Q: Question 2(d)How is the refractive index of a material related to? [2](i) real and apparent depth?(ii) velocity of light in vacuum or air and the velocity of light in a given medium?

(i) Refractive index (μ) = Real depthApparent depth\dfrac{\text{Real depth}}{\text{Apparent depth}}Apparent depthReal depth(ii) Refractive index (μ) = Speed of light in vacuum or airSpeed of light in given medium \dfrac{\text{Speed of light in vacuum or air}}{\text{\text{Speed of light in given medium }}}Speed of light in given medium Speed of light in vacuum or air

Question 1

Question 1(d)If the power of a motor be 100 kW, at what speed can it raise a load of 50,000 N? [2]

Accepted Answer

Given, power P = 100 kW = 100 x 103 W, force F = 50,000 NPower P = Force (F) x Speed (v)Substituting the values we get,100×103=50,000×v⇒v=100×10350,000⇒v=2ms−1100 \times 10^3 = 50,000 \times v \[0.5em] \Rightarrow v = \dfrac{100 \times 10^3 }{50,000} \[0.5em] \Rightarrow v = 2 m s^{-1}100×103=50,000×v⇒v=50,000100×103​⇒v=2ms−1

Question 2

Question 2(c)A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C, when the final temperature recorded is 20°C. Find the specific heat capacity of the solid. (Specific heat capacity of water = 4.2 J/g°C). [2]

Accepted Answer

Given,Mass of solid m1 = 50 g,temperature of solid t1 = 150°C,mass of water m2 = 100 gtemperature of water t2 = 11°C ,temperature of mixture t = 20°CLet specific heat capacity of the solid be c1Heat lost by the solid Q1 = m1 c1 (t1 - t)= 50 x c1 x (150 - 20) = 6500 c1 JHeat gained by water = m2 c2 (t - t2 )= 100 x 4.2 x (20 - 11) = 3780 JIf there is no loss of heat, by the principle of calorimetryHeat lost by the solid = Heat gained by the water6500 c1 = 3780c1 = 37806500\dfrac{3780}{6500}65003780​ = 0.582 Jg-1° C-1

Question 3

Question 2(d)How is the refractive index of a material related to? [2](i) real and apparent depth?(ii) velocity of light in vacuum or air and the velocity of light in a given medium?

Accepted Answer

(i) Refractive index (μ) = Real depthApparent depth\dfrac{	ext{Real depth}}{	ext{Apparent depth}}Apparent depthReal depth​(ii) Refractive index (μ) = Speed of light in vacuum or airSpeed of light in given medium \dfrac{	ext{Speed of light in vacuum or air}}{	ext{	ext{Speed of light in given medium }}}Speed of light in given medium Speed of light in vacuum or air​

Question 4

Question 2(e)State the conditions required for total internal reflection of light to take place. [2]

Accepted Answer

The conditions required for total internal reflection are:The light must travel from a denser medium to a rarer medium.The angle of incidence of light in the denser medium must be greater than the critical angle for that pair of media.

Question 5

Question 3(a)Draw a ray diagram to show the refraction of a monochromatic ray through a prism when it suffers minimum deviation. [2]

Accepted Answer

The ray diagram below shows the refraction of a monochromatic ray of light through a prism in the position of minimum deviation. The refracted ray QR is parallel to the base BC of the prism or angle of incidence (i) = angle of emergence (e).

Question 6

Question 3(c)An enemy plane is at a distance of 300 km from a radar. In how much time the radar will be able to detect the plane? Take velocity of radio waves as 3 x 108 m s-1. [2]

Accepted Answer

Given,Distance d = 300 km,Velocity V = 3 x 108 m s-1If it takes time t for the signal to go up to the enemy plane and come back, thenV=2dt⇒t=2dV=2×(300×103m)3×108ms−1=2×10−3 sV = \dfrac{2d}{t} \[0.5em] \Rightarrow t = \dfrac{2d}{V} \[0.5em] = \dfrac{2 \times (300 \times 10^3 m)}{3 \times 10^8 m s ^{-1}} \[0.5em] = 2 \times 10^{-3} \text{ s}V=t2d​⇒t=V2d​=3×108ms−12×(300×103m)​=2×10−3 sHence, the radar will detect the enemy plane in 2 x 10-3 s

Question 7

Question 3(d)How is the frequency of a stretched string related to: [2](i) it's length?(ii) it's tension?

Accepted Answer

(i) Frequency is inversely proportional to the length of the string i.e., f ∝ 1l\dfrac{1}{l}l1​(ii) Frequency is directly proportional to the square root of the tension i.e., f ∝ √T

Question 8

Question 4(a)An electric bulb of resistance 500 Ω, draws a current of 0.4 A. Calculate the power of the bulb and the potential difference at it's end. [2]

Accepted Answer

Given,R = 500 Ω,I = 0.4 APower of bulb P = I2 R= (0.4)2 x 500 = 80 WPotential difference at the ends of the bulb V = IR = 0.4 x 500 = 200 V

Question 9

Question 4(b)State two causes of energy loss in a transformer. [2]

Accepted Answer

The two causes of energy loss in a transformer are :Heating loss in the wire of primary and secondary coils.Hysteresis loss in the core of the transformer due to it's magnetization and demagnetization in each cycle.

Question 10

Question 4(d)State two factors upon which the rate of emission of thermions depends. [2]

Accepted Answer

The two factors upon which the rate of emission of thermions depends are :The temperature of the emitter.The surface area of the emitter.

Question 11

Question 4(e)When does the nucleus of an atom tend to be radioactive? [2]

Accepted Answer

The nucleus of an atom tends to be radioactive when the number of neutrons inside it becomes more than the number of protons in it.

Question 12

Question 5(a)A uniform half metre rule balances horizontally on a knife edge at 29 cm mark when a weight of 20 gf is suspended from one end. [3](i) Draw a diagram of the arrangement.(ii) What is the weight of the half metre rule?

Accepted Answer

(i) Diagram of the arrangement is shown below:(ii) If W gf is the weight of half meter rule, by the principle of moments,W×(29−25)=20×(50−29)⇒W×4=20×21⇒W=20×214⇒W=105 gfW \times (29 - 25) = 20 \times (50 - 29) \[0.5em] \Rightarrow W \times 4 = 20 \times 21 \[0.5em] \Rightarrow W = \dfrac{20 \times 21}{4} \[0.5em] \Rightarrow W = 105 \text{ gf}W×(29−25)=20×(50−29)⇒W×4=20×21⇒W=420×21​⇒W=105 gf

Question 13

Question 5(b)(i) A boy uses a single fixed pulley to lift a load of 50 kgf to some height. Another boy uses a single movable pulley to lift the same load to the same height. Compare the effort applied by them. Give a reason to support your answer. [3](ii) How does uniform circular motion differs from uniform linear motion?(iii) Name the process used for producing electricity using nuclear energy.

Accepted Answer

(i) From relation,M.A.=LE	ext{M.A.} = \dfrac{	ext{L}}{	ext{E}}M.A.=EL​we can say that M.A. and effort are inversely related.Therefore, we get,Single Fixed Pulley EffortSingle Movable Pulley Effort=21\dfrac{	ext{Single Fixed Pulley Effort}}{	ext{Single Movable Pulley Effort}} = \dfrac{2}{1}Single Movable Pulley EffortSingle Fixed Pulley Effort​=12​The reason is that the mechanical advantage of a single movable pulley is 2 while that of a single fixed pulley is 1.(ii) The difference between uniform circular motion and uniform linear motion is as follows:Uniform circular motionUniform linear motionIn a uniform circular motion the velocity is variable (although the speed is uniform), so it is an accelerated motion.In uniform linear motion the speed and velocity are constant and acceleration is zero i.e., the uniform linear motion is an unaccelerated motion.(ii) Nuclear fission

Question 14

Question 5(c)A pulley system with V.R. = 4 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort required is 50 kgf in the downward direction. (g = 10 N kg-1) [4]Calculate:(i) Distance moved by the effort.(ii) Work done by the effort.(iii) M.A. of the pulley system.(iv) Efficiency of the pulley system.

Accepted Answer

Given,V.R. = 4L = 175 kgfdL = 15 mE = 50 kgf(i) V.R. = dEdL\dfrac{d_E}{d_L}dL​dE​​Substituting the values we get,4=dE15⇒dE=15×4⇒dE=60 m4 = \dfrac{d_E}{15} \[0.5em] \Rightarrow d_E = 15 \times 4 \[0.5em] \Rightarrow d_E = 60 \text{ m}4=15dE​​⇒dE​=15×4⇒dE​=60 m(ii) Work done by the effort = E x dE     = 50 x 10 x 60     = 3 x 104(iii) M.A. = LE\dfrac{L}{E}EL​Substituting the values we get,M.A.=17550=3.5\text {M.A.} = \dfrac{175}{50} = 3.5M.A.=50175​=3.5(iv) Efficiency η = M.A.V.R.\dfrac{M.A.}{V.R.}V.R.M.A.​Substituting the values we get,η=3.54=0.875∴η=87.5η = \dfrac{3.5}{4} = 0.875 \[0.5em] \therefore η = 87.5 %η=43.5​=0.875∴η=87.5

Question 15

Question 6(b)Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C. [3]Specific heat capacity of calorimeter = 0.4 J g-1 °C-1Specific heat capacity of water = 4.2 Jg-1°C-1Latent heat capacity of ice = 330 Jg-1

Accepted Answer

Given,Mass of water m1 = 150 gMass of calorimeter m2 = 50 gTemperature of water t1 = 32°CFinal temperature t = 5°CLet m be the mass of ice addedHeat given by calorimeter Q1 = m2 c2 x (t1 - t)= 50 x 0.4 x (32 - 5) = 540 JHeat given by water Q2 = m1 c1 x (t1 - t)= 150 x 4.2 x (32 - 5) = 17010 JHeat taken by ice to melt at 0°C = m L= m x 330 Jg-1Heat energy taken by the melted ice to to raise it's temperature from 0°C to 5 °C= m x 4.2 x (5 - 0) = 21m JTotal heat taken by ice = 330m + 21m = 351 m JBy principle of calorimeter, if there is no loss of heat.351 m = 540 + 17010m = 17550351\dfrac{17550}{351}35117550​ = 50 g

Question 16

Question 6(c)(i) Name the radiations which are absorbed by greenhouse gases in the earth’s atmosphere. [4](ii) A radiation X is focused by a particular device on the bulb of a thermometer and mercury in the thermometer shows a rapid increase. Name the radiation X.(iii) Name two factors on which the heat energy liberated by a body depends.

Accepted Answer

(i) The green house gases in the earth's atmosphere absorb the low energy infrared radiations.(ii) The radiation X is Infrared radiation.(iii) The two factors are :Material of the bodyTemperature of the body.

Question 17

Question 7(a)A lens forms an upright and diminished image of an object when the object is placed at the focal point of the given lens. [3](i) Name the lens.(ii) Draw a ray diagram to show the image formation.

Accepted Answer

(i) The lens is concave(ii) The ray diagram is shown below :

Question 18

Question 8(c)(i) A wire of length 80 cm has a frequency of 256 Hz. Calculate the length of a similar wire under similar tension, which will have frequency 1024 Hz. [4](ii) A certain sound has a frequency of 256 hertz and a wavelength of 1.3 m.Calculate the speed with which this sound travels.What difference would be felt by a listener between the above sound and another sound travelling at the same speed, but of wavelength 2.6 m?

Accepted Answer

(i) Given,l1 = 80 cm, f1 = 256 Hz, f2 = 1024 Hz, l2 = ?Since, f ∝ 1l\dfrac{1}{l}l1​Therefore, f1l1 = f2l2orf2=f1l1f2=256×801024=20 cmf_2 = \dfrac{f_1l_1}{f_2} \[0.5em] = \dfrac{256 \times 80}{1024} \[0.5em] = 20 \text{ cm}f2​=f2​f1​l1​​=1024256×80​=20 cm(ii) Given, f = 256 Hz, λ = 1.3 m(a) V = fλ = 256 x 1.3 = 332.8 m s-1(b) Speed of second sound (V') = 332.8 m s-1Wavelength of second sound (λ') = 2.6 mf=V′λ′=332.82.6=128 Hz\text{f} = \dfrac{V'}{λ'} \[0.5em] = \dfrac{332.8}{2.6} \[0.5em] = 128 \text{ Hz}f=λ′V′​=2.6332.8​=128 HzAs frequency of second sound is less as compared to the first sound, hence, to the listener, the first sound of wavelength 1.3 m will appear to be shriller than the second sound of wavelength 2.6 m.

Question 19

Question 9(c)(i) Explain the meaning of the statement 'current rating of a fuse is 5A'. [4](ii) In the transmission of power, the voltage of power generated at the generating stations is stepped up from 11 kV to 132 kV before it is transmitted. Why?

Accepted Answer

(i) Current rating of a fuse is 5A, implies that if the current in a circuit having this fuse in the live wire exceeds 5A, it blows off and the circuit becomes incomplete without damaging the appliance connected in that circuit.(ii) The power generated at the generating station is stepped up from 11 kV to 132 kV before transmission so as to reduce the magnitude of current to 112\dfrac{1}{12}121​th of it's value so that the heating loss of power in the line wires used for transmission are reduced [since heat produced ∝ (current)2]

Question 20

Question 10(c)(i) What are free electrons? [4](ii) Why do they not leave the metal surface on their own?(iii) How can they be made to leave the metal surface? (State any two ways)

Accepted Answer

(i) The loosely bound outer most electrons of atoms in a metal which leave their atoms and become free to move inside the metal are called the free electrons.(ii) The free electrons can move in a random manner inside the metal but they do not have sufficient kinetic energy to leave the metal surface.(iii) The free electrons can be made to leave the metal surface by imparting energy from outside either :by heating in thermionic emission.by making the ultraviolet radiation incident on it in photoelectric effect.

Solved 2017 Question Paper ICSE Class 10 Physics

Section I 40 Marks

Section Ii 40 Marks