Q: Question 1(b)Two bodies A and B have masses in the ratio 5:1 and their kinetic energies are in the ratio 125:9. Find the ratio of their velocities. [2]

Given,mAmB=51KAKB=1259\dfrac{m_A}{m_B} = \dfrac{5}{1} \\[0.5em] \dfrac{K_A}{K_B} = \dfrac{125}{9}mBmA=15KBKA=9125Since,KAKB=12mAvA212mBvB2=(mAmB)×(vA2vB2)\dfrac{K_A}{K_B} = \dfrac{\dfrac{1}{2}m_Av_A^2}{\dfrac{1}{2}m_Bv_B^2} \\[0.5em] = \Big(\dfrac{m_A}{m_B}\Big) \times \Big(\dfrac{v_A^2}{v_B^2}\Big)KBKA=21mBvB221mAvA2=(mBmA)×(vB2vA2)Substituting the values we get,1259=51×vA2vB2⇒vA2vB2=125×19×5⇒vAvB=259=53\dfrac{125}{9} = \dfrac{5}{1} \times \dfrac{v_A^2}{v_B^2} \\[0.5em] \Rightarrow \dfrac{v_A^2}{v_B^2} = \dfrac{125 \times 1}{9 \times 5} \\[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \sqrt{\dfrac{25}{9}} = \dfrac{5}{3} \\[0.5em]9125=15×vB2vA2⇒vB2vA2=9×5125×1⇒vBvA=925=35Hence, ratio between velocities of A and B is 5 : 3

Q: Question 2(b)(i) What is the relation between the refractive index of water with respect to air (aμw) and the refractive index of air with respect to water (wμa)(ii) If the refractive index of water with respect to air (aμw) is 53\dfrac{5}{3}35. Calculate the refractive index of air with respect to water (wμa). [2]

(i) aμw = 1wμa\dfrac{1}{_wμ_a}wμa1 i.e., they are inverse of each other.(ii) Given, aμw = 53\dfrac{5}{3}35Therefore, wμa = 35\dfrac{3}{5}53 = 0.6

Question 1

Question 1(b)Two bodies A and B have masses in the ratio 5:1 and their kinetic energies are in the ratio 125:9. Find the ratio of their velocities. [2]

Accepted Answer

Given,mAmB=51KAKB=1259\dfrac{m_A}{m_B} = \dfrac{5}{1} \[0.5em] \dfrac{K_A}{K_B} = \dfrac{125}{9}mB​mA​​=15​KB​KA​​=9125​Since,KAKB=12mAvA212mBvB2=(mAmB)×(vA2vB2)\dfrac{K_A}{K_B} = \dfrac{\dfrac{1}{2}m_Av_A^2}{\dfrac{1}{2}m_Bv_B^2} \[0.5em] = \Big(\dfrac{m_A}{m_B}\Big) \times \Big(\dfrac{v_A^2}{v_B^2}\Big)KB​KA​​=21​mB​vB2​21​mA​vA2​​=(mB​mA​​)×(vB2​vA2​​)Substituting the values we get,1259=51×vA2vB2⇒vA2vB2=125×19×5⇒vAvB=259=53\dfrac{125}{9} = \dfrac{5}{1} \times \dfrac{v_A^2}{v_B^2} \[0.5em] \Rightarrow \dfrac{v_A^2}{v_B^2} = \dfrac{125 \times 1}{9 \times 5} \[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \sqrt{\dfrac{25}{9}} = \dfrac{5}{3} \[0.5em]9125​=15​×vB2​vA2​​⇒vB2​vA2​​=9×5125×1​⇒vB​vA​​=925​​=35​Hence, ratio between velocities of A and B is 5 : 3

Question 2

Question 2(b)(i) What is the relation between the refractive index of water with respect to air (aμw) and the refractive index of air with respect to water (wμa)(ii) If the refractive index of water with respect to air (aμw) is 53\dfrac{5}{3}35​. Calculate the refractive index of air with respect to water (wμa). [2]

Accepted Answer

(i) aμw = 1wμa\dfrac{1}{_wμ_a}w​μa​1​ i.e., they are inverse of each other.(ii) Given, aμw = 53\dfrac{5}{3}35​Therefore, wμa = 35\dfrac{3}{5}53​ = 0.6

Question 3

Question 2(e)The diagram below shows a light source P embedded in a rectangular glass block ABCD of critical angle 42°. Complete the path of the ray PQ till it emerges out of the block. [Write necessary angles.] [2]

Accepted Answer

The path of ray is as shown:

Question 4

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

Accepted Answer

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 Rs then,Rs = 5 + 4 = 9 ΩThen, in second part there is parallel combination of 3 Ω and Rs (9 Ω ), if equivalent resistance is Rp1Rp=13+19⇒1Rp=3+19⇒1Rp=49⇒Rp=94⇒Rp=2.25Ω\dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{1}{9} \[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{3 + 1}{9} \[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{4}{9} \[0.5em] \Rightarrow R_p = \dfrac{9}{4} \[0.5em] \Rightarrow R_p = 2.25 ΩRp​1​=31​+91​⇒Rp​1​=93+1​⇒Rp​1​=94​⇒Rp​=49​⇒Rp​=2.25ΩIn the third part, 8 Ω and Rp ( 2.25 Ω ) are in series. Therefore, the equivalent resistance between AB i.e., RAB is8 + 2.25 = 10.25 Ω .Hence, Total resistance between A and B is 10.25 Ω

Question 5

Question 4(d)(i) Why is a nuclear fusion reaction called a thermo nuclear reaction? [2](ii) Complete the reaction:3He2 + 2H1 ⟶ 4He2 + .......... + Energy

Accepted Answer

(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

Question 6

Question 4(e)State two ways to increase the speed of rotation of a D.C. motor. [2]

Accepted Answer

The two ways of increasing the speed of a D.C. motor areBy increasing the strength of current through the coil.By increasing the number of turns in the coil.

Question 7

Question 5(a)A body of mass 10 kg is kept at a height of 5 m. It is allowed to fall and reach the ground. [3](i) What is the total mechanical energy possessed by the body at the height of 2 m assuming it is a frictionless medium?(ii) What is the kinetic energy possessed by the body just before hitting the ground? Take g = 10 m s-2.

Accepted Answer

(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-1Mechanical energy at 2m height = P.E. + K.E. = mgh + 12\dfrac{1}{2}21​ mv2= (10 x 10 x 2) + (12\dfrac{1}{2}21​ x 10 x 60)= 200 + 300= 500 JHence, 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 JHence, kinetic energy possessed by the body just before hitting the ground = 500 J

Question 8

Question 5(c)The diagram below shows a pulley arrangement : [4](i) Copy the diagram and mark the direction of tension on each strand of the string.(ii) What is the velocity ratio of the arrangement?(iii) If the tension acting on the string is T, then what is the relationship between T and effort E?(iv) If the free end of the string moves through a distance x, find the distance by which the load is raised.

Accepted Answer

(i) The direction of tension (T) is marked in the diagram:(ii) V.R. = Distance moved by effortDistance moved by load\dfrac{\text{Distance moved by effort}}{\text{Distance moved by load}}Distance moved by loadDistance moved by effort​ = 2(iii) Effort E balances the tensions therefore, T = E(iv) From the relation,V.R. = Distance moved by effortDistance moved by load\dfrac{\text{Distance moved by effort}}{\text{Distance moved by load}}Distance moved by loadDistance moved by effort​ = 2If the free end of the string moves through a distance x then distance moved by load = ?Substituting the values we get,2=xDistance moved by load⇒Distance moved by load=x22 = \dfrac{\text{x}}{\text{Distance moved by load}} \[0.5em] \Rightarrow \text{Distance moved by load} = \dfrac{x}{2}2=Distance moved by loadx​⇒Distance moved by load=2x​

Question 9

Question 6(a)How does the angle of deviation formed by a prism change with the increase in the angle of incidence? Draw a graph showing the variation in the angle of deviation with the angle of incidence at a prism surface. [3]

Accepted Answer

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:

Question 10

Question 6(b)A virtual, diminished image is formed when an object is placed between the optical centre and the principal focus of a lens. [3](i) Name the type of lens which forms the above image.(ii) Draw a ray diagram to show the formation of the image with the above stated characteristics.

Accepted Answer

(i) The lens is concave in nature.(ii) Ray diagram showing the formation of the image is given below:

Question 11

Question 6(c)An object is placed at a distance 24 cm in front of a convex lens of focal length 8 cm. [4](i) What is the nature of the image so formed ?(ii) Calculate the distance of the image from the lens.(iii) Calculate the magnification of the image.

Accepted Answer

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,1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​−u1​=f1​Substituting the values we get,1v+124=181v=18−1241v=3−1241v=2241v=112⇒v=12cm\dfrac{1}{v} + \dfrac{1}{24} = \dfrac{1}{8} \[0.5em] \dfrac{1}{v} = \dfrac{1}{8} - \dfrac{1}{24} \[0.5em] \dfrac{1}{v} = \dfrac{3 - 1}{24} \[0.5em] \dfrac{1}{v} = \dfrac{2}{24} \[0.5em] \dfrac{1}{v} = \dfrac{1}{12} \[0.5em] \Rightarrow v = 12 \text{cm}v1​+241​=81​v1​=81​−241​v1​=243−1​v1​=242​v1​=121​⇒v=12cmHence, the distance of the image from the lens = 12 cm(iii) From relation, m = vu\dfrac{v}{u}uv​Substituting the values we get,m=12−24⇒m=−0.5m = \dfrac{12}{-24} \[0.5em] \Rightarrow m = -0.5m=−2412​⇒m=−0.5-ve sign of magnification tells us that the image is inverted.Hence, magnification = -0.5

Question 12

Question 7(b)A pendulum has a frequency of 4 vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears the echo from the cliff after 6 vibrations of the pendulum. If the velocity of sound in air is 340 m s-1, find the distance between the cliff and the observer. [3]

Accepted Answer

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 = 64\dfrac{6}{4}46​ = 1.5 s.If d is the distance between the cliff and the observer, then velocity of sound V = 2dt\dfrac{2d}{t}t2d​Substituting the values in the equation above:340=2d1.5⇒d=340×1.52=255 cm340 = \dfrac{2d}{1.5} \[0.5em] \Rightarrow d = \dfrac{340 \times 1.5}{2} = 255 \text{ cm}340=1.52d​⇒d=2340×1.5​=255 cmHence, distance between the cliff and the observer = 255 cm

Question 13

Question 7(c)Two pendulums C and D are suspended from a wire as shown in the given figure. Pendulum C is made to oscillate by displacing it from it's mean position. It is seen that D also starts oscillating. [4](i) Name the type of oscillation, C will execute.(ii) Name the type of oscillation, D will execute.(iii) If the length of D is made equal to C, then what difference will you notice in the oscillations of D?(iv) What is the name of the phenomenon when the length of D is made equal to C?

Accepted Answer

(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.

Question 14

Question 8(b)The diagram below shows a dual control switch circuit connected to a bulb. [3](i) Copy the diagram and complete it so that the bulb is switched ON.(ii) Out of A & B which one is the live wire and which one is the neutral wire?

Accepted Answer

(i) The following figure shows the completed diagram:(ii) Wire A is the live wire and wire B is the neutral wire.

Question 15

Question 8(c)The diagram shows a circuit with the key k open. Calculate: [4](i) the resistance of the circuit when the key k is open.(ii) the current drawn from the cell when the key k is open.(iii) the resistance of the circuit when the key k is closed.(iv) the current drawn from the cell when the key k is closed.

Accepted Answer

(i) When k is open, resistance is 5 + 0.5 = 5.5 Ω(ii) From relation, V = IRSubstituting the values in the formula we get,3.3=I×5.5⇒I=3.35.5⇒I=0.6 A3.3 = \text{I} \times 5.5 \[0.5em] \Rightarrow \text{I} = \dfrac{3.3}{5.5} \[0.5em] \Rightarrow \text{I} = 0.6 \text{ A}3.3=I×5.5⇒I=5.53.3​⇒I=0.6 AHence, 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,1Rp=15+151Rp=1+15⇒1Rp=25⇒Rp=2.5Ω\dfrac{1}{R_p} = \dfrac{1}{5} + \dfrac{1}{5} \[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 1}{5} \[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{2}{5} \[0.5em] \Rightarrow R_p = 2.5 ΩRp​1​=51​+51​Rp​1​=51+1​⇒Rp​1​=52​⇒Rp​=2.5ΩSo now the resistance of circuit is 2.5 + 0.5 = 3 Ω(iv) Current drawn I = VR\dfrac{\text{V}}{\text{R}}RV​Substituting the values we get,I=3.33⇒I=1.1 A\text{I} = \dfrac{3.3}{3} \[0.5em] \Rightarrow \text{I} = 1.1 \text{ A}I=33.3​⇒I=1.1 A

Question 16

Question 9(a)(i) Define Calorimetry. [3](ii) Name the material used for making a Calorimeter.(iii) Why is a Calorimeter made-up of thin sheets of the above material answered in (ii) ?

Accepted Answer

(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.

Question 17

Question 9(b)The melting point of naphthalene is 80°C and the room temperature is 30°C. A sample of liquid naphthalene at 100°C is cooled down to the room temperature. Draw a temperature time graph to represent this cooling. In the graph, mark the region which corresponds to the freezing process. [3]

Accepted Answer

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

Question 18

Question 9(c)104 g of water at 30°C is taken in a calorimeter made of copper of mass 42 g. When a certain mass of ice at 0°C is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10°C. Find the mass of ice added.[Specific heat capacity of water = 4.2 J g-1 °C-1; Specific latent heat of fusion of ice = 336 J g-1; Specific heat capacity of copper = 0.4 J g-1 °C-1 ] [4]

Accepted Answer

Given,Mass of water = 104 gTemperature of water = 30°CFinal temperature = 10°CMass of calorimeter = 42 gTemperature of ice = 0°CThe heat given by water = m x s x tSubstituting the values we get,The heat given by water = 104 x 4.2 x (30 - 10) = 8736 JThe heat given by calorimeter = m x s x tSubstituting the values we get,The heat given by calorimeter = 42 x 0.4 x (30 - 10) = 336 JLet mass of ice added be m gmHeat taken by ice to melt = mL = 336m JHeat taken by ice-water to reach the mixture temperature = mst = m x 4.2 x (10 - 0) = 42m JTherefore, from principle of calorimetry, heat taken = heat given.i.e., 336 m + 42 m = 8736 + 336Therefore, m = 9072378\dfrac{9072}{378}3789072​ = 24 gHence, mass of ice added = 24 g

Question 19

Question 10(b)(i) Define nuclear fission. [3](ii) Rewrite and complete the following nuclear reaction by filling in the atomic number of Ba and mass number of Kr :92235U_{92}^{235}	ext{U}92235​U + 01n_{0}^{1}	ext{n}01​n ⟶ ...144Ba_{...}^{144}	ext{Ba}...144​Ba + 36...Kr_{36}^{...}	ext{Kr}36...​Kr + 3 01n_{0}^{1}	ext{n}01​n + Energy

Accepted Answer

(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) 92235U_{92}^{235}	ext{U}92235​U + 01n_{0}^{1}	ext{n}01​n ⟶ 56144Ba_{56}^{144}	ext{Ba}56144​Ba + 3689Kr_{36}^{89}	ext{Kr}3689​Kr + 3 01n_{0}^{1}	ext{n}01​n + Energy

Question 20

Question 10(c)The diagram below shows a magnetic needle kept just below the conductor AB which is kept in North South direction. [4](i) In which direction does the needle deflect when the key is closed ?(ii) Why is the deflection produced?(iii) What will be the change in the deflection if the magnetic needle is taken just above the conductor AB ?(iv) Name one device which works on this principle.

Accepted Answer

(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.

Solved 2019 Question Paper ICSE Class 10 Physics

Section I 40 Marks

Section Ii 40 Marks