ICSE Class 10 Physics — 2020 Paper

(i) The moment of force is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation.

(ii) The S.I. and C.G.S. unit of moment of force are related as follows —

1 N m = 10⁵ dyne x 10² cm = 10⁷dyne cm

One kilowatt hour is the electrical energy consumed by an electrical appliance of power 1 kilowatt when it is used for 1 hour.

1 kilowatt hour = 1 kilowatt x 1 hour
= 1000 J s^-1 x 3600 s = 3.6 x 10⁶ J

The work done is zero because the centripetal force on the body at any instant is directed towards the centre of the circular path and the displacement at that instant is along the tangent to the circular path, i.e., normal to the direction of force on the body. Since, θ = 90° , cos 90° = 0. Therefore, work done is zero.

(i) The class of the lever shown in the diagram is class III.

(ii) By applying effort at a larger distance from the fulcrum i.e., shifting effort E towards load L, thereby increasing the length of effort arm.

(i) Chemical energy changes into electrical energy — In a battery when current is drawn from it in the external circuit.

(ii) Electrical energy changes into sound energy — In a loud speaker, electrical signals vibrate the diaphragm of loud speaker to produce sound.

Let the work done by both cranes be 'W'

Given,

Time taken by crane A to do work, t₁ = 5 s

Time taken by crane B to do work, t₂ = 2 s

Power of crane A, P_A = $\frac{W}{t_1}$

Power of crane B, P_B = $\frac{W}{t_2}$

$$\begin{aligned}\frac{P_A}{P_B} = \frac{Wt_2}{Wt_1} \\[0.5em] \frac{P_A}{P_B} = \frac{t_2}{t_1} \\[0.5em] \frac{P_A}{P_B} = \frac{2}{5}\end{aligned}$$

P_A : P_B = 2 : 5

Below is the ray diagram showing the path of the ray till it emerges out of the slab:

Below ray diagram shows the path of ray AB till it emerges out of the prism:

(i) In order to get an enlarged real image, the object should be placed between F₁ (or f) and 2F₁ (or 2f) distance i.e., 2F₁ > u > F₁.

(ii) In order to get an enlarged virtual image, the object should be placed between F₁ and optical centre i.e., u < F₁

Given,

apparent depth = 2.7 m

Refractive index of the water = $\frac{4}{3}$

Refractive index of the liquid = $\frac{\text{Actual depth}}{\text{Apparent depth}}$

Substituting the values we get,

$$\begin{aligned}\frac{4}{3} = \frac{\text{Actual depth}}{2.7} \\[0.5em] \Rightarrow \text{Actual depth} = \frac{4}{3} \times 2.7 = 3.6 \text{ m}\end{aligned}$$

Hence, actual depth = 3.6 m

(i) The speed is same for both colours in vacuum because velocity of light of all colors in vacuum or air remains the same and it changes only when the medium changes.

(ii) The light of red colour has greater speed in glass as it has a larger wavelength.

Below is the graph between displacement from mean position and time for a body executing free vibration in a vacuum:

Given,

wavelength = 0.4 m

speed of sound in water = 1400 m s^-1

As we know,

frequency of wave = $\frac{\text{Speed}}{\text{Wavelength}}$

Substituting the values we get,

$$\begin{aligned}\text{frequency of wave} = \frac{1400}{0.4} \\[0.5em] \text{frequency of wave} = 3500 \text{ Hz} = 3.5 \text{ kHz} \\[0.5em]\end{aligned}$$

Hence, we get the frequency of wave = 3500 Hz and this lies in the audible range of frequency (i.e., 20 Hz to 20,000 Hz), hence it is audible.

The stone lying in the sun get heated up much more than water lying for the same duration of time because the specific heat capacity of water is much more than that of stone. With the same exposure to sun, stone acquires higher temperature due to it's low specific heat capacity.

It is not advisable to use a piece of copper wire as fuse wire in an electric circuit because due to it's high conductivity, low specific resistance and high melting point, copper does not melt when heated up and thus does not break the flow of current in the circuit. Thus, the purpose of fuse wire will not get fulfilled.

This circuit consists of two parts. In the first part, the resistances 3 ohm and 6 ohm are connected in parallel. If R_p is the equivalent resistance then

$$\begin{aligned}\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} \\[0.5em] \frac{1}{R_p} = \frac{2 + 1}{6} \\[0.5em] \frac{1}{R_p} = \frac{3}{6} \\[0.5em] \frac{1}{R_p} = \frac{1}{2} \\[0.5em] \Rightarrow R_p = 2 \space Ω\end{aligned}$$

Therefore, R_p = 2 Ω

In the second part, R_p (i.e., 2 ohm) and 5 ohm are in series. If the equivalent resistance = R_s the we get,

R_s = 2 + 5 = 7 Ω

Hence, equivalent resistance across A and B = 7 Ω

Let mass of block P = m_p and mass of block Q = m_Q

Given,

$\frac{m_p}{m_Q}$ = $\frac{2}{1}$

Q = m x s x ∆t

Since, heat supplied (∆Q) and rise in temperature (∆t) is same for both block P and Q,

Therefore,

$$\begin{aligned}m_p \times s_p \times ∆t = m_Q \times s_Q \times ∆t \\[0.5em] m_p \times s_p = m_Q \times s_Q \\[0.5em] \frac{s_p}{s_Q} = \frac{m_Q}{m_p} = \frac{1}{2}\end{aligned}$$

Therefore, s_P : s_Q = 1 : 2

(i) In order to experience zero force, the conductor should be parallel to the magnetic field i.e., the angle between the length of the conductor and the magnetic field should be zero.

(ii) In order to experience maximum force, the conductor should be kept perpendicular to the magnetic field i.e.,the angle between the length of the conductor and the magnetic field should be 90°.

The emission of alpha and beta particle is represented below:

$$\begin{aligned}_{84}^{202}\text{X} \longrightarrow \space _{82}^{198}\text{Z} + \space _{2}^{4}\text{He} \text{ (α particle emission)} \\[2em] _{82}^{198}\text{Z} \longrightarrow \space _{83}^{198}\text{Y} + \space _{-1}^{\space \space \space 0}e \text{ (β particle emission)}\end{aligned}$$

∴ a = 83 and b = 198

(i) The magnetic polarity of the loop that faces us is south pole as the current is in clockwise direction.

(ii) We can increase the strength of the magnetic field produced by this loop by decreasing the value of variable resistance which in turn will increase the current I in the circuit.

Given,

h = 5 m, m = 200 g = 0.2 kg, g = 10 ms^-2

(i) Potential energy U_B at B is given by

U_B = m x g x h

Substituting the values we get,

$$\begin{aligned}U_B = 0.2 \times 10 \times 5 \\[0.5em] \Rightarrow U_B = 10 J\end{aligned}$$

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

(ii) 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.

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

Therefore,

$$\begin{aligned}\frac{1}{2} \times m \times (v_A)^{2} = U_B \\[0.5em] 0.5 \times 0.2 \times (v_A)^{2} = 10 \\[0.5em] (v_A)^{2} = \frac{10}{0.1} \\[0.5em] \Rightarrow v_A = \sqrt{100} \\[0.5em] \Rightarrow v_A = 10 \text{ m s}^{-1}\end{aligned}$$

(i) Centripetal force always acts towards the centre of circle along it's radius. It is this force which compels the body to move along the circular path. A centrifugal force appears to act away (outward) from the centre of circular path. Thus, the two forces act in opposite directions.

(ii) No, the centrifugal force is not the force of reaction of centripetal force.

(iii) Centripetal force and centrifugal force have equal magnitudes.

(i) Labelled diagram of the system with direction and points of application of load and effort marked is shown below:

(ii) Increase in weight of movable block does not affect velocity ratio. Therefore, V.R. = 4.

(i) The object appears taller to the diver.

(ii) Below diagram shows the path of two rays AC & AD:

The path of the ray AB through the glass prism in PQR till it emerges out of the prism is shown below:

Note that the total internal reflection takes place at C.

(i) The lens used is convex lens.

(ii) Given, f = +20 cm, v = +60 cm, u = ?

From lens formula,

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

Substituting the values in the formula,

$$\begin{aligned}\frac{1}{20} = \frac{1}{60} - \frac{1}{u} \\[0.5em] \Rightarrow \frac{1}{u} = \frac{1}{60} - \frac{1}{20} \\[0.5em] \Rightarrow \frac{1}{u} = \frac{1 - 3}{60} \\[0.5em] \Rightarrow \frac{1}{u} = \frac{- 2}{60} \\[0.5em] \Rightarrow u = - 30 cm \\[0.5em]\end{aligned}$$

Hence, the lens is 30 cm in front of the object

(iii) Magnification m = $\frac{v}{u}$

Substituting the values in the formula above we get,

$m = \frac{60}{- 30} = -2$

Thus, magnification = 2 times

(i) Clouds are dust particles and water molecules suspended in the atmosphere near the earth of size bigger than the wavelength of visible light. Thus, they scatter all colours of incident white light from the sun to the same extent and hence when scattered light reaches our eyes, the clouds appear white.

(ii) The air molecules in the atmosphere scatter the blue end of spectrum having shorter wavelengths much more than colours on the red end. In the absence of direct light from sun, the light reaching our eyes is the scattered light which is rich in blue colour. Hence, light from all directions of sky being rich in blue colour makes the sky appear blue.

(i) The system which enables us to locate underwater objects by transmitting ultrasonic waves and detecting the reflecting impulse is called sound navigation and ranging or SONAR. Here, ultrasonic waves from a ship are transmitted in all directions. The reflected waves received in the receiver gives the distance and position of the object (e,g, submarine etc.) in water.

(ii) The acoustically measurable quantity related to pitch is frequency and to loudness is intensity of sound.

(i) When the stem of the tuning fork is placed in contact with a table top, it sets the table surface in small vibrations but the large number of air molecules in contact with the table surface also start vibrating, thereby producing a loud sound.

(ii) Let the man be standing at distance d from the cliff and v be the speed of sound in air.

For first echo, t₁ = $\frac{2d}{v}$ = 3.5 s    [Eqn i ]

and for second echo, t₂ = $\frac{2(d - 84)}{v}$ = 3 s    [Eqn ii]

Dividing eqn (ii) by eqn (i)

$$\begin{aligned}\frac{\frac{2(d - 84)}{v}}{\frac{2d}{v}} = \frac{3}{3.5} \\[0.5em] \Rightarrow \frac{d - 84}{d} = \frac{3}{3.5} \\[0.5em] \Rightarrow 3.5 (d - 84) = 3d \\[0.5em] \Rightarrow (3.5d) - (3.5 \times 84) = 3d \\[0.5em] \Rightarrow (3.5d) - 3d = 294 \\[0.5em] \Rightarrow 0.5 d = 294 \\[0.5em] \Rightarrow d = \frac{294}{0.5} \\[0.5em] \Rightarrow d = 588 m\end{aligned}$$

Hence, the distance of the cliff from the initial position of the man = 588 m

(i) The core is made up of soft iron.

(ii) The completed diagram of the transformer is shown below:

(i) A superconductor is a substance of zero resistance (or infinite conductance) at a very low temperature.

(ii) For the given appliance, W = 110 W, V = 220 V

$\therefore \text{Current I} = \frac{W}{V} = \frac{110}{220} = 0.5 \text{ A}$

Hence, current drawn by the appliance = 0.5 A

(iii) The resistance of semi conductors such as carbon, silicon, etc., decreases with increase in temperature.

(i) The equivalent resistance R_p of resistances 3 Ω and 1.5 Ω joined in parallel is given by

$$\begin{aligned}\frac{1}{R_p} = \frac{1}{3} + \frac{2}{3} \space \space \big[\because 1.5 Ω = \frac{3}{2} Ω\big] \\[0.5em] \Rightarrow \frac{1}{R_p} = \frac{3}{3} \\[0.5em] \Rightarrow R_p = 1 Ω\end{aligned}$$

From, V = IR

the potential drop across this combination = 0.3 x 1 = 0.3 V

$$\begin{aligned}\therefore \text{Current through 3 Ω resistance} = \frac{\text{Voltage}}{\text{Resistance}} \\[0.5em] = \frac{0.3}{3} = 0.1 \text{ A}\end{aligned}$$

(ii) Total resistance of the circuit = R_p + 4 + r = (5 + r) Ω

Current in the circuit,

$$\begin{aligned}0.3 = \frac{1.8}{5 + r} \\[0.5em] 0.3 \times (5 + r) = 1.8 \\[0.5em] (0.3 \times 5) + (0.3 \times r) = 1.8 \\[0.5em] 1.5 + (0.3 \times r) = 1.8 \\[0.5em] 0.3 \times r = 1.8 - 1.5 \\[0.5em] r = \frac{0.3}{0.3} = 1 Ω\end{aligned}$$

∴ Internal resistance r = 1 Ω

(i) The heat required to raise the temperature of a substance or body through 1 °C (or 1 K) is called it's heat capacity.

(ii) S.I. unit of heat capacity is J °C^-1 or J K^-1

(iii) Specific heat capacity = $\frac{\text{Heat capacity}}{\text{Mass of body or substance}}$

(i) In part CD, the rate of rise in temperature is slow as the substance in this phase has high specific heat capacity while temperature rises rapidly in part AB (so higher slope) as the substance has lower specific heat capacity in this phase. The different specific heat capacities are attributed to different phases of state (i.e. solid and liquid phase).

(ii) Boiling point of substance is t₂ and melting point of substance is t₁

Let final temperature of water be T °C.

Heat energy taken by the ice to melt at 0 °C

Q₁ = mass of ice x specific latent heat of fusion of ice
= 60 x 336 = 20,160 J

Heat energy taken by the melted ice to raise it's temperature from 0 °C to T °C.

Q₂ = mass of ice x specific heat capacity of water x rise in temperature
= 60 x 4.2 x (T - 0) = 252T J

Total heat energy taken by ice = Q₁ + Q₂
= (20,160 + 252 T) J

Heat energy given by water in fall of it's temperature from 50 °C to T °C
= mass of water x specific heat capacity of water x fall in temperature
= 140 x 4.2 x (50 - T)
= 588 (50 - T) J

If there is no heat loss,

heat energy given by water = total heat energy taken by ice

Substituting the values we get,

$$\begin{aligned}588 (50 - T) = 20,160 + 252 T \\[0.5em] (588 \times 50) - 588 T = 20,160 + 252 T \\[0.5em] 29,400 - 588 T = 20,160 + 252 T \\[0.5em] 29,400 - 20,160 = 588 T + 252 T \\[0.5em] 9240 = 840 T \\[0.5em] \Rightarrow T = \frac{9240}{840} \\[0.5em] \Rightarrow T = 11 °C\end{aligned}$$

Hence, T = 11 °C

(i) Below is the labeled diagram of a d.c. motor:

(ii) It is used to produce rotational motion in appliances like fan, mixer, grinder, washing machine etc.

(i) The difference between nuclear fusion and nuclear fission is as follows :

(ii) The nuclear waste should be enclosed in thick casks and buried in deep underground places far from populated areas as they can be harmful.

(i) After the emission of an alpha particle from A, the resulting nucleus B has 2 protons and 2 neutrons less than the part nucleus (i.e., A), therefore, the nucleus B has :
No. of proton = 82
No of neutron = 126

(ii) After the emission of a beta particle from B, the resulting nucleus C 1 proton more and 1 neutron less than the part nucleus (i.e., B) therefore, the nucleus C has :
No. of proton = 83
No. of neutron = 125

(iii) Mass number of the nucleus A = (84 + 128 ) = 212.

(iv) No, the composition of a nucleus does not change if it emits gamma radiation.