ICSE Class 10 Physics — 2025 Paper

have only rotational motion

Reason — Forces are equal and opposite, but not along same straight line.

So, the forces will form a couple and the body will have rotational motion only.

moment of couple

Reason — A couple has both magnitude and direction, making it a vector quantity. Its direction is perpendicular to the plane in which the two forces lie. The magnitude is calculated by multiplying one of the forces with the perpendicular distance between their lines of action.

chemical → heat + light

Reason — When a candle burns, chemical energy stored in the wax is transformed into heat and light energy through a combustion reaction.

Frequency of the wave

Reason — Frequency of wave depends on the source and not on the medium through which it propagates.

Gamma-rays, visible light, microwaves

Reason — Wavelength of gamma rays is less than 0.01 nm, wavelength of visible light ranges from 400 to 800 nm and of microwave ranges from 1 mm to 1 m.

Mechanical advantage

Reason — Load vs effort graph is a straight line and slope of the graph = Load/effort = Mechanical advantage.

beyond 2F

Reason — Here, real image size < object size and this happens when the object is placed beyond 2F.

60°

Reason — As, the ray is normally incident on face AB so it passes undeviated. Angle of incidence on face AC is 60˚. The reflected ray is incident normally on face BC and passes undeviated.

Now, from the above diagram,

∠NRS = 60˚

∴ ∠SRC = 30˚

∠ARQ = 30˚

∴ ∠CRT = 30˚

RT is the original path of the incident ray.

So, total deviation = ∠SRT = 30˚ + 30˚ = 60˚

V = ε – Ir

Reason — : The circuit is :

When current flows,

ε = V + Ir

∴ V = ε – Ir

0.5 J g⁻¹ K⁻¹

Reason — As,

$\text {Specific heat capacity} = \frac {\text {Thermal capacity}}{\text {mass}} = \frac{2.5\ \text J \text K⁻¹}{5\ \text g} = 0.5\ \text J\ \text g⁻¹\ \text K⁻¹$

(A) is true but (R) is false.

Explanation:

Assertion (A) is true because as the water column increases, air column decreases, frequency increases.

Reason (R) is false because the frequency is directly proportional to the length of water column.

parallel in both the circuits

Reason — Circuit Y may be redrawn as :

Both the resistors are connected across points A and B. Hence, those are connected in parallel.

Circuit Z may be redrawn as :

Both the resistors are connected across points A and B. Hence, those are connected in parallel.

Two groups to the left of P

Reason — The reaction is :

$$\begin{aligned}_{\text Z}^{\text A}\text{P}\ \longrightarrow \\ _{\text Z - 2}^{\text A - 4}\text{Q}\ + \\ _{2}^{4}\text{He}\end{aligned}$$

In periodic table elements are arranged in ascending order of atomic number and the atomic number of Q is 2 less than that of P. Hence, position of Q will 2 groups left of P.

Lead

Reason — The substance whose specific heat capacity is the lowest will attain the highest temperature.

anticlockwise when magnet is above the ring and clockwise when the magnet is below the ring

Reason

Magnet above the ring (approaching it)

• Flux change The north pole of the magnet faces the ring, so magnetic field lines enter the ring downwards. As the magnet falls closer, the downward flux through the ring increases.

• Opposition required To oppose this increase in downward flux, the ring must create an upward field (its own north pole must face the magnet).

• Current direction Using the right-hand rule, a coil whose top acts as a north pole carries anticlockwise current when viewed from above.

Result: Anticlockwise current while the magnet is above the ring.

Magnet inside and leaving the ring (moving away below it)

• Flux change Once the magnet has passed through, the same downward field lines are now moving away, so the downward flux decreases.

• Opposition required The ring tries to maintain the original flux by producing more downward field—i.e., its top must now behave like a south pole.

• Current direction A coil whose top is a south pole carries clockwise current when viewed from above.

Result: Clockwise current while the magnet is below the ring.

(a) In uniform circular motion the centrifugal force acts away from the centre.

(b) Refractive index of a medium is independent of angle of incidence.

(c) Heat absorbed during change of phase depends on mass of the substance.

(d) Emf of a cell is equal to the terminal voltage when the cell is in open circuit.

(e) In a step-up transformer the turns ratio is more than 1.

(f) The nuclear radiation with lowest ionising power is γ.

(a) R is the most probable position of its centre of gravity.

(b) When a body is freely suspended from a point, it comes to rest (i.e. stops oscillating) in such a position that its centre of gravity lies vertically below the point of suspension.
So, in the figure, the string from the branch passes vertically through point R, hence R is the most probable position of the kite’s centre of gravity.

(a) As PR is the time period of the vibration and reduction of time period means the increase of frequency. Also, Frequency is proportional to square root of the tension.

So, tension is to be increased to increase the frequency i.e., to reduce the time period.

(b) Since, pitch depends on frequency, pitch will be affected by reduction of PR.

This ray will not undergo total internal reflection because for total internal reflection, the light ray should travel from denser to rarer medium but in the given situation light ray travels from rarer to denser medium since refractive index of glass is more than that of water. So, there will be no total internal reflection.

(a) Earth wire is connected to the metal body of electrical appliances and according to new convention the colour of earth wire is green or yellow.

(b) Live wire is connected to the switch of an appliance and according to new convention the colour of live wire is brown.

(a) Alternating current produces a varying magnetic field when it flows through a conductor because alternating current changes its magnitude as well as direction periodically and thus, it provides a varying current and hence varying magnetic field as magnetic field is directly proportional to the applied current.

(b) 50 Hz.

Given,

Mass of paraffin (m) = 200 g

Specific latent heat of fusion of paraffin (L) = 146 J g⁻¹

Let, required heat be Q.

Then,

$\text Q = \text {mL} = 200 \times 146 = 29200\ \text J = 29.2\ \text {kJ}$

(a) Since, looking from the F side current is flowing in anti-clockwise direction and hence, north pole developed at that end and south pole at the T end.

The labelled diagram is given below :

(b)

(a) As current flows for a long period of time, heat is generated and temperature increases. So, the resistance of metallic wire increases as resistance is directly proportional to the temperature of a conductor.

(b) As temperature increases, resistivity of metallic wire increases.

(a) As, Curium has 96 protons and 247 – 96 = 151 neutrons and point P represents 96 protons and 151 neutrons. Hence, P point indicates element Curium.

(b) After the radioactive decay of 1 α and 2 β particles, the atomic number and mass number of daughter nucleus will be 243 and 96 respectively. So, it will have 243 – 96 = 147 neutrons.

As, point R represents 96 protons and 147 neutrons. Hence R point indicates the daughter nucleus.

(c) Mass number of daughter nucleus = Mass number of Curium – Mass number of α particle = 247 – 4 = 243

(a)

From the above diagram, it is visible that inside the prism, the angles of incidence of rays I, J and H are respectively 30˚, 60˚ and 30˚. Hence only the ray J will suffer total internal reflection.

(b)

(a) Given,

Refractive index of glass = 1.5

Apparent depth of air bubble = 4 cm

As,

$$\begin{aligned}\text {Refractive index of glass} = \frac{\text {Real depth of air bubble}}{\text {Apparent depth of air bubble}} \\[1em] ⇒ \text {Real depth of air bubble} = \text {Refractive index of glass}\times \text {Apparent depth of air bubble} \\[1em] = 1.5 \times 4 = 6 \text { cm}\end{aligned}$$

So, actual depth of the air bubble from the surface AB is 6 cm.

(b) As,

$$\begin{aligned}\text {Refractive index of glass} = \frac{\text {Real depth of air bubble}}{\text {Apparent depth of air bubble}} \\[1em] ⇒ \text {Real depth of air bubble} = \text {Refractive index of glass}\times \text {Apparent depth of air bubble} \\[1em]\end{aligned}$$

Now, apparent depth will increase, if refractive index decreases and as wavelength increases, refractive index decreases. As wavelength of yellow light is greater than wavelength of blue light so, refractive index for yellow light is smaller than refractive index of blue light.

Hence, for yellow light apparent depth will be greater.

(c) When the block is turned upside down, then the real depth of the bubble is 15 – 6 = 9 cm

And,

$$\begin{aligned}\text {Apparent depth of air bubble} = \frac{\text {Real depth of air bubble}}{\text {Refractive index of glass}}=\frac{9}{1.5} =6 \text { cm}\\[1 em]\end{aligned}$$

So, the statement is false.

(a) Ray diagram showing the formation of the image when the object is placed at 2F position of a convex lens is shown below:

(b) If we ONLY replace the original lens with another lens of greater focal length then object distance (u) which was at 2F₁ earlier, now comes between F₁ and 2F₁.

And

When an object is placed between F₁ and 2F₁ of the convex lens then it produces real, inverted and magnified image beyond 2F₂ behind the lens.

Hence, if we, ONLY replace the original lens with another lens of a greater focal length then the size of the image increases.

(a) Given,

Object distance (u) = -45 cm

Image distance (v) = -30 cm

From lens formula,

$$\begin{aligned}\frac{1}{\text f} = \frac{1}{\text v} - \frac{1}{\text u} \\[1em] \frac{1}{\text f} = \frac{1}{-30}-\frac{1}{(-45)}\\[1em] \frac{1}{\text f}=-\frac{1}{30}+\frac{1}{45}\\[1em] \frac{1}{\text f} = \frac{-3+2}{90}\\[1em] \frac{1}{\text f} = -\frac{1}{90} \\[1 em] \text f = -\frac{90}{1}\\[1 em] \text f= - 90 \text { cm}\end{aligned}$$

(a) Angle of incidence for both the rays being same, the directions of both the rays remain same. As the refracted ray corresponding to RS bends towards the normal is more than that corresponding to ray PQ hence, the lateral displacement of the emergent ray corresponding to incident ray RS will also be more.

Diagram shows, x₁ and x₂ are the lateral shifts of the emergent rays corresponding to incident rays PQ and RS.

So, here x₂ > x₁

(b) Since, the refracted ray corresponding to incident ray PQ bends less towards the normal, hence this travels with greater speed.

(c) Speed corresponding to incident ray PQ is more than the speed corresponding to incident ray RS. So, wavelength corresponding to incident ray PQ is more than the wavelength corresponding to incident ray RS. So, RS will scatter more in atmosphere.

(a)

1. For ultraviolet radiation, a quartz prism is used since, ordinary glass absorbs UV radiation.
2. Infra red radiation.
3. Visible light radiation (Red, Yellow, Green)

(b) All electromagnetic radiations travels with the same speed ie., 3 × 10⁸ ms^-1.

The diagram of the arrangement is shown below :

Given,

Weight at the end X (W₁) = 2 N

Weight at the end Y (W₂) = 5 N

Distance of X from the knife (R₁) = 70 cm

Distance of Y from the knife (R₂) = 100 cm - 70 cm = 30 cm

Let,

W = weight of the scale

Now,

Anticlockwise moment (due to 2N) = R₁ x W₁ = 2 × 70 = 140 Nm

Clockwise moment (due to 5N) = R₂ x W₂ = 5 × 30 = 150 Nm

Anticlockwise moment (due to weight of the scale) = W × 20 = 20 W

For equilibrium,

Anticlockwise moment of force about knife = Clockwise moment of force about knife

$$\begin{aligned}150 = 140 + 20\text { W} \\[1em] 20 \text W = 150 - 140 = 10 \\[1em] \text W = \frac{10}{20} = 0.5\ \text N\end{aligned}$$

So, weight of the scale is 0.5 N.

(a) All are class I levers since, the fulcrum is in between the load and effort.

(b) As,

$\text {Mechanical Advantage} = \frac{\text {Effort Arm}}{\text {Load Arm}}$

For X, effort arm is greater than the load arm. Hence, it has the maximum mechanical advantage.

(a) Given,

Speed at Q (v) = 10 ms^-1

Then,

Kinetic energy at Q = $\frac{1}{2}$ mv² = $\frac{1}{2}$ × 40 × 10² = 2000 J

(b) If the height of P is H, then :

Potential energy at P = mgH = 40 x 10 x H = 400H

As, friction is negligible so total mechanical energy is conserved.

From law of conservation of total mechanical energy,

Potential energy at P = Kinetic energy at Q

⇒ 400H = 2000 J

⇒ H = $\frac{2000}{400}$ = 5 m

(c) Given,

Height of point R (h) = 3 m

Energy lost from Q to R through friction = 500 J

Now

Potential energy at R = mgh = 40 × 10 × 3 = 1200 J

So, from law of conservation of energy

Kinetic energy at R + Potential energy at R + Energy lost from Q to R through friction = Potential energy at P

⇒ Kinetic energy at R = Potential energy at P - Potential energy at R - Energy lost from Q to R through friction

⇒ Kinetic energy at R = 2000 - 1200 - 500 = 300 J

The block and tackle system of pulleys with velocity ratio equal to 3 is shown below :

Given,

Time taken = 1 minute 40 seconds = 100 s

Speed of sound = 1500 ms^-1)

So,

$$\begin{aligned}\text {Distance of the obstacle} = \frac {\text {Speed of sound} \times \text {Time taken}}{2} \\[1em] =\frac{1500 \times 100}{2} = \frac{150000}{2} \\[1em] = 75000\ \text m = 75\ \text {km}\end{aligned}$$

(a) The phenomenon responsible for falling off the paper rider is resonance.
Resonance is a special case of forced vibrations. When the frequency of the externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance.

(b) A thicker wire will vibrate at a lower resonant frequency than a thinner wire if the length, tension, and material are unchanged. This is because the thicker wire has greater mass per unit length, which lowers its natural frequency of vibration. Therefore, the frequency will be less than 300 Hz.

(a) X is a company fuse or pole fuse. It does not have a fixed current rating. It depends on the load capacity of the connection.

(b) Z can connect & disconnected live and neutral line simultaneously. But the switch which is used to operate domestic appliances connect & disconnected live line only.

(c) Unit is kWh.

(a) Nuclear fission.

(b) X = 236

(c) As, the mass number and atomic number is conserved in the reaction then

From conservation of mass number

Let, mass number of Y be y.

mass number of neutron + mass number of $_{92}^{235}\text{U}$ = 3 x mass number of Y + mass number of $_{36}^{89}\text{Kr}$ + mass number of $_{56}^{144}\text{Ba}$

$$\begin{aligned}1 + 235 = 3 \times \text y + 89 + 144 \\[1em] ⇒ 236 = 3 \text y + 233 \\[1em] ⇒ 3 \text y = 236 - 233 = 3 \\[1em] ⇒ \text y = \frac{3}{3}=1 \\[1em]\end{aligned}$$

So, mass number of Y is 3.

From conservation of atomic number

Let, atomic number of Y be z.

atomic number of neutron + atomic number of $_{92}^{235}\text{U}$ = atomic number of Y + atomic number of $_{36}^{89}\text{Kr}$ + atomic number of $_{56}^{144}\text{Ba}$

$$\begin{aligned}⇒ 92 + 0 = \text z + 36 + 56 \\[1em] ⇒ 92 = \text z + 92 \\[1em] ⇒ \text z = 92 - 92 = 0 \\[1em]\end{aligned}$$

As mass number of Y is 1 and it's atomic number is zero which is a property of a neutron.

Hence, Y is a neutron.

Given,

Battery emf (ε) = 12 V

Voltage rating of eac bulb (V) = 12 V

Power rating of eac bulb (P) = 18 W

(a)

1. Let resistance of each bulb be R.

Then,

$$\begin{aligned}\text P = \frac {\text V^2}{\text R} \\[1 em] ⇒ \text R = \frac {\text V^2}{\text P} = \frac {12^2}{18} = \frac {144}{18}\\[1 em] ⇒ \text R = 8\ \text Ω\end{aligned}$$

2. As, B₂ and B₃ are in parallel combination then

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R_1} + \frac{1}{\text R_2} \\[1 em]\end{aligned}$$

Here, R₁ = R₂ = 8 Ω

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

Now this combination is in series with B₁ then

R_S = R_P + R = 4 + 8 = 12 Ω

So,

$$\begin{aligned}\text {Current drawn} = \frac{\text {Battery emf}}{\text {Total resistance}} = \frac{\text ε}{\text R_\text S} \\[1 em] =\frac{12}{12} = 1\ \text A\end{aligned}$$

Hence, current drawn from the battery is 1 A.

(b) When B₃ is removed, the circuit becomes :

Now, the equivalent resistance = 8 + 8 = 16 Ω.

Since, the resistance increases, current in the circuit decreases.

Hence, brightness of B₁ decreases.

Given,

For ice

Mass of ice (m_i) = 30 g

Initial temperature (T_i) = 0°C

Final temperature (T_f) = 20°C

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

For water

Initial temperature (T_i) = 70°C

Final temperature (T_f) = 20°C

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

Let,

Mass of water = (m_w)

Now,

Heat lost by water = m_wc_w△t

= m_w × 4.2 × (70 - 20)

= m_w × 4.2 × 50 = 210m_w

Heat gained by ice = m_iL_i + m_ic_w△t

= 30 × 336 + 30 × 4.2 × (20-0)

= 10080 + 2520 = 12600 J

By principle of calorimetry,

Heat lost by water = Heat gained by ice

⟹ 210m_w = 12600

⟹ m_w = $\frac{12600}{210}$ = 60 g

So, the mass of required water is 60 g.

(a) Given,

Mass of X = mass of Y = m = 1 g

Change in temperature of X (△t_X) = 10°C

Change in temperature of Y (△t_Y) = 40°C

Let,

Amount of heat = Q

Specific heat capacity of X = c_X

Specific heat capacity of X = c_Y

Now,

For X,

Q = m × c_X × Δt_X = 1 × c_X × 10 = 10c_X

For Y,

Q = m × c_Y × Δt_Y = 1 × c_Y × 40 = 40c_Y

⟹ 10c_X = 40c_Y

⟹ c_X = 4c_Y

Hence, material X has higher specific heat capacity.

(b) Material chosen for calorimeter should have low specific heat capacity. So, here Y material should be used to make a calorimeter.

(c) False as the specific heat capacity of a substance does not remain the same when it changes state and it is phase dependant.

(a) The statement is true. as the magnetic field is upward to downward direction and the direction of current is from left to right so, the rod PQ will experience a force towards C. So, PQ will move towards C.

(b) Fleming’s left hand rule.

(c) If the replaced copper rod has same length but greater cross-sectional area then the resistance of this rod will be lower than the previous rod. So, the current through the new rod will be more than the previous one. Hence, the force experienced will also be more.

(d) As force experienced is directly proportional to the applied current so if current increases, then force obviously increases.