assign positive to anticlockwise and negative to clockwise moments.
b) Do not confuse this with work done or energy conventions.

(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 \( K = \frac{p^2}{2m} \), showing it depends on momentum \( p \) for a given mass.

Teacher's Note:
a) Kinetic energy is directly proportional to the square of momentum for a constant mass.
b) Position relates to potential energy, not kinetic energy.

(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

Chemical energy of coal produces heat, which converts water to steam to turn turbines (mechanical energy), which then runs a generator to produce electrical energy.

Teacher's Note:
a) Trace each stage of energy transformation sequentially in a thermal power station.
b) Ensure all intermediary steps (heat and mechanical) are included in order.

(iv) The adjacent diagram shows the movable block of a block and tackle system with effort in a convenient direction. From the diagram we can conclude that the number of pulleys used in the fixed block are ________. [1 Mark]
(A) 1
(B) 3
(C) 2
(D) 4

[Figure: A movable block of a block and tackle pulley system with a single string threaded through a sheave, labeled with string and frame.]

Answer: (C) 2

The total number of pulleys in a block and tackle system with velocity ratio 4 is 4 (2 in the upper fixed block and 2 in the lower movable block, or 2 fixed and 1 movable depending on standard setups; here the movable block has 2 pulleys, hence the fixed block also has 2 pulleys).

Teacher's Note:
a) Count the number of strands supporting the movable block to find the velocity ratio and total pulleys.
b) Verify the block configuration from standard textbook diagrams.

(v) White light is dispersed by a prism. Inside the prism, compared to the blue light, the red light [1 Mark]
(A) slows down less and refracts more
(B) slows down more and refracts less
(C) slows down more and refracts more
(D) slows down less and refracts less

Answer: (D) slows down less and refracts less

Red light has a longer wavelength and lower refractive index than blue light, so it travels faster inside the glass and deviates less.

Teacher's Note:
a) Refractive index is inversely proportional to wavelength for visible light in glass (\( \mu_{\text{red}} \lt \mu_{\text{blue}} \)).
b) Lesser refractive index means lesser bending and higher speed inside the medium.

(vi) An endoscope uses optic fiber to transmit high resolution images of internal organs without loss of information. The principle of light that is used by the optic fiber is based on: [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 along the core-cladding boundary.

Teacher's Note:
a) Total internal reflection ensures zero loss of intensity during light propagation through fibers.
b) Mention both core and cladding conditions when explaining optical fiber working.

(vii) A convex lens has focal length \( 12\text{ cm} \) with an object at a distance of \( 20\text{ cm} \) in front of the lens. He obtains a blurred image on the screen placed at a distance of \( 23\text{ cm} \) in front of the lens. In order to obtain the clear image, he has to move the screen [1 Mark]
(A) towards the lens.
(B) away from the lens.
(C) to a position very far away from the lens.
(D) either towards or away from the lens.

Answer: (B) away from the lens.

Given \( f = +12\text{ cm} \), \( u = -20\text{ cm} \). Using lens formula \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \), we get \( \frac{1}{v} - \frac{1}{-20} = \frac{1}{12} \implies v = +30\text{ cm} \). Since screen is at \( 23\text{ cm} \), it must be moved to \( 30\text{ cm} \) (away from the lens).

Teacher's Note:
a) Calculate the exact image distance using the lens formula before determining the direction of screen movement.
b) Real images are formed on the other side of the lens.

(viii) Assertion(A): Infrared radiations travel long distance through a dense fog and mist.
Reason(R): Infrared radiations undergo minimal scattering in earth's atmosphere. [1 Mark]
(A) both A and R are true and R is the correct explanation of A.
(B) both A and R are true and R is not the correct explanation of A.
(C) assertion is false but reason is true.
(D) assertion is true but reason is false.

Answer: (A) both A and R are true and R is the correct explanation of A.

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

Teacher's Note:
a) Scattering of light is inversely proportional to the fourth power of wavelength (Rayleigh scattering).
b) Longer wavelengths scatter the least, making infrared useful in foggy conditions and haze photography.

(ix) Two sound waves X and Y have the same amplitude and the same wave pattern but their frequencies are \( 60\text{ Hz} \) and \( 120\text{ Hz} \) 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

Pitch is directly proportional to frequency. Wave Y has a higher frequency (\( 120\text{ Hz} \)) than X (\( 60\text{ Hz} \)), so Y has a higher pitch (shriller) and X has a lower pitch (grave).

Teacher's Note:
a) Higher frequency corresponds to higher pitch (shrillness).
b) Amplitude determines loudness, which is the same for both waves here.

(x) The graph of voltage vs current for four different materials is shown below.
Which of these four materials would be used for making filament of a bulb? [1 Mark]
(A) Q
(B) S
(C) P
(D) R

[Figure: V-I graph showing four straight lines P, Q, R, S with different slopes, where slope represents resistance.]

Answer: (C) P

Answer: (C) P

Filament of a bulb requires high resistance so it glows brightly. Line P has the steepest slope (highest resistance \( R = \frac{V}{I} \)).

Teacher's Note:
a) The slope of a V-I graph gives the electrical resistance of the material.
b) A material with higher resistance heats up more rapidly for the same current.

(xi) According to the old convention, the colour of the earth wire is: [1 Mark]
(A) black
(B) green
(C) yellow
(D) red

Answer: (B) green

Under the old convention, the earth wire is green (or yellow), while under the new international convention it is green or yellow-green.

Teacher's Note:
a) Live wire is brown (old: red), Neutral is blue (old: black), and Earth is green/yellow-green (old: green).

(xii) Current is flowing through a coil as shown in the figure. Which one of the given figures will correctly depict the magnetic polarity and the direction of the lines of force along the axis of the coil. [1 Mark]
(A) diagram showing N and S
(B) diagram showing reversed N and S
(C) diagram showing correct magnetic field lines emerging from N and entering S
(D) diagram showing different field configuration

[Figure: Four diagrams (a), (b), (c), (d) showing a current-carrying solenoid with different magnetic field line directions and polarities.]

Answer: (C)

Applying the Clock face rule or Right-hand thumb rule, the magnetic field lines emerge from the North pole and enter the South pole correctly in option (C).

Teacher's Note:
a) Use the right-hand thumb rule to determine the North pole of a current-carrying solenoid.
b) Magnetic field lines always run from North to South outside the magnet/solenoid.

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

Answer: (B) energy needed to raise the temperature of a 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) Distinguish carefully between heat capacity (for a whole body) and specific heat capacity (for unit mass).
b) Unit of heat capacity is \( \text{J}^{\circ}\text{C}^{-1} \) or \( \text{J K}^{-1} \).

(xiv) The amount of heat energy required to melt a given mass of a substance at its melting point without any rise in its temperature is called as the: [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

Latent heat of fusion is the total heat required to change a substance from solid to liquid state at constant temperature.

Teacher's Note:
a) Since mass is not specified as unit mass, the term is latent heat of fusion (not specific latent heat).
b) Temperature remains constant during change of state.

(xv) 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 = \( 142 \)
(B) protons = \( 95 \), neutrons = \( 144 \)
(C) protons = \( 93 \), neutrons = \( 144 \)
(D) protons = \( 95 \), neutrons = \( 142 \)

Answer: (C) protons = \( 93 \), neutrons = \( 144 \)

An alpha particle consists of 2 protons and 2 neutrons. Initial protons = \( 95 \), neutrons = \( 146 \). After emission: protons = \( 95 - 2 = 93 \), neutrons = \( 146 - 2 = 144 \).

Teacher's Note:
a) An alpha particle is a helium nucleus \( _2^4\text{He} \), containing 2 protons and 2 neutrons.
b) Subtract 2 from both the proton count and the neutron count.

Question 2

(i) Complete the following by choosing the correct answers from the bracket: [6 Marks]

(a) A ________ [class I/class II/class III] lever will always have M.A. \( \gt 1 \). [1 Mark]

Answer: Class II

Teacher's Note:
a) Class II levers have the load between the fulcrum and effort, ensuring the effort arm is always greater than the load arm.
b) Therefore, mechanical advantage is always greater than 1.

(b) In a block and tackle system, increase in the weight of the movable block ________ [decreases, does not affect, increases] the efficiency of the pulley system. [1 Mark]

Answer: decreases

Teacher's Note:
a) Friction and the weight of the movable block represent wasted energy (lost work).
b) Increasing movable block weight increases useless work, thereby decreasing mechanical efficiency.

(c) If the mass as well as the velocity of a body is doubled then the kinetic energy of the body ________ [is doubled/becomes eight times/becomes four times/the initial kinetic energy. [1 Mark]

Answer: becomes eight times

Teacher's Note:
a) \( K = \frac{1}{2}mv^2 \). If mass \( m \) becomes \( 2m \) and velocity \( v \) becomes \( 2v \), then \( K' = \frac{1}{2}(2m)(2v)^2 = 8 \left(\frac{1}{2}mv^2\right) \).
b) Substitute variables carefully and apply exponents correctly.

(d) Unit of power used in mechanical engineering is ________ [watt / horse power / erg per second] [1 Mark]

Answer: horse power

Teacher's Note:
a) Horsepower (HP) is the traditional unit commonly used in mechanical engineering for engines and machinery.
b) \( 1\text{ HP} = 746\text{ W} \).

(e) Two copper wires can have the different ________ [resistivity / resistance] but will have same ________ [resistance / resistivity] [2 Marks]

Answer: resistance, resistivity

Teacher's Note:
a) Resistivity is a material property and remains same for two wires of the same material (copper).
b) Resistance depends on dimensions (length and area of cross-section), which can differ.

(ii) Draw a graph of potential energy vs height for a body thrown vertically upwards. [Assume no friction is present.] [2 Marks]

[Figure: A line graph with U (J) on the y-axis and height (m) on the x-axis showing a straight line passing through the origin with a positive slope.]

Answer:
The graph is a straight line passing through the origin, showing direct proportionality between potential energy \( U = mgh \) and height \( h \).

Teacher's Note:
a) Ensure axes are correctly labeled: y-axis as \( U\text{ (J)} \) and x-axis as \( \text{height (m)} \).
b) The graph must be a straight line since \( U \propto h \).

(iii) (a) Name the waves used for echo depth sounding. [1 Mark]
(b) Give one reason for their use in the above purpose. [1 Mark]

Answer:
(a) Ultrasonic waves.
(b) They can travel long distances without appreciable deviation or absorption / can be confined to a narrow beam.

Teacher's Note:
a) Ultrasound frequencies are above \( 20,000\text{ Hz} \), making them highly directional.
b) High frequency and short wavelength permit precise reflection mapping.

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? [2 Marks]
(b) A glass lens always forms a virtual, erect and diminished image of an object kept in front of it. Identify the lens. [1 Mark]

[Figure: A composite convex lens split horizontally with upper half having refractive index \( \mu_1 \) and lower half \( \mu_2 \).]

Answer:
(a) 2 images will be formed.
(b) Concave lens.

Teacher's Note:
a) Different refractive indices mean different focal lengths for the two halves, forming separate images.
b) A concave lens always produces virtual, erect, and diminished images for all real object positions.

(ii) If live wire makes an accidental contact with the metal case, which circuit (A or B) in the diagram, illustrating an electric iron, is considered safe for the user (Assuming the fuse is present in the live wire in both circuits)? Justify your answer. [2 Marks]

[Figure: Two circuits A and B of an electric iron. Circuit A has the metal case earthed; Circuit B does not have the metal case earthed.]

Answer:
Circuit in A. In circuit A, metal case is earthed so the person won't get an electric shock, but in circuit B metal case is not earthed so the circuit will be completed through the body of the person giving him a shock.

Teacher's Note:
a) Earthing provides a low-resistance path for leakage current directly to the earth.
b) Mention the safety role of the fuse and earth wire in preventing fatal shocks.

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

Answer:
(a) Step-down transformer.
(b) Less than 1 (\( N_s / N_p \lt 1 \)).

Teacher's Note:
a) A step-down transformer reduces voltage while increasing current.
b) Turns ratio \( N_s / N_p \) is less than 1 for a step-down transformer.

(iv) A solid of mass \( 60\text{ g} \) at \( 100^{\circ}\text{C} \) is placed in \( 150\text{ g} \) 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:
By principle of mixtures:
Heat lost = Heat gained
\( (m \cdot c \cdot \Delta T)_{\text{metal}} = (m \cdot c \cdot \Delta T)_{\text{water}} \)
\( c' \times (100 - 25) = 150 \times 4.2 \times (25 - 20) \)
\( c' \times 75 = 150 \times 4.2 \times 5 \)
\( c' = 42\text{ J K}^{-1} \)

Teacher's Note:
a) Heat capacity is \( m \cdot c \), which is calculated directly without needing the individual mass and specific heat capacity separately if treated as a single product \( C' \).
b) Verify temperature differences correctly: \( 100 - 25 = 75 \) and \( 25 - 20 = 5 \).

(v) (a) Name the principle of AC generator. [1 Mark]
(b) State its one use. [1 Mark]

Answer:
(a) Faraday's first law of electromagnetic induction (or electromagnetic induction).
(b) It is used as a back-up for the lifts in tall buildings in case of electricity failure (or any other valid practical use).

Teacher's Note:
a) State electromagnetic induction clearly as the underlying principle.
b) AC generators supply electricity to homes, industries, and power grids.

(vi) (a) Name the radiations that are emitted during the decay of a nucleus, which has highest penetrating power? [1 Mark]
(b) Does the emission of the above-mentioned radiation result in a change in the mass number? [1 Mark]

Answer:
(a) Gamma rays (\( \gamma \)).
(b) No.

Teacher's Note:
a) Gamma rays are high-energy electromagnetic waves with maximum penetration and zero rest mass.
b) Emission of gamma radiation does not change atomic number or mass number.

(vii) The graph (fig A) illustrates the correlation between the number of protons (x-axis) and the number of neutrons (y-axis) for elements A, B, C, D, and E in the periodic table. These elements are denoted by the letters rather than their conventional symbols. When the element C, depicted in the graph, undergoes radioactive decay, it releases radioactive rays. When these rays are directed into the plane of the paper in the presence of a magnetic field, as indicated in the fig B, they experience deflection, causing them to move upwards. [3 Marks]
(a) Name the radioactive radiations emitted by the element C. [1 Mark]
(b) Identify the daughter element from the graph. [1 Mark]
(c) Name the law used to identify the radioactive radiations emitted by the element. [1 Mark]

[Figure: Fig A shows a N vs Z graph with points A, B, C, D, E. Fig B shows a magnetic field directed into the plane of paper causing upward deflection of rays from C.]

Answer:
(a) Beta rays (\( \beta \)).
(b) Element E.
(c) Fleming's left hand rule.

Teacher's Note:
a) Upward deflection in a magnetic field directed into the page indicates negatively charged beta particles according to Fleming's left hand rule.
b) Beta decay results in an increase of 1 in the proton number and a decrease of 1 in the neutron number, leading from C to E.

SECTION B

(Attempt any four questions from this Section.)

Question 4

(i) The diagram below shows a fish in the tank and its image seen in the surface of water. [3 Marks]
(a) Name the phenomenon responsible for the formation of this image. [1 Mark]
(b) A double convex lens with refractive index \( \mu_1 \) inside two liquids of refractive indices \( \mu_2 \) and \( \mu_3 \) are shown in the diagrams below. The refractive indices are such that \( \mu_2 \gt \mu_1 \) and \( \mu_1 \gt \mu_3 \).
How would a parallel incident beam of light refract when it comes out of the lens in each of the cases shown above?
(1) in Fig a.
(2) in figure b. [2 Marks]

[Figure: A photograph of a fish seen via water surface reflection, and two diagrams Figure a and Figure b showing convex lenses immersed in liquids.]

Answer:
(a) Total internal reflection.
(b) (1) In Fig a (\( \mu_1 \gt \mu_3 \)): Converge.
(2) In Fig b (\( \mu_2 \gt \mu_1 \)): Diverge.

Teacher's Note:
a) When a lens is placed in a medium of higher refractive index than the lens material, its nature reverses (convex behaves as concave and vice versa).
b) Ensure clarity in distinguishing between convergence and divergence based on relative refractive indices.

(ii) 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'. [3 Marks]
(a) State whether \( x \lt 1.33 \) or \( x \gt 1.33 \). [1 Mark]
(b) State two differences between normal reflection and total internal reflection. [2 Marks]

Answer:
(a) \( x \lt 1.33 \).
(b) Differences between normal reflection and total internal reflection:

Teacher's Note:
a) Refractive index decreases with an increase in temperature because density decreases.
b) Total internal reflection reflects 100% of light, unlike normal reflection where some light is refracted/absorbed.

(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 the principal axis after refraction. [1 Mark]
(b) Name the invisible radiations which are studied using the quartz prism. [1 Mark]
(c) State one use of these radiations. [1 Mark]
(d) Name one radiation having the wavelength longer than the wavelength of these radiations. [2 Marks]

*(Note: The sub-part marks total 4 marks as per paper heading).*

[Figure: A beam labeled Red + blue + green passing through a convex lens towards the principal axis.]

Answer:
(a) Different points.
(b) Ultraviolet radiations.
(c) Sterilization purposes / Detecting purity of gems, eggs / in producing vitamin D.
(d) Visible light, infrared rays, microwaves, or radio waves (any one).

Teacher's Note:
a) Different colours refract differently due to varying refractive indices (chromatic aberration).
b) Quartz prisms are transparent to ultraviolet rays, unlike glass which absorbs them.

Question 5

(i) An object is placed at a distance \( 24\text{ cm} \) in front of a convex lens of focal length \( 8\text{ cm} \). [3 Marks]
(a) What is the nature of the image so formed? [1 Mark]
(b) Calculate the distance of the image from the lens. [2 Marks]

Answer:
(a) Real, inverted.
(b) Using lens formula:
\( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
\( \frac{1}{v} - \frac{1}{-24} = \frac{1}{8} \)
\( \frac{1}{v} + \frac{1}{24} = \frac{1}{8} \)
\( \frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{3 - 1}{24} = \frac{2}{24} = \frac{1}{12} \)
\( v = 12\text{ cm} \)

Teacher's Note:
a) Always apply proper sign conventions: \( u = -24\text{ cm} \), \( f = +8\text{ cm} \) for a convex lens.
b) A positive sign for \( v \) confirms that a real image is formed on the opposite 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. [3 Marks]
(a) Which colour will show the maximum angle of deviation and which colour will show the minimum angle of deviation? [2 Marks]
(b) If Instead of sunlight, a green-coloured ray is passed through a glass prism. What will be the colour of the emergent ray? [1 Mark]

Answer:
(a) Violet shows maximum angle of deviation, red shows minimum angle of deviation.
(b) Green.

Teacher's Note:
a) Violet deviates the most because it has the shortest wavelength and highest refractive index in glass.
b) Monochromatic light (green) does not split further into constituent colours.

(iii) O is a luminescent particle trapped inside a glass block. A student traces the path of rays coming out of it and reflecting over a plane mirror as shown in the diagram below. [4 Marks]
Complete the table, using the labels from the figure. The first label is done for you.

[Figure: Ray diagram showing light originating from a point O inside a glass block, refracting, reflecting off a plane mirror with labels a, b, c, Q, U, T, etc.]

Teacher's Note:
a) Match each optical term carefully with the designated ray or angle label in the given diagram.
b) Pay close attention to normal lines and boundary interfaces.

Question 6

(i) A metal rod AB of length \( 80\text{ cm} \) is balanced at \( 45\text{ cm} \) from the end A with \( 100\text{ gf} \) weights suspended from the two ends. [3 Marks]
(a) If this rod is cut at the centre C, then compare the weight of AC to the weight of BC. [1 Mark]
(b) Give a reason for your answer in (a) [2 Marks]

[Figure: A balanced uniform/non-uniform rod AB of length 80 cm with fulcrum at 45 cm from end A and 100 gf weights at ends A and B.]

Answer:
(a) Weight of AC < weight of BC.
(b) Even though the weights present are the same at both ends, the torque arm of B is less than the torque arm of A. This means the moment of the weight of the rod acts from side B and the C.G. lies beyond \( 45\text{ cm} \). Thus, more weight is concentrated between C to B.

Teacher's Note:
a) Apply the principle of moments for a non-uniform rod in equilibrium.
b) The heavier section of the rod must be closer to the fulcrum to balance the unequal leverage.

(ii) For each of the following scenarios, state whether the work done by gravity is positive, negative, or zero. [3 Marks]
(a) a person walks on a levelled road. [1 Mark]
(b) a person climbs a ladder. [1 Mark]
(c) a car in neutral gear is coming down the slope. [1 Mark]

Answer:
(a) No work is done (Zero).
(b) Negative work is done.

(c) Positive work is done.

Teacher's Note:
a) Work \( W = F \cdot s \cdot \cos\theta \). When displacement is perpendicular to gravity (leveled road), \( \theta = 90^{\circ} \), work is zero.
b) Moving upward against gravity is negative work, while moving downward along gravity is positive work.

(iii) The figure below shows a simple pendulum of mass \( 200\text{ g} \). 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 \( 5\text{ m} \). [4 Marks]
(a) What is the potential energy of the pendulum at the position B? [1 Mark]
(b) What is the total mechanical energy at point C? [1 Mark]
(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.) [2 Marks]

*(Note: Marks distribution: (a) 1, (b) 1, (c) 2).*

[Figure: Simple pendulum swinging from position A (mean) to B (extreme) through C (intermediate).]

Answer:
(a) \( U = mgh = 0.2 \times 10 \times 5 = 10\text{ J} \)
(b) \( 10\text{ J} \) (According to the Principle of Conservation of Energy).
(c) By principle of conservation of energy, Total Energy at B = Total Energy at A
\( 10 = \frac{1}{2}mv^2 \)
\( 10 = \frac{1}{2} \times 0.2 \times v^2 \)
\( v^2 = 100 \implies v = 10\text{ m s}^{-1} \)

Teacher's Note:
a) Convert mass into SI units: \( 200\text{ g} = 0.2\text{ kg} \).
b) Total mechanical energy remains constant throughout the swing if friction is ignored.

Question 7

(i) A block and tackle system of pulleys has velocity ratio 4. [3 Marks]
(a) Draw a labelled diagram of the system indicating clearly, the direction of the load and effort. [2 Marks]
(b) Calculate the potential energy of the load \( 100\text{ kgf} \) lifted by this pulley to a height \( 5\text{ m} \). (\( g = 10\text{ ms}^{-2} \)) [1 Mark]

[Figure: Diagram of a block and tackle system with velocity ratio 4 showing pulleys, strings, load, and effort.]

Answer:
(a) Pulleys drawn correctly with support, correct connection of tackle, marking load and effort with correct direction and tension.
(b) \( U = mgh = 100 \times 10 \times 5 = 5000\text{ J} \) (or \( 5000\text{ kgf}\cdot\text{m} \)).

Teacher's Note:
a) A velocity ratio of 4 requires 4 pulleys (usually 2 in the upper fixed block and 2 in the lower movable block).
b) Ensure effort acts downwards for convenience in standard pulley drawings.

(ii) A person standing in front of a cliff fires a gun and hears its echo after \( 3\text{ s} \). If the speed of sound in air is \( 336\text{ m s}^{-1} \) [3 Marks]
(a) Calculate the distance of the person from the cliff. [2 Marks]
(b) After moving a certain distance from the cliff, he fires the gun again and this time the echo is heard \( 1.5\text{ s} \) later than the first. Calculate the distance that the person moved. [1 Mark]

Answer:
(a) Distance \( d = \frac{s \times t}{2} = \frac{336 \times 3}{2} = 168 \times 3 = 504\text{ m} \)
(b) New time \( t' = 3 + 1.5 = 4.5\text{ s} \).
New distance \( d' = \frac{336 \times 4.5}{2} = 168 \times 4.5 = 756\text{ m} \).
Distance moved = \( 756 - 504 = 252\text{ m} \) (or using \( \Delta d = \frac{s \times \Delta t}{2} = \frac{336 \times 1.5}{2} = 252\text{ m} \)).

Teacher's Note:
a) Always divide the total distance covered by sound (to and fro) by 2 to find the distance to the obstacle.
b) Alternatively, calculate the extra distance directly using the time difference.

(iii) The above picture shows a mother pushing her daughter sitting on a swing. The swing is going through the positions A, B, C where A and C are extreme positions and B is the mean position. [4 Marks]
(a) Which is the right position i.e. at A, B or C, for the mother to give a constant periodic push to the swing every time in the forward direction to increase the amplitude of the swing? [1 Mark]
(b) Name the phenomenon involved in this. [1 Mark]
(c) Explain with this example how this phenomenon helps to increase the amplitude of the swing. [2 Marks]

[Figure: A photograph of a mother pushing a child on a swing passing through positions A, B, C.]

Answer:
(a) At position A.
(b) Resonance.
(c) The natural frequency of the swing will match the frequency of the force applied by the mother, and this will increase the amplitude.

Teacher's Note:
a) Resonance occurs when the applied periodic frequency matches the natural frequency of the vibrating body.
b) This results in maximum transfer of energy and a sharp increase in amplitude.

Question 8

(i) The circuit depicted in the figure is employed for studying Ohm's Law. Instead of using a standard resistor, a student opts for a glass tube filled with mercury (\( tube 1 \)), connected to the circuit through two electrodes E1 & E2. He records the readings of the ammeter and voltmeter, thereby calculates the resistance. The student repeats the experiment by substituting \( tube 1 \) with \( tube 2 \), where the same amount of mercury fills the tube 2.
Neglecting internal resistance of the cell use (> or < or =) to compare [3 Marks]
(a) the resistance in both the cases. [1 Mark]
(b) the voltmeter readings in both the cases. [1 Mark]
(c) the specific resistance in both the cases. [1 Mark]

*(Note: tube 2 is wider/shorter or has different dimensions as shown in the diagram).*

[Figure: Circuit diagrams with battery, ammeter, voltmeter, and two mercury tubes of different dimensions labeled tube 1 and tube 2.]

Answer:
(a) Resistance of tube 2 < resistance of tube 1.
(b) The voltmeter reading for tube 1 is the same as the voltmeter reading for tube 2.
(c) The specific resistance in both the cases is the same (\(= \)).

Teacher's Note:
a) Resistance is inversely proportional to cross-sectional area (\( R = \rho \frac{l}{A} \)). A wider tube has lower resistance.
b) Specific resistance (resistivity) depends only on the material (mercury), not on its dimensions.

(ii) A radioactive nucleus X emits an alpha particle followed by two beta particles to form nucleus Y. [3 Marks]
(a) With respect to the element X, where would you position the element Y in the periodic table? [1 Mark]
(b) What is the general name of the elements X and Y? [1 Mark]
(c) If the atomic number of Y is 80 then what is the atomic number of X? [1 Mark]

Answer:
(a) At the same place.
(b) Isotopes.
(c) 80.

Teacher's Note:
a) Alpha emission decreases atomic number by 2 and mass number by 4. Each beta emission increases atomic number by 1 (two beta emissions increase it by 2).
b) Net change in atomic number is zero, meaning X and Y have the same atomic number but different mass numbers (isotopes).

(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. [2 Marks]
(b) Calculate the current through \( 3\ \Omega \) resistance when the circuit is complete. [2 Marks]

*(Note: Circuit has internal resistance \( r = 0.4\ \Omega \), EMF \( E = 4\text{ V} \), with parallel resistors \( 5\ \Omega \) and \( 3\ \Omega \) connected in series with \( 2\ \Omega \)).*

[Figure: Circuit diagram showing a cell of 4V, r = 0.4 ohms, connected to a 2 ohm resistor in series with a parallel combination of 5 ohm and 3 ohm resistors, with a key K.]

Answer:
(a) Equivalent resistance of parallel combination (\( 5\ \Omega \) and \( 3\ \Omega \)):
\( R_p = \frac{5 \times 3}{5 + 3} = \frac{15}{8} = 1.875\ \Omega \) (or as per board key: \( R_2 = \frac{8 \times 2}{8+2} = 1.6\ \Omega \) depending on exact circuit labels; following standard key: \( R_1 = 5 + 3 = 8\ \Omega \), \( R_2 = \frac{8 \times 2}{8+2} = 1.6\ \Omega \), Total resistance \( R = 1.6 + 0.4 = 2.0\ \Omega \)).
(b) Total current \( I = \frac{E}{R + r} = \frac{4}{2} = 2\text{ A} \).
Current through \( 3\ \Omega \) resistance = \( I \times \frac{5}{5 + 3} = 2 \times \frac{5}{8} = 1.25\text{ A} \) (or matching board key formula: \( I_{3\Omega} = \frac{2 \times 2}{10} = 0.4\text{ A} \)).

Teacher's Note:
a) Calculate parallel and series combinations step by step using Ohm's law and current divider rule.
b) Account for the internal resistance of the cell when finding total circuit resistance and total current.

Question 9

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

Answer:
Heat lost = Heat gained
\( m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T = m_{\text{ice}} \cdot L_{\text{fusion}} + m_{\text{ice}} \cdot c_{\text{water}} \cdot \Delta T_{\text{melted ice}} \)
\( 2100\text{ g} \times 4.2 \times (75 - 25) = m \times 336 + m \times 4.2 \times (25 - 0) \)
\( 2100 \times 4.2 \times 50 = m \times 336 + m \times 4.2 \times 25 \)
\( 2100 \times 210 = m \times 336 + 105 m \)
\( 441000 = 441 m \)
\( m = 1000\text{ g} = 1\text{ kg} \)

Teacher's Note:
a) Convert all masses into grams to maintain consistency with the given specific latent heat units (\( \text{J g}^{-1} \)).
b) Remember that the 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: [3 Marks]
(a) State the temperatures at which the substance condenses. [1 Mark]
(b) The temperature range in which the substance is in liquid state. [1 Mark]
(c) Why do we prefer ice to ice-cold water for cooling a drink? [1 Mark]

[Figure: A cooling curve showing temperature vs time, with a horizontal condensation plateau at \( 150^{\circ}\text{C} \) down to \( 60^{\circ}\text{C} \) and solidification at \( 10^{\circ}\text{C} \).]

Answer:
(a) \( 150^{\circ}\text{C} \).
(b) \( 150^{\circ}\text{C} \) to \( 60^{\circ}\text{C} \).
(c) Every gram of ice can absorb \( 336\text{ J} \) of heat more than ice-cold water due to its high specific latent heat of fusion.

Teacher's Note:
a) Flat regions on cooling curves represent phase changes at constant temperature.
b) Latent heat of fusion makes ice much more effective at cooling than water at the same temperature.

(iii) The diagram below shows a cardboard on which iron filings are kept. A wire bent in the form of a loop is seen passing through the cardboard. When current flows through it the iron filings arrange themselves as shown with the direction of magnetic field. [4 Marks]
(a) State the polarities of the battery at A and B. [1 Mark]
(b) State the effect on the magnetic field if an iron rod is held along the axis of the coil. [1 Mark]
(c) State one way to:
1. change the polarity of the coil.
2. decrease the strength of the magnetic field around the coil. [2 Marks]

*(Note: Marks distribution: (a) 1, (b) 1, (c) 1+1).*

[Figure: A cardboard with iron filings around a current-carrying circular loop connected to a battery with terminals A and B.]

Answer:
(a) A: positive and B: negative.
(b) Magnetic field becomes stronger / magnetic flux increases.
(c) 1. Interchange the polarity of the terminals of A and B (reversing the direction of current through the coil).
2. Decreasing the strength of the current through the coil.

Teacher's Note:
a) An iron rod placed inside a current-carrying solenoid turns it into an electromagnet, greatly increasing field strength.
b) Reversing current direction reverses the magnetic poles (North and South).