ICSE Class 10 Physics (Science Paper – 1) Solved Paper 2026

(b) Its acceleration must be zero.

Explanation:

A body in dynamic equilibrium moves with uniform velocity. Therefore, its acceleration is zero.

(c) Light energy to chemical energy.

Explanation:

During photosynthesis, plants convert light energy from the Sun into chemical energy stored in food.

(c) 3

Explanation:

When the effort is applied upwards, one end of the rope is fixed to the movable block. The number of supporting rope segments is three, so the velocity ratio is 3.

(d) \frac{\sin 60^\circ}{\sin 45^\circ}

Explanation:

The angle of incidence = 60° and

the angle of refraction = 45°.

Therefore, refractive index of B with respect to A = \frac{\sin 60^\circ}{\sin 45^\circ}.

(a) Infrared radiations.

Explanation:

Infrared radiations lie beyond the red end of the visible spectrum and produce a strong heating effect.

(d) Total mechanical energy remains constant.

Explanation:

In the absence of friction, kinetic energy and potential energy interchange while their sum remains constant.

(d) 15 cm.

Explanation:

For a concave lens, the virtual image is always formed between the optical centre and the principal focus. Hence its distance cannot exceed the focal length.

(d) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.

Explanation:

Blue light is scattered more strongly because scattering varies inversely as the fourth power of wavelength. The stated refractive-index relation does not explain this scattering.

(b) L2

Explanation:

L2 is connected in a separate parallel branch. Its failure does not break the paths through the other lamps.

(b) B

Explanation:

For the same volume of water, the jar with the smallest radius has the greatest water height and therefore the shortest air column. A shorter air column produces a higher frequency and a shriller sound.

(d) Remains the same.

Explanation:

Specific heat capacity is a property of the material and does not depend on its dimensions.

(b) \frac{\mathrm{R}_1\mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}

Explanation:

For two parallel resistors, \frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}.

(a) Only compass 1.

Explanation:

For current into the plane of the paper (×), the right-hand thumb rule gives a clockwise magnetic field.

Compass 1 is tangent to the field and points in the correct direction, whereas compass 2 points opposite to the field direction.

Therefore, only compass 1 is correctly aligned.

(b) B

Explanation:

For equal masses heated at the same rate, the substance taking the longest time absorbs the greatest latent heat.

(c) 7

Explanation:

Mass number = number of protons + number of neutrons = 3 + 4 = 7.

(a) tangential to the path

(b) increases

(c) convection

(d) waveform

(e) superconductor

(f) Alpha

The position of the centre of gravity of a body depends mainly on two factors:

1. Shape of the body
2. Distribution of mass in the body

Explanation:

The position of the centre of gravity depends on the shape of the body and on how its mass is distributed. For example, two rods may have the same shape and size, but if one rod is uniform and the other is heavier at one end, their centres of gravity will be at different positions.

(a) The potato was closer to her hand in Case 1.

(b) The cutter acts as a Class II lever, in which the load lies between the fulcrum and the effort. When the potato is placed closer to the hand applying effort, it is farther from the fulcrum, so the load arm becomes longer. For a fixed effort arm, a longer load arm requires greater effort to cut. Since E1 > E2, the potato must have been closer to her hand in Case 1.

(a) The lens used is a convex lens.

(b) If the object is placed between F and 2F, the magnification will be more than 1.

Explanation:

(a) The graph shows that for an object placed in front of the lens, a real image is formed on the other side and the image distance changes according to the position of the object. This behaviour is shown by a convex lens.

(b) For a convex lens, when the object is placed between F and 2F, the image is formed beyond 2F. The image is real, inverted and magnified. Therefore, the magnification is more than 1.

(a) The voltmeter reading decreases.

(b) The voltmeter reading is given by: V = ε – Ir

where I is the current flowing in the circuit, ε is the emf of the cell and r is the internal resistance of the cell.

When the rheostat is adjusted to increase the current in the circuit, the current drawn from the cell becomes larger. Due to the internal resistance of the cell, a greater potential drop occurs inside the cell.

As a result, the terminal voltage of the cell decreases. Since the voltmeter measures this terminal voltage across the cell, its reading decreases.

(a) The periodic vibrations of a body in the absence of any external force on it are called natural (or free) vibrations.

(b) In natural vibrations, the amplitude remains constant because there is no loss of energy, while in damped vibrations the amplitude gradually decreases with time due to energy loss caused by the presence of resistive force.

Given,

Thermal capacity = 40 J K⁻¹

Heat absorbed = 800 J

Let the rise in temperature be T.

Temperature rise = \frac{\text{Heat absorbed}}{\text{Thermal capacity}}

T = \frac{800}{40} = 20 K

Since a rise of 1 K = 1°C, the rise in temperature is also 20°C.

Hence, the rise in the temperature of the metal piece is 20°C.

(a) Soft iron core

(b) Carbon brushes

(a) The angle of incidence is maximum for prism B.

(b) The angle of incidence is minimum for prism A.

(c) The angle of incidence is equal to the angle of emergence for prism C.

Explanation:

For identical prisms, a larger angle of incidence produces a refracted ray inside the prism making a larger angle with the normal. Hence, diagram B corresponds to maximum incidence, while diagram A corresponds to minimum incidence.

When the angle of incidence is equal to the angle of emergence, the path of the ray through the prism is symmetrical, and the refracted ray inside the prism becomes parallel to the base. This is shown in diagram C.

(a) At the first surface, from air to glass:

\frac{\sin \mathrm{i_1}}{\sin \mathrm{r_1}} = μ

At the second surface, from glass to air:

\frac{\sin \mathrm{i_2}}{\sin \mathrm{r_2}}=\frac{1}{\text{μ}}

Therefore, the product is \mu\times\frac{1}{\text{μ}}=1.

(b) The opposite surfaces are not parallel.

(c) If the opposite surfaces were parallel, the angle of refraction at the first surface would equal the angle of incidence at the second surface. Since these angles are not equal, the surfaces cannot be parallel.

(a) Given, refractive index of the glass block = 1.8 and speed of light in air = 3 × 10⁸ m s⁻¹.

Speed of light in glass = \frac{3\times10^8}{1.8} = 1.67 × 10⁸ m s⁻¹.

(b) If the width of the glass block is doubled, the speed of light does not change because it depends on the refractive index of the material, not its thickness.

Therefore, the speed remains 1.67 × 10⁸ m s⁻¹.

(a) Ultraviolet (UV) radiation

(b) The completed diagram is shown below

Explanation:

The ray AB enters the prism normally at point B, so it passes straight into the prism without bending. It then strikes the slant face with an angle of incidence of 60°. Since 60° is greater than the critical angle of 42°, the ray undergoes total internal reflection. It then reaches the opposite face with an angle of incidence of 30°, which is less than the critical angle, and refracts out of the prism.

Initially, the real image is the same size, so magnification = −1 and u = −v.

From the lens formula:

\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{2}{\mathrm{v}}

Therefore, v = 2f.

After moving the object 12 cm closer, u′ = u + 12 and magnification = −3. Hence, v′ = 3v − 36.

For the final position:

\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v'}}-\frac{1}{\mathrm{u'}}=\frac{4}{\mathrm{v'}}

Therefore, f = v′/4 = (3v − 36)/4.

Using v = 2f:

f = \frac{6\mathrm{f}-36}{4}

Hence, f = 18 cm.

(a) False

Explanation:

After the hailstorm, ice absorbs the heat energy required for melting from the surroundings, so the temperature of the surroundings falls further.

(b) Making the base heavier increases its heat capacity because heat capacity depends on mass.

(c) The principle of calorimetry states that when a hot body is brought in contact with a cold body, heat lost by the hot body equals heat gained by the cold body, provided there is no heat loss to the environment.

(a) The freezing temperature is 20°C because the graph becomes horizontal at this temperature.

(b) The phase change in region QR is freezing, where the liquid changes into solid.

(c) The substance liberates heat faster in the solid state. The graph after point R has a steeper slope, indicating a faster decrease in temperature than in the liquid state before point Q.

(a) Force F1 produces the negative moment because it tends to rotate the wheel clockwise.

(b) No, the wheel is not in equilibrium.

(c) For equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments about the centre O. Although F1 and F2 have equal magnitudes, their perpendicular distances from the centre are different. Therefore, their moments are not equal and opposite, the net moment is not zero, and the wheel is not in equilibrium.

(a) electron volt (eV)

(b) Class I lever

(c) Work done is W = Force × Displacement × cos θ.

In Case 1, force is perpendicular to displacement, so work done is zero.

In Case 2, force acts along displacement, so work done is maximum.

In Case 3, θ < 90°, so the work done is greater than Case 1 but less than Case 2.

Therefore, the ascending order is: Case 1 < Case 3 < Case 2.

Given,  Mass (m) = 20 g = 0.02 kg,

Initial height (h₁) = 45 m,

Rebound height (h₂) = 40 m and g = 10 m s⁻².

(a) Initial potential energy = mgh₁ = 0.02 × 10 × 45 = 9 J.

(b) Just before hitting the ground:

\frac{1}{2}\mathrm{mv^2}=9

v = \sqrt{900} = 30 m s⁻¹.

(c) Kinetic energy after rebound = 0.02 × 10 × 40 = 8 J.

Loss in kinetic energy = 9 − 8 = 1 J.

Given,

Load = 30 kgf

(a) Mechanical advantage (MA) of a machine is:

\mathrm{MA}=\frac{\mathrm{Load}}{\mathrm{Effort}}

For a single fixed pulley, MA = 1.

\mathrm{E_1}=\frac{\mathrm{Load}}{\mathrm{MA}}=\frac{30}{1}=30\ \mathrm{kgf}

For a single movable pulley, MA = 2.

\mathrm{E_2}=\frac{\mathrm{Load}}{\mathrm{MA}}=\frac{30}{2}=15\ \mathrm{kgf}

Ratio of efforts = E1 : E2 = 30 : 15 = 2 : 1.

Hence, the ratio of the efforts in the two cases is 2 : 1.

(b) Potential energy gained = Load × Height raised.

If the effort displacement is the same:

• In a fixed pulley, the load rises through the same distance as the effort.
• In a movable pulley, the load rises through half the distance moved by the effort.

Let the load in the fixed-pulley case rise through height h. The load in the movable-pulley case rises through h/2.

Ratio of potential energies:

\frac{30\times\mathrm{h}}{30\times\frac{\mathrm{h}}{2}}=\frac{2}{1}=2:1

Hence, the ratio of the potential energies gained by the loads is 2 : 1.

(c) In an ideal machine, efficiency = 100% for both pulleys.

Therefore, the ratio of their efficiencies is 1 : 1.

(a) The ruler produces natural vibrations.

(b) The frequency increases because pushing the ruler further inside decreases the length of its free vibrating part. A shorter vibrating length oscillates more rapidly.

Given

• Distance between A and B = 170 m
• Time taken by the person B to hear the echo = 3 s
• Speed of sound in air = 340 ms^-1

(a) (1) Distance of B from the cliff = \frac{340\times3}{2} = 510 m.

(a) (2) Minimum time = \frac{170}{340} = 0.5 s.

(b) The echo is softer (less loud) than the original sound due to the decrease in amplitude of the wave.

(a) Resistance of bulb A = \frac{40^2}{160} = 10 Ω.

Resistance of bulb B = \frac{40^2}{40} = 40 Ω.

Since the bulbs are in series, potential difference divides in the ratio of resistance.

V1 : V2 = 10 : 40 = 1 : 4.

(b) False

Explanation:

If bulb A fuses, the circuit becomes open and the current becomes zero.

(a) The live red wire is connected to the earth terminal, and the green earth wire is connected to the live terminal. The live and earth wires should be interchanged.

(b) The appliance’s outer metallic body may become live, creating a risk of electric shock.

(c) No, the appliance will not function because the wiring connections are incorrect.

(a) When the switch is open, the middle connection does not join the two branches. Therefore, the circuit has two separate branches in parallel. In each branch, the 12 Ω and 6 Ω resistors are in series.

For the top branch:

R1 = 12 + 6 = 18 Ω

For the bottom branch:

R2 = 12 + 6 = 18 Ω

Since R1 and R2 are in parallel:

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

or, \frac{1}{\mathrm{R}}=\frac{1}{18}+\frac{1}{18}=\frac{2}{18}=\frac{1}{9}

Therefore, R = 9 Ω.

Hence, the resistance across AB when switch S is open is 9 Ω.

(b) When switch S is closed, the midpoints are connected. The circuit is rearranged into two parallel combinations connected in series.

For the left combination containing 12 Ω and 6 Ω:

\frac{1}{\mathrm{R_1}}=\frac{1}{12}+\frac{1}{6}=\frac{1}{12}+\frac{2}{12}=\frac{3}{12}=\frac{1}{4}

Therefore, R₁ = 4 Ω.

Similarly, for the right combination:

\frac{1}{\mathrm{R_2}}=\frac{1}{12}+\frac{1}{6}=\frac{1}{12}+\frac{2}{12}=\frac{3}{12}=\frac{1}{4}

Therefore, R₂ = 4 Ω.

These two combinations are in series:

R = R₁ + R₂ = 4 + 4 = 8 Ω.

Hence, the resistance across AB when switch S is closed is 8 Ω.

Given, power (P) = 1100 W, voltage (V) = 220 V and time (t) = 5 h.

(a) Operating current = \frac{\text{P}}{\text{V}}  = \frac{1100}{220} = 5 A.

The fuse rating should be slightly higher than the operating current. Therefore, the minimum suitable fuse rating is 6 A.

(b) Energy consumed = Power × time

= 1100 × 5

= 5500 Wh = 5.5 kWh.

(c) Cost of energy consumption = 5.5 × ₹ 10 = ₹ 55.

The galvanometer deflection depends on the rate of change of magnetic flux and hence on the relative motion between the magnet and coil.

(a) The magnet and coil move in the same direction at the same speed, so there is no relative motion. No current is induced and the galvanometer shows no deflection.

(b) The magnet and coil move in opposite directions, so their relative speed is 10 m s⁻¹. The direction of deflection remains to the right, but its magnitude increases.

(a) (i) Lead is used in the lining of the special aprons.

(ii) It is preferred because it is a dense metal and absorbs harmful nuclear radiations effectively.

(b) (1) The emitted radiation is beta (β) radiation.

(2) Radioactive decay equation:

{}^{24}_{11}\mathrm{Na}\longrightarrow{}^{24}_{12}\mathrm{Mg}+{}^{0}_{-1}\text{β}

In beta decay, the atomic number increases by 1 while the mass number remains unchanged.