Q: Question 2(c)(i) State the relation between the critical angle and the absolute refractive index of a medium.(ii) Which colour of light has a higher critical angle? Red light or Green light. [2]

(i) μ = 1sin C\dfrac{1}{\text{sin C}}sin C1 where μ is the refractive index of the medium and C is the critical angle.(ii) The critical angle of Red light is higher than that of green light.

Question 1

Question 1(d)(i) Why is the ratio of the velocities of light of wavelengths 4000 Å and 8000 Å in vacuum 1 : 1 ?(ii) Which of the above wavelengths has a higher frequency? [2]

Accepted Answer

(i) In vacuum the light of each wavelength travels with the same speed i.e., the speed of light of wavelength of 4000 Å is same as that of light of wavelength 8000 Å.(ii) The light of wavelength 4000 Å will have higher frequency.

Question 2

Question 1(e)(i) Why is the motion of a body moving with a constant speed around a circular path said to be accelerated? [2](ii) Name the unit of physical quantity obtained by the formula 2KV2\dfrac{2K}{V^2}V22K​.Where K: kinetic energy, V: Linear Velocity.

Accepted Answer

(i) The motion of a body moving with a constant speed around a circular path is said to be accelerated because as the direction keeps changing the velocity changes and therefore it's an accelerated motion.(ii) 2KV2\dfrac{2K}{V^2}V22K​ = 2×(12×mV2)V2\dfrac{2 	imes (\dfrac{1}{2} 	imes mV^2)}{V^2}V22×(21​×mV2)​ = m.Therefore, the quantity, 2KV2\dfrac{2K}{V^2}V22K​ will have the unit of mass m which is kilogram (kg).

Question 3

Question 2(c)(i) State the relation between the critical angle and the absolute refractive index of a medium.(ii) Which colour of light has a higher critical angle? Red light or Green light. [2]

Accepted Answer

(i) μ = 1sin C\dfrac{1}{	ext{sin C}}sin C1​ where μ is the refractive index of the medium and C is the critical angle.(ii) The critical angle of Red light is higher than that of green light.

Question 4

Question 3(c)(i) What do you understand by the term nuclear fusion? [2](ii) Nuclear power plants use the nuclear fission reaction to produce electricity. What is the advantage of producing electricity by fusion reaction?

Accepted Answer

(i) The combining of two or more light nuclei at very high temperature and pressure, to form a heavy nucleus is called nuclear fusion. In this process, huge amount of energy is released.(ii) The energy released in nuclear fusion will be more than in nuclear fission for the same mass.

Question 5

Question 4(d)(i) What are isobars?(ii) Give one example of isobars. [2]

Accepted Answer

(i) Isobars are the atoms of same mass number but different atomic number.(ii) 1123Na_{11}^{23}	ext{Na}1123​Na and 1223Mg_{12}^{23}	ext{Mg}1223​Mg are isobars.

Question 6

Question 4(e)State any two advantages of electromagnets over permanent magnets. [2]

Accepted Answer

The two advantages of electromagnets over permanent magnets are :An electromagnet can be made to produce a strong magnetic field (increasing the number of turns in the coil).The polarity of the electromagnet can be reversed.

Question 7

Question 5(a)(i) Derive a relationship between S.I. and C.G.S. unit of work. [3](ii) A force acts on a body and displaces it by a distance S in a direction at an angle θ with the direction of force. What should be the value of θ to get the maximum positive work?

Accepted Answer

S.I. unit of work is joule and C.G.S. unit of work is erg.Since, work = Force x displacement1 joule = 1 newton x 1 metre= 105 dyne x 100 cm= 107 dyne x cm= 107 erg1 joule = 107 erg(ii) Work W = F.S. cos θFor maximum work, cos θ = 1 or θ = 0°

Question 8

Question 5(b)A half metre rod is pivoted at the centre with two weights of 20 gf and 12 gf suspended at a perpendicular distance of 6 cm and 10 cm from the pivot respectively as shown below. [3](i) Which of the two forces acting on the rigid rod causes clockwise moment?(ii) Is the rod in equilibrium?(iii) The direction of 20 gf force is reversed. What is the magnitude of the resultant moment of the forces on the rod?

Accepted Answer

(i) The force 12 gf rotates the rod clockwise.(ii) Since, the clockwise moment (= 20 gf x 6 cm = 120 gf cm) is equal to the anticlockwise moment (= 12 gf x 10 cm = 120 gf cm), hence the rod is in equilibrium.(iii) On reversing the direction of force, 20 gf will also produce clockwise moment, total clockwise moment will then be (20 gf x 6 cm + 12 gf x 10 cm) = 240 gf cm

Question 9

Question 5(c)(i) Draw a diagram to show a block and tackle pulley system having a velocity ratio of 3 marking the direction of load (L), effort (E) and tension (T). [4](ii) The pulley system drawn lifts a load of 150 N when an effort of 60 N is applied. Find it's mechanical advantage.(iii) Is the above pulley system an ideal machine or not?

Accepted Answer

(i) The block and tackle pulley system with velocity ratio 3 is shown below in which the direction of load (L), effort (E) and tension (T) has been marked:(ii) Given,L = 150 N,E = 60 NMechanical advantage M.A. = 𝐿E\dfrac{𝐿}{E}EL​ = 15060\dfrac{150}{60}60150​ = 2.5(iii) This pulley system is not ideal.

Question 10

Question 6(c)An object is placed at a distance of 12 cm from a convex lens of focal length 8 cm. [4]Find:(i) the position of the image(ii) nature of the image

Accepted Answer

(i) Given,u = 12 cm (negative),f = 8 cm (positive)From lens formula1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}v1​−u1​=f1​Substituting the values we get,1v−1−12=18⇒1v+112=18⇒1v=18−112⇒1v=3−224⇒1v=124⇒v=24 cm\dfrac{1}{v} - \dfrac{1}{-12} = \dfrac{1}{8} \[0.5em] \Rightarrow \dfrac{1}{v} + \dfrac{1}{12} = \dfrac{1}{8} \[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{8} - \dfrac{1}{12} \[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{3 - 2}{24} \[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{24} \[0.5em] \Rightarrow v = 24 \text{ cm}\[0.5em]v1​−−121​=81​⇒v1​+121​=81​⇒v1​=81​−121​⇒v1​=243−2​⇒v1​=241​⇒v=24 cmThe image will be formed at a distance 24 cm behind the lens.(iii) The image will be real, inverted and magnified.

Question 11

Question 7(a)Draw the diagram of a right angled isosceles prism which is used to make an inverted image erect. [3]

Accepted Answer

Below is the diagram of a right angled isosceles prism which is used to make an inverted image erect:

Question 12

Question 7(c)A person is standing at the sea shore. An observer on the ship which is anchored in between a vertical cliff and the person on the shore fires a gun. The person on the shore hears two sounds, 2 seconds and 3 seconds after seeing the smoke of the fired gun. If the speed of sound in the air is 320 ms-1 then calculate: [4](i) the distance between the observer on the ship and the person on the shore.(ii) the distance between the cliff and the observer on the ship.

Accepted Answer

Let d1 be the distance between the observer on ship and person on shore. The first sound heard by the person at shore after t1 = 2 s will be the direct sound of fire.From relation v = dt\dfrac{d}{t}td​,320 = d12\dfrac{d_1}{2}2d1​​ or d1 = 640 m.(ii) The second sound heard by the person at shore will be the echo heard after reflection from the cliff. Since, sound takes 2 s to reach from the observer to the person, so the time taken by the echo of fire to reach the observer will be t2 = 3 - 2 = 1 s.Let d2 be the distance between the observer and the cliff. Then 2d2t2\dfrac{2d_2}{t_2}t2​2d2​​ = vOr d2 = vt22\dfrac{vt_2}{2}2vt2​​ = 320×12\dfrac{320 	imes 1}{2}2320×1​ = 160 m.

Question 13

Question 8(a)(i) A fuse is rated 8A. Can it be used with an electrical appliance rated 5 KW, 200 V? Give a reason. [3](ii) Name two safety devices which are connected to the live wire of a household electric circuit.

Accepted Answer

Given,P = 5 KW = 5000 W,V = 200 VCurrent drawn i = PV\dfrac{P}{V}VP​ = 5000W200V\dfrac{5000 W}{200 V}200V5000W​ = 25 ASince, the appliance when in use will draw 25 A current, the fuse rated 8 A cannot be used because it will blow off as the current will exceed 8 A and the circuit wil break.(ii) Safety devices which are connected to the live wire of a household electric circuit are switch and fuse.

Question 14

Question 8(b)(i) Find the equivalent resistance between A and B. [3](ii) State whether the resistivity of a wire changes with the change in the thickness of the wire.

Accepted Answer

(i) In figure, a parallel combination of resistors 6 Ω and 3 Ω are connected in series with another parallel combination of resistors 4 Ω and 12 Ω between the terminals A and B.In the first part, let the equivalent resistance of parallel combination of resistors 3 Ω and 6 Ω be R1.1R1=13+16⇒1R1=2+16⇒1R1=36⇒1R1=12⇒R1=2Ω\dfrac{1}{R_1} = \dfrac{1}{3} + \dfrac{1}{6} \[0.5em] \Rightarrow \dfrac{1}{R_1} = \dfrac{2 + 1}{6} \[0.5em] \Rightarrow \dfrac{1}{R_1} = \dfrac{3}{6} \[0.5em] \Rightarrow \dfrac{1}{R_1} = \dfrac{1}{2} \[0.5em] \Rightarrow R_1 = 2 ΩR1​1​=31​+61​⇒R1​1​=62+1​⇒R1​1​=63​⇒R1​1​=21​⇒R1​=2ΩIn the second part, 4 Ω and 12 Ω are in parallel connection. If the equivalent resistance is R2 then1R2=14+112⇒1R2=3+112⇒1R2=412⇒1R2=13⇒R2=3Ω\dfrac{1}{R_2} = \dfrac{1}{4} + \dfrac{1}{12} \[0.5em] \Rightarrow \dfrac{1}{R_2} = \dfrac{3 + 1}{12} \[0.5em] \Rightarrow \dfrac{1}{R_2} = \dfrac{4}{12} \[0.5em] \Rightarrow \dfrac{1}{R_2} = \dfrac{1}{3} \[0.5em] \Rightarrow R_2 = 3 ΩR2​1​=41​+121​⇒R2​1​=123+1​⇒R2​1​=124​⇒R2​1​=31​⇒R2​=3ΩIn the third part R1 and R2 are in series. Therefore we get,2 + 3 = 5 ΩHence, equivalent resistance between A and B = 5 Ω.(ii) The resistivity of wire does not change with the change in the thickness of the wire.

Question 15

Question 8(c)An electric iron is rated 220V, 2 kW. [4](i) If the iron is used for 2h daily, find the cost of running it for one week if it costs ₹4.25 per kWh.(ii) Why is the fuse absolutely necessary in a power circuit?

Accepted Answer

Given,power P = 2 kW,time t = 2h,number of days = 7 and cost = ₹4.25 per kWhElectrical energy consumed = Power x total time= 2 kW x (2 x 7) h= 28 kWhCost = 28 x 4.25 = ₹119∴ Cost of running the iron for a week is ₹119(ii) The appliances connected to the power circuit like air conditioner, geyser, washing machine, etc. consume heavy current. The fuse is absolutely necessary in a power circuit because if somehow the circuit is overloaded (i.e., the total power of appliances connected in the circuit exceeds the power limit of the circuit) an excessive current will flow in the circuit and the line wire or appliances may get damaged. Moreover, due to the excessive heat produced by such high current, the wires may get overheated causing a fire. The use of fuse in power circuit will safe guard against any such mishap.

Question 16

Question 9(b)(i) State two differences between "Heat Capacity" and "Specific Heat Capacity". [3](ii) Give a mathematical relation between Heat Capacity and Specific Heat Capacity.

Accepted Answer

(i) Difference between "Heat Capacity" and "Specific Heat Capacity" are as follows :Heat CapacitySpecific Heat CapacityIt is the amount of heat energy required to raise the temperature of entire body by 1 °C.It is the amount of heat energy required to raise the temperature of unit mass of the body by 1 °C.The S.I. unit is JK-1S.I. unit is J kg-1K-1.(ii) Heat Capacity = Mass x Specific Heat Capacity

Question 17

Question 9(c)The temperature of 170 g of water at 50°C is lowered to 5°C by adding a certain amount of ice to it. Find the mass of ice added.Given: Specific heat capacity of water = 4200 J kg-1 ⁰C-1 and Specific latent heat of ice = 336000 J kg-1 [4]

Accepted Answer

Given,mass of water m = 170 g = 0.17 kginitial temperature = 50°Cfinal temperature = 5°Cfall in temperature = (50 - 5) = 45°C = 45 KHeat lost by water = m x c x Δt= 0.17 x 4200 x 45= 3.213 x 104If m' kg ice is added, heat gained by it to melt to 0°C = m'L= m' x 3.36 x 105Heat gained by it to raise temperature by 5°C = m' x C x Δt= m' x 4200 x 5= m' x 2.1 x 104 JTotal heat gained by ice = (m' x 3.36 x 105) + (m' x 2.1 x 104)= m' x 3.57 x 105 JBy the principle of method of mixtures:heat lost by water = heat gained by ice3.213×104=m′×3.57×105m′=3.213×1043.57×105m′=0.09 kg=90 g3.213 \times 10^4 = \text{m}' \times 3.57 \times 10^5 \[0.5em] \text{m}' = \dfrac{3.213 \times 10^4}{3.57 \times 10^5} \[0.5em] \text{m}' = 0.09 \text{ kg} = 90 \text{ g}3.213×104=m′×3.57×105m′=3.57×1053.213×104​m′=0.09 kg=90 gHence, mass of ice added = 90 g

Question 18

Question 10(a)The diagram shows a coil wound around a U shape soft iron bar AB. [3](i) What is the polarity induced at the ends A and B when the switch is pressed?(ii) Suggest one way to strengthen the magnetic field in the electromagnet.(iii) What will be the polarities at A & B if the direction of current is reversed in the circuit?

Accepted Answer

(i) The polarity induced at the end A is North and at the end B is South.(ii) To strengthen the magnetic field in the electromagnet, the current flowing in it is increased.(iii) On reversing the direction of current, the end A will be south pole and the end B will be the north pole.

Question 19

Question 10(b)The ore of Uranium found in nature contains 92238U_{92}^{238}	ext{U}92238​U and 92235U_{92}^{235}	ext{U}92235​U. Although both the isotopes are fissionable, it is found out experimentally that one of the two isotopes is more easily fissionable. [3](i) Name the isotope of Uranium which is easily fissionable.(ii) Give a reason for your answer.(iii) Write a nuclear reaction when Uranium 238 emits an alpha particle to form a Thorium (Th) nucleus.

Accepted Answer

(i) 92235U_{92}^{235}	ext{U}92235​U is easily fissionable.(ii) The reason is that the fission of 92238U_{92}^{238}	ext{U}92238​U nucleus is possible only by the fast neutrons, while the fission of 92235U_{92}^{235}	ext{U}92235​U nucleus can be done by both slow as well as fast neutrons.(iii) 92238U_{92}^{238}	ext{U}92238​U ⟶ 90234Th_{90}^{234}	ext{Th}90234​Th + 24He_{2}^{4}	ext{He}24​He

Question 20

Question 10(c)Radiations given out from a source when subjected to an electric field in a direction perpendicular to their path are shown in the diagram. The arrows show the path of the radiation A, B and C. Answer the following question in terms of A, B and C. [4](i) Name the radiation B which is unaffected by the electrostatic field.(ii) Why does the radiation C deflect more than A?(iii) Which among the three causes the least biological damage externally.(iv) Name the radiation which is used in carbon dating.

Accepted Answer

(i) The radiation B i.e., γ radiation is unaffected by the electrostatic field.(ii) The radiation C is the β radiation which is lighter than radiation A (α radiation).(iii) The radiation A (i.e., α) causes the least biological damage.(iv) The radiation C (i.e., β radiation) is used in carbon dating.

Solved 2018 Question Paper ICSE Class 10 Physics

Section I 40 Marks

Section Ii 40 Marks