ICSE Class 10 Physics (Science Paper – 1) Solved Paper 2025

(b) have only rotational motion

Explanation:

Forces are equal and opposite, but not along the same straight line. Therefore, the forces form a couple and the body has rotational motion only.

(d) moment of couple

Explanation:

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. Its magnitude is calculated by multiplying one of the forces by the perpendicular distance between their lines of action.

(c) chemical → heat + light

Explanation:

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

(b) Frequency of the wave

Explanation:

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

(a) Gamma-rays, visible light, microwaves

Explanation:

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.

(a) Mechanical advantage

Explanation:

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

(b) beyond 2F

Explanation:

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

(b) 60°

Explanation:

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°

(b) V = ε – Ir

Explanation:

The circuit is :

When current flows,

ε = V + Ir

∴ V = ε – Ir

(d) 0.5 J g⁻¹ K⁻¹

Explanation:

As,

Specific heat capacity = \frac{\text{Thermal capacity}}{\text{mass}}

= \frac{2.5}{5} = 0.5 J g⁻¹ K⁻¹

(c) (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.

(b) parallel in both the circuits

Explanation:

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.

(d) Two groups to the left of P

Explanation:

The reaction is :

{}^{A}_{Z}\mathrm{P}\longrightarrow{}^{A-4}_{Z-2}\mathrm{Q}+{}^{4}_{2}\mathrm{He}

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.

(d) Lead

Explanation:

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

(c) anticlockwise when magnet is above the ring and clockwise when the magnet is below the ring

Explanation:

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.

Explanation:

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 of paraffin (L) = 146 J g⁻¹

Let, required heat be Q.

Then,

Q = mL

= 200 × 146

= 29200 J

= 29.2 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) Two magnetic lines of force around FT clearly indicating the direction

(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, point P 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 96 and 243 respectively. So, it will have 243 − 96 = 147 neutrons.

As point R represents 96 protons and 147 neutrons, point R 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 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) Complete path

(a) Given,

Refractive index of glass (μ) = 1.5

Apparent depth of air bubble = 4 cm

Refractive index of glass (μ) = \frac{ \text{Real depth of air bubble}}{ \text{Apparent depth of air bubble}}

Real depth of air bubble = 1.5 × 4 = 6 cm.

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

(b) Apparent depth will increase if refractive index decreases. As wavelength increases, refractive index decreases. The wavelength of yellow light is greater than that of blue light, so the apparent depth will be greater for yellow light.

(c) When the block is turned upside down,

the real depth of the bubble is 15 − 6 = 9 cm.

Apparent depth = 9 ÷ 1.5 = 6 cm.

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:

(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.

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

Hence, the size of the image increases.

Given,

Object distance (u) = −45 cm

Image distance (v) = −30 cm

From lens formula,

\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}

or, \frac{1}{\mathrm{f}}=\frac{1}{-30}-\frac{1}{-45}

or, \frac{1}{\mathrm{f}}=-\frac{1}{30}+\frac{1}{45}=-\frac{1}{90}

or, f = − 90 cm

(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 more than that corresponding to ray PQ, the lateral displacement of the emergent ray corresponding to incident ray RS will also be more.

Here, x₂ > x₁.

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

(c) The speed and wavelength corresponding to incident ray PQ are greater than those corresponding to incident ray RS. So, RS will scatter more in the atmosphere.

(a) (1) Ultraviolet radiation. A quartz prism is used since ordinary glass absorbs UV radiation.

(2) Infrared radiation.

(3) Visible light radiation (red, yellow and green).

(b) All electromagnetic radiations travel with the same speed, i.e. 3 × 10⁸ m s⁻¹ in vacuum.

The diagram of the arrangement is shown below.

Given,

Weight at end X = 2 N

Weight at end Y = 5 N

Distance of X from the knife edge = 70 cm

Distance of Y from the knife edge = 100 − 70 = 30 cm

Let the weight of the scale be W. Its weight acts at the 50 cm mark, 20 cm from the knife edge.

For equilibrium,

Anticlockwise moment = Clockwise moment

2 × 70 + W × 20 = 5 × 30

140 + 20W = 150

20W = 10

W = 0.5 N

Therefore, the weight of the ruler is 0.5 N.

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

(b) Mechanical advantage = \text{Effort arm} \over \text{Load arm}

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

(a) Given, speed at Q (v) = 10 m s⁻¹

Kinetic energy at Q = \frac{1}{2} \times40 \times10^2 = 2000 J

(b) If the height of P is H, then potential energy at P = 40 × 10 × H = 400H.

As friction is negligible, total mechanical energy is conserved.

400H = 2000

H = 5 m

(c) Height of point R = 3 m and energy lost from Q to R = 500 J.

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

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 m s⁻¹

Distance of the obstacle = \frac{\text{Speed of sound}\times\text{Time taken}}{2}

= \frac{1500\times100}{2}

= 75,000 m

= 75 km

(a) The phenomenon is resonance. Resonance is a special case of forced vibrations. When the frequency of an externally applied periodic force equals the natural frequency of a body, the body vibrates with a large amplitude.

(b) The frequency f is less than 300 Hz. A thicker wire has greater mass per unit length and therefore a lower natural frequency when its length, tension and material remain unchanged.

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

(b) Z connects and disconnects the live and neutral lines simultaneously, whereas a switch used for a domestic appliance connects and disconnects only the live line.

(c) The unit is kWh.

(a) Nuclear fission.

(b) X = 236.

(c) Conservation of mass number gives:

1 + 235 = 89 + 144 + 3y

236 = 233 + 3y, so y = 1.

Conservation of atomic number gives:

0 + 92 = 36 + 56 + z, so z = 0.

Therefore, Y has mass number 1 and atomic number 0. Hence, Y is a neutron.

Given,

Battery emf (ε) = 12 V

Voltage rating of each bulb (V) = 12 V

Power rating of each bulb (P) = 18 W

(a) 1. Let the resistance of each bulb be R.

Then, \mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}

⇒ \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{12^2}{18}=\frac{144}{18}

⇒ R = 8 Ω

2. Since B₂ and B₃ are connected in parallel:

\frac{1}{\mathrm{R_P}}=\frac{1}{\mathrm{R_1}}+\frac{1}{\mathrm{R_2}}

Here, R1 = R2 = 8 Ω

⇒ \frac{1}{\mathrm{R_P}}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}

⇒ R_P = 4 Ω

This parallel combination is in series with B1.

Total resistance, R_S = R_P + R = 4 + 8 = 12 Ω

Therefore,

Current drawn = \frac{\text{Battery emf}}{\text{Total resistance}}=\frac{12}{12} = 1 A

Hence, the current drawn from the battery is 1 A.

(b) When B₃ is removed, B₁ and B₂ remain in series.

Equivalent resistance = 8 + 8 = 16 Ω

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

Hence, the brightness of B₁ decreases.

Given, mass of ice = 30 g.

Heat gained by ice = heat required to melt the ice + heat required to raise the temperature of the melted ice to 20°C.

= 30 × 336 + 30 × 4.2 × 20

= 10080 + 2520 = 12600 J

Let the mass of water be m g.

Heat lost by water = m × 4.2 × (70 − 20) = 210m J

By the principle of calorimetry:

210m = 12600

m = 60 g

Therefore, the required mass of water is 60 g.

(a) For the same heat supplied to equal masses:

Q = mcΔT

For X, Q = 1 × cX × 10 = 10cX.

For Y, Q = 1 × cY × 40 = 40cY.

Therefore, 10cX = 40cY, so cX = 4cY. Material X has the higher specific heat capacity.

(b) Material Y should be used to make a calorimeter because a calorimeter should have low specific heat capacity.

(c) False. Specific heat capacity generally changes with the physical state of a substance.

(a) True. The magnetic field is directed downwards and the current is from left to right, so rod PQ experiences a force towards C.

(b) Fleming’s left-hand rule.

(c) The force experienced by the rod will increase.

(d) A rod with a greater cross-sectional area has lower resistance. Therefore, a larger current flows through it. Since magnetic force is directly proportional to current, the force increases.