Question 1

Question 2(c)Complete the path of the monochromatic light ray AB incident on the surface PQ of the equilateral glass prism PQR till it emerges out of the prism due to refraction. [2]

Accepted Answer

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

Question 2

Question 2(e)A pond appears to be 2.7 m deep. If the refractive index of water is 4/3, find the actual depth of the pond. [2]

Accepted Answer

Given,apparent depth = 2.7 mRefractive index of the water = 43\dfrac{4}{3}34​Refractive index of the liquid = Actual depthApparent depth\dfrac{\text{Actual depth}}{\text{Apparent depth}}Apparent depthActual depth​Substituting the values we get,43=Actual depth2.7⇒Actual depth=43×2.7=3.6 m\dfrac{4}{3} = \dfrac{\text{Actual depth}}{2.7} \[0.5em] \Rightarrow \text{Actual depth} = \dfrac{4}{3} \times 2.7 = 3.6 \text{ m}34​=2.7Actual depth​⇒Actual depth=34​×2.7=3.6 mHence, actual depth = 3.6 m

Question 3

Question 3(c)A sound wave travelling in water has wavelength 0.4 m. Is this wave audible in air? (The speed of sound in water = 1400 ms-1) [2]

Accepted Answer

Given,wavelength = 0.4 mspeed of sound in water = 1400 m s-1As we know,frequency of wave = SpeedWavelength\dfrac{\text{Speed}}{\text{Wavelength}}WavelengthSpeed​Substituting the values we get,frequency of wave=14000.4frequency of wave=3500 Hz=3.5 kHz\text{frequency of wave} = \dfrac{1400}{0.4} \[0.5em] \text{frequency of wave} = 3500 \text{ Hz} = 3.5 \text{ kHz} \[0.5em]frequency of wave=0.41400​frequency of wave=3500 Hz=3.5 kHzHence, 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.

Question 4

Question 4(a)Calculate the total resistance across AB: [2]

Accepted Answer

This circuit consists of two parts. In the first part, the resistances 3 ohm and 6 ohm are connected in parallel. If Rp is the equivalent resistance then1Rp=13+161Rp=2+161Rp=361Rp=12⇒Rp=2 Ω\dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{1}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{2 + 1}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{3}{6} \[0.5em] \dfrac{1}{R_p} = \dfrac{1}{2} \[0.5em] \Rightarrow R_p = 2 \space ΩRp​1​=31​+61​Rp​1​=62+1​Rp​1​=63​Rp​1​=21​⇒Rp​=2 ΩTherefore, Rp = 2 ΩIn the second part, Rp (i.e., 2 ohm) and 5 ohm are in series. If the equivalent resistance = Rs the we get,Rs = 2 + 5 = 7 ΩHence, equivalent resistance across A and B = 7 Ω

Question 5

Question 4(b)Two metallic blocks P and Q having masses in ratio 2 : 1 are supplied with the same amount of heat. If their temperatures rise by same degree, compare their specific heat capacities. [2]

Accepted Answer

Let mass of block P = mp and mass of block Q = mQGiven,mpmQ\dfrac{m_p}{m_Q}mQ​mp​​ = 21\dfrac{2}{1}12​Q = m x s x ∆tSince, heat supplied (∆Q) and rise in temperature (∆t) is same for both block P and Q,Therefore,mp×sp×∆t=mQ×sQ×∆tmp×sp=mQ×sQspsQ=mQmp=12m_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] \dfrac{s_p}{s_Q} = \dfrac{m_Q}{m_p} = \dfrac{1}{2}mp​×sp​×∆t=mQ​×sQ​×∆tmp​×sp​=mQ​×sQ​sQ​sp​​=mp​mQ​​=21​Therefore, sP : sQ = 1 : 2

Question 6

Question 4(d)A nucleus 84X202 of an element emits an alpha particle followed by a beta particle. The final nucleus is aYb. Find a and b. [2]

Accepted Answer

The emission of alpha and beta particle is represented below:84202X⟶ 82198Z+ 24He (α particle emission)82198Z⟶ 83198Y+ −1   0e (β particle emission)_{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)}84202​X⟶ 82198​Z+ 24​He (α particle emission)82198​Z⟶ 83198​Y+ −1   0​e (β particle emission)∴ a = 83 and b = 198

Question 7

Question 4(e)The diagram alongside shows a loop of wire carrying current I: [2](i) What is the magnetic polarity of the loop that faces us?(ii) With respect to the diagram how can we increase the strength of the magnetic field produced by this loop?

Accepted Answer

(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.

Question 8

Question 5(a)The figure alongside shows a simple pendulum of mass 200 g. It is displaced from the mean position A to the extreme position B. The potential energy at the position A is zero. At the position B the pendulum bob is raised by 5 m. [3](i) What is the potential energy of the pendulum at the position B?(ii) What is the total mechanical energy at point C?(iii) What is the speed of the bob at the position A when released from B?(Take g = 10 ms-2 and there is no loss of energy.)

Accepted Answer

Given,h = 5 m, m = 200 g = 0.2 kg, g = 10 ms-2(i) Potential energy UB at B is given byUB = m x g x hSubstituting the values we get,UB=0.2×10×5⇒UB=10JU_B = 0.2 \times 10 \times 5 \[0.5em] \Rightarrow U_B = 10 JUB​=0.2×10×5⇒UB​=10JHence, the potential energy of the pendulum at the position B = 10 J(ii) Total mechanical energy at point C = 10 JThe 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,12×m×(vA)2=UB0.5×0.2×(vA)2=10(vA)2=100.1⇒vA=100⇒vA=10 m s−1\dfrac{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} = \dfrac{10}{0.1} \[0.5em] \Rightarrow v_A = \sqrt{100} \[0.5em] \Rightarrow v_A = 10 \text{ m s}^{-1}21​×m×(vA​)2=UB​0.5×0.2×(vA​)2=10(vA​)2=0.110​⇒vA​=100​⇒vA​=10 m s−1

Question 9

Question 5(b)(i) With reference to the direction of action, how does a centripetal force differ from a centrifugal force during uniform circular motion? [3](ii) Is centrifugal force the force of reaction of centripetal force?(iii) Compare the magnitudes of centripetal and centrifugal force.

Accepted Answer

(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.

Question 10

Question 5(c)A block and tackle system of pulleys has velocity ratio 4. [4](i) Draw a neat, labelled diagram of the system indicating clearly the points of application and direction of load and effort.(ii) What will be it's V.R. if the weight of the movable block is doubled?

Accepted Answer

(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.

Question 11

Question 6(a)A diver in water looks obliquely at an object AB in air. [3](i) Does the object appear taller, shorter or of the same size to the diver?(ii) Show the path of two rays AC & AD starting from the tip of the object as it travels towards the diver in water and hence obtain the image of the object.

Accepted Answer

(i) The object appears taller to the diver.(ii) Below diagram shows the path of two rays AC & AD:

Question 12

Question 6(b)Complete the path of the ray AB through the glass prism in PQR till it emerges out of the prism. Given the critical angle of the glass as 42°. [3]

Accepted Answer

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.

Question 13

Question 6(c)A lens of focal length 20 cm forms an inverted image at a distance 60 cm from the lens. [4](i) Identify the lens.(ii) How far is the lens present in front of the object?(iii) Calculate the magnification of the image.

Accepted Answer

(i) The lens used is convex lens.(ii) Given, f = +20 cm, v = +60 cm, u = ?From lens formula,1f\dfrac{1}{f}f1​ = 1v\dfrac{1}{v}v1​ - 1u\dfrac{1}{u}u1​Substituting the values in the formula,120=160−1u⇒1u=160−120⇒1u=1−360⇒1u=−260⇒u=−30cm\dfrac{1}{20} = \dfrac{1}{60} - \dfrac{1}{u} \[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{1}{60} - \dfrac{1}{20} \[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{1 - 3}{60} \[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{- 2}{60} \[0.5em] \Rightarrow u = - 30 cm \[0.5em]201​=601​−u1​⇒u1​=601​−201​⇒u1​=601−3​⇒u1​=60−2​⇒u=−30cmHence, the lens is 30 cm in front of the object(iii) Magnification m = vu\dfrac{v}{u}uv​Substituting the values in the formula above we get,m=60−30=−2m = \dfrac{60}{- 30} = -2m=−3060​=−2Thus, magnification = 2 times

Question 14

Question 7(a)Give reasons for the following: [3]During the day:(i) Clouds appear white.(ii) Sky appears blue.

Accepted Answer

(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.

Question 15

Question 7(c)(i) When a tuning fork [vibrating] is held close to ear, one hears a faint hum. The same [vibrating tuning fork] is held such that it's stem is in contact with the table surface, then one hears a loud sound. Explain. [4](ii) A man standing in front of a vertical cliff fires a gun. He hears the echo after 3.5 seconds. On moving closer to the cliff by 84 m, he hears the echo after 3 seconds. Calculate the distance of the cliff from the initial position of the man.

Accepted Answer

(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 = 2dv\dfrac{2d}{v}v2d​ = 3.5 s    [Eqn i ]and for second echo, t2 = 2(d−84)v\dfrac{2(d - 84)}{v}v2(d−84)​ = 3 s    [Eqn ii]Dividing eqn (ii) by eqn (i)2(d−84)v2dv=33.5⇒d−84d=33.5⇒3.5(d−84)=3d⇒(3.5d)−(3.5×84)=3d⇒(3.5d)−3d=294⇒0.5d=294⇒d=2940.5⇒d=588m\dfrac{\dfrac{2(d - 84)}{v}}{\dfrac{2d}{v}} = \dfrac{3}{3.5} \[0.5em] \Rightarrow \dfrac{d - 84}{d} = \dfrac{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 = \dfrac{294}{0.5} \[0.5em] \Rightarrow d = 588 mv2d​v2(d−84)​​=3.53​⇒dd−84​=3.53​⇒3.5(d−84)=3d⇒(3.5d)−(3.5×84)=3d⇒(3.5d)−3d=294⇒0.5d=294⇒d=0.5294​⇒d=588mHence, the distance of the cliff from the initial position of the man = 588 m

Question 16

Question 8(b)(i) What are superconductors? [3](ii) Calculate the current drawn by an appliance rated 110 W, 220 V when connected across 220 V supply.(iii) Name a substance whose resistance decreases with the increase in temperature.

Accepted Answer

(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∴Current I=WV=110220=0.5 A	herefore 	ext{Current I} = \dfrac{W}{V} = \dfrac{110}{220} = 0.5 	ext{ A}∴Current I=VW​=220110​=0.5 AHence, current drawn by the appliance = 0.5 A(iii) The resistance of semi conductors such as carbon, silicon, etc., decreases with increase in temperature.

Question 17

Question 8(c)The diagram alongside shows three resistors connected across a cell of e.m.f. 1.8 V and internal resistance r. Calculate: [4](i) Current through 3 Ω resistor.(ii) The internal resistance r.

Accepted Answer

(i) The equivalent resistance Rp of resistances 3 Ω and 1.5 Ω joined in parallel is given by1Rp=13+23  [∵1.5Ω=32Ω]⇒1Rp=33⇒Rp=1Ω\dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{2}{3} \space \space \big[\because 1.5 Ω = \dfrac{3}{2} Ω\big] \[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{3}{3} \[0.5em] \Rightarrow R_p = 1 ΩRp​1​=31​+32​  [∵1.5Ω=23​Ω]⇒Rp​1​=33​⇒Rp​=1ΩFrom, V = IRthe potential drop across this combination = 0.3 x 1 = 0.3 V∴Current through 3 Ω resistance=VoltageResistance=0.33=0.1 A\therefore \text{Current through 3 Ω resistance} = \dfrac{\text{Voltage}}{\text{Resistance}} \[0.5em] = \dfrac{0.3}{3} = 0.1 \text{ A}∴Current through 3 Ω resistance=ResistanceVoltage​=30.3​=0.1 A(ii) Total resistance of the circuit = Rp + 4 + r = (5 + r) ΩCurrent in the circuit,0.3=1.85+r0.3×(5+r)=1.8(0.3×5)+(0.3×r)=1.81.5+(0.3×r)=1.80.3×r=1.8−1.5r=0.30.3=1Ω0.3 = \dfrac{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 = \dfrac{0.3}{0.3} = 1 Ω0.3=5+r1.8​0.3×(5+r)=1.8(0.3×5)+(0.3×r)=1.81.5+(0.3×r)=1.80.3×r=1.8−1.5r=0.30.3​=1Ω∴ Internal resistance r = 1 Ω

Question 18

Question 9(b)The diagram below shows the change of phases of a substance on a temperature vs time graph on heating the substance at a constant rate.[3](i) Why is the slope of CD less than slope of AB?(ii) What is the boiling and melting point of the substance?

Accepted Answer

(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

Question 19

Question 9(c)A piece of ice of mass 60 g is dropped into 140 g of water at 50 °C. [4]Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding).Specific heat capacity of water (c) = 4.2 Jg-1k-1,Specific latent heat of fusion of ice (L) = 336 Jg-1

Accepted Answer

Let final temperature of water be T °C.Heat energy taken by the ice to melt at 0 °CQ1 = mass of ice x specific latent heat of fusion of ice= 60 x 336 = 20,160 JHeat 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 JTotal heat energy taken by ice = Q1 + Q2= (20,160 + 252 T) JHeat 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) JIf there is no heat loss,heat energy given by water = total heat energy taken by iceSubstituting the values we get,588(50−T)=20,160+252T(588×50)−588T=20,160+252T29,400−588T=20,160+252T29,400−20,160=588T+252T9240=840T⇒T=9240840⇒T=11°C588 (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 = \dfrac{9240}{840} \[0.5em] \Rightarrow T = 11 °C588(50−T)=20,160+252T(588×50)−588T=20,160+252T29,400−588T=20,160+252T29,400−20,160=588T+252T9240=840T⇒T=8409240​⇒T=11°CHence, T = 11 °C

Question 20

Question 10(c)An atomic nucleus A is composed of 84 protons and 128 neutrons. The nucleus A emits an alpha particle and is transformed into a nucleus B. [4](i) What is the composition of B?(ii) The nucleus B emits a beta particle and is transformed into a nucleus C. What is the composition of C?(iii) What is mass number of the nucleus A?(iv) Does the composition of C change if it emits gamma radiations?

Accepted Answer

(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 = 82No 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 = 83No. 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.

Solved 2020 Question Paper ICSE Class 10 Physics

Section I 40 Marks

Section Ii 40 Marks