ICSE Class 10 Physics — 2019 Paper

(i) A claw hammer belongs to class I lever.

(ii) Another example of class I lever is shears (used to cut metal sheets)

Given,

$$\begin{aligned}\frac{m_A}{m_B} = \frac{5}{1} \\[0.5em] \frac{K_A}{K_B} = \frac{125}{9}\end{aligned}$$

Since,

$$\begin{aligned}\frac{K_A}{K_B} = \frac{\frac{1}{2}m_Av_A^2}{\frac{1}{2}m_Bv_B^2} \\[0.5em] = \Big(\frac{m_A}{m_B}\Big) \times \Big(\frac{v_A^2}{v_B^2}\Big)\end{aligned}$$

Substituting the values we get,

$$\begin{aligned}\frac{125}{9} = \frac{5}{1} \times \frac{v_A^2}{v_B^2} \\[0.5em] \Rightarrow \frac{v_A^2}{v_B^2} = \frac{125 \times 1}{9 \times 5} \\[0.5em] \Rightarrow \frac{v_A}{v_B} = \sqrt{\frac{25}{9}} = \frac{5}{3} \\[0.5em]\end{aligned}$$

Hence, ratio between velocities of A and B is 5 : 3

(i) Heat energy is measured in calories.

(ii) 1 cal = 4.2 J where joule is the S.I. unit of heat energy.

(i) Two equal and opposite parallel forces, not acting along the same line, form a couple.

(ii) The S.I. unit of moment of couple is Newton x meter or Nm.

(i) Critical angle is the angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90°.

(ii) The refractive index of the medium affects the critical angle of a given medium.

(i) Infra-Red radiation.

(ii) Infra-Red radiation has larger wavelength hence, it is scattered least among visible radiations and thus have deep penetration even in fog.

(i) _aμ_w = $\frac{1}{_wμ_a}$ i.e., they are inverse of each other.

(ii) Given, _aμ_w = $\frac{5}{3}$

Therefore, _wμ_a = $\frac{3}{5}$ = 0.6

Substance B is a good conductor of heat energy. For the same heat energy and same mass, the rise in temperature of B will be more hence, B is a good conductor of heat.

When the holes near the mouth are closed, the effective length of air column increases so the pitch of sound decreases as the frequency of note produced is inversely proportional to the length of air column.

The path of ray is as shown:

(i) Focal length of lens increases.

(ii) A thin lens has greater focal length as compared to a thick lens.

(i) Intensity is the measure of loudness.
Loudness ∝ (amplitude) ²

$$\begin{aligned}\therefore \frac{L_1}{L_2} = \Big(\frac{a_1}{a_2}\Big)^2 \\[0.5em] \Rightarrow \frac{L_1}{L_2} = \Big(\frac{1}{3}\Big)^2 \\[0.5em] \Rightarrow \frac{L_1}{L_2} = \frac{1}{9}\end{aligned}$$

Therefore, Ratio of intensities is 1:9

(ii) Ratio of frequencies is 1:1 (same frequency as pitch is same for both waves).

(i) On increasing the temperature of a metal it's specific resistance increases.

(ii) On increasing the temperature of a semiconductor, it's specific resistance decreases.

(i) Resonant vibrations are vibrations of high amplitude which are produced when the natural frequency of vibrations of the vibrating body is equal to the frequency of forced vibration imposed on it.

(ii) Quality (or timbre) is the characteristic of sound that makes it possible to recognize a person by his voice without seeing him.

No, hydrogen nucleus is just a proton which obviously cannot give out an alpha particle made of two protons and two neutrons.

The given circuit consists of three parts.

In the first part 5 Ω and 4 Ω are in series and if the equivalent resistance is given by R_s then,

R_s = 5 + 4 = 9 Ω

Then, in second part there is parallel combination of 3 Ω and R_s (9 Ω ), if equivalent resistance is R_p

$$\begin{aligned}\frac{1}{R_p} = \frac{1}{3} + \frac{1}{9} \\[0.5em] \Rightarrow \frac{1}{R_p} = \frac{3 + 1}{9} \\[0.5em] \Rightarrow \frac{1}{R_p} = \frac{4}{9} \\[0.5em] \Rightarrow R_p = \frac{9}{4} \\[0.5em] \Rightarrow R_p = 2.25 Ω\end{aligned}$$

In the third part, 8 Ω and R_p ( 2.25 Ω ) are in series. Therefore, the equivalent resistance between AB i.e., R_AB is

8 + 2.25 = 10.25 Ω .

Hence, Total resistance between A and B is 10.25 Ω

(i) The specific heat capacity of a substance changes when it's state changes.

(ii) The specific heat capacity of water is 4.2 J g^-1°C^-1 while that of ice (frozen water) is 2.1 J g^-1°C^-1

x < y

Since, the coil A has less number of turns, therefore the e.m.f. generated in this coil will be less, hence the galvanometer x connected to this coil shows less deflection i.e., x < y.

(i) A nuclear fusion reaction is called a thermo nuclear reaction because it takes place at a temperature of nearly 10⁷ K (10 million).

(ii) ³He₂ + ²H₁ ⟶ ⁴He₂ + ¹H₁ + Energy

The two ways of increasing the speed of a D.C. motor are

1. By increasing the strength of current through the coil.
2. By increasing the number of turns in the coil.

(i) The speed of the body when at a height of 2m above the ground i.e., after falling through 3 m is :

V² = U² + 2gh = (0) + (2 x 10 x 3)
= 60

∴ V = √(60) m s^-1

Mechanical energy at 2m height = P.E. + K.E. = mgh + $\frac{1}{2}$ mv²
= (10 x 10 x 2) + ($\frac{1}{2}$ x 10 x 60)
= 200 + 300
= 500 J

Hence, total mechanical energy possessed by the body at the height of 2 m = 500 J

(ii) The kinetic energy on reaching the ground will be same as the potential energy at it's highest point from which it started falling.

Mechanical energy = Initial potential energy

= m x g x h = 10 x 10 x 5 = 500 J

Hence, kinetic energy possessed by the body just before hitting the ground = 500 J

(i) Let W gf be the weight of metre rule which acts downwards from 50 cm mark.

In equilibrium position, taking moments about fulcrum.

Anticlockwise moment = 40 x (30 - 5) = 1000 gf cm

Clockwise moment = W (50 - 30) = 20 W gf cm

∴ Using principle of moments:
20 W = 1000

⇒ W = $\frac{1000}{20}$ = 50 gf

∴ Weight of the metre scale = 50 gf

(ii) In order to keep the ruler in equilibrium, F is shifted towards 0 cm mark.

(i) The direction of tension (T) is marked in the diagram:

(ii) V.R. = $\frac{\text{Distance moved by effort}}{\text{Distance moved by load}}$ = 2

(iii) Effort E balances the tensions therefore, T = E

(iv) From the relation,

V.R. = $\frac{\text{Distance moved by effort}}{\text{Distance moved by load}}$ = 2

If the free end of the string moves through a distance x then distance moved by load = ?

Substituting the values we get,

$$\begin{aligned}2 = \frac{\text{x}}{\text{Distance moved by load}} \\[0.5em] \Rightarrow \text{Distance moved by load} = \frac{x}{2}\end{aligned}$$

The angle of deviation (δ) produced by a prism first decreases with the increase in angle of incidence (i) and reaches to a minimum value at a certain angle of incidence. With further increase in angle of incidence, angle of deviation also increases.

The graph below shows the variation of angle of deviation with the angle of incidence:

(i) The lens is concave in nature.

(ii) Ray diagram showing the formation of the image is given below:

Given u = 24 cm (-ve), f = 8 cm (+ve)

(i) Image formed is real, inverted and diminished.

(ii) Let v be the distance of image from optical centre. From the relation,

$$\begin{aligned}\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\[0.5em]\end{aligned}$$

Substituting the values we get,

$$\begin{aligned}\frac{1}{v} + \frac{1}{24} = \frac{1}{8} \\[0.5em] \frac{1}{v} = \frac{1}{8} - \frac{1}{24} \\[0.5em] \frac{1}{v} = \frac{3 - 1}{24} \\[0.5em] \frac{1}{v} = \frac{2}{24} \\[0.5em] \frac{1}{v} = \frac{1}{12} \\[0.5em] \Rightarrow v = 12 \text{cm}\end{aligned}$$

Hence, the distance of the image from the lens = 12 cm

(iii) From relation, m = $\frac{v}{u}$

Substituting the values we get,

$$\begin{aligned}m = \frac{12}{-24} \\[0.5em] \Rightarrow m = -0.5\end{aligned}$$

-ve sign of magnification tells us that the image is inverted.

Hence, magnification = -0.5

(i) Loudness

(ii) Base drum has a greater surface area compared to the side drums so amplitude of vibration of base drum is greater than amplitude of vibration of side drums. Hence Loudness of base drum is increased compared to the side drums and it can be distinctly heard from a distance.

Echo is heard after 6 vibrations of pendulum. Since frequency of pendulum is 4 vib per sec, so the time taken for the echo to be heard is t = $\frac{6}{4}$ = 1.5 s.

If d is the distance between the cliff and the observer, then velocity of sound V = $\frac{2d}{t}$

Substituting the values in the equation above:

$$\begin{aligned}340 = \frac{2d}{1.5} \\[0.5em] \Rightarrow d = \frac{340 \times 1.5}{2} = 255 \text{ cm}\end{aligned}$$

Hence, distance between the cliff and the observer = 255 cm

(i) C executes free vibrations (damped if medium is present).

(ii) D executes forced oscillation.

(iii) If the length of D is made equal to C, then the amplitude of vibration of pendulum D increases substantially.

(iv) The phenomenon is called resonance.

(i) The advantage of connecting electrical appliances in parallel combination is that each appliance works independently i.e., any change in one appliance does not affect the other appliance(s).

(ii) The fuse wire should have high specific resistance and low melting point.

(iii) The earth wire in a power circuit is connected to the metallic body of the appliance

(i) The following figure shows the completed diagram:

(ii) Wire A is the live wire and wire B is the neutral wire.

(i) When k is open, resistance is 5 + 0.5 = 5.5 Ω

(ii) From relation, V = IR

Substituting the values in the formula we get,

$$\begin{aligned}3.3 = \text{I} \times 5.5 \\[0.5em] \Rightarrow \text{I} = \frac{3.3}{5.5} \\[0.5em] \Rightarrow \text{I} = 0.6 \text{ A}\end{aligned}$$

Hence, current drawn from the cell when the key k is open = 0.6 A.

(iii) When k is closed, the resistance 2 Ω and 3 Ω are in series, therefore their total resistance is 2 Ω + 3 Ω = 5 Ω . Now since, 5 Ω resistance is in parallel to it,

Therefore,

$$\begin{aligned}\frac{1}{R_p} = \frac{1}{5} + \frac{1}{5} \\[0.5em] \frac{1}{R_p} = \frac{1 + 1}{5} \\[0.5em] \Rightarrow \frac{1}{R_p} = \frac{2}{5} \\[0.5em] \Rightarrow R_p = 2.5 Ω\end{aligned}$$

So now the resistance of circuit is 2.5 + 0.5 = 3 Ω

(iv) Current drawn I = $\frac{\text{V}}{\text{R}}$

Substituting the values we get,

$$\begin{aligned}\text{I} = \frac{3.3}{3} \\[0.5em] \Rightarrow \text{I} = 1.1 \text{ A}\end{aligned}$$

(i) The science of measuring the amount of heat exchanged by the system with it's surrounding is called calorimetry.

(ii) A copper sheet is used to make a calorimeter.

(iii) It is because Copper is a good conductor of heat so it soon acquires the temperature of it's contents and a thin Copper sheet has low heat capacity, so it takes only a small amount of heat from it's contents. Hence, a Calorimeter is made-up of thin sheets of Copper.

The temperature-time graph representing the cooling of naphthalene is shown below:

Given,

Mass of water = 104 g

Temperature of water = 30°C

Final temperature = 10°C

Mass of calorimeter = 42 g

Temperature of ice = 0°C

The heat given by water = m x s x t

Substituting the values we get,

The heat given by water = 104 x 4.2 x (30 - 10) = 8736 J

The heat given by calorimeter = m x s x t

Substituting the values we get,

The heat given by calorimeter = 42 x 0.4 x (30 - 10) = 336 J

Let mass of ice added be m gm

Heat taken by ice to melt = mL = 336m J

Heat taken by ice-water to reach the mixture temperature = mst = m x 4.2 x (10 - 0) = 42m J

Therefore, from principle of calorimetry, heat taken = heat given.

i.e., 336 m + 42 m = 8736 + 336

Therefore, m = $\frac{9072}{378}$ = 24 g

Hence, mass of ice added = 24 g

Labelled diagram of an AC generator is as shown:

(i) The process of splitting of a heavy nucleus into two lighter nuclei of nearly same mass, when bombarded with slow neutrons, is called nuclear fission.

(ii) $_{92}^{235}\text{U}$ + $_{0}^{1}\text{n}$ ⟶ $_{56}^{144}\text{Ba}$ + $_{36}^{89}\text{Kr}$ + 3 $_{0}^{1}\text{n}$ + Energy

(i) North pole of needle deflects towards east.

(ii) On passing current through the wire AB, a magnetic field is produced around the wire which aligns the magnetic needle in it's direction.

(iii) The direction of deflection gets reversed i.e., now it deflects towards west.

(iv) An electric bell works on this principle.