ICSE Class 10 Physics — 2025 Specimen

positive

Reason — Conventionally, if the effect on the body is to turn it anticlockwise, the moment of couple is called anticlockwise moment and it is taken positive, while if the effect on the body is to turn it clockwise, the moment of couple is called clockwise moment and it is taken negative.

momentum

Reason — As,

$\text {Kinetic Energy} = \frac{1}{2}\text m \text v^2$

And,

Momentum (p) = mv

Then,

$\text {Kinetic Energy} = \frac{1}{2\text m}\text m^2 \text v^2 = \frac{p^2}{2\text m}$

So, the kinetic energy of a given body depends on its momentum.

chemical to heat to mechanical to electrical

Reason — For burning of coal in a thermoelectric station, the chemical energy of coal on burning changes into the heat energy of steam then in a steam engine, the heat energy of steam changes into the kinetic energy of the turbine in the form of rotational kinetic energy due to which it rotates the armature of the generator connected to it and thus kinetic energy gets transformed into electrical energy.

2

Reason — The diagram shows that the load is supported by two segment of strings, hence the number of pulleys used in the fixed block are two.

slows down less and refracts less

Reason — In a prism, the refractive index of a given transparent medium decreases with the increase in the wavelength of light. Thus, for visible light the refractive index of the material of a prism is greater for blue colour and minimum for red colour i.e., speed of red colour is greater than speed of blue colour. Consequently, a given prism deviates blue light more as compared to red light.

total internal reflection

Reason — Total internal reflection is the principle that allows light to propagate through the core of an optical fiber by continuously reflecting off the inner walls of the fiber due to the angle at which it strikes those walls.

away from the lens.

Reason —

Given,

Focal length (f) = +12 cm

Object distance (u) = -20 cm

Blurred image distance (v') = +23 cm

Let, clear image distance be v.

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} – \frac{1}{\text u} \\[1 em] \frac{1}{\text v} = \frac{1}{\text f} + \frac{1}{\text u} \\[1 em] \frac{1}{\text v}=\frac{1}{12} + \frac{1}{(-20)}\\[1 em] \frac{1}{\text v}=\frac{1}{12} - \frac{1}{20}\\[1 em] \frac{1}{\text v}=\frac{5-3}{60 }= \frac{2}{60}\\[1 em] \text v = \frac{60}{2} = +30 \text { cm}\end{aligned}$$

Now,

Clear image distance (v) - Blurred image distance (v') = 30 - 23 = 7 cm

So, in order to obtain the clear image, he has to move the screen 7 cm away from the lens.

both A and R are true and R is the correct explanation of A.

Explanation:

Assertion (A) is true : Infrared radiations can travel longer distances through fog and mist because they are less affected by scattering compared to visible light.

Reason (R) is true : Infrared radiations have longer wavelengths, and scattering is inversely proportional to the wavelength. Hence, infrared rays undergo minimal scattering.

Since minimal scattering allows infrared rays to penetrate dense fog and mist more effectively, the reason correctly explains the assertion.

X will be grave and Y will be shriller

Reason — The shrillness of a sound wave is directly proportional to its frequency, meaning that higher frequencies result in shriller sounds and lower frequencies result in grave sound.

P

Reason — The filament of a bulb should produce a high heating effect so its resistance should be high. From the graph, slope of line for substance P is the greatest hence, resistance of P is highest. So, substance P is most suitable for making the filament of a bulb.

green

Reason — In the old colour convention, green colour is used for earth wire.

Reason — As current is flowing in clockwise direction in the coil and according to clock rule if the current in wire around that face is in anticlockwise direction, the face has the north polarity, while if the current at that face is in clockwise direction, the face has the south polarity.

energy needed to raise the temperature of a body by 1°C

Reason — The heat capacity of a body is the amount of heat energy required to raise its temperature by 1°C (or 1 K).

latent heat of fusion

Reason — Heat energy absorbed in change of phase that is not externally manifested by any rise in temperature is called the latent heat of fusion.

protons = 93, neutrons = 144

Reason — An alpha particle consists of two protons and two neutrons. Emission of alpha particle reduces the number of protons by 2 and also the number of neutrons by 2. Hence, number of protons = 95 - 2 = 93 and number of neutrons = 146 - 2 = 144.

(a) A class II lever will always have M.A. > 1.

(b) In a block and tackle system, increase in the weight of the movable block decreases the efficiency of the pulley system.

(c) If the mass as well as the velocity of a body is doubled then the kinetic energy of the body becomes eight times the initial kinetic energy.

(d) Unit of power used in mechanical engineering is horse power.

(e) Two copper wires can have the different resistance but will have same resistivity.

The graph of potential energy vs height for a body thrown vertically upwards is shown below where u(J) represents potential energy in joules and h(m) refers to height in metres.

(a) Ultrasonic waves.

(b) Ultrasonic waves are used for echo depth sounding because they travel long distance without deviation and is not easily absorbed by the medium or can be confined to a narrow beam.

(a) The two materials bend the incoming rays by different amounts and produce two focal points. Hence, two images are formed.

(b) Concave lens because irrespective of the position of the object, the image formed by a thin concave (or divergent) lens is always virtual, upright, diminished, and it is situated on the side of the object between the focus and the lens.

Hence, 2 images will be formed.

(b) Concave lens because irrespective of the position of the object, the image formed by a thin concave (or divergent) lens is always virtual, upright, diminished, and it is situated on the side of the object between the focus and the lens.

The electric iron, is considered safe for the user in circuit A because in this circuit 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.

(a) Step down transformer since it is used to change a high alternating voltage to a low alternating voltage.

(b) Turns ratio is n < 1, i.e., number of turns in the secondary are less than the number of turns in the primary.

Given,

Mass of solid (m_s) = 60 g

Mass of water (m_w) = 150 g

Fall in temperature of solid (△t_s) = (100 - 25) = 75°C

Rise in temperature of water (△t_w) = (25 - 20) = 5°C

Specific heat capacity of water (c_w) = 4.2 Jg⁻¹K⁻¹

Let specific heat capacity of solid be c_s.

Then,

Heat energy given by solid = m_sc_s△t_s = 60 x c x 75 = 4500 x c …..(1)

Heat energy taken by water = m_wc_w△t_w = 150 × 4.2 × 5 = 3150 …..(2)

Assuming that there is no loss of heat energy,

Heat energy given by solid = Heat energy taken by water.

Equating equations 1 & 2, we get,

4500 × c_s = 3150

$\text c_\text s = \frac{3150}{4500}$

c_s = 0.7 J g⁻¹ K⁻¹

Now,

Heat capacity of solid = mass x specific heat capacity = 60 x 0.7 = 42 J K⁻¹

Hence, Heat capacity of solid = 42 J K⁻¹.

(a) Faraday's first law or principle of electromagnetic induction.

(b) It is used as a back-up for the lifts in tall buildings in case of electricity failure.

(a) Gamma (γ) radiation.

(b) No, the emission of gamma radiation does not result in a change in the mass number as they are a form of electromagnetic radiation, essentially high-energy photons, and they have no mass or charge. Since they are massless, their emission does not affect the number of protons or neutrons in the nucleus, and therefore the mass number remains unchanged.

(a) Beta radiations.

(b) The daughter element is E because emission of beta particle reduces one neutron from C and increases it's one proton so the element with one neutron less than C and one excess proton will be the daughter element which is clearly E as seen from the graph.

(c) Fleming's left hand rule.

(a) AS, the refractive index of a medium decreases with increase in temperature, so when the temperature of water is increased by 40 °C then it refractive index will decreases and hence new refractive index will be smaller than 1.33.

Hence, X < 1.33.

(b)

(a) The ray will pass through different points on principal axis after refraction through the convex lens as the refractive index of a medium decreases with the increase in wavelength so red colour will deviate the least, followed by green and then blue will deviate the most.

(b) Ultraviolet radiations.

(c) Ultraviolet radiations are used for sterilizing air, surgical equipment, etc.

(d) Infrared radiation.

(a) As, the object distance (= 24 cm) is greater than the focal length (= 8 cm) of the convex lens which means the object placed beyond it's focus so the lens will form a real and inverted image.

(b)

Given,

Object distance (u) = - 24 cm

Focal length (f) = +8 cm

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1 em] \frac{1}{\text v} = \frac{1}{\text f} + \frac{1}{\text u} \\[1 em] \frac{1}{\text v} = \frac{1}{8}+\frac{1}{(-24)} \\ [1 em] \frac{1}{\text v} =\frac{1}{8}-\frac{1}{24} \\[1 em] \frac{1}{\text v} = \frac{3-1}{24}\\[1 em] \frac{1}{\text v} = \frac{2}{24} \\[1 em] \text v = \frac{24}{2} \\[1 em] \text v =12 \text { cm}\end{aligned}$$

∴ The image is formed at a distance of 12 cm behind the lens.

(a) As the angle of deviation depends directly on the refractive index of a given transparent medium which is itself inversely proportional to the wavelength of light so the angle of deviation decreases with the increase in the wavelength of light. Thus, for visible light the angle of deviation of a prism is maximum for violet colour and minimum for red colour.

(b) A green-coloured ray is a single colour light and hence a prism can not split a single light so the emergent ray will be of green colour.

Given,

Weight at the end A (W_A) = 100 gf

Weight at the end B (W_B) = 100 gf

Length of the rod (L) = 80 cm

Distance of A from the fulcrum (l_A) = 45 cm

Distance of B from the fulcrum (l_B) = 80 cm - 45 cm = 35 cm

Let,

Weight of the rod AC is (W_AC) and distance of centre of gravity of the rod AC from the fulcrum is (l_AC).

And

Weight of the rod BC is (W_BC) and distance of centre of gravity of the rod BC from the fulcrum is (l_BC).

Here,

(l_AC) = 45 - 20 = 25 cm,

(l_BC) = 60 - 45 = 15 cm,

Now,

Anticlockwise moment (due to weight of 100 gf at end A and weight of rod AC) = l_A x W_A + l_AC x W_AC = 45 x 100 + 25 x W_AC

Clockwise moment (due to weight of 100 gf at end B and weight of rod BC) = l_B x W_B + l_BC x W_BC = 35 x 100 + 15 x W_BC

As system is balanced then for equilibrium,

Anticlockwise moment of force about the fulcrum = Clockwise moment of force about the fulcrum

45 x 100 + 25 x W_AC = 35 x 100 + 15 x W_BC

⇒ 4500 + 25W_AC = 3500 + 15W_BC

⇒ 15W_BC - 25W_AC = 4500 - 3500 = 1000

⇒ 3W_BC - 5W_AC = 200

⇒ 3W_BC - 5W_AC > 0

⇒ 3W_BC > 5W_AC

$\Rightarrow \frac{\text W_\text {BC}}{\text W_\text {AC}} \gt \frac{5}{3} \gt 1$

⇒ W_BC > W_AC

Hence, weight of AC < weight of BC.

(b) Even though the weights present are the same at both ends and 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. Thus, more weight is concentrated between C to B.

(a) Zero because when a person walks on a levelled, no work is done against the force of gravity as the displacement of the person is normal to the direction of force of gravity which is vertically downwards.

(b) Negative because when a person climbs a ladder (upward), the displacement covered by the person is opposite to the direction of force of gravity (downwards), so the work done on the person by the force of gravity is negative.

(c) Positive because when a car in neutral gear is coming down the slope, the direction of displacement covered by it is parallel to the direction of force of gravity (downwards), so the work done on the car by the force of gravity is positive.

Given,

h = 5 m,

m = 200 g = 0.2 kg,

g = 10 ms^-2

(a) Potential energy U_B at B is given by

U_B = m x g x h

Substituting the values we get,

U_B = 0.2 x 10 x 5

⇒ U_B = 10 J

Hence, the potential energy of the pendulum at the position B = 10 J.

(b) Total mechanical energy at point C = 10 J

The total mechanical energy is same at all points of the path due to conservation of mechanical energy.

(c) At A, bob has only kinetic energy which is equal to potential energy at B,

Therefore,

$$\begin{aligned}\frac {1}{2}\text m \text v_\text A^2 = \text U_\text B \\[1 em] ⇒ 0.5 \times 0.2 \times \text v_\text A^2 = 10 \\[1 em] ⇒ \text v_\text A^2 = \frac{10}{0.1} \\[1 em] ⇒ \text v_\text A = \sqrt {100} = 10\ \text m \text s^{-1}\end{aligned}$$

(a)The labelled diagram of the system indicating clearly, the direction of the load and effort is shown below :

(b) Given,

m = 100 kg

h = 5 m

g = 10 ms^-2

Then,

Potential Energy (U) = mgh = 100 × 10 × 5 = 5000 J

(a) Given,

Time taken to hear the echo = 3s

Speed of sound = 336 ms^-1

So,

$$\begin{aligned}\text {Distance of the person from the cliff} = \frac {\text {Speed of sound} \times \text {Time taken}}{2} \\[1em] =\frac{336 \times 3}{2} = \frac{1008}{2} \\[1em] = 504\ \text m\end{aligned}$$

So, distance of the person from the cliff is 504 m.

(b) Given,

Time taken to hear the new echo = 3 s + 1.5 s = 4.5 s

So,

$$\begin{aligned}\text {New distance of the person from the cliff} = \frac {\text {Speed of sound} \times \text {Time taken}}{2} \\[1em] =\frac{336 \times 4.5}{2} = \frac{1512}{2} \\[1em] = 756\ \text m\end{aligned}$$

Distance moved by the person = 756 - 504 = 252 m

So, distance moved by the person is 252 m.

(a) At position A, the mother should give a constant periodic push to the swing every time in the forward direction to increase the amplitude of the swing because it is an extreme position where speed of the swing is zero and direction of motion changes from backward direction to forward so at this position direction of motion and direction of push are in parallel while for C it is opposite and B speed is maximum on either direction which can collide with mother and hurt her.

(b) Resonance phenomenon.

(c) The natural frequency of the swing will match the frequency of the force applied by mother this will increase the amplitude.

(a) As the resistance depends directly on length and inversely on area of the resistor and since here tube 2 is thicker and shorter than tube 1 thus, resistance of the tube 2 < resistance of the tube 1.

(b) The voltmeter reading for tube 1 is the same as the voltmeter reading for tube 2 because according to Ohm's law potential difference across a resistor depends only on the current applied not on the resistance of the resistor.

(c) As the specific resistance of a resistor is independent of its dimensions but depends only on the contained material and since here both tubes contain only mercury thus, the specific resistance in both the cases is the same.

(a) $$\begin{aligned}_{\text Z}^{\text A}\text{X} \longrightarrow \\ _{\text Z-2}^{\text A - 4}\text{X}_1 + _{2}^{4}\text{He}\ \longrightarrow \\ _{\text Z-1}^{\text A - 4}\text{X}_2 + \\ _{-1}^{0}\text{e}\ \longrightarrow \\ _{\text Z}^{\text A - 4}\text{Y} + \\ _{-1}^{0}\text{e}\end{aligned}$$

We can see from the above that X and Y have same atomic number. Hence, X and Y will occupy the same position in the periodic table

(b) Isotope.

(c) When the atomic number of Y is 80 and X and Y are isotopes then atomic number of X will also be 80.

(a) 5Ω and 3Ω are in series so their equivalent resistance (R_s) = 5 + 3 = 8Ω

Now 2Ω and R_s are in parallel.

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R_\text s} + \frac{1}{2} \\[1 em] ⇒ \frac{1}{\text R_\text P} = \frac{1}{8} + \frac{1}{2} = \frac{1+4}{8}\\[1 em] ⇒ \text R_\text P = \frac{8}{5} = 1.6\ \text Ω\end{aligned}$$

From circuit diagram,

Internal Resistance = 0.4 Ω

Total resistance of circuit = 1.6 + 0.4 = 2 Ω

(b) Current drawn from the battery :

$$\begin{aligned}\text I = \frac{\text {emf}}{\text {Total resistance}} \\[1em] ⇒ \text I = \frac{4}{2} = 2\ \text A\end{aligned}$$

Now, the current I divides in 2 parts.

Let current across 2Ω be I₁ and across the series combination of 5Ω and 3Ω be I₂.

I = I₁ + I₂ = 2 A ..............(1)

Terminal Voltage of cell = V = IR_p = 2 x 1.6 = 3.2 V

∴ P.d. across 2 Ω resistor = 3.2 V

I₁ = $\frac{3.2}{2}$ = 1.6 A

Substituting value of I₁ in Eq 1,

2 = 1.6 + I₂

⇒ I₂ = 2 - 1.6 = 0.4 A

Hence, current across 3Ω = 0.4 A

Given,

For Water

Mass of water (m_w) = 2.1 kg = 2100 g

Initial temperature (T_i) = 75°C

Final temperature (T_f) = 25°C

Specific heat capacity of water (c_w) = 4.2 J g⁻¹ °C⁻¹

For ice

Initial temperature (T_i) = 0°C

Final temperature (T_f) = 25°C

Specific latent heat of fusion (L_i) = 336 J g⁻¹

Let,

Mass of ice = (m_i)

Now,

Heat lost by water = m_wc_w△t

= 2100 × 4.2 × (75 - 25)

= 2100 × 4.2 × 50 = 441000 J = 441 kJ

Heat gained by ice = m_iL_i + m_ic_w△t

= m_i × 336 + m_i × 4.2 × (25-0)

= 336m_i + m_i × 4.2 × 25

= 336m_i + 105m_i = 441m_i

By principle of calorimetry,

Heat lost by water = Heat gained by ice

⟹ 441000 = 441 m_i

⟹ m_i = $\frac{441000}{441}$ = 1000 g = 1 kg

So, the mass of required ice is 1 kg.

(a) Substance condenses at 150° C.

(b) 150 to 60°C.

(c) 1 g of ice at 0° C takes 336 J of heat energy from the drink to melt into water at 0° C. Thus, the drink liberates an additional 336 J of heat energy to 1 g ice at 0° C than to 1 g ice-cold water at 0° C. Therefore, cooling produced by 1 g ice at 0° C is much more than that by 1 g water at 0° C.

(a) Polarity of A is positive and B is negative.

Reason — Since at the left side of the loop, the lines of force are anti-clockwise, so, by right hand thumb rule we find that the current is going out of terminal A hence, it is positive. Similarly, at the right side of the loop, the lines of force are clockwise, so, terminal B is negative.

(b) The strength of the magnetic field increases when an iron rod is held along the axis of the coil.

(c)

1. By reversing the direction of the current.
2. By decreasing the current.