Q: Question 4(b)An element ZSA decays to 85R222 after emitting 2 α particles and 1 β particle. Find the atomic number and atomic mass of the element S. [2]

85R222_{85}\text{R}^{222}85R222 before the emission of β particle, would have atomic number 84, and before 2 α emissions, atomic number would have been 88. Hence, atomic number of X = 88.85R222_{85}\text{R}^{222}85R222 before the emission of β particle, would have mass number 222, and before 2 α emissions, mass number would have been 230. Hence, mass number of X = 230.

Question 1

Question 1(b)A boy weighing 40 kgf climbs up a stair of 30 steps each 20 cm high in 4 minute and a girl weighing 30 kgf does the same in 3 minutes. Compare :-(i) The work done by them.(ii) The power developed by them. [2]

Accepted Answer

(i) The work done by them = ?While climbing both the girl and boy have to do work against the force of gravity.Force of gravity on boy = F1 = m1g = 40 x 10 = 400Force of gravity on girl = F2 = m2g = 30 x 10 = 300Total height climbed uph = number of steps x height of each step= 30 x 20 = 600 cm = 6 mWork done by the boy = W1 = F1 x h = 400 x 6 = 2400 JWork done by the girl = W2 = F2 x h = 300 x 6 = 1800 JTherefore, W1 : W2 = 2400 : 1800 = 4 : 3(ii) The power developed by them = ?P1P2=W1t1W2t2=W1W2×t2t1=43×34=1:1\dfrac{P_1}{P_2} = \dfrac{\dfrac{W_1}{t_1}}{{\dfrac{W_2}{t_2}}} \[0.5em] = \dfrac{W_1}{W_2} \times \dfrac{t_2}{t_1} \[0.5em] = \dfrac{4}{3} \times \dfrac{3}{4} \[0.5em] = 1 : 1P2​P1​​=t2​W2​​t1​W1​​​=W2​W1​​×t1​t2​​=34​×43​=1:1

Question 2

Question 1(d)Calculate the mass of ice required to lower the temperature of 300 g of water at 40°C to water at 0°C. (Specific latent heat of ice = 336 J g-1, Specific heat capacity of water = 4.2 J g-1 °C-1) [2]

Accepted Answer

Given,mass of water m = 300 ginitial temperature = 40°Cfinal temperature = 0°CTherefore, fall in temperature = 40 - 0 = 40 °CHeat lost by water = m x c x Δt= 300 x 4.2 x 40 = 5.04 x 104 JIf m' g ice is added, heat gained by it to melt to 0 °C = m'L = m' x 336 JBy the principle of method of mixture,heat lost by water = heat gained by iceTherefore, 5.04 x 104 = m' x 336m' = 5.04×104336\dfrac{5.04 	imes 10^{4}}{336}3365.04×104​ = 150 g

Question 3

Question 2(b)A boy uses blue colour of light to find the refractive index of glass. He then repeats the experiment using red colour of light. Will the refractive index be the same or different in the two cases? Give a reason to support your answer. [2]

Accepted Answer

The speed of blue light in glass is less that of red light.i.e. cb < crWe know that,μ=Speed of light in vacuum or air (c) Speed of light in glass (V)μ = \dfrac{\text{Speed of light in vacuum or air (c) }}{\text {Speed of light in glass (V)}} \[0.5em]μ=Speed of light in glass (V)Speed of light in vacuum or air (c) ​Hence, the refractive index of blue light is greater than that of red light.i.e. μb > μr.Hence, the refractive index of red and blue light in glass will be different.

Question 4

Question 2(c)Copy the diagram given below and complete the path of light ray till it emerges out of the prism. The critical angle of glass is 42°. In your diagram mark the angles wherever necessary. [2]

Accepted Answer

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

Question 5

Question 2(e)The ratio of amplitude of two waves is 3 : 4. What is the ratio of their :(i) loudness?(ii) frequencies? [2]

Accepted Answer

Given, a1 : a2 = 3 : 4(i) Since, loudness ∝ (amplitude)2Therefore,L1L2=(a1a2)2=(34)2=916\dfrac{L_1}{L_2} = \Big(\dfrac{a_1}{a_2}\Big)^2 = \Big(\dfrac{3}{4}\Big)^2 = \dfrac{9}{16}L2​L1​​=(a2​a1​​)2=(43​)2=169​(ii) The frequency of sound does not depend on it's amplitude, so the ratio of frequency f1 : f2 = 1 : 1

Question 6

Question 3(c)The V-I graph for a series combination and for a parallel combination of two resistors is shown in the figure below. Which of the two A or B, represents the parallel combination? Give a reason for your answer. [2]

Accepted Answer

The slope of V-I graph gives the resistance. Since the straight line A is less steeper than B, so the straight line A represents small resistance. In parallel combination, the equivalent resistance is less than in series combination so A represents the parallel combination.

Question 7

Question 3(d)A music system draws a current of 400 mA when connected to a 12 V battery.(i) What is the resistance of the music system?(ii) The music system is left playing for several hours and finally the battery voltage drops and the music system stops playing when the current drops to 320 mA. At what battery voltage does the music system stop playing? [2]

Accepted Answer

Given,I = 400 mA = 400 x 10-3 = 0.4 AV = 12 VFrom Ohm's law,V = IRSubstituting the values in the formula we get,12=0.4×R⇒R=120.4⇒R=30Ω12 = 0.4 \times R \[0.5em] \Rightarrow R = \dfrac{12}{0.4} \[0.5em] \Rightarrow R = 30 Ω12=0.4×R⇒R=0.412​⇒R=30ΩHence, the resistance of the music system = 30 Ω(b) Given,I = 320 mA = 320 x 10-3 A = 0.32 AR = 30 ΩFrom Ohm's law,V = IRSubstituting the values in the formula we get,V = 0.32 x 30 = 9.6VHence, the battery voltage when the music system stops playing = 9.6 V

Question 8

Question 4(a)State the characteristics required of a good thermion emitter. [2]

Accepted Answer

The characteristics required of a good thermion emitter are :The work function of the substance should be low so that the electrons may be emitted from it even when it is not related to a high temperature.The melting point of the substance should be high so that it may not melt when it is heated to the temperature required for thermionic emitter.

Question 9

Question 4(b)An element ZSA decays to 85R222 after emitting 2 α particles and 1 β particle. Find the atomic number and atomic mass of the element S. [2]

Accepted Answer

85R222_{85}	ext{R}^{222}85​R222 before the emission of β particle, would have atomic number 84, and before 2 α emissions, atomic number would have been 88. Hence, atomic number of X = 88.85R222_{85}	ext{R}^{222}85​R222 before the emission of β particle, would have mass number 222, and before 2 α emissions, mass number would have been 230. Hence, mass number of X = 230.

Question 10

Question 4(d)State the characteristics required in a material to be used as an effective fuse wire. [2]

Accepted Answer

The characteristics required in a material to be used as an effective fuse wire are :The melting point should be low,The specific resistance should be high.

Question 11

Question 4(e)Which coil of a step up transformer is made thicker and why? [2]

Accepted Answer

Primary Coil is made thicker than the secondary coil in a step up transformer. As current in primary coil is more than in the secondary coil so the wire in the primary coil is kept thicker.The use of thicker wire in a coil reduces it's resistance and therefore, reduces the loss of energy as heat in that coil.

Question 12

Question 5(a)A stone of mass 'm' is rotated in a circular path with a uniform speed by tying a strong string with the help of your hand. Answer the following questions:(i) Is the stone moving with a uniform or variable speed?(ii) Is the stone moving with a uniform acceleration? In which direction does the acceleration act?(iii) What kind of force acts on the hand and state it's direction? [3]

Accepted Answer

(i) The stone is moving with uniform speed.(ii) Yes, the stone is moving with a uniform acceleration. The acceleration acts towards the centre.(iii) Centrifugal force acts on the hand. It is directed radially outwards away from the centre.

Question 13

Question 5(c)(i) An effort is applied on the bigger wheel of a gear having 32 teeth. It is used to turn a wheel of 8 teeth. Where is it used?(ii) A pulley system has three pulleys. A load of 120 N is overcome by applying an effort of 50 N. Calculate the Mechanical Advantage and Efficiency of this system. [4]

Accepted Answer

(i) In a gear system, the wheel on which effort is applied is called driving gear and other one is called driven gear. If number of teeth are more in driving gear than the driven gear, then we gain in speed.(ii) Given, Load = 120 N and E = 50 N,From relation,M.A.=LE\text{M.A.} = \dfrac{\text{L}}{\text{E}}M.A.=EL​Substituting the values in the formula above we get,M.A.=12050⇒M.A.=2.4\text{M.A.} = \dfrac{120}{50} \[0.5em] \Rightarrow \text{M.A.}= 2.4M.A.=50120​⇒M.A.=2.4V.R. = 3Efficiency (η)=M.A.V.R.×100\text{Efficiency (η)} = \dfrac{\text{M.A.}}{\text{V.R.}} \times 100%Efficiency (η)=V.R.M.A.​×100Substituting the values in the formula above we get,η=2.43%η=80%\text{η} = \dfrac{2.4}{3}\% \[0.5em] \text{η} = 80\%η=32.4​%η=80%

Question 14

Question 6(b)Some ice is heated at a constant rate, and it's temperature is recorded after every few seconds, till steam is formed at 100°C. Draw a temperature time graph to represent the change. Label the two phase changes in your graph. [3]

Accepted Answer

Temperature-time graph representing the change of ice to steam is shown below:

Question 15

Question 6(c)A copper vessel of mass 100 g contains 150 g of water at 50 °C. How much ice is needed to cool it to 5 °C?Given : Specific heat capacity of copper = 0.4 J g-1 ⁰C-1Specific heat capacity of water = 4.2 J g-1 ⁰C-1Specific latent heat of fusion of ice = 336 J g-1 [4]

Accepted Answer

Heat energy imparted by vessel = 100 x 0.4 x (50 - 5) = 1800 JHeat energy imparted by water = 150 x 4.2 x (50 - 5) = 28,350 JLet m g of ice be used, thenHeat energy taken by ice to melt = m x 336 JHeat energy taken by melted ice to raise it's temperature from 0 °C to 5 °C = m x 4.2 x (5 - 0) = 21m JBy law of conservation of energyheat energy imparted by vessel and water = heat energy taken by ice and melted ice1800 + 28,350 = 336m + 21m357 m = 30,150m = 30150357\dfrac{30150}{357}35730150​ = 84.45 g

Question 16

Question 7(c)What do you understand by the term 'Scattering of light'? Which colour of white light is scattered the least and why? [4]

Accepted Answer

Scattering is the process of absorption and then re-emission of light energy by the dust particles and air molecules present in the atmosphere.Red colour of white light is scattered the least because wavelength of red light (λ = 8000 Å) is longest and intensity of scattered light (I) ∝ 1λ4\dfrac{1}{λ^4}λ41​.

Question 17

Question 8(b)(i) What is an echo?(ii) State two conditions for an echo to take place. [3]

Accepted Answer

(i) The sound heard after reflection from a distinct obstacle (such as a cliff, a hillside etc.) after the original sound has ceased is called an echo.(ii) Conditions for an echo to take place are :The size of the reflector must be large enough as compared to the wavelength of the sound wave.The intensity of sound should be such that the reflected sound reaching the ear is sufficiently loud to be audible.

Question 18

Question 8(c)(i) Name the phenomenon involved in tuning a radio set to a particular station.(ii) Define the phenomenon named by you in part (i) above.(iii) What do you understand by loudness of sound?(iv) In which units is the loudness of sound measured? [4]

Accepted Answer

(i) Resonance.(ii) Resonance is a special case of forced vibration. When the frequency of the externally applied periodic force on a body is equal to it's natural frequency, the body begins to vibrate with an increased amplitude. This phenomenon is known as resonance.(iii) Loudness is the characteristic by virtue of which a loud sound can be distinguished from a faint one, both having the same pitch and quality.(iv) The unit of loudness is phon. The level of sound is expressed in decibel (dB).

Question 19

Question 9(c)A battery of e.m.f. 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series. [4]Find:(i) Current in the circuit.(ii) The terminal voltage of the cell.(iii) The potential difference across 6 Ω Resistor.(iv) Electrical energy spent per minute in 4 Ω Resistor.

Accepted Answer

Given,RA = 4 ΩRB = 6 Ωe.m.f. = 12 Vr = 2 Ω(i) Two resistors ( 4 Ω and 6 Ω) are connected in series, so Rs = 4 + 6 = 10 ΩFrom relation,I = 𝐸𝑅+𝑟\dfrac{𝐸}{𝑅 + 𝑟}R+rE​Substituting the values we get,I=1210+2I=1212⇒I=1AI = \dfrac{12}{10 + 2} \[0.5em] I = \dfrac{12}{12} \[0.5em] \Rightarrow I = 1 AI=10+212​I=1212​⇒I=1A(ii) Terminal voltage of cell V = ε - IrSubstituting the values we get,V=12−(1×2)V=12−2⇒V=10VV = 12 - (1 \times 2) \[0.5em] V = 12 - 2 \[0.5em] \Rightarrow V = 10 VV=12−(1×2)V=12−2⇒V=10V(iii) P.d. across 6 Ω resistor = I R6ΩSubstituting the values we get,V6Ω = 1 x 6 = 6 V(iv) Electrical energy spent per minute (60 s) in 4 Ω Resistor = I2RAt = 1 x 6 = 6 V

Question 20

Question 10(c)(i) Represent the change in the nucleus of a radioactive element when a β particle is emitted.(ii) What is the name given to elements with same mass number and different atomic number?(iii) Under which conditions does the nucleus of an atom tend to be radioactive? [4]

Accepted Answer

(i) If a radioactive nucleus P with mass number A and atomic number Z emits a β particle to form a daughter nucleus Q with mass number A and atomic number Z + 1, the change can be represented as follows:ZAP⟶Z+1     AQ+ −1   0eParentDaughterBetanucleusnucleusparticle\begin{matrix} _{Z}^{A}	ext{P} & \longrightarrow & _{Z+1}^{\space \space \space \space \space A}	ext{Q} & + & \space _{-1}^{\space \space \space 0}	ext{e} \ \footnotesize{	ext{Parent}} & & \footnotesize{	ext{Daughter}} & & \footnotesize{	ext{Beta}} \ \footnotesize{	ext{nucleus}} & & \footnotesize{	ext{nucleus}} & & \footnotesize{	ext{particle}} \end{matrix}ZA​PParentnucleus​⟶​Z+1     A​QDaughternucleus​+​ −1   0​eBetaparticle​ii) Elements with same mass number A, but different atomic number Z are called isobars.iii) The nucleus of an atom tend to be radioactive when the number of neutrons is more than the number of protons. The nucleus thus becomes unstable and in order to gain stability it tends to emit α, β, γ radiations.

Solved 2016 Question Paper ICSE Class 10 Physics

Section I 40 Marks

Section Ii 40 Marks