ICSE Class 10 Physics Sample Paper 2026 with Solutions

SECTION A (40 Marks)

(Attempt all questions from this Section.)

Question 1 [15 Marks]

Choose the correct answers to the questions from the given options. (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. The moment of couple is taken as: [1 Mark]
(A) positive
(B) negative
(C) maximum
(D) zero

Answer: (A) positive

By sign convention, anticlockwise rotation produced by a moment of couple is considered positive.

Teacher's Note:
a) Remember that anticlockwise moments are taken as positive and clockwise moments as negative.
b) Students often confuse this with direction conventions in mathematics coordinate geometry.

(ii) The kinetic energy of a given body depends on the: [1 Mark]
(A) position
(B) centre of gravity
(C) momentum
(D) displacement

Answer: (C) momentum

Kinetic energy \( K = \frac{p^2}{2m} \), showing its dependence on momentum \( p \) for a given mass.

Teacher's Note:
a) Kinetic energy is related to momentum by the formula \( K = \frac{p^2}{2m} \).
b) Do not confuse kinetic energy with potential energy, which depends on position.

(iii) During power production in a coal-based thermoelectric power plant, the correct sequence of energy conversions taking place is: [1 Mark]
(A) heat \(\rightarrow\) mechanical \(\rightarrow\) chemical
(B) heat \(\rightarrow\) mechanical \(\rightarrow\) electrical
(C) chemical \(\rightarrow\) heat \(\rightarrow\) light
(D) heat \(\rightarrow\) chemical \(\rightarrow\) electrical

Answer: (B) heat \(\rightarrow\) mechanical \(\rightarrow\) electrical

Coal burning produces heat energy, which converts water to steam to rotate turbines (mechanical energy), which then generate electrical energy.

Teacher's Note:
a) Chemical energy of coal first changes to heat energy upon burning.
b) Note that the question asks the sequence starting from heat production during power generation: heat to mechanical (turbine) to electrical (generator).

(iv) Anita used a single movable pulley to lift a bucket of water from a well. She lubricates the pulley. Which of the following statements is true regarding the performance of the pulley used? [1 Mark]
(A) Mechanical Advantage decreases and efficiency increases.
(B) Velocity Ratio increases and efficiency decreases.
(C) Mechanical Advantage remains unchanged and efficiency increases.
(D) Velocity Ratio remains unchanged and efficiency increases.

Answer: (D) Velocity Ratio remains unchanged and efficiency increases.

Lubrication reduces friction, lowering resistive forces, which increases efficiency and mechanical advantage, while velocity ratio remains constant.

Teacher's Note:
a) Velocity ratio depends only on the geometry of the system and remains constant.
b) Lubrication reduces friction, thus increasing efficiency and mechanical advantage.

(v) Inside the prism, during the dispersion of white light, compared to blue, 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 higher speed in glass (less slowing down) and a lower refractive index, resulting in less deviation (refraction) than blue light.

Teacher's Note:
a) Red light has the maximum wavelength and minimum refractive index in glass.
b) Consequently, red light travels faster than blue light inside the prism and deviates the least.

(vi) When objects are viewed through the rising heat of a campfire they appear to shimmer. The optical phenomenon responsible for this effect is: [1 Mark]
(A) refraction
(B) reflection
(C) scattering
(D) total internal reflection

Answer: (A) refraction

Varying temperatures of air near a fire create layers of different densities and refractive indices, causing continuous refraction.

Teacher's Note:
a) Shimmering is caused by refraction of light through layers of air at changing temperatures.
b) Similar atmospheric refraction causes twinkling of stars.

(vii) A convex lens with a focal length of 12 cm has an object at a distance of 20 cm in front of the lens. A blurred image is obtained on the screen placed at a distance of 23 cm in front of the lens. In order to obtain a clear image, the screen has to be moved: [1 Mark]
(A) towards the lens.
(B) away from the lens.
(C) to a position very far from the lens.
(D) either towards or away from the lens.

Answer: (B) away from the lens.

Using lens formula, actual image distance \( v = 30\text{ cm} \). Since screen is at 23 cm, it must be moved away to 30 cm.

Teacher's Note:
a) Calculate the correct image distance using \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \), which yields \( v = +30\text{ cm} \).
b) Since the screen is currently at 23 cm, it needs to be moved further away to 30 cm.

(viii) Assertion(A): Soldiers avoid firing at a target in foggy weather conditions.
Reason(R): In foggy weather, light gets scattered by tiny water droplets, reducing visibility. [1 Mark]
(A) (A) is true but (R) is false.
(B) (A) is false but (R) is true.
(C) Both (A) and (R) are true and (R) is the correct explanation of (A).
(D) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

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

Scattering by water droplets reduces visibility, making targets unclear, which justifies the assertion.

Teacher's Note:
a) Fog consists of tiny water droplets that scatter light in all directions.
b) This scattering blurs targets, making both assertion and reason correct with proper causal connection.

(ix) Two sound waves X and Y have the same amplitude and the 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.

Lower frequency (60 Hz) corresponds to a grave (low pitched) sound, and higher frequency (120 Hz) corresponds to a shrill sound.

Teacher's Note:
a) Pitch is directly proportional to frequency.
b) Wave X with 60 Hz has lower pitch (grave) and wave Y with 120 Hz has higher pitch (shrill).

(x) In the circuit given below, identify the correct relation between the currents flowing through the 2\(\Omega\), 3\(\Omega\), and 5\(\Omega\) resistors: [1 Mark]
[Figure: Circuit diagram with a 12V battery connected in parallel to a branch containing 3\(\Omega\) and 2\(\Omega\) resistors in series, and another parallel branch containing a 5\(\Omega\) resistor.]
(A) current through 2\(\Omega\) > current through 3\(\Omega$
(B) current through 5\(\Omega\) < current through 3\(\Omega$
(C) current through 2\(\Omega\) = current through 5\(\Omega$
(D) current through 5\(\Omega\) > current through 3\(\Omega$

Answer: (C) current through 2\(\Omega\) = current through 5\(\Omega$

The 2\(\Omega\) and 3\(\Omega\) resistors are in series, so the same current flows through both. The official key specifies current through 2\(\Omega\) equals current through 5\(\Omega\); let us check circuit branch currents: total resistance of left branch is \( 2 + 3 = 5\Omega \), same as the right branch resistor (5\(\Omega\)), hence currents are equal.

Teacher's Note:
a) Resistors of 2\(\Omega\) and 3\(\Omega\) are connected in series, carrying identical current.
b) Since both branches have an equivalent resistance of 5\(\Omega$, the current divides equally between them.

(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

Per old and new conventions, earth wire insulation is green or yellow-green.

Teacher's Note:
a) Earth wire colour convention is green or yellow.
b) Neutral is black (old) or blue (new), and live is red (old) or brown (new).

(xii) For an ideal step up transformer: [1 Mark]
(A) \( \frac{\text{Voltage}_{\text{primary}}}{\text{Voltage}_{\text{secondary}}} > 1$
(B) \( \frac{\text{Current}_{\text{primary}}}{\text{Current}_{\text{secondary}}} < 1$
(C) \( \frac{\text{Number of turns}_{\text{primary}}}{\text{Number of turns}_{\text{secondary}}} = 1$
(D) \( \frac{\text{Power}_{\text{primary}}}{\text{Power}_{\text{secondary}}} = 1$

Answer: (D) \( \frac{\text{Power}_{\text{primary}}}{\text{Power}_{\text{secondary}}} = 1$ (the marking scheme also accepts option (D))

For an ideal transformer, input power equals output power, so their ratio is 1.

Teacher's Note:
a) An ideal transformer has 100% efficiency, meaning input power equals output power.
b) Therefore, the power ratio between primary and secondary is always 1.

(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 K.

Teacher's Note:
a) Heat capacity is a body-specific quantity, unlike specific heat capacity which is material-specific.
b) Read carefully to distinguish between heat capacity (whole body) and specific heat capacity (unit mass).

(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 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

Total heat required to melt a given mass without temperature change is the latent heat of fusion.

Teacher's Note:
a) The term "given mass" without unit mass specification refers to total latent heat.
b) If "unit mass" were specified, it would be specific latent heat of fusion.

(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. Subtracting these: 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 initial proton and neutron counts to find the remaining particles.

Question 2

(i) Complete the following by choosing the correct answers from the bracket: [6 Marks]
(a) A ___________ [Class II/Class III] lever will always have M.A. > 1.
(b) A boy uses a GPS device to locate his missing friend in a crowded area; the system primarily uses __________ [ultraviolet waves / microwaves] to track the location.
(c) Unit of specific heat capacity is _______. [\( \text{kg}^2\text{m}^2\text{s}^{-2}\text{K}^{-1} \) / \( \text{m}^2\text{s}^{-2}\text{K}^{-1} \)]
(d) The threshold of hearing is __________. [0dB, 20Hz]
(e) Two copper wires can have different resistivity if they have different __________. [lengths / temperatures]
(f) The reaction responsible for the production of energy in the sun is __________. [Nuclear Fusion / Nuclear Fission]

Answer:
(a) Class II
(b) microwaves
(c) \( \text{m}^2\text{s}^{-2}\text{K}^{-1} \) (Note: Joules per kg Kelvin simplifies to \( \text{m}^2\text{s}^{-2}\text{K}^{-1} \))
(d) 0dB
(e) temperatures
(f) Nuclear fusion

Teacher's Note:
a) Class II levers have load between fulcrum and effort, giving MA always greater than 1.
b) Resistivity of a metallic conductor depends on its temperature, not its dimensions.

(ii) Match the movement of the body part in Column A to the class of lever in Column B. [2 Marks]

Answer:
(a) Nodding head \(\rightarrow\) Class I
(b) Lifting body weight on your toes \(\rightarrow\) Class II

Teacher's Note:
a) In nodding the head, the fulcrum lies between the load and effort (Class I).
b) When lifting on toes, the load is in between the fulcrum at toes and effort at the heel (Class II).

(iii) (a) Name the wave used for echo depth sounding.
(b) Give one reason why the waves mentioned in (a) is used for the above purpose. [2 Marks]

Answer:
(a) Ultrasonic waves.
(b) They travel long distances without deviation / not easily absorbed by the medium OR can be confined to a narrow beam.

Teacher's Note:
a) Ultrasonic waves have very high frequency and short wavelength.
b) Their directional property allows them to travel as a focused beam without significant diffraction.

Question 3

(i) (a) Refer to the diagram given below. A lens is made of two materials of different refractive indices (\(\mu_1\), \(\mu_2\)) as 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: Diagram of a lens divided horizontally into two halves with top refractive index \(\mu_1\) and bottom \(\mu_2\), intersected by a principal axis.]

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

Teacher's Note:
a) Different refractive indices for each half create different focal lengths, forming two separate images.
b) A concave lens always diverges light, producing virtual, erect, and diminished images.

(ii) The image given below displays kilowatt-hour meter readings recorded at two distinct points in time.
Assuming a continuously running 2000W air conditioner as the sole electrical device in use, calculate the time interval, in hours, between these two meter readings. [2 Marks]
[Figure: Two energy meter displays showing 00856 kWh and 01567 kWh respectively.]

Answer:
Energy spent = \( 1567 - 856 = 711\text{ kWh} \)
Time in hours = \( \frac{711\text{ kWh}}{2\text{ kW}} = 355.5\text{ hours} \)

Teacher's Note:
a) Energy consumed is the difference between final and initial meter readings.
b) Time is calculated by dividing total energy (kWh) by power in kilowatts (2 kW).

(iii) ABCDE is a regular pentagon with its centre of gravity at O. What will be the most probable position (W, X, Y, Z or O) of the new centre of gravity:
(a) if a piece of clay is attached at point A?
(b) if the pentagon is cut along the line PQ? (Of the remaining part DCBQPE) [2 Marks]
[Figure: Regular pentagon ABCDE with vertices labelled, center O, and internal points W, X, Y, Z marked near different sides.]

Answer:
(a) X
(b) Z

Teacher's Note:
a) Adding mass at point A shifts the centre of gravity towards vertex A (position X).
b) Removing the lower apex portion shifts the balance point upwards towards Z.

(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 the solid.
[sp. heat capacity of water = 4.2 J g\(^{-1}\) K\(^{-1}\)] [2 Marks]

Answer:
By the principle of mixtures:
Heat lost by solid = Heat gained by water
\( (m \times c \times \Delta T)_{\text{solid}} = (m \times c \times \Delta T)_{\text{water}} \)
Note: Heat capacity \( C' = m \times c \).
\( C' \times (100 - 25) = 150 \times 4.2 \times (25 - 20) \)
\( C' \times 75 = 150 \times 4.2 \times 5 \)
\( C' = \frac{3150}{75} = 42\text{ J K}^{-1} \)

Teacher's Note:
a) Heat capacity is the product of mass and specific heat capacity (\( m \times c \)).
b) Ensure temperature differences are computed accurately for both cooling and heating components.

(v) The diagram given below shows a copper wire wound around a U-shape soft iron bar. An iron pin is brought near the arrangement. First Source P and then Source Q are connected across AB, each operating independently.
(a) State True or False: Source P as well as Source Q, when connected across AB, can attract the iron pin.
(b) Justify your answer to (a) with a suitable reason. [2 Marks]
[Figure: U-shaped iron core with wire wound around it, connected to Source P (AC symbol) and Source Q (DC symbol).]

Answer:
(a) True
(b) Both AC and DC currents produce a magnetic field around the electromagnet, which magnetizes the soft iron core to attract the iron pin.

Teacher's Note:
a) Both alternating current (AC) and direct current (DC) create magnetic fields when passing through a coil.
b) The soft iron core gets magnetized regardless of the current type, attracting the iron pin.

(vi) (a) Name the radiation that is emitted during the decay of a nucleus that has the highest penetrating power.
(b) Does the emission of the above-mentioned radiation result in a change in the mass number? [2 Marks]

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

Teacher's Note:
a) Gamma rays are high-energy electromagnetic waves with maximum penetration power.
b) Since they are photons, their emission does not change the atomic number or mass number.

(vii) Advanced optical sensors in air-to-air missiles use fiber optic cables to transmit light signals with minimal loss. This relies on a physical phenomenon that confines light within the fibers, making the system very dependable for guiding the missile precisely.
(a) Name the optical phenomenon that allows light signals to remain confined within the fiber optic cables during transmission.
(b) Explain the two main conditions necessary for this phenomenon to occur. [3 Marks]

Answer:
(a) Total internal reflection.
(b) 1. Light must travel from a denser medium to a rarer medium.
2. The angle of incidence in the denser medium must be greater than the critical angle for that pair of media.

Teacher's Note:
a) Total internal reflection ensures zero light loss inside fiber optic cables.
b) Both conditions (denser to rarer medium, and incidence angle exceeding critical angle) must be explicitly stated to secure full marks.

SECTION B (40 Marks)

(Attempt any four questions from this Section.)

Question 4

(i) The diagram below shows a fish in the tank and its image seen on the surface of water.
(a) Name the phenomenon responsible for the formation of this image.
(b) A double convex lens with refractive index \(\mu_1\) is placed inside two liquids of refractive indices \(\mu_2\) and \(\mu_3\) as shown in the diagrams below. The refractive indices are such that \(\mu_2 > \mu_1\) and \(\mu_1 > \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 Figure a.
(2) in Figure b. [3 Marks]
[Figure: Two diagrams showing a double convex lens inside Liquid 1 (\(\mu_3\)) in Figure a and Liquid 2 (\(\mu_2\)) in Figure b.]

Answer:
(a) Total internal reflection.
(b) (1) In Figure a (\(\mu_1 > \mu_3$): The lens behaves as a converging lens, so the parallel beam will converge.
(2) In Figure b (\(\mu_2 > \mu_1$): The surrounding medium has a higher refractive index than the lens, so the lens behaves as a diverging lens, making the parallel beam diverge.

Teacher's Note:
a) A fish's image visible from above on the water surface is caused by total internal reflection.
b) Lens behavior depends on the relative refractive index of the lens material compared to the surrounding medium.

(ii) A scientist lowers a metallic ruler vertically into a transparent oil tank. The ruler touches an object placed at the bottom of the tank and gets wet up to the 25 cm mark. If the refractive index of the glycerin is 1.25:
(a) up to which mark will the ruler get wet, if the scientist lowers it up to the image of the object?
(b) how will this length in (a) change if another liquid of \(\mu > 1.25\) is used? [3 Marks]

Answer:
(a) Real depth = 25 cm, Refractive index = 1.25
Apparent depth = \( \frac{\text{Real depth}}{\text{Refractive index}} = \frac{25}{1.25} = 20\text{ cm} \).
(b) The wet length will decrease further because apparent depth decreases with an increase in refractive index.

Teacher's Note:
a) Apparent depth formula is \( \text{Apparent Depth} = \frac{\text{Real Depth}}{\mu} \).
b) Higher refractive index causes greater bending of light, raising the image closer to the surface and decreasing the apparent depth.

(iii) (a) A mixture of red, blue, and green light rays is passed through a convex lens, as illustrated in the diagram below. State whether the ray passes through a single point or through different points on the principal axis after refraction.
(b) Name the invisible radiation which is studied using the quartz prism.
(c) State one use of the radiation mentioned by you in (b) above.
(d) Name one type of radiation with a wavelength greater than that of the radiation mentioned by you in (b) above. [4 Marks]
[Figure: Convex lens refracting a combined beam of red, blue, and green light.]

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

Answer:

Teacher's Note:
a) Chromatic aberration occurs because different colours have different wavelengths and thus different focal lengths in a lens.
b) Quartz prisms are transparent to ultraviolet rays, making them ideal for UV spectrum analysis.

Question 5

(i) An object is placed at a distance of 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:
(a) Real and inverted.
(b) Using lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
Given \( u = -24\text{ cm} \), \( f = 8\text{ cm} \)
\( \frac{1}{v} - \frac{1}{-24} = \frac{1}{8} \)
\( \frac{1}{v} + \frac{1}{24} = \frac{1}{8} \)
\( \frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{3 - 1}{24} = \frac{2}{24} = \frac{1}{12} \)
\( v = +12\text{ cm} \)

Teacher's Note:
a) Apply standard sign conventions strictly: object distance \( u \) is always negative, focal length \( f \) for a convex lens is positive.
b) A positive image distance \( v \) confirms that a real and inverted image is formed on the opposite side of the lens.

(ii) The diagram below shows a cooling curve for a substance X:
(a) State the temperatures at which the substance condenses.
(b) Mention the temperature range in which the substance is in its liquid state.
(c) State True or False: The amount of heat released when a substance is cooled by 10\(^{\circ}\text{C}$ in its liquid state is greater than the heat released when it is cooled by the same amount in its solid state. [3 Marks]
[Figure: Cooling curve graph plotting temperature versus time, showing plateaus at 150\(^{\circ}\text{C}$ and 10\(^{\circ}\text{C}$ and sloping regions.]

Answer:
(a) 150\(^{\circ}\text{C}$
(b) 150\(^{\circ}\text{C}$ to 60\(^{\circ}\text{C}$ (Note: condensation happens at 150\(^{\circ}\text{C}$, freezing at 10\(^{\circ}\text{C}$, and liquid state lies between condensation and freezing temperatures: 150\(^{\circ}\text{C}$ to 60\(^{\circ}\text{C}\))
(c) True.

Teacher's Note:
a) Horizontal portions of a cooling curve represent change of state at constant temperature.
b) Specific heat capacity of the liquid phase is generally greater than that of the solid phase, resulting in more heat release for the same temperature drop.

(iii) In an experiment to measure the temperature of the flame of a Bunsen burner, a lump of copper of mass 0.12 kg is heated on the flame for a long time. The copper then is quickly transferred into a beaker of negligible heat capacity containing 0.84 kg of water and the temperature of the water rose from 15 \(^{\circ}\text{C}$ to 35 \(^{\circ}\text{C}$. Calculate the temperature of the flame.
[Given sp. Heat capacity of copper = 0.4 Jg\(^{-1}\) \(^{\circ}\text{C}\)\(^{-1}\), Sp. Heat capacity of water = 4.2 Jg\(^{-1}\) \(^{\circ}\text{C}\)\(^{-1}\).] [4 Marks]

Answer:
Let flame temperature be \( t \).
Mass of copper \( m_{\text{cu}} = 0.12\text{ kg} = 120\text{ g} \)
Mass of water \( m_w = 0.84\text{ kg} = 840\text{ g} \)
Heat lost by copper = Heat gained by water
\( m_{\text{cu}} \times c_{\text{cu}} \times (t - 35) = m_w \times c_w \times (35 - 15) \)
\( 120 \times 0.4 \times (t - 35) = 840 \times 4.2 \times (35 - 15) \)
\( 48 \times (t - 35) = 840 \times 4.2 \times 20 \)
\( 48 \times (t - 35) = 70560 \)
\( t - 35 = \frac{70560}{48} = 1470 \)
\( t = 1470 + 35 = 1505 \(^{\circ}\text{C}$

Teacher's Note:
a) Ensure mass units are consistent (convert kg to grams or vice-versa) before applying the principle of mixtures.
b) Heat lost by hot copper equals heat gained by cold water up to the final equilibrium temperature.

Question 6

(i) A metal rod AB of length 80 cm is balanced at 45 cm from the end A with 100 gf weights suspended from the two ends.
(a) If this rod is cut at the centre C, then compare the weight of AC to the weight of BC. (Use >, < or =)
(b) Give a reason for your answer in (a). [3 Marks]
[Figure: Rod AB of length 80 cm, fulcrum at C (45 cm from A), with 100 gf weights at ends A and B.]

Answer:
(a) Weight of AC < weight of BC
(b) Even though equal weights are suspended at both ends, the torque arm of B is less than the torque arm of A, meaning the centre of gravity of the rod lies beyond 40 cm towards B, concentrating more weight in section BC.

Teacher's Note:
a) For the rod to balance at 45 cm (which is off-center since total length is 80 cm), the heavier portion must be on the shorter side.
b) Thus, section BC has more mass and weight than section AC.

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

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

Teacher's Note:
a) Work by gravity is zero when displacement is perpendicular to gravitational force (walking on level road).
b) Work is negative when moving upwards against gravity, and positive when moving downwards in the direction of gravity.

(iii) A, B, C and D are four points on a hemispherical cup placed inverted on the ground. Diameter BC = 360 cm and AE = R/3 (R is the radius of the cup). A small spherical mass 500 g at rest at the point A, slides down along the smooth surface of the cup. Assuming that there is no loss of energy, calculate its:
(a) Potential Energy at A relative to B.
(b) Speed at the point B (lowest point).
(c) Kinetic Energy at D (\(g = 10\text{ ms}^{-2}\)). [4 Marks]
[Figure: Inverted hemispherical cup with diameter BC = 360 cm, height AE = R/3, and point D marked on the curve.]

Answer:
Given diameter BC = 360 cm, so radius \( R = 180\text{ cm} = 1.8\text{ m} \).
Height of A above B: \( h = AE = \frac{R}{3} = \frac{1.8}{3} = 0.6\text{ m} \).
(a) Potential Energy at A: \( \text{PE} = mgh = 0.5 \times 10 \times 1.8 = 9\text{ J} \). (Note: height of A above B is \( R = 1.8\text{ m} \), so \( \text{PE} = 0.5 \times 10 \times 1.8 = 9\text{ J} \)).
(b) Speed at B: Using conservation of energy, \( \text{PE at A} = \text{KE at B} \)
\( mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1.8} = \sqrt{36} = 6\text{ ms}^{-1} \).
(c) At point D, height above B is \( h_D = \frac{2R}{3} = \frac{2 \times 1.8}{3} = 1.2\text{ m} \).
\( \text{PE at D} = mgh_D = 0.5 \times 10 \times 1.2 = 6\text{ J} \).
\( \text{KE at D} = \text{Total Energy} - \text{PE at D} = 9 - 6 = 3\text{ J} \).

Teacher's Note:
a) Total mechanical energy remains conserved throughout the motion since friction is absent.
b) Pay careful attention to height measurements relative to the reference base level B.

Question 7

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

Answer:
(a) [Figure: Diagram showing a block and tackle pulley system with 2 pulleys in the upper fixed block and 2 in the lower movable block, total 4 strands supporting the load, showing load downwards, effort downwards, and tensions T in strands.]
(b) Potential Energy \( U = mgh = 100\text{ kg} \times 10\text{ ms}^{-2} \times 5\text{ m} = 5000\text{ J} \).

Teacher's Note:
a) A velocity ratio of 4 requires a total of 4 pulleys (usually 2 in the upper block and 2 in the lower block).
b) Potential energy gained equals the work done against gravity on the load.

(ii) A person standing in front of a cliff fires a gun and hears its echo after 3s. 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 the distance that the person has moved. [3 Marks]

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

Teacher's Note:
a) Echo formula accounts for the round trip of sound: \( d = \frac{v \times t}{2} \).
b) Add the time increment to the initial time to find the new total time for the second echo.

(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.
(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?
(b) Name the phenomenon involved in this.
(c) On the basis of this example, explain how this phenomenon helps to increase the amplitude of the swing. [4 Marks]
[Figure: Mother pushing a girl on a swing passing positions A, B, C.]

Answer:
(a) At A
(b) Resonance
(c) When the frequency of the periodic push applied by the mother matches the natural frequency of the swing, resonance occurs, transferring maximum energy and significantly increasing the amplitude of oscillation.

Teacher's Note:
a) Pushing at the extreme position A at regular intervals ensures energy is supplied in phase with the motion.
b) This is a classic example of resonance in mechanical systems.

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 the following:
(a) Resistance in both cases.
(b) Voltmeter readings in both cases.
(c) Specific resistance in both cases. [3 Marks]
[Figure: Circuit diagram for Ohm's law with voltmeter, ammeter, battery, and two glass tubes filled with mercury of different dimensions labelled tube 1 and tube 2.]

Answer:
(a) Resistance of tube 2 < Resistance of tube 1 (assuming tube 2 has a larger cross-sectional area or shorter length as standard in such comparison problems).
(b) The voltmeter reading for tube 1 is the same as the voltmeter reading for tube 2 (since terminal voltage of the ideal source remains constant across parallel branches/components).
(c) The specific resistance in both cases is the same.

Teacher's Note:
a) Resistance depends on physical dimensions (\( R = \rho \frac{l}{A} \)).
b) Specific resistance (resistivity) is a material property and remains constant for mercury regardless of container dimensions.

(ii) An appliance with a metal covering, rated at 2 kW, 220 V, is to be connected in a circuit. Given below are four diagrams (P, Q, R & S) depicting different circuit configurations,
(a) Identify the safest circuit.
(b) Write two reasons, supported by mathematical calculations, where applicable, to justify your choice. [3 Marks]
[Figure: Four circuit diagrams P, Q, R, and S showing appliances with varying fuse ratings and earthing connections.]

Answer:
(a) S
(b) 1. The metal body of the appliance is properly earthed.
2. Current rating calculation: \( I = \frac{P}{V} = \frac{2000\text{ W}}{220\text{ V}} = 9.09\text{ A} \), so a 10A fuse is appropriate and safe.

Teacher's Note:
a) Safety requires both proper earthing of the metallic casing and an appropriate fuse rating slightly above operating current.
b) Operating current is approximately 9.09 A, so a 10A fuse provides adequate protection without frequent blowing.

(iii) A nichrome wire X with length (\(l\)) & cross-sectional area (\(A\)) is connected to a 10 V source and another nichrome wire Y with length (\(2l\)) & cross-sectional area (\(A/2\)), is connected to a 20 V source.
(a) Compare the resistances of wires X and Y. [Given that the resistivity of nichrome is (\(\rho\)).]
(b) Compare the electrical power consumed by each wire.
(c) Compare the masses of these wires. (Given that the density of nichrome is \(d\).)
(d) State True or False: Wire X and wire Y both show the same rise in temperature in the same time. [4 Marks]

Answer:
(a) \( R_X = \rho \frac{l}{A} \), \( R_Y = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R_X \). Ratio \( R_X : R_Y = 1 : 4 \).
(b) \( P_X = \frac{V_X^2}{R_X} = \frac{10^2}{R} = \frac{100}{R} \). For wire Y, \( P_Y = \frac{V_Y^2}{R_Y} = \frac{20^2}{4R} = \frac{400}{4R} = \frac{100}{R} \). Ratio \( P_X : P_Y = 1 : 1 \).
(c) Mass = Volume \(\times\) Density = \( (\text{Area} \times \text{length}) \times d \).
Mass of X: \( m_X = A \times l \times d \).
Mass of Y: \( m_Y = \frac{A}{2} \times 2l \times d = A \times l \times d \).
Ratio \( m_X : m_Y = 1 : 1 \).
(d) True. Heat produced \( Q = P \times t \). Since both power and time are equal, heat generated is equal, leading to the same temperature rise.

Teacher's Note:
a) Resistance formula is \( R = \rho \frac{l}{A} \); doubling length and halving area increases resistance by a factor of 4.
b) Power calculations using \( P = \frac{V^2}{R} \) show that adjusting voltage compensates for the resistance change, resulting in equal power consumption.

Question 9

(i) Three bulbs of powers \(P_1\), \(P_2\) and \(P_3\) (\(P_1 < P_2 < P_3\)) are connected in a certain way that \(P_3\) glows brightest.
(a) What type of connection exists between these bulbs?
(b) Compare the voltage across these bulbs. (Use >, < or =)
(c) Will the circuit still function if one of the bulbs is fused? [3 Marks]

Answer:
(a) Parallel connection.
(b) \( V_1 = V_2 = V_3 $
(c) Yes.

Teacher's Note:
a) In a parallel connection, every appliance experiences the same potential difference.
b) Bulb with highest power (\(P_3\) lowest resistance) glows brightest, and fusing one bulb does not affect the others.

(ii) The given diagram shows the output of an AC generator. If the speed of the generator coil is doubled, then:
(a) what is the effect on the physical quantity indicated by ‘a’?
(b) what is the effect on the physical quantity indicated as ‘b’?
(c) give reason for your answer in (b). [3 Marks]
[Figure: Sinusoidal AC output waveform with peak amplitude labelled 'a' and time period of one full cycle labelled 'b'.]

Answer:
(a) It will be doubled.
(b) It will be halved.
(c) Doubling the rotational speed halves the time taken to complete one full cycle (frequency is doubled, so time period \( b \) is halved).

Teacher's Note:
a) Peak induced emf (\( a \)) is proportional to rotational speed (\( NAB\omega \)), so doubling speed doubles the peak amplitude.
b) Time period (\( b \)) is inversely proportional to frequency, hence it gets halved when speed is doubled.

(iii) 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.
(a) Identify the radioactive radiation emitted when element C decays into element E. Represent this using a nuclear reaction.
(b) What is the special name given to elements D and E?
(c) If element C transforms into element B by emitting a radioactive ray, how will this ray behave in an electric field? [4 Marks]
[Figure: Neutron-proton scatter plot showing points A, B, C, D, and E with C and D at 147 neutrons and 94/95 protons.]

Answer:
(a) Beta particle (\(\beta\)-emission): \( _{92}^{238}\text{C} \rightarrow _{93}^{238}\text{E} + _{-1}^0\beta $
(b) Isotopes.
(c) It will shift towards the positive plate (if alpha particle) or negative plate depending on the particle emitted; for alpha emission it shifts towards the negative plate.

Teacher's Note:
a) Beta decay involves a neutron converting into a proton, increasing atomic number by 1 while mass number remains constant.
b) Elements with the same number of protons (atomic number) but different numbers of neutrons are called isotopes.