Physics for You - January 2017

Volume 25 Managing Editor Mahabir Singh Editor Anil Ahlawat (BE, MBA)

No. 1

January 2017

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CONTENTS

Class 11 NEET | JEE Essentials

8

Ace Your Way CBSE

23

MPP-7

34

Brain Map

38

Class 12 Brain Map

39

NEET | JEE Essentials

40

Ace Your Way CBSE

59

Key Concept

69

MPP-7

73

Competition Edge Physics Musing Problem Set 42

76

JEE Main Practice Paper

78

Physics Musing Solution Set 41

85

You ask We Answer

87

At a Glance

88

Crossword

89

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PHYSICS FOR YOU | JANUARY ‘17

7





    

 Heat and Temperature •

Heat is a form of energy that is transferred between two bodies or between adjacent part of a body or between a system and its environment at different temperatures. When some heat is given to a body and its state does not change, the temperature of the body rises and if heat is taken from a body its temperature falls i.e., temperature can be regarded as the effect of cause, heat. Temperature is that property of a body which helps us to decide the degree of its hotness. Zeroth law of thermodynamics : If two systems A and B are each in thermal equilibrium with a third system C, then A and B will be in thermal equilibrium with each other. Scales of Temperature : Centigrade or Celsius (°C), Fahrenheit (°F) and Kelvin (K), are commonly used temperature scales.  Relation between these scales

•

• •

•

TC TF − 32 TK − 273 = = 5 9 5 Although the temperature of a body can be raised without limit, it cannot be lowered without limit and theoretically limiting low temperature is taken to be zero on the Kelvin scale.

•

8

PHYSICS FOR YOU | JANUARY ‘17

 



 





















 

•

•

•

  

  

Mathematically, heat supplied to the body, or heat released by the body. ΔQ = msΔT Where, m = masss of the body, s = specific heat of the body and ΔT = change in temperature The amount of heat required to change the state of a body of mass m from solid to liquid at melting point of the solid or from liquid to gas at boiling point of the liquid is ΔQ = mL, where L is the latent heat of the substance. ΔQ , if the substance As ΔQ = msΔT ⇒ s = mΔT undergoes the change of state which occurs at ΔQ constant temperature (ΔT = 0), then s = = ∞. 0

•

•

Thus the specific heat of a substance when it melts or boils at constant temperature is infinite. Thermal or heat capacity of a body, H = ms  Thermal capacity of the body and its water equivalent are numerically equal.  If thermal capacity of a body is expressed in terms of mass of water, it is called water equivalent of the body. Heating curve of a body (state changes from solid to liquid and liquid to gas)          

•

•







    



     





 



When two substances at different temperatures are mixed together, heat flows from the substance at higher temperature to the substance at lower temperature till a common temperature is reached. In this process, heat lost by one substance = heat gained by the other substance, i.e., m1 × s1 (ΔT1) = m2 × s2 (ΔT2) Joule’s mechanical equivalent of heat J=



mechanical work done (W ) heat energy produced (H )

= 4.2 J cal–1 = 4.2 × 107 ergs cal–1 If T is rise in temperature of a body of mass m on falling through a height h, then mgh gh or T = smT Js When a block of ice of mass m melts on falling through a height h, then J=



J= 

10

•

 

        



 Thermal Expansion

•

Thermal expansion in solids is of three types  Coefficient of linear expansion Increase in length α= Original length × Rise in temperature L − L0 α= T or LT = L0 (1 + αΔT ) L0 × ΔT  Coefficient of area expansion Increase in area β= Original area × Rise in temperature A − A0 β= T or AT = A0 (1 + βΔT ) A0 × ΔT  Coefficient of volume expansion Increase in volume γ= Original volume × Rise in temperature V − V0 γ= T or VT = V0 (1 + γΔT ) V0 × ΔT  The three coefficients of thermal expansion are β γ related as α = = . 2 3 Effect of thermal expansion  With increase in temperature volume of substance increase while mass remains constant, therefore density should decrease. ρ ρ′ = ≈ ρ(1 – γ·Δθ) (if γ·Δθ f or immersed fraction will increase. When a solid whose density is more than the density of liquid is immersed completely, then upthrust will act on 100% volume of solid and apparent weight appears less than the actual weight. Wapp = W – FB

Here, FB = Vsρl g With increase in temperature, Vs will increase and ρl will decrease, while g will remain unchanged. Therefore upthrust may increase, decrease or remain same.  With increase in temperature, length of pendulum will increase. Therefore time period will increase. A pendulum clock will become slow and it loses the time. At some higher temperature, 1 ⎛ 1 ⎞ T ′ = T(1 + αΔθ)2 or T ′ ≈ T ⎜1 + αΔθ ⎟ (if α Δθ T2)

( Σx i ) ⎛ xi ⎞ ⎜⎝ Σ K ⎟⎠ i

•

Conducting slabs in parallel



⎛T −T ⎞ Q = (K1 A1 + K 2 A2 + K 3 A3 ) ⎜ 1 2 ⎟ t ⎝ x ⎠ Equivalent thermal conductivity K A + K 2 + A2 + K 3 A3 K1, A1 K= 1 1 A1 + A2 + A3 T1

•

K2, A2

T2

(ΣK i Ai ) K3, A3 In general, K = x ΣAi Time taken in growth of ice layer (from thickness ρL 2 x1 to x2) on water surface is t = (x 2 – x21); 2KT T = temperature of atmosphere. 

 Heat Transfer •

Equivalent thermal conductivity (K)

The time interval to change the thickness from 0 to x, from x to 2x and so on will be in the ratio Δt1 : Δt2 : Δt3 : : 1 : 3 : 5.

 Radiation •

•

• • •

•

• •

All objects emit radiations simply because their temperature is above absolute zero, and all objects absorb some of radiations. The intensity of radiation is inversely proportional to the square of distance of point of observation form the source (i.e., I ∝ 1/d2). When a body is heated, all radiations having wavelengths from zero to infinity are emitted. Radiations of longer wavelengths are predominant at lower temperature. energy absorbed Absorptive power a = energy incident – a < 1 for ordinary body – a = 1 for perfectly black body Spectral absorptive power aλ = absorptive power of wavelength λ. – aλ < 1 for ordinary body – aλ = 1 for perfectly black body Emissive power : Energy radiated per unit area per unit time is called emissive power of a body. Stefan’s law : Emissive power of a body is given by, E = eσT4 Here e = emissivity e ≤ 1 and e = 1 for a perfectly black body

Total energy radiated by a body, Q = eσT 4At Here, A = surface area of body, T = temperature of body, t = time σ = Stefan’s constant.  Net emissive power of a body E = emissive power – absorptive power = eσ(T4 – T04) Kirchhoff ’s law : If different bodies (including a perfectly black body) are kept at same temperature, then emissive power is proportional to the absorptive power. e e λ ∝ aλ or λ = constant aλ 

or



⎛ eλ ⎞ ⎛e ⎞ ⎛e ⎞ =⎜ λ ⎟ =⎜ λ ⎟ ⎜⎝ a ⎟⎠ ⎝ aλ ⎠ Body-2 ⎝ aλ ⎠ Perfectly black body λ Body-1

Good absorbers of a particular wavelength λ are also good emitters of same wavelength.  At a given temperature, ratio of eλ and aλ for any body is constant. This ratio is equal to eλ of perfectly black body at that temperature. Wien’s Displacement law 

•



⎡ T1 + T2 ⎤ ⎛ T1 − T2 ⎞ ⎜⎝ t ⎟⎠ = k ⎢ 2 − T0 ⎥ . ⎣ ⎦

L 

 L

•

L

1 λ m ∝ or λ mT = Constant = Wien’s constant (b) T Here, b = 2.89 × 10–3 mK Further, area of this graph will give total emissive power which is proportional to T 4. Cooling of a body by radiation  Rate of cooling dT eAσ 4 dT − = (T − T04 ) or − ∝ (T 4 − T04 ) dt ms dt

Newton’s law of cooling : If temperature difference of a body with atmosphere is small, then rate of cooling is proportional to the temperature difference. If body cools by radiation according to this law, then temperature of body decreases exponentially. In the figure, Ti = initial temperature of   body T0 = temperature of  surrounding.  Temperature at any time t can be written as, T = T0 + (Ti − T0 )e − kt ; k = constant If body is cooling according to this law then to find temperature of a body at any time t, we will have to calculate e −αt . To avoid this, you can use a shortcut approximate formula given below

If ln(T – T0) be plotted against t, then the equation assumes the form y = mx + c; where m = –k and C = ln(T1 – T0)

ln( – 0)

•







This is a straight line with negative slope. If Qemission > Qabsorption → temperature of body decreases and consequently the body appears colder. If Qemission < Qabsorption → temperature of body increases and it appears hotter. If Qemission = Qabsorption → temperature of body remains constant (thermal equilibrium).

•

• •

 Gas Law’s Name of law Boyle’s law

Charle’s law

14

Constant terms Basic concept (i) Mass of gas PV = constant (ii) Temperature V ∝1/P P1V1 = P2V2 (i) Mass of gas (ii) Pressure

PHYSICS FOR YOU | JANUARY ‘17

Graph 



 



V/T = constant; V ∝T V1 V2 = T1 T2



Vt =V0(1 + αt)

(α → volume expansion coefficient = 1/273 °C–1)

 slope = 0A (K)

–273°C

(°C)

Gay-Lussac’s law (i) Mass of gas (ii)Volume

P/T = constant; P ∝T P1 P2 = T1 T2 Pt =P0(1 + βt)

•

•

Avogadro’s law : At same temperature and pressure equal volumes of all gases contains equal number of molecules. N1 = N2 ; if P, V and T are same. Dalton’s law : According to this law, the pressure exerted by a mixture of several gases equals the sum of the pressure exerted by each component of gas present in the mixture i.e., Pmix = P1 + P2 + P3 ...... ⎛ RT ⎞ P=⎜ n ⇒ Pmix ∝ n ⎝ V ⎟⎠



slope = 0B (K)

 (°C)

–273°C

(β → pressure expansion coefficient = 1/273 °C–1) 

Average speed vav =

•

RT (n + n + n + .......) V 1 2 3 Grahm’s law of diffusion : According to this law, at same temperature and pressure, the rate of diffusion of gases inversely proportional to the square root of the density of gas i.e., 1 Rate of diffusion rd ∝ ρ 1 Also, vrms ∝ so vrms ∝ rd ; ρ

8P 8 RT 8 kT = = πρ π M π m

Kinetic energy of gas (internal energy) 

Translatory kinetic energy 1 3 2 ET = Nm vrms = PV 2 2 Total internal energy of an ideal gas is kinetic.



Energy per unit volume or energy density (EV)





Total Energy

E 1 2 3 = ρvrms = P Volume V 2 2 Molar K.E. or Mean Molar K.E. (E) : K.E. of N molecules 3 3 (R = Nk) E = RT = NkT 2 2 Molecular kinetic energy or mean molecular K.E.(E ) : K.E. of a gas molecule EV =

∴ Pmix = •



E=

=

E 3 RT 3 or E = ⇒ E = kT N 2 N 2

 Kinetic Theory of an Ideal Gas

 Degree of Freedom

•

•

The number of independent ways in which a molecule or an atom can exhibit motion is called its degrees of freedom.

•

The number of independent coordinates required to specify the dynamical state of a system is called its degrees of freedom.

•

The degrees of freedom are of three types :

•

Pressure of an ideal gas inside the container 1 mN 2 1 2 P= vrms = ρvrms 3 V 3 Where, m = mass of each molecule, N = total number of molecules, V = volume of container or total volume of gas, ρ = density of gas, vrms = root mean square speed of the gas. Various types of speeds of gas molecules 

Root mean square speed, 3P 3RT 3kT = = ρ M m Here, M = molar mass of the gas



Translational degree of freedom : Maximum three degree of freedom are there corresponding to translational motion.



Rotational degree of freedom : The number of degrees of freedom in this case depends on the structure of the molecule.



Vibrational degree of freedom : It exhibits at high temperature.

vrms =



Most probable speed vmp =

2P 2RT 2kT = = ρ M m

PHYSICS FOR YOU | JANUARY ‘17

15

•

Degree of freedom for different gases depends on atomicity of gas. Atomicity of gas

Translational

Rotational

3

0

Monatomic e.g., He, Ar, Ne, Ideal gas etc

Vibrational Total 0

Graphically

3







Diatomic e.g., H2, O2, Cl2, N2 etc

3

2

0

5





Triatomic or Polyatomic

(linear) e.g., CO2, C2H2

3

2

2

7

 

O=C=O





Triangular (non-linear) e.g., H2O, O3 etc

•

3

Relation between degree of freedom and specific heat of gas.  Energy of one mole of gas related with each degree of freedom = RT/2  Energy of one mole of gas related with all degrees of freedom = fRT/2  Energy of each molecule of gas related with all degrees of freedom = fkT/2

3

 Mixture of Non Reactive Gases • •

n = n1 + n2 U = U1 + U2

•

CV =

f ΔU R ⎞ ⎛f ; CV = R = ; CP = CV + R = ⎜ + 1⎟ R ⎝2 ⎠ ΔT 2 γ −1 CP γR 2 2 CP = ;γ= = 1+ ; γ = 1+ CV f f γ −1

•

•

•

Van der waal’s equation : ⎛ n2 a ⎞ P + (V − nb) = nRT [for n - mole] ⎜⎝ V 2 ⎟⎠ where a and b are Van der waal’s constant Critical parameter :(Pc , Vc , Tc) a 8 a Pc = , Tc = , V = 3nb 2 27 Rb c 27b Approximate ideal gas law PV = nRT = NkT

16

PHYSICS FOR YOU | JANUARY ‘17

n1CV1 + n2CV2

P = P1 + P2 ΔU = ΔU1 + ΔU2 CP =

n1CP1 + n2CP2

n1 + n2 n1 + n2 n1 n2 CP n γ= or = + CV γ − 1 γ1 − 1 γ 2 − 1 n M +n M M= 1 1 2 2 n1 + n2

= CV + R

 Thermodynamic System and Process •

 Real Gas Equation •



Here, n = number of moles, N = number of molecules, k = R/NA

•

CV =

6



Internal energy of one mole of ideal gas (total KE) U = fRT/2 (ΔU = fRΔT/2)



0

•

•

A thermodynamic system can be described by specifying its pressure (P), volume (V), temperature (T), internal energy (U) and the number of moles (n). The relation between the thermodynamic variables (P, V, T) of the system is called equation of state.  For n moles of an ideal gas, equation of state is PV = nRT Thermodynamics system may be of three types :  Open system : It exchanges both energy and matter with the surroundings.  Closed system : It exchanges only energy (not matter) with the surroundings.



Isolated system : It exchanges neither energy nor matter with the surroundings.

 Work Done •

Mathematical method : ΔW = PΔV ⇒

V2

W = ∫ PdV V1

If P constant, W = P(V2 – V1) = nR(T2 – T1)  If V constant, W = 0  If T constant, W = 2.303 n RT log10(V2/ V1) = 2.303 nRT log10(P1/P2)  If Q constant, W = nR(T1 – T2)/γ – 1 = (P1V1 – P2V2)/γ – 1   Graphical method :   Work done = Area  enclosed between P - V 

 curve on V axis   Sign concept for work done : — If V ↑ ⇒ dV = + ve ⇒ expansion of gas ΔW = (+ve) ⇒ work done by the system. 

•

If V ↓ ⇒ dV = – ve ⇒ Compression of gas ΔW = (–ve) ⇒ work done on the system.  Sign concept for heat : — If heat given to system or heat absorbed by the system ΔQ = +ve — If heat rejected by system or heat evolved by the system ΔQ = –ve  Sign concept for internal energy : Obtained by difference of (ΔQ – ΔW) — If dU (+ve) ⇒ then U ↑ — If dU (–ve) ⇒ then U ↓ First law of thermodynamic is based on energy conservation, ΔQ = ΔW + dU Heat and work both are path dependent so they called unexact differential parameter. Internal energy is a point function or state function, Internal energy only depends on initial and final state of system so it is called exact differential parameter. —

• • •

•

Different types of thermodynamics processes Basic Point Isochoric Isobaric Isothermal Constant Volume Pressure Temperature parameter Equation of state P ∝ T PV = constant V∝T Work done

ΔW = zero

ΔW = P Δ V = nRΔT

ΔW = 2.303nRT log10

Adiabatic Heat PV γ = constant, TV γ–l = constant P1–γTγ = constant, Pρ–γ = constant, −PV V2 nR(T1 − T2 ) PV = 1 1 2 2 ΔW = V1 γ −1 γ −1

Applied to first ΔQ ≠ 0, ΔU = 0, ΔQ = ΔW ΔQ = dU law of thermodyΔW ≠ 0, = nCVΔT namics ΔU ≠ 0 ΔQ = μCPΔT, dU = μCVΔT  P-V diagram Q  o  

ΔQ = 0, ΔW= –ΔU = –nCVΔT



  



Slope of PV graph Slope = dP/dV = ∞ Specific condition Closed container







Slope = dP/dV = 0 Constant pressure (Open container)

Q



Slope = dP/dV = –P/V

Slope = dP/dV = –γP/V

(i) Conductive vessel (ii) slowly process

(i) Non-conductive vessel/insulated vessel (ii) Rapidly/fast/sudden process

 Second Law of Thermodynamics •



Kelvin Plank’s statement : It state that in cyclic process total heat can not be converted into mechanical work.

•

•

Claussius statement : It is impossible to have net heat flow from a low temperature body to a high temperature body. Carnot’s theorem : This theorem states that PHYSICS FOR YOU | JANUARY ‘17

17

•

efficiency of any irreversible heat engine can not be greater than or equal to efficiency of a reversible heat engine provided both work between same heat source and sink. Heat engine : Main elements of heat engine  Heat source at temperature T1 = Heat reservoir  Working substance  Sink at low temperature T2 = Cold reservoir 
  
 
   

•

⎛ Q ⎞ ⎛ T ⎞ η = ⎜1 − 2 ⎟ × 100% = ⎜1 − 2 ⎟ × 100% ⎝ Q1 ⎠ ⎝ T1 ⎠ Refrigerator : It is just opposite to a heat engine.  In refrigerator, heat is absorbed from a cold body and some external work is to be done on refrigerant and the total heat is given out at higher temperature source.  The coefficient of performance (C.O.P.) is reciprocal of efficiency and for refrigerator it is better to work out with its C.O.P.

  
!"    

 
         


  

   

Q1 = Q2 + W W Q1 − Q2 T1 − T2 Efficiency η = = = Q1 Q1 T1

18

PHYSICS FOR YOU | JANUARY ‘17

C.O.P.(β) =

Q2 (Heat absorbed from cold body)

W (Mechanical work on refrigerator) Q2 T2 C.O.P. = = Q1 − Q2 T1 − T2 ””

1. A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in the same thermal contact with the two reservoir in time t ? Q Q Q (a) (b) (c) 2 Q (d) 4 16 2 2. A piece of blue glass heated to a high temperature and a piece of red glass at room temperature, are taken inside a dimly lit room, then (a) the blue piece will look blue and red will look as usual (b) red looks brighter red and blue looks ordinary blue (c) blue shines like brighter red compared to the red piece (d) both the pieces will look equally red. 3. A steel scale measures the length of a copper rod as 80 cm when both are at 20°C, the calibration temperature for the scale. The scale read for the length of the rod when both are at 40°C is (Given, α for steel = 11 × 10–6 °C–1 and α for copper = 17 × 10–6 °C–1.) (a) 97.096 cm (b) 80.0096 cm (c) 89.0096 cm (d) 92.23 cm 4. If the ratio of specific heat of a gas at constant pressure to that at constant volume is γ, the change in internal energy of a mass of the gas, when the volume changes from V to 2 V at constant pressure P, is PV R γPV (a) (b) PV (c) (d) (γ − 1) (γ − 1) (γ − 1) 5. A reversible engine converts (1/6)th of the heat into work. When the temperature of the sink is reduced by 62 K, the efficiency of the engine is doubled. The temperatures of the source and sink respectively are (a) 95°C and 47°C (b) 80°C and 37°C (c) 99°C and 37°C (d) 90°C and 47°C 6. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats γ. It is moving with speed v and is suddenly brought to

rest. Assuming no heat is lost to the surroundings, its temperature increases by (γ − 1) (γ − 1) Mv 2 (b) Mv 2 (a) 2 (γ + 2) R 2γ R γ Mv 2 (γ − 1) Mv 2 (c) (d) 2R 2R 7. The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21 × 10–21 J and 484 m s–1 respectively. The corresponding values at 600 K are nearly (a) 12.42 × 10–21 J and 968 m s–1 (b) 8.78 × 10–21 J and 684 m s–1 (c) 6.21 × 10–21 J and 968 m s–1 (d) 12.42 × 10–21 J and 684 m s–1 8. The Wien’s displacement law express relation between (a) wavelength corresponding to maximum energy and temperature (b) radiation energy and wavelength (c) temperature and wavelength (d) colour of light and temperature. 9. At what temperature is (1/2)kBT equal to minimum rotational energy permitted for a hydrogen molecule? (a) 87 K (b) 92 K (c) 82 K (d) 98 K 10. A polyatomic gas with n degrees of freedom has a mean energy per molecule given by nkT nkT nkT 3kT (b) (c) (d) (a) N 2 2 2N 11. 10 g of ice cubes at 0ºC are released in a tumbler (water equivalent 55 g) at 40ºC. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L = 80 cal g–1) (a) 31ºC (b) 22ºC (c) 19ºC (d) 15ºC 12. A piece of lead is dropped from an aeroplane at a height of 200 m. If 60% of the striking energy is converted into heat, then the rise in temperature is (specific heat for lead is 127.7 J kg–1 K–1) (a) 9.2 K (b) 9.8 K (c) 12.7 K (d) 11.4 K 13. The molecules of a given mass of a gas have rms velocity of 200 m s–1 at 27°C and 1.0 × 105 N m–2 pressure. When the temperature and pressure of the PHYSICS FOR YOU | JANUARY ‘17

19

gas are respectively, 127°C and 0.05 × 105 N m–2, the rms velocity of its molecules in m s–1 is (a)

100 2 100 (b) 3 3

(c) 100 2 (d)

400

3 [NEET Phase I 2016]

14. Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100 °C, while the other one is at 0 °C. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is (a) 50 °C (b) more than 50 °C (c) less than 50 °C but greater than 0 °C [NEET Phase II 2016] (d) 0 °C 15. A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be 7 3 4 (a) T (b) T (c) T (d) T 4 2 3 [NEET Phase II 2016] 16. The temperature inside a refrigerator is t2 °C and the room temperature is t1 °C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be t1 t + 273 (a) (b) 1 t1 − t2 t1 − t2 t2 + 273 t1 + t2 (c) (d) t1 − t2 t1 + 273 [NEET Phase II 2016] 17. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4 s a day if the temperature is 20 °C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively (a) 25°C; α = 1.85 × 10–5 °C–1 (b) 60°C; α = 1.85 × 10–4 °C–1 (c) 30°C; α = 1.85 × 10–3 °C–1 (d) 55°C; α = 1.85 × 10–2 °C–1 [JEE Main Offline 2016] 18. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn = constant, then n is given by (Here CP and CV are molar specific 20

heat at constant pressure and constant volume, respectively) C − CP C (a) n = P (b) n = C − CV CV C −C C − CV (c) n = P (d) n = C − CV C − CP

PHYSICS FOR YOU | JANUARY ‘17

[JEE Main Offline 2016] 19. ‘n’ moles of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of the gas during the process will be 9P0V0 3P0V0 9P0V0 9P0V0 (b) (c) (d) (a) 4nR 2nR 2nR nR [JEE Main Offline 2016] 20. 200 g water is heated from 40° C to 60°C. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J kg–1 K–1) (a) 167.4 kJ (b) 8.4 kJ (c) 4.2 kJ (d) 16.7 kJ [JEE Main Online 2016] 1.

2.

3.

4.

SOLUTIONS (b) : A′ = A/4 and l′ = 4l Q′ A′ l 1 1 1 Q = = ⋅ = , ∴ Q′ = Q A l ′ 4 4 16 16 (c) : According to Stefan’s law, E ∝ T4 As the temperature of blue glass is more than that of red glass, so it will appear brighter than red glass. (b) : 1 cm length of steel scale at 40°C = 1 cm + 1 × (11 × 10–6)(40 – 20) cm = 1.00022 cm Length of copper rod at 40°C = 80 cm + (80 × 17 × 10–6) (40 – 20) cm = 80.0272 cm Number of division on the steel scale 80.0272 cm ≈ 80.0096 = 1.00022 cm Length of the rod = 80.0096 cm CP − CV CP (a) : As =γ ∴ =γ–1 CV CV C − CV R = or CV = P γ −1 γ −1 nRΔT Now ΔU = nCVΔT= (∵ PΔV = nRΔT) γ −1 P ΔV P (2V − V ) PV ⇒ ΔU = = = γ −1 γ −1 γ −1

T 1 T 6 5. (c) : As η = 1 − 2 , = 1 − 2 ; T1 = T2 T1 6 T1 5 T2′ 1 1 Also, η′ = 2 × = = 1 − T1′ 6 3 or

6.

7.

8.

9.

(T − 62) 1 =1− 2 T1 3

or

(T − 62) 1 =1− 2 (6/5)T2 3

Hence, T2 = 310 K = 37°C 6 6 Also, T1 = T2 = × 310 K = 372 K = 99°C 5 5 1 (d) : Loss in KE of the gas = mv2 2 Heat gained by gas = nCVΔT 1 2 m R ∴ mv = nCVΔT = ⋅ ΔT 2 M γ −1 (γ − 1) Mv 2 or ΔT = 2R (d) : Average translational energy ∝ temperature. Thus, when temperature is doubled, average translational energy becomes double, i.e., 12.42 × 10–21 J. Further, as vrms ∝ T ; when T becomes 2 times of its previous value, vrms becomes 2 times of its previous value, i.e., 2 × 484 m s–1 = 684 m s–1 (a) : Wien’s displacement law states that the product of absolute temperature and the wavelength at which the emissive power is maximum is constant i.e. λmax T = constant. Therefore it expresses relation between wavelength corresponding to maximum energy and temperature. 1 (a) : Kinetic energy of rotation = Iω2 2 (I ω)2 L2 = = (as L = Iω) 2I 2I 1 For kBT to be equal to minimum rotational energy, 2 L2 L2 1 kBT = min or T = min ...(i) kB I 2 2I From quantum mechanics, h ≈ 10 −34 kg m2 s −1 L2min = ...(ii) 2π In case of hydrogen molecule I = 2mR2, m = 1.67 × 10–27 kg, R ≈ 5 × 10–11 m Thus, I = 2(1.67 × 10–27)(5 × 10–11)2 kg m2 = 8.3 × 10–48 kg m2 ...(iii) From eqns. (i), (ii) and (iii), T=

(10 −34 )2 (1.38 × 10 −23 )(8.3 × 10 −48 )

≈ 87 K

10. (c) : According to law of equipartition of energy, 1 the energy per degree of freedom is kT . For a 2 polyatomic gas with n degrees of freedom, the mean 1 energy per molecule = nkT . 2 11. (b) : Let the final temperature be T. Heat required by ice = mL + m × s × (T – 0) = 10 × 80 + 10 × 1 × T Heat lost by water = 55 × (40 – T) By using law of calorimetry, heat gained = heat lost; 800 + 10T = 55 × (40 – T) ⇒ T = 21.54ºC = 22ºC 12. (a) : We are given, h = 200 m, c = 127.7 J kg–1 K–1 Let m be the mass of the piece of lead. Potential energy of the lead piece = mgh Since 60% of the potential energy is converted into heat, heat produced = (60/100) × mgh = 0.6 mgh. If ΔT is the rise in temperature, then heat gained by the piece = mcΔT Assuming that there is no loss of heat, heat gained = heat produced or mcΔT = 0.6 mgh 0.6 gh 0.6 × 9.8 × 200 = = 9. 2 K or ΔT = c 127.7 13. (d) 14. (b) : Since, heat capacity of material increases with increase in temperature so, body at 100 °C has more heat capacity than body at 0 °C. Hence, final common temperature of the system will be closer to 100 °C. ∴ Tc > 50 °C 15. (b) : According to Newton’s law of cooling, dT = K (T − Ts) dt dT dT For two cases, 1 = K (T1 − Ts) and 2 = K (T2 − Ts) dt dt 3T + 2T Here, Ts = T, T1 = = 2. 5 T 2 dT1 3T − 2T T and = = dt 10 10 dT 2T + T ′ 2T − T ′ T2 = and 2 = dt 2 10 T So, ...(i) = K (2.5T − T ) 10 2T − T ′ ⎛ 2T + T ′ ⎞ ...(ii) =K⎜ −T ⎟ ⎝ 2 ⎠ 10 PHYSICS FOR YOU | JANUARY ‘17

21

Dividing eqn. (i) by eqn. (ii), we get T (2.5T − T ) = 2T − T ′ ⎛ 2T + T ′ ⎞ −T ⎟ ⎜⎝ ⎠ 2 3 T′ = 3(2T – T′) or, 4T′ = 6T ∴ T′ = T 2 16. (b) : Temperature inside refrigerator = t2 °C Room temperature = t1 °C For refrigerator, Heat given to higher temperature (Q1 ) T = 1 Heat taken from lower temperature (Q2 ) T2 Q1 t1 + 273 = Q2 t2 + 273 Q1 t + 273 t −t W = 1 or, = 1 2 ⇒ Q1 − W t 2 + 273 Q1 t1 + 273 The amount of heat delivered to the room for each joule of electrical energy (W = 1 J) t + 273 Q1 = 1 t1 − t2 17. (a) : Time period of the pendulum clock at temperature θ is given by 1

l (1 + αθ) l l Tθ = 2 π θ = 2 π 0 = 2 π 0 (1 + αθ) 2 g g g ⎛ 1 ⎞ Tθ ≈ T0 ⎜1 + αθ ⎟ ⎝ 2 ⎠

...(i)

Assume pendulum clock gives correct time at temperature θ0 ∴

⎛ 1 ⎞ Tθ = T0 ⎜1 + αθ0 ⎟ 0 ⎝ 2 ⎠

...(ii)

At θ = 40°C > θ0 as clock loses time. ⎛ 1 ⎞ T40 = T0 ⎜1 + α × 40 ⎟ ⎝ 2 ⎠ At θ = 20°C < θ0 as clock gains time. ⎛ 1 ⎞ T20 = T0 ⎜1 + α × 20 ⎟ ⎝ 2 ⎠

...(iii)

...(iv)

From equations (ii) and (iii), we get T40 − Tθ 1 0 = α(40 − θ0 ) T0 2 or 12 s = α(40 – θ0) (12 h) From equations (ii) and (iv), we get Tθ − T20 1 0 = α(θ0 − 20) T0 2 or, 4 s = α(θ0 – 20)(12 h) 22

PHYSICS FOR YOU | JANUARY ‘17

..(v)

...(vi)

From equations (v) and (vi), we get 3(θ0 – 20) = (40 –θ0) 3θ0 + θ0 = 40 + 60 100 θ0 = = 25°C 4 From equation (vi), 4 s = α(25 – 20)(12 × 3600 s) 4 α= = 1.85 × 10−5 °C −1 5 × 12 × 3600 18. (b) : Here, PVn = constant or, PnVn–1 dV + Vn dP = 0 or, nPdV = – V dP Also, from ideal gas equation PV = nRT PdV + VdP = nR dT or PdV – nPdV = nRdT nRdT or, PdV = (1 − n) Also, dQ = dU + dW ⇒ nC dT = nCVdT + PdV nRdT nCdT = nCV dT + (1 − n) R R or, C = CV + or, (1− n) = C − CV (1 − n) C − (CV + R) C − CP R or, n = 1 − = = C − CV C − CV C − CV 19. (a) : Equation of line AB is given by y −y y − y1 = 2 1 (x − x1 ) x2 − x1 2P − P P − P0 = 0 0 (V − 2V0 ) V0 − 2V0 P0 P or P = − V + 3P0 or, PV = − 0 V 2 + 3P0V V0 V0 P0 2 or nRT = − V + 3P0V V0 ⎞ 1 ⎛ P0 2 or T = − V + 3P0V ⎟ ...(i) ⎜ nR ⎝ V0 ⎠ For maximum value of T, dT = 0 dV P 3 or − 0 (2V ) + 3P0 = 0 ∴ V = V0 V0 2 So, from equation (i) ⎞ 9 P0V0 1 ⎛ P0 9 2 9 Tmax = − × V + PV = nR ⎜⎝ V0 4 0 2 0 0 ⎟⎠ 4 nR 20. (d) : For isochoric process, ΔU = Q = ms ΔT Here, m = 200 g = 0.2 kg, s = 4184 J kg–1 K–1 ΔT = 60°C – 40°C = 20 °C = 20 K ∴ ΔU = 0.2 × 4184 × 20 = 16736 J = 16.7 kJ ””

CLASS XI Series 7

 Oscillations and Waves

Time Allowed : 3 hours Maximum Marks : 70

GENERAL INSTRUCTIONS (i)

All questions are compulsory.

(ii)

Q. no. 1 to 5 are very short answer questions and carry 1 mark each.

(iii) Q. no. 6 to 10 are short answer questions and carry 2 marks each. (iv)

Q. no. 11 to 22 are also short answer questions and carry 3 marks each.

(v)

Q. no. 23 is a value based question and carries 4 marks.

(vi)

Q. no. 24 to 26 are long answer questions and carry 5 marks each.

(vii) Use log tables if necessary, use of calculator is not allowed.

SECTION-A

1. What change in mass is required to double the frequency of a harmonic oscillator? 2. Why is a loud sound heard at resonance?

vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

3. If an explosion takes place at the bottom of a lake, will the shock waves in water be longitudinal or transverse?

9. What are the differences between stationary waves and progressive waves?

4. How does the frequency of a tuning fork change, when the temperature is increased?

OR Why is a tuning fork used as a standard oscillator? On what factors does the pitch of a tuning fork depend?

5. The length of a string tied to two rigid supports is 40 cm. What is the maximum wavelength of the stationary wave produced in it? SECTION-B

6. Show that the motion of a particle represented by y = sin ωt – cos ωt is simple harmonic with a time period of 2π/ω. 7. The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon? 8. A wire stretched between two rigid supports

10. A body of mass m is situated in a potential field U(x) = U0 (1 – cos αx) where U0 and α are constants. Find the time period of small oscillations. SECTION-C

11. Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to mass m at either end. Each end of the spring in figure (b) is stretched by the same force F. PHYSICS FOR YOU | JANUARY ‘17

23

(i) is independent of pressure, (ii) increases with temperature, (iii) increases with humidity.



17. State the principle of superposition of waves. Distinguish between conditions for the production of stationary waves and beats.

  





 



(i) What is the maximum extension of the spring in the two cases? (ii) If the mass in figure (a) and the two masses in figure (b) are released, what is the period of oscillation in each case? 12. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 m s–1). 13. A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. The speed of sound in still air can be taken as 340 m s–1. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s–1? (b) recedes from the platform with a speed of 10 m s–1? (ii) What is the speed of sound in each case? 14. Two simple harmonic motions are represented by the equations: π y1 = 10 sin (12t + 1), y2 = 5(sin 3πt + 3 cos 3πt) 4 Here y1 and y2 are in cm and t is in second. Find the ratio of their amplitudes. What are time periods of the two motions? 15. A simple pendulum is hung in a stationary lift and its periodic time is T. What will be the effect on its periodic time T if (i) the lift goes up with uniform velocity v, (ii) the lift goes up with uniform acceleration a, and (iii) the lift comes down with uniform acceleration a? 16. Use the formula v = of sound in air 24

γP to explain why the speed ρ

PHYSICS FOR YOU | JANUARY ‘17

18. Show that for small oscillations the motion of a simple pendulum is simple harmonic. Derive an expression for its time period. Does it depend on the mass of the bob? 19. On a quiet day, two persons A and B, each sounding a note of frequency 580 Hz, are standing a few metres apart. Calculate the number of beats heard by each in one second when A moves towards B with a velocity of 4 m s–1. (Speed of sound in air = 330 m s–1.) 20. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body? 21. The patterns of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is 360 m s–1 and their frequencies are 256 Hz.    



 

  



 


 


 


 


(i) Calculate the time at which the second curve is plotted. (ii) Mark nodes and antinodes on the curve. (iii) Calculate the distance between A′ and C′. OR A horizontal spring block system of mass M executes simple harmonic motion. When the block is passing through its equilibrium position, an object of mass m is put on it and the two move together. Find the new amplitude and frequency of vibration. 22. Discuss first three modes of vibration of a closed organ pipe.

SECTION-D

23. Rohit was a good football player. But since last few days he was getting pain in his stomach. His parents took him to a doctor who examined him and asked him to get an ultrasound done to detect the exact cause. Rohit was afraid of ultrasound scanner and refused to get it done. His parents made him understand that the scanner uses ultrasonic rays which go inside and detect any problem inside the body. So he got it done and the scanner showed that he has small tumour in his stomach and that has to be operated as early as possible. Doctor operated him off the tumour and after a month he became fine again. Answer the following questions based on above information : (i) What are the values shown by Rohit's parents? (ii) On which principle does the ultrasonic scanner work? (iii) If the ultrasound uses the operating frequency of 4.2 MHz, the speed of sound in the tissue is 1.7 km s–1. What is the wavelength of the sound in tissue? SECTION-E

24. A cylindrical piece of cork of base area A and height h floats in a liquid of density ρ1. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T = 2π

hρ ρ1 g

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid). OR A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. 25. Find the total energy of the particle executing SHM and show graphically the variation of potential energy and kinetic energy with displacement in SHM. OR Explain the formation of beats analytically. Prove that the beat frequency is equal to the difference in frequencies of the two superposing waves.

26. Derive Newton's formula for the speed of sound in a gas. Why and what correction was applied by Laplace in this formula? OR Explain why (or how) : (i) in a sound wave, a displacement node is a pressure antinode and vice versa, (ii) bats can ascertain distances, directions, nature and sizes of the obstacles without any eyes, (iii) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (iv) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (v) the shape of a pulse gets distorted during propagation in a dispersive medium. SOLUTIONS

1. The time period (T) of a harmonic oscillator (a mass attached to a spring) is given by T = 2π m/k , where k is the force constant of the spring. If υ is the frequency of the harmonic oscillator, υ=

1 1 k = or υ ∝ 1/ m T 2π m

Thus, to double the frequency of the oscillator, the mass should be reduced to (1/4)th of its original value. 2. At resonance, a compression falls on a compression and a rarefaction falls on a rarefaction. On account of this, the amplitude of the vibrating particles increases. Since the intensity of sound is directly proportional to the square of the amplitude of the vibrating particles, hence maximum sound is heard at resonance position. 3. An explosion in a lake generates shock waves in water thereby resulting in a tremendous increase in pressure in the medium (water). A shock wave is thus a longitudinal wave travelling at a speed which is greater than that of a longitudinal wave of ordinary intensity. 4. As the temperature increases, the length of the prong of the tuning fork increases. This increases the wavelength of the stationary waves set up in the tuning fork. As frequency, υ ∝ 1/λ, so frequency of the tuning fork decreases. PHYSICS FOR YOU | JANUARY ‘17

25

5. When the string vibrates in one segment, L =

λ 2

∴ λ = 2 L = 2 × 40 cm = 80 cm 6. Given, y = sin ωt – cos ωt ⎛ 1 ⎞ 1 sin ωt − cos ωt ⎟ = 2⎜ ⎝ 2 ⎠ 2 π π ⎛ ⎞ = 2 ⎜ cos sin ωt − sin cos ωt ⎟ ⎝ ⎠ 4 4 π⎞ ⎛ ∴ y = 2 sin ⎜ ωt − ⎟ ⎝ 4⎠ Hence, (sin ωt – cos ωt) represents SHM. Again, π⎞ π ⎛ ⎛ ⎞ y = 2 sin ⎜ ωt − ⎟ = 2 sin ⎜ ωt − + 2π⎟ ⎝ ⎝ ⎠ 4⎠ 4 ⎡ 2π π ⎤ = 2 sin ⎢ω(t + ) − ⎥ 4⎦ ω ⎣

Hence, time period of SHM is 2π/ω. 7. For second’s pendulum, T = 2 s l l l 1 ⇒ 2 = 2π ⇒ = = constant g g g π2 l l ∴ moon = earth g moon g earth

∵ T = 2π

⎞ ⎛g 1 1 ⇒ lmoon = learth ⎜ moon ⎟ = 1 × = m 6 6 g ⎝ earth ⎠

8. Here, υ = 45 Hz, M = 3.5 × 10–2 kg ; mass/length = μ = 4.0 × 10–2 kg m–1 ∴

l=

M 3.5 × 10 −2 7 = m = μ 4.0 × 10 −2 8

As wire vibrates in its fundamental mode λ 7 =l= 2 8

∴ λ=

7 m = 1.75 m 4

The speed of the transverse wave, v = υλ = 45 × 1.75 = 78.75 m s–1 (b) As, v = T

μ

T = v2 × μ = (78.75)2 × 4.0 × 10–2 = 248.06 N 9. Difference between stationary and progressive waves (i)

26

Stationary waves The disturbance remains confined to a particular region, and there is no onward motion.

Progressive waves The disturbance travels forward, being handed over from one particle to the neighbouring particles.

PHYSICS FOR YOU | JANUARY ‘17

(ii)

There is no transfer of energy in the medium.

Energy is transferred in the medium along the waves. (iii) The amplitude of The amplitude of vibration of particles vibration of each varies from zero at particle is same. nodes to maximum at antinodes.

OR When a tuning fork is struck lightly against a rubber pad, it produces only fundamental tone. If it is struck forcefully, it produces overtones which soon dies out. So a tuning fork can be used as a source of standard frequency. Factors on which the pitch of a tuning fork depends : (i) It is inversely proportional to the square of the length of its prongs. (ii) It is directly proportional to the thickness of the fork. (iii) It is directly proportional to the square root of the Young's modulus of elasticity of its material. (iv) It is inversely proportional to the square root of the density of its material. Hence low frequency tuning forks are long and thin while high frequency tuning forks are short and thick. 10. Given, U(x) = U0(1 – cos αx) Differentiating both sides with respect to x dU (x ) = U 0 (0 + α sin αx ) = U 0α sin αx dx dU (x ) ∴ F=− = −U 0α sin αx dx For small oscillations, sinθ ≈ θ ⇒ sin αx≈ αx ∴ F = –U0 α(αx) = –U0α2x Also, F = –kx From equations (i) and (ii) k = U0α2 m m = 2π T = 2π Thus, k U 0α2

...(i) ...(ii)

11. (i) Maximum extension of the spring : (a) Suppose the maximum extension produced in the spring is y. Then, F=ky (in magnitude) or y = F/k (b) In this case, force F on each mass acts as the force of reaction developed due to force F on

the other mass. Therefore, in this case also, maximum extension is given by y = F/k (ii) Period of oscillation: If T1 is the time period in case (a), then m T1 = 2π k In case (b), the time period of oscillation of a two body oscillator (two bodies of mass m1 and m2 connected at the ends of a spring of spring constant k) is given by μ T2 = 2π k where μ is called the reduced mass of the system defined as m1m2 μ= m1 + m2 In the present case, m1 = m2 = m. So, m×m m μ= = m+m 2 m/2 m = 2π k 2k 12. Here, L = 20 cm = 0.2 m, υn = 430 Hz, v = 340 m s–1 The frequency of nth normal mode of vibration of closed pipe is v 340 υn = (2n − 1) ∴ 430 = (2n − 1) 4L 4 × 0.2 430 × 4 × 0.2 or 2n − 1 = = 1.01 340 2n = 2.01 or n ≈ 1 Hence, it will excite the lst normal mode of vibration. In a pipe, open at both ends, we have v n × 340 430 × 2 × 0.2 = 430 ∴ n = υn = n × ⇒ = 0.5 340 2L 2 × 0.2 As n has to be an integer, therefore, open organ pipe cannot be in resonance with the source.

Thus, T2= 2π

13. Here, frequency of source of sound, υ = 400 Hz; speed of sound, v = 340 m s–1 speed of source, vs = 10 m s–1 (i) (a) When the train approaches the platform, the apparent frequency as heard by the observer on the platform will be 340 v υ′ = υ= × 400 340 − 10 v − vs 340 = × 400 = 412.1 Hz 330 (b) When the train recedes from the platform, the apparent frequency as heard by the observer will be :

340 v 340 υ= × 400 = × 400 340 + 10 v + vs 350 = 388.6 Hz 389 Hz (ii) The speed of sound in each case remains same i.e., 340 m s–1. υ′ =

π π 14. y1 = 10 sin (12t + 1) = 10 sin ⎛⎜ 3πt + ⎞⎟ ...(i) ⎝ 4 4⎠ y2 = 5(sin 3πt + 3 cos 3πt) ⎛ ⎞ = 10 ⎜ sin 3πt × 1 + cos 3πt × 3 ⎟ ⎝ 2 ⎠ 2 π π⎞ ⎛ = 10 ⎜ sin 3πt cos + cos 3πt sin ⎟ ⎝ 3 3⎠ π ⎛ ⎞ or y2 = 10 sin ⎜ 3πt + ⎟ ...(ii) ⎝ 3⎠ The general equation for SHM is y = A sin (ωt + φ0 ) = A sin ⎛⎜ 2π t + φ0 ⎞⎟ ...(iii) ⎝T ⎠ Comparing equations (i) and (ii) with (iii), we get 2π 2π A1 = 10 cm, A2 = 10 cm, = = 3π T1 T2 2 A1 ∴ = 1; T1 = T2 = s 3 A2

15. (i) When the lift goes up a a figure (a) with uniform a velocity v, tension in  
   the string, T ′ = mg. The value of g remains    unaffected.    The period T remains same as that in stationary lift, l i.e., T = 2π g (ii) When the lift goes up with acceleration a as shown in figure (b), the net upward force on the bob is T′ – mg = ma ∴ T′ = m(g + a) The effective value of g is (g + a) and the time period is T1 = 2π

l g +a

Clearly, T1 < T, i.e., time period decreases. (iii) When lift comes down with acceleration a figure (c), the net downward force on the bob is mg – T′ = ma ∴ T′ = m(g – a) The effective value of g becomes (g – a) and the time period is l T2 = 2π g −a Clearly, T2 > T, i.e., time period increases. PHYSICS FOR YOU | JANUARY ‘17

27

16. (i) Effect of pressure: The speed of sound in a gas is given by, v =

γP ρ

At constant temperature, PV = constant; Pm = constant ρ P Since m is constant, so = constant ρ i.e., when pressure changes, density also changes P in the same ratio so that the factor remains ρ unchanged. Hence the pressure has no effect on the speed of sound in a gas for a given temperature. (ii) Effect of temperature : We know that nRT PV = nRT or P = V γP γnRT γRT Also v = = = M ρ ρV where M = molecular weight of the gas As γ, R and M are constants, so v ∝ T , i.e., velocity of sound in a gas is directly proportional to the square root of its temperature, hence we conclude that the velocity of sound in air increases with increase in temperature. (iii) Effect of humidity: As v = γP , i.e., v ∝ 1 ρ ρ The density of water vapours is less than that of dry air. Since the speed of sound is inversely proportional to the square root of density, so sound travels faster in moist air than in dry air. 17. Principle of superposition of waves states that when a number of waves travel through a medium simultaneously, the resultant displacement of any particle of the medium at any given time is equal to the algebraic sum of the displacements due to the individual waves. Mathematically, y = y1 + y2 + y3 + ..... + yn (i) When two waves of same frequency moving with the same speed in the opposite directions in a medium superpose on each other, they produce stationary waves. (ii) When two waves of slightly different frequencies moving with the same speed in the same direction in a medium superpose on each other they produce beats. 18. Suppose at any instant during oscillation, the bob of a simple pendulum lies at position A when its displacement is OA = x and the thread makes angle θ 28

PHYSICS FOR YOU | JANUARY ‘17



   with the vertical. The forces acting on the bob are Q (i) Weight mg of the  bob acting vertically  downwards.    Q (ii) Tension T along the  Q   Q   Q string. Thus, the restoring force is  F = –mg sin θ 3 5 ⎛ ⎞ = −mg ⎜ θ − θ + θ − .....⎟ 3! 5! ⎝ ⎠ However, if θ is so small that its higher powers can be neglected, then, F = –mg θ If l is the length of the simple pendulum, then x arc θ(rad) = = radius l x ∴ F = −mg l mg g x or a = − x = − ω2 x or ma= − l l Hence for small oscillations, the motion of the bob is simple harmonic. Its time period is 2π 2π l or T = 2π T= = ω g g /l Obviously, the time period of a simple pendulum depends on its length l and acceleration due to gravity g but is independent of the mass m of the bob.

19. In one case, A can be regarded as a source of sound moving towards observer B. vs = + 4 m s–1, vo = 0 v − vo 330 − 0 ∴ υ′ = × 580 ×υ= 330 − 4 v − vs 330 = × 580 = 587 Hz 326 





 

 


In another case, A can be regarded as observer moving towards stationary source B.  



 

 


vo = –4 m s–1, vs = 0 v − vo 330 + 4 ∴ υ″ = × 580 ×υ= 330 − 0 v − vs 334 = × 580 = 587 Hz 330 Number of beats heard per second by A = υ″ – υ = 587 – 580 = 7 Number of beats heard per second by B = υ′ – υ = 587 – 580 = 7

v = Velocity of mass M when passing through mean position. Maximum kinetic energy = Total energy 1 1 or Mv 2 = kA2 2 2 k A M When mass m is put on the system,

∴

v=

total mass = (M + m). If v′ is the velocity of the combination in equilibrium position, then by the conservation of linear momentum, Mv Mv = (M + m)v′ or v′ = M+m If A′ is the new amplitude, then 1 1 ( M + m)v ′2 = kA′2 2 2

M +m × v′ = k

A′ =

M +m Mv × k M +m

M +m M k × × A= k M+m M

=

New frequency, υ′ =

M A M +m

1 k 2π M + m



22. (i) First mode of vibration : In this mode of vibration, there is only one node at the closed end and one antinode at the open end. If L is the length of the organ pipe, then 

  



 


L= Frequency, υ1 =



λ1 4

or λ1 = 4L

v v = υ (say) = λ1 4 L

This frequency is called first harmonic or fundamental frequency. (ii) Second mode of vibration : In this mode of vibration, there is one node and one antinode between a node at the closed end and an antinode at the open end. 

21. (i) Here, v = 360 m s–1, υ = 256 Hz v 360 λ= = = 1.406 m υ 256 λ 1.406 AA′ = = = 0.3516 m 4 4 Time (t) at which the second curve is plotted AA ′ 0.3516 = = 9.8 × 10 −4 s v 360 (ii) Nodes : A, B, C, D, E Antinodes : A′, C ′ (iii) Distance between A′ and C ′ = λ = 1.406 m = 1.41 m OR Original frequency, 1 k υ= 2π M Let A =Initial amplitude of oscillation

or





 

 

 






L= Frequency,

3λ 2 4

or λ2 =

4L 3

v 3v =3υ = λ2 4 L This frequency is called first overtone or third harmonic. (iii) Third mode of vibration : In this mode of vibration, there are two nodes and two antinodes between a node at the closed end and an antinode at the open end.

υ2 =



20. Here, Maximum mass, m = 50 kg, Maximum extension, y = 20 – 0 = 20 cm = 0.2 m Maximum force, F = mg = 50 × 9.8 = 490 N ∴ Spring constant, F 490 = 2450 N m −1 k= = y 0. 2 When a body of mass M is suspended from the spring balance, it oscillates with a period of 0.6 s. M M ∵ Time period, T = 2π or T 2 = 4 π2 k k T 2 k (0.6)2 × 2450 ∴ M= 2 = 4π 4 × (3.14)2 M = 22.36 kg ∴ Weight of the body, W = Mg = 22.36 × 9.8 = 219.1 N

  


 





 

 

  



∴

5λ L= 3 4

or λ3 =

4L 5

PHYSICS FOR YOU | JANUARY ‘17

29

v 5v =5υ = λ3 4 L Hence different frequencies produced in a closed organ pipe are in the ratio 1 : 3 : 5 : 7 : ...... i.e., only odd harmonics are present in a closed organ pipe.

Frequency, υ3 =

23. (i) Sense of responsibility, concern for their son and awareness about technology. (ii) An ultrasonic scanner works on the principle of reflection of ultrasonic waves from a region where there is a change in tissue density. −1 1.7 × 103 m s −1 v 1.7 km s (iii) Wavelength, λ = = = υ 4.2 MHz 4.2 × 106 s −1 = 404.76 μm 24. Let X be the equilibrium position of a cylinder floating in a given liquid. Assume V = volume of cork = Ah m = mass of cork = Ahρ.   

 





   

l = length of the cylindrical piece of cork dipped in the liquid upto point P in position X. W = weight of the cylindrical cork. W1 = weight of the liquid displaced by the cork. W= mg = (Ahρ)g ∴ W1 = Area of cross-section of cork × length of cylinder dipped in liquid × density of liquid × g = Alρ1g ∴ According to the law of flotation, W = W1 or Ahρg = Alρ1g lρ or h= 1 ρ Let the cylinder be pushed into the liquid through a small distance y from equilibrium. Since W = W1 then the restoring force acting on the cylinder is given by F = – weight of the liquid displaced by the length y of the cylindrical cork = −( Ay )ρ1 × g = −( Aρ1 g ) y = –ky where k = Aρ1g is the force constant. If a be the acceleration produced in the cylindrical piece of the cork, then F F − Aρ1 gy = = a= mass of cork m Ahρ ρ1 g y a=− or hρ 30

PHYSICS FOR YOU | JANUARY ‘17

Now as the acceleration of the cylindrical cork is directly proportional to its displacement from equilibrium position and acts towards the equilibrium position, so the motion of the bob is simple harmonic having time period (T) given by y displacement = 2π acceleration ρ1 g y hρ hρ T = 2π ρ1 g T = 2π

or

OR Let the displacement of the particle at any time t be represented by x = A cos(ωt + φ0) ...(i) where A = amplitude, φ0 = initial phase If v be the velocity of the particle at time t, Then dx d = v= ⎡ Acos(ωt + φ0 )⎤⎦ dt dt ⎣ = −Aωsin(ωt + φ0 ) ...(ii) At t = 0, x = x0 and v = v0 By putting t = 0 in equations (i) and (ii), we get x0 = Acosφ0, v0 = –Aωsinφ0 or or

v0 = −ω ( A sin φ0 )2 = −ω A2 (1 − cos2 φ0 ) = −ω A2 − x02

...(iii)

Equation (iii) shows that initial velocity is negative. Squaring on both sides of equation (iii), we get v2 v02 = ω2 ( A2 − x02 ); A2 − x02 = 02 ω 2⎞ 2 ⎛ v v A2 = x02 + 02 ; A = ⎜ x02 + 02 ⎟ ⎝ ω ω ⎠ 25. The energy of a harmonic oscillator is partly kinetic and partly potential. When a body is displaced from its equilibrium position by doing work upon it, it acquires potential energy. When the body is released, it begins to move back to equilibrium position, thus acquires kinetic energy. At any instant, the displacement of a particle executing SHM is given by x = A cos(ωt + φ0) dx ∴ Velocity, v = = –ω A sin(ωt + φ0) dt Hence, kinetic energy of the particle at any time t is given by 1 1 K = mv2 = mω2 A2 sin2(ωt + φ0) 2 2

But A2 sin2(ωt + φ0) = A2[1 – cos2(ωt + φ0)] = A2 – A2 cos2 (ωt + φ0) = A2 – x2 1 1 or K = mω2 (A2 – x2) = k(A2 – x2) 2 2 When the displacement of a particle from its equilibrium position is x, the restoring force acting on it is F = –kx If we displace the particle further through a small distance dx, then work done against the restoring force is given by dW = –Fdx = + kxdx The total work done in moving the particle from mean position (x = 0) to displacement x is given by x

x

⎡ 2⎤ 1 W = ∫ dW = ∫ kxdx = k ⎢ x ⎥ = kx 2 ⎣ 2 ⎦0 2 0 The work done against the restoring force is stored as the potential energy of the particle. Hence potential energy of a particle at displacement x is given by 1 1 1 U = kx2 = mω2x2 = mω2 A2 cos2(ωt + φ0) 2 2 2 At any displacement x, the total energy of a harmonic oscillator is given by 1 1 E = K + U = k(A2 – x2) + kx2 2 2 1 2 1 or E = kA = mω2A2 = 2π2mυ2A2 2 2 (∵ ω = 2πυ)

Thus the total mechanical energy of a harmonic oscillator is independent of time or displacement. At the mean position, x = 0 1 1 Kinetic energy, K = k(A2 – 02) = kA2 2 2 1 Potential energy, U = k(02) = 0 2 Hence at the mean  position, particle has 



 only kinetic energy.  At the extreme positions, x=±A  Kinetic energy, 1  = –   = + K = k(A2 – A2) = 0 2 1 Potential energy, U = kA2 2 Hence at the two extreme positions particle has only potential energy.

OR Consider two harmonic waves of frequencies υ1 and υ2 (υ1 being slightly greater than υ2) and each of amplitude A travelling in a medium in the same direction. The displacements due to the two waves at a given observation point may be represented by y1 = A sinω1 t = A sin2π υ1 t y2 = A sinω2 t = A sin2π υ2 t By the principle of superposition, the resultant displacement at the given point will be y = y1 + y2 = A sin2π υ1 t + A sin2π υ2 t ⎛ υ −υ ⎞ ⎛υ +υ ⎞ = 2 A cos 2π ⎜ 1 2 ⎟ t ⋅ sin 2π ⎜ 1 2 ⎟ t ⎝ 2 ⎠ ⎝ 2 ⎠ If we write υ −υ υ +υ υmod = 1 2 and υav = 1 2 2 2 then y = 2A cos (2π υmod t) sin (2π υav t)

or

y = R sin(2π υav t)

where R = 2A cos(2π υmod t) is the amplitude of the resultant wave. The amplitude R of the resultant wave will be maximum, when cos 2 π υmod t = ± 1 or 2 π υmod t = nπ where n = 0, 1, 2, ... π(υ1 – υ2)t = nπ 1 2 n = 0, , , ..... or t = υ1 − υ2 υ1 − υ2 υ1 − υ2 ∴ Time interval between two successive maxima 1 = υ1 − υ2 Similarly, the amplitude R will be minimum, when cos 2π υmod t = 0 or 2π υmod t = (2n + 1)π/2 where n = 0, 1, 2, ...

or

π(υ1 – υ2)t = (2n + 1)π/2 (2n + 1) 1 3 5 or t = = , , , ... 2(υ1 − υ2 ) υ1 − υ2 2(υ1 − υ2 ) (2 υ1 − υ2 ) ∴ The time interval between successive minima 1 = υ1 − υ2 Clearly, both maxima and minima of intensity occur alternately. Hence the time interval between two successive beats 1 tbeat = υ1 − υ2 The number of beats produced per second is called beat frequency. or

PHYSICS FOR YOU | JANUARY ‘17

31

υbeat =

1 t beat

or υbeat = υ1 – υ2

26. Newton assumed that sound waves travel through a gas under isothermal conditions. Thus the temperature of gas remains constant. If Biso is the bulk modulus of the gas at constant temperature, then the speed of sound in the gas will be v=

Biso ρ

For an isothermal change, PV = constant (Boyle's law) Differentiating both sides, we get PdV + VdP = 0 VdP dP or P = − =− dV dV /V Change in pressure (dP ) = = Biso Change in volume (dV )/Original volume (V ) Hence the Newton's formula for the speed of sound in a gas is P v= ρ Speed of sound in air at STP, 1.013 × 105

280 m s–1 1.293 This value is about 16% less than the experimental value (331 m s–1) of the speed of sound in air at STP. Hence Newton's formula is not acceptable. The French scientist Laplace pointed out that sound travels through a gas under adiabatic conditions not under isothermal conditions. So, when sound travels through a gas, the temperature does not remain constant. The pressure and volume variations are adiabatic. If Badia is the adiabatic bulk modulus of the gas, then the formula for the speed of sound in the gas would be v=

v=

Badia ρ

For an adiabatic change, PVγ = constant Differentiating both sides, we get PγVγ – 1 dV + Vγ dP = 0 or PγVγ – 1 dV = –Vγ dP −dP γP= = Badia dV /V where γ = CP/CV, is the ratio of two specific heats. Hence the Laplace formula for the speed of sound in a gas is 32

PHYSICS FOR YOU | JANUARY ‘17

v=

γP ρ

This modification of Newton's formula is known as Laplace correction. For air γ = 7/5, so speed of sound in air at STP will be P 7 v= γ = × 280 m s–1 = 331.3 m s–1 ρ 5 OR (i) In a sound wave, a node is a point where the displacement is zero as here a compression and a rarefaction meet and the pressure is maximum, so it is also called pressure antinode. While an antinode is a point where the amplitude displacement is maximum but pressure is minimum. So this point is also called pressure node. Hence displacement node is a pressure antinode and displacement antinode is pressure node. (ii) Bats emit ultrasonic waves of large frequencies (small wavelength) when they fly. These ultrasonic waves are received by them after reflection from the obstacle. Their ears are so sensitive and trained that they can not only get the information of the distance of the obstacle but also the nature of the reflecting surface. (iii) The quality of the sound produced by an instrument depends upon the number of overtones. Since the number of overtones is different in the cases of sounds produced by violin and sitar therefore we can distinguish through them. (iv) Solids possess both the volume elasticity and the shear elasticity. Therefore they can support both longitudinal and transverse waves. On the other hand, gases have only the volume elasticity and no shear elasticity, so only longitudinal waves can propagate in gases. (v) A sound pulse is a combination of waves of different wavelengths. In a dispersive medium, the waves of different wavelengths travel with different speeds in different directions, i.e., with different velocities. So the shape of the pulse gets distorted, i.e., a plane wavefront in a non-dispersive medium does not remain a plane wavefront in a dispersive medium. ””



Class XI

T

his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.

Thermodynamics Kinetic Theory of Gases Total Marks : 120

Time Taken : 60 min

NEET / AIIMS / PMTs Only One Option Correct Type

(c)

1. Let v , vrms and vp respectively denote the mean speed, the root mean square speed and the most probable speed of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m. Then, (a) No molecules can have speed greater than vrms. v (b) No molecule can have speed less than p . 2 (c) v p < v > vrms . (d) The average kinetic energy of a molecule is 3 2 mv . 4 p 2. Figure shows graph of   pressure versus density for an  ideal gas at two temperatures T1 and T2. R

(a) T1 > T2 (c) T1 < T2

(b) T1 = T2 (d) None of these

3. A cyclic process is shown on the P–T diagram. Which of the curves show the same process on a V–T diagram?













(a)





34









(b)







PHYSICS FOR YOU | JANUARY ‘17















 



(d)



  

 

 





4. A Carnot engine is designed to operate between 480 K and 300 K. If the engine actually produces 1.2 J of mechanical energy per calorie of heat absorbed, then the ratio of actual efficiency to theoretical efficiency is (a) 16/21 (b) 21/16 (c) 5/16 (d) 16/5 5. A gas at pressure P is adiabatically compressed so that its density becomes twice that of initial value. Given that γ = CP /CV = 7/5, what will be the final pressure of the gas? (a) 2 P

(b) 1.4 P

(c) 2.64 P (d) P

6. The root mean square velocity of the molecules in a sample of helium is (5/7)th that of the molecules in a sample of hydrogen. If the temperature of the hydrogen gas is 0 °C, that of helium sample is about (a) 0 °C (b) 4 °C (c) 273 °C (d) 100 °C 7. One mole of an ideal gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is (Given the gas constant R = 8.3 J mol–1 K–1) (a) 198.7 J (b) 290 J (c) 215.3 J (d) 124 J

8. At what frequency would the wavelength of sound be of the order of the mean free path in nitrogen at 1.02 atm pressure and 18.0 °C? Take the diameter of the nitrogen molecule to be 315 pm and speed of sound to be 343 m s–1. (a) 5.36 × 109 Hz (b) 7.38 × 108 Hz 8 (c) 2.88 × 10 Hz (d) 3.88 × 109 Hz 9. A thermodynamic process is    shown in the figure. The pressure and volumes corresponding to some points in the figure are    PA = 3 × 104 Pa; VA = 2 × 10–3 m3;  PB = 8 × 104 Pa; VD = 5 × 10–3 m3. In the process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be (a) 560 J (b) 800 J (c) 600 J (d) 640 J 10. The internal energy of a monatomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross section 8.5 cm2. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas. If the temperature rises through 2 °C, find the distance moved by the piston. Atmospheric pressure = 100 kPa. (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm 11. A gas is expanded from volume V0 to 2V0 under three different processes is shown in figure. Process 1 is isobaric, process 2 is isothermal and process 3 is adiabatic. Let ΔU1, ΔU2 and ΔU3 be the change in internal energy of the gas in these three processes. Then 





  

(a) ΔU1 > ΔU2 > ΔU3 (c) ΔU2 < ΔU1 < ΔU3





(b) ΔU1 < ΔU2 < ΔU3 (d) ΔU2 < ΔU3 < ΔU1

12. If the pressure of n mole of an ideal gas varies according to the law, P = P0 – aV2, where P0 and a are constant, the highest temperature of the gas attained is

(a)

P0 ⎛ P0 ⎞1/2 ⎜ ⎟ nR ⎝ 3a ⎠

(b)

2P0 ⎛ P0 ⎞1/2 ⎜ ⎟ nR ⎝ 3a ⎠

(c)

2 P0 ⎛ P0 ⎞1/2 ⎜ ⎟ 3 nR ⎝ 3a ⎠

(d)

P0 ⎛ P0 ⎞1/2 ⎜ ⎟ 3 nR ⎝ 3a ⎠

Assertion & Reason Type

Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 13. Assertion : All molecular motion ceases at –273°C. Reason : Temperature below –273°C cannot be attained. 14. Assertion : A refrigerator transfers heat from lower temperature to higher temperature. Reason : Heat cannot be transferred from lower temperature to higher temperature normally. 15. Assertion : The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume. Reason : The molecules of a gas collide with each other and the velocities of the molecules change due to collision. JEE MAIN / JEE ADVANCED / PETs Only One Option Correct Type

16. Two identical containers A and B have frictionless pistons. They contain the same volume of an ideal gas at the same temperature. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. If the changes in the pressure in A and B are found to be ΔP and 1.5 ΔP respectively then (a) 4mA = 9mB (b) 2mA = 3mB (c) 3mA = 2mB (d) 9mA = 4mB 17. A smooth vertical tube having two different cross sections is open from both the ends but closed by two sliding pistons as shown in figure and tied with an inextensible string. One mole of an ideal gas is enclosed between the piston. PHYSICS FOR YOU | JANUARY ‘17



35

The difference in cross-sectional areas of the two pistons is given ΔS. The masses of pistons are m1 and m2 for larger and smaller one, respectively. Find the temperature by which the gas is raised so that the pistons will be displaced by a distance l. Take atmospheric pressure equal to P0. l (a) [P0ΔS + (m1 + m2)g] R l (b) [P0ΔS + (m1 – m2)g] R P0 Δ S l (m1 − m2 ) gl (c) (d) R R 18. Figure shows a process  ABCA performed on an  ideal gas. The net heat   given to the system during   the process will be    V2 ⎞ ⎛ (a) nR ⎜T2 ln − (T2 − T1 ) ⎟ V1 ⎠ ⎝ V ⎞ ⎛ (b) nR ⎜T2 ln 2 − (T1 + T2 ) ⎟ V1 ⎠ ⎝ V ⎞ ⎛ (c) nR ⎜T1 ln 1 + (T1 − T2 ) ⎟ V2 ⎠ ⎝ V ⎞ ⎛ (d) nR ⎜T1 ln 1 − (T2 − T1 ) ⎟ V2 ⎠ ⎝ 19. Assume that the temperature remains essentially constant in the upper part of the atmosphere. The mean molecular weight of air is M. An expression for the variation in pressure in the upper atmosphere with height h is (P0 is the pressure at h = 0) (a) P = 2P0 e–Mgh/RT

(b) P = P0 e–Mgh/RT

(c) P = P0 e–Mgh/2RT

(d) P = P0 e–2Mgh/RT

More than One Options Correct Type

20. A gas may expand either adiabatically or isothermally. A number of P–V curves are drawn for the two processes over different ranges of pressure and volume at different temperatures. It will be found that (a) two adiabatic curves do not intersect (b) two isothermal curves do not intersect (c) an adiabatic curve and an isothermal curve may intersect (d) the magnitude of the slope of an adiabatic curve is greater than the magnitude of the slope of an isothermal curve for the same value of pressure and volume. 36

PHYSICS FOR YOU | JANUARY ‘17

21. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass m. The piston and the cylinder have equal cross-sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from the equilibrium position and released. Then (Assuming that the system is completely isolated from its surrounding.) (a) Piston will execute SHM. (b) Motion of the piston will be periodic only. 2 Aγ P0 . (c) Frequency of motion is π mV0 (d) Frequency of motion is

A γ P0 . 2 π mV0

22. An ideal gas whose adiabatic exponent equals γ is expanded according to the law P = αV, where α is a constant. The initial volume of the gas is equal to V0. As a result of expansion the volume of the gas increases η times. Then (a) the increment of the internal energy of the gas (η2 − 1) is α V02 . ( γ − 1) (b) the work performed by the gas is 0.5 α V20 (η2 – 1). (c) the molar heat capacity of the gas in the process ( γ + 1) R . ( γ − 1) (d) Both (b) and (c). 23. The speeds of ten particles in m s–1 are 0, 1.0, 2.0, 3.0, 3.0, 3.0, 4.0, 4.0, 5.0 and 6.0. (a) The average speed is 3.1 m s–1. (b) The root mean square speed is 3.75 m s–1. (c) The most probable speed of these particle is 3 m s–1. (d) Mean square speed is 17 m2 s–2. Integer Answer Type

24. A vessel has 6 g of hydrogen at pressure P and temperature 500 K. A small hole is made in it so that hydrogen leaks out. If the final pressure is P/2 and the temperature falls to 300 K, the mass of hydrogen (in g) that leaks out is 25. A vessel contains a mixture of 1 mole of oxygen and 2 moles of nitrogen at 300 K. Find the ratio of average rotational kinetic energies per O2 molecule to per N2 molecule.

26. n moles of a gas in a cylinder  under a piston are transferred  infinitely slowly from a state with a volume of V0 and a  pressure 3P0 to another state with 3V0 and a pressure P0    as shown in figure. If the maximum temperature that the gas will reach x P0V0 , what is the value of x? in this process is nR Comprehension Type



A container of volume 4V0 made of a perfectly non-conducting material is divided into two equal parts by a fixed rigid wall whose lower half is non-conducting and upper half is purely conducting. The right side of the wall is divided into equal parts (initially) by means of a massless non-conducting piston free to move as shown. Section A contains 2 mol of a gas while the section B and C contain 1 mol each of the same gas (γ = 1.5) at pressure P0. The heater in left part is switched on till the final pressure in section C becomes 125/27 P0. 









 



Matrix Match Type  29. One mole of a monatomic ideal   gas is taken along two cyclic processes E → F → G → E and     E → F → H → E   as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. Match the paths in column I with the magnitudes of the work done in column II. Column I Column II (A) G → E (P) 160 P0V0 ln2 (B) G → H (Q) 36 P0V0 (C) F → H (R) 24 P0V0 (D) F → G (S) 31 P0V0 A B C D (a) Q R S P (b) P Q S R (c) S Q P R (d) S R Q P

30. Heat given to process is positive, match the following option of column I with the corresponding option of column II.



27. Final temperature in part C is PV 5P0V0 (b) (a) 0 0 R 3R P0V0 5P0V0 (c) (d) R 3R 28. The heat supplied by the heater is 368 113 P0V0 (b) PV (a) 9 5 0 0 316 405 (c) P0V0 (d) PV 9 8 0 0

Column I (A) JK (B) KL (C) LM (D) MJ A B (a) Q, S S (b) P Q, R (c) Q P, S (d) P Q

 



















Column II (P) ΔW > 0 (Q) ΔQ < 0 (R) ΔW < 0 (S) ΔQ > 0 C D P Q Q P, R S Q, R P, S R ””

Keys are published in this issue. Search now! 

Check your score! If your score is > 90%

EXCELLENT WORK !

You are well prepared to take the challenge of final exam.

No. of questions attempted

……

90-75%

GOOD WORK !

You can score good in the final exam.

No. of questions correct

……

74-60%

SATISFACTORY !

You need to score more next time.

Marks scored in percentage

……

< 60%

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

PHYSICS FOR YOU | JANUARY ‘17

37

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 Energy Bands in Solids • •





•

•

Range of energy possessed by electron in a solid is known as energy bands. There are two types of energy band Valence band : Range of energy possessed by valence electron is known as valence band. These electrons are bounded and not responsible for flow of current. Conduction band : Range of energy possessed by free electron is known as conduction band. These electrons are responsible for flow of current.



 

Width of forbidden energy gap depends upon the nature of substance.

 
   


If width is more then valence electrons are strongly attached with nucleus. As temperature increases, forbidden energy gap decreases (very slightly).

According to energy band theory there are three types of solids. Property

40

Forbidden energy gap (Eg) : It is the energy gap between the bottom of the conduction band and top of the valence band. No electron exist in this gap. 

 



Conductors

Semiconductors

Insulators

1.

Electrical conductivity Very high, and its value 102 – 108 S m–1

Between those of conductors Negligible, and insulators 105 – 10–6 S m–1 10–11 – 10–19 S m–1

2.

Resistivity value

Between those of conductors Very high, and insulators 10–5 – 106 Ω m 1011 – 1019 Ω m

3.

Energy gap and its Zero or very small value

More than that in conductors Very large but less than that in insulators For diamond, For Ge, Eg = 0.72 eV; for Si, Eg Eg = 7 eV = 1.1 eV; for GaAs, Eg = 1.3 eV

4.

Current carriers

Free electrons and holes

and

its Negligible 10–2 – 10–8 Ω m

Free electrons

PHYSICS FOR YOU | JANUARY ‘17

Free electrons

5.

Number of current carriers (electrons or holes) at ordinary temperature 6. Temperature coefficient of resistivity (α) 7. Effect of temperature on conductivity 8. On increasing temperature the number of current carriers 9. On mixing impurities their resistance 10. Current flow in these takes place 11. Examples

Very high

Very low

Negligible

Positive

Negative

Negative

Conductivity decreases

Conductivity increases

Unaffected

Increases

Conductivity increases Increases

Increases

Decreases

Remains unchanged

Easily

Very slow

Does not take place

Cu, Ag, Au, Na, Pt, Hg etc.

Ge, Si, Ga, As etc.

Wood, plastic, mica, diamond, glass etc.

 Classification of Semiconductor • •

•

A semiconductor in its pure state is called intrinsic semiconductor. A semiconductor doped with suitable impurity to increase its conductivity, is called extrinsic semiconductor. On the basis of doped impurity extrinsic semiconductors are of two types  n-type semiconductor : Extrinsic semiconductor doped with pentavalent impurity like As, Sb, Bi, etc. in which negatively

1.



charged electrons works as charge carriers, is called n-type semiconductor. Every pentavalent impurity atom donate one electron in the crystal, therefore it is called a donor atom. p-type semiconductor : Extrinsic semiconductor doped with trivalent impurity like Al, B, etc. in which positively charged holes work as charge carriers, is called p-type semiconductor. Every trivalent impurity atom have a tendency to accept one electron, therefore it is called an acceptor atom.

Intrinsic Semiconductors

n-type Semiconductor

p-type Semiconductor





  




 
 

  



 








 
  
  




 


  
 

3.

Current due to electrons and holes

Mainly due to electrons

Mainly due to holes

4.

ne = nh

nh > ne (nA nh)

5.

I = Ie + Ih

I

I

6.

Entirely neutral

Entirely neutral

Ie

Ih

Entirely neutral PHYSICS FOR YOU | JANUARY ‘17

41

•

•

•

•

•

In a doped semiconductor ne nh = n2i where ne and nh are the number density of electrons and holes and ni is number density of intrinsic carriers, i.e., electrons or holes. Electrical conductivity of extrinsic semiconductor is given by σ = 1/ρ = e(neμe + nh μh) Where ρ is resistivity, μe and μh are mobility of electrons and holes respectively.

Diffusion current : Because of high carrier concentration difference, holes from p-side try to diffuse n-side, produces diffusion current. Drift current : Because of barrier potential, minority charge carriers force to move and such movement produces drift current.

• • •

•

 



Forward-biased 

– – – – 

 



Effective barrier potential decreases. Depletion width decreases. Low resistance offered at junction. High current flows through the circuit. Reverse-biased  

– – – –



Effective barrier potential increases. Depletion width increases. High resistance offered at junction. Low current flows through the circuit. PHYSICS FOR YOU | JANUARY ‘17

$ '  "

 

 (



  

  

 

 
   &
             (      %  
     !    "#

     $      "

 

   




I-V characteristics



Unbiased 

It is a basic unit of all semiconductor devices. p(n) region has majority carrier holes (electrons) and minority carrier electrons (holes). The layer of immobile positive and negative ions, which have no free electrons and holes is called depletion layer.



Biasing of p-n junction 

42

 p -n Junction



Current flowing through the diode,

I = I0 [e(eV /nkT ) − 1]

•

where I0 is reverse saturation current and V is potential difference across the diode, ⎧1 for Ge n = constant = ⎨ ⎩2 for Si ΔV  Dynamic resistance of p-n junction, rd = ΔI V  Static or dc resistance, r = z dc 1 For an ideal diode,  In reverse bias  In forward bias ON switch r=0

OFF switch r=∞

PHYSICS FOR YOU | JANUARY ‘17

43

 Diode as a Rectifier •

•

Peak current through the load εm Im = (r f + RL ) I  Ripple factor = ac = 0.48 I dc Pdc  Efficiency = = 81.2% Pac (double the efficiency of the half wave rectifier)  Output frequency = 2 × Input frequency The pulsating dc obtained in both the rectifiers can be further rectified by using filter circuits. 

It is a device which is used for converting alternating current into direct current. There are two types of rectifiers . Half wave rectifier Junction diode  

+  Output voltage

•

–

 Special Purpose Diode Input voltage



Output voltage

• Time



Time

Zener diode : It is heavily doped p-n junction and operated in reverse bias.  Zener voltage remains constant, even though current through the zener diode varies over a wide range.    
 



Output frequency = Input frequency  Peak current through load εm Im = (r f + RL ) rf = forward resistance of the diode ; RL = load resistance ; εm = peak of ac voltage.  Ripple factor rms value of ac component = = 1.21 dc component in the rectifier circcuit  Efficiency dc output power of half wave rectifier = = 40.6% ac power supplied to the reectifier 

•

 


 


 

    



Zener diode as a voltage regulator : Any increase (decrease) in the input voltage results in, increase (decrease) of voltage drop across Rs without any change in voltage across the zener diode.   
    

Full wave rectifier



   
 





 

•

44

PHYSICS FOR YOU | JANUARY ‘17

   
 

Photodiode : It is a special type p-n junction diode fabricated with a transparent window to allow light to fall on the diode. It is operated under reverse bias.  When it is illuminated with light of photon energy greater than the energy gap of the semiconductor, electron-hole pairs are generated in near depletion region, hence reverse current increases.



•

generates emf when solar radiation falls on the p-n junction.  It converts solar energy into electrical energy.  It works on the same principle (photovoltaic effect) as the photodiode, except that no external bias is applied and the junction area is kept large.

The symbol of a photodiode is shown in the figure.

Light emitting diode : It is a heavily doped p-n junction which emits spontaneous radiation under forward bias.  It converts electrical energy into light energy.  The symbol of a LED is shown in the figure.

 Junction Transistor •

•

The I-V characteristics of a LED is similar to that of Si junction diode. But the threshold voltages are much higher and slightly different for each colour. The reverse breakdown voltages of LEDs are very low, typically around 5 V.  The semiconductor used for fabrication of visible LEDs must have minimum band gap of 1.8 eV.  The compound semiconductor gallium arsenide phosphide (GaAsP) is used for making LEDs of different colours.  GaAs is used for making infrared LED. Solar cell : It is basically a p-n junction which 

•

A semiconductor device which is obtained by growing either a very thin layer of n-type (p-type) crystal between two much thicker p-type (n-type) layers. It is a current operated device, i.e., action of transistor is controlled by the motion of charge carriers, i.e., current.  Emitter : Supply a large number of majority charge carriers for the flow of current through the transistor. It is heavily doped and medium in size.  Base : Control the flow of majority charge carriers from emitter to collector. It is lightly doped and very thin in size.  Collector : Collects a major portion of the majority carriers supplied by emitters for the circuit operation. It is moderately doped and has large size.

   

  











  



     





 









    

  

Bas ic Transistor Configur ation

C ommon Emi tter (CE)     

 

C ommon Base (CB)     

 







C ommon Collector(CC)     









 





PHYSICS FOR YOU | JANUARY ‘17

45

•

 Action : In general (active region), emitter base is forward biased and base collector is reverse biased  IE = IB + IC , (IC >> IB)





•

•

 



Transistor characteristics : n-p-n transistor as CE mode 

Input characteristic Input resistance,

⎛ ΔV ⎞ ri = ⎜ BE ⎟ ⎝ ΔI B ⎠V

CE



Power gain = current gain × voltage gain R ⎛ R ⎞ = βAV = β β C = β2 C ⎜⎝ R ⎟⎠ RB B Transistor as a switch : Operated in cut off region or saturation region.  VCC = ICRC + VCE

 VCE = VCC – ICRC When Vi = 0  or < 0.7 V, IB = 0  Hence IC = 0    ∴ VCE = VCC  (open circuit (switch))   When Vi > 0.7 V, then this is similar to a closed circuit (switch). Transistor as an oscillator : An oscillator works on positive feedback. A  Voltage gain with feedback A f = 1 − Aβ A → open voltage gain 



•

Output characteristic



•

⎛ ΔV ⎞ Output resistance, ro = ⎜ CE ⎟ ⎝ ΔIC ⎠ I B Transistor parameters : α and β  Current amplification factor, I I ⎛ ΔI ⎞ βac = ⎜ C ⎟ and βdc = C ; α dc = C , IB IE ⎝ ΔI B ⎠VCE

•

Relation between α and β, α β β= and α = 1− α 1+ β Transistor as an amplifier (CE-configuration) : To use transistor as an amplifier it is being operated somewhere in the middle of its active region. V0 IC RC R = =β C  Voltage gain, AV = Vi I B RB RB 

 

46







 

PHYSICS FOR YOU | JANUARY ‘17



 



 



   




 Logic Gates



•

β → feedback factor Barkhausen criterion for sustained oscillation, Aβ = 1



•

A logic gate is a digital circuit which is based on certain logical relationship between the input and the output voltage of the circuit. The operation of a logic gate is indicated in a table, known as truth table. This table contains all possible combination of inputs and corresponding outputs.

Name of gate OR

Symbol

Truth table A B Y 0

0

0

0

1

1

1

0

1

1

1

1

Boolean expression Y=A+B

JANUARY ’17





JANUARY ’17

AND

Y=A·B

•

Transmitter : A transmitter processes the incoming message signal so as to make it suitable for transmission through a channel and subsequent reception. Receiver : Extract the desired message signals from the received signals at the channel output. Attenuation : The loss of strength of a signal while propagating through a medium. Amplification : Process of increasing the strength of a signal using an amplifier. Range : The largest distance between a source and a destination up to which the signal is received with sufficient strength. Bandwidth : The frequency range over which an equipment operates or the portion of spectrum occupied by the signal. Modulation : The process of superposing a low frequency message signal on a high frequency wave (carries wave). Demodulation : The process of retrieval of information from the carrier wave at the receiver. Repeater : Picks up the signal from the transmitter, amplifies and retransmits it to the receiver sometimes with a change in carrier frequency.

A

B Y

0

0 0

0

1 0

1

0 0

1

1 1

A

Y

0

1

1 A

0 B Y

0

0 1

0

1 1

1

0 1

1 A

1 0 B Y

0

0 1

0

1 0

1

0 0

1

1 0

XOR

A

B Y

(also

0

0 0

called exclusive OR gate)

 Propagation of Electromagnetic Waves

0

1 1

•

1

0 1

1

1 0

A

B Y

0

0 1

0

1 0

1

0 0

1

1 1

NOT

NAND

NOR

XNOR

• • Y=A

• Y = A⋅B

•

•

Y = A+B

•

• • Y = A⋅B + A⋅B

Y = A⋅B + A⋅B



 Basic Terminology used in Electronic Communication Systems • •

•

Ground wave propagation : The radiowaves from the transmitting antenna propagate along the surface of the earth, to reach the receiving antenna.

Transducer : It is a device that converts one form of energy into another form. Signal : Information converted in electrical form and suitable for transmission is called a signal. Signal can be either analog or digital. Noise : The unwanted signals that tend to disturb the transmission and processing of the message signals in a communication system.

It is suitable for low frequency and medium frequency (i.e., less than 2 MHz)  Uses : Local broadcasting (ship communication, radio navigation). Sky wave propagation : The radiowaves from the transmitting antenna propagate through sky to reach the receiving antenna, after the reflection through ionosphere.  The ionospheric layer acts as a reflector for a 

•

PHYSICS FOR YOU | JANUARY ‘17

51



•

certain range of frequencies (3 to 30 MHz). Critical frequency : Maximum frequency of radiowave reflected from ionosphere. υC = 9 Nmax

Here, Nmax = Maximum electron density of ionosphere. Space wave (Line of Sight) propagation  The radiowaves from the transmitting antenna propagate along the space surrounding the earth to reach the receiving antenna either directly or after reflection from the ground.  Range of a transmitting antenna dT = 2Rh T

 Modulation •

•

•

•



•

Range of space wave propagation on earth’s surface = Maximum line of sight distance between two antennas (receiving and transmitting) dM = dT + dR = 2RhT + 2RhR

Amplitude modulation The modulated signal cm(t) can be written as Am sinωmt) sin ωct Ac = Acsinωct + μAc sinωmt sinωct

cm(t) = Ac(1 +

52

PHYSICS FOR YOU | JANUARY ‘17

Low frequencies cannot be transmitted to long distances. Therefore, they are superimposed on a high frequency carrier signal by a process known as modulation. In modulation, some characteristic of the carrier signal like amplitude, frequency or phase varies in accordance with the modulating or message signal. Correspondingly, they are called Amplitude Modulated (AM), Frequency Modulated (FM) or Phase Modulated (PM) waves. Pulse modulation could be classified as : Pulse Width Modulation (PAM), Pulse Duration Modulation (PDM) or Pulse Width Modulation (PWM) and Pulse Position Modulated (PPM). Need for modulation  Size of antenna or aerial ≈ wavelength of the signal (atleast λ/4 in dimension) 2  Effective power radiated by an antenna ∝(l/λ) . For a good transmission we need high powers, i.e., λ should be small. up of signals from different  Mixing transmitters.

= Acsinωct +

μAc μA cos(ωc – ωm)t − c cos(ωc + ωm)t 2 2

  



•

The total variation in frequency from the minimum to the maximum is called the carrier swing. Carrier swing = 2 Δυ

•

Frequency modulated wave consists of an infinite number of side frequency components on each side of the carrier frequency υc, υc ± υm, υc ± 2υm,.......

Am Ac μ ≤ 1 to avoid distortion. Modulation index in terms of amplitude of AM wave, A − Amin μ = max Amax + Amin

 Optical Fibres

Power in AM wave

•

Modulation index, μ =

A2 For carrier wave, Pc = c , R is the resistance of 2R antenna 2

2

⎛ μAc ⎞ 1 ⎛ μAc ⎞ 1 μ2 = + P ⎝ 2 ⎟⎠ 2R ⎜⎝ 2 ⎟⎠ 2R 2 c

For side bands, Ps = ⎜

•

2⎞ ⎛ Total power of AM wave, Pt = Pc ⎜1 + μ ⎟ 2 ⎠ ⎝

•





Fraction of total power carried by side bands, Ps μ2 = Pt (2 + μ2 ) If several modulating signals are present in AM

•

Frequency modulation  Modulation factor (index), μf =

For TIR in an optical fibre, i > θc where critical angle θc is given by μ12 − μ22 μ2 sin θc = or cos θc = μ1 μ1

2 2 2 wave then μ total = μ1 + μ2 + μ3 + .....

•

Optical fibres have an inner core of refractive index μ1 and a cladding of refractive index μ2 such that μ2 < μ1. The light is restricted to travel within the fibre by the process of total internal reflection. Optical fibres are used for transmitting optical frequencies.

Acceptance angle for optical fibre is θa where sin θa =

μ12 − μ22

μ0 Numerical aperture, NA = μ0 sin θa = μ12 − μ22

Maximum frequency deviation Δυ = Modulation frequency υm

Here Δυ = υmax – υc = υc – υmin

Here, μ0 is refractive index of surroundings. Numerical aperture is measure of light gathering power of the fibre. ””

   


 !

   


 !

   
!

 !

  

  



 
 
 

  

   


 

 
 


 

PHYSICS FOR YOU | JANUARY ‘17

53

1. The saturation current density of a p-n junction germanium diode is 250 mA m–2 at 300 K. The voltage that would have to be applied across the junction to cause a forward current density of 105 A m–2 to flow is (a) 0.33 V (b) 0.50 V (c) 0.66 V (d) 0.99 V.

6. A carrier wave is modulated by a number of sine waves with modulation indices m1, m2, m3.... . The total modulation index (m) of the wave is (a) m1 + m2 + m3 + ... (b) m1 – m2 + m3 + ...

2. In figure, assuming the diodes to be ideal,

m12 + m22 + m32 + ... . n 7. Choose only false statement from the following. (a) Substances with energy gap of the order of 10 eV are insulators. (b) The conductivity of semiconductor increases with increase in temperature. (c) In conductors, the valence and conduction bands may overlap. (d) The resistivity of a semiconductor increases with increase in temperature.

 







 

(a) D1 is forward biased and D2 is reverse biased and hence current flows from A to B (b) D2 is forward biased and D1 is reverse biased and hence no current flows from B to A and vice versa (c) D1 and D2 are both forward biased and hence current flows from A to B (d) D1 and D2 are both reverse biased and hence no current flows from A to B and vice versa. 3. A transmitter radiates 10 kW power with only carrier and 11.2 kW power with modulated carrier. The depth of modulation is (a) 24.5% (b) 36.5% (c) 49.0% (d) 53.4%. 4. What is the ratio of critical frequency for reflection of radiowaves from E, F1 and F2 layers in ionosphere if their electron densities are 2 × 1011, 3 × 1011 and 8 × 1011 m–3 respectively? (a) 1 : 1.5 : 3 (b) 1 : 1.22 : 2 (c) 2 : 3 : 8 (d) 1 : 1.5 : 4 5. In an n-p-n transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, (a) the emitter current will be 9 mA (b) the base current will be 1 μA (c) the emitter current will be 11 mA (d) the base current will be –1 mA. 54

PHYSICS FOR YOU | JANUARY ‘17

(c)

m12 + m22 + m32 + ...

(d)

8. A transmitting antenna of height h and the receiving antenna of height 45 m are separated by a distance of 40 km for satisfactory communication in line-of-sight mode. Then the value of h is (given radius of the earth is 6400 km) (a) 15 m (b) 20 m (c) 30 m (d) 25 m. 9. The output current of an 80% modulating amplitude modulated generator is 1.8 A. To what value will the current rise if the generator is additionally modulated by another audiowave of modulation index 0.6 ? (a) 1.71 A (b) 1.81 A (c) 1.91 A (d) 2.01 A 10. Figure shows a modified biasing circuit where the base resistor RB is connected to the collector instead of connecting it to the battery VCC. The value of IB is not given by  (VCC − IC RC ) − VBE (a) I B =  (RC + RB ) (b)

IB =

(VCE − VBE ) (RC + RB )

(c) VCE = VCC – (IC + IB)RC (d) VCC = VCE – (IC + IB)RB.



PHYSICS FOR YOU | JANUARY ‘17

55

11. A ground receiving station receiving a signal at 6 MHz transmitted from a ground transmitter at a height of 500 m located at a distance of 100 km. If radius of the earth is 6 . 4 × 106 m; maximum number density of electrons in ionosphere is 1012 m–3, the signal is coming via (a) ground wave (b) space wave (c) sky wave (d) none of these.

of the circuit is 192 Ω, the voltage gain and the power gain of the amplifier will respectively be (a) 4, 4 (b) 4, 3.69 (c) 4, 3.84 (d) 3.69, 3.84. [NEET Phase I 2016] 17. Identify the semiconductor devices whose characteristics are given below, in the order (i), (ii), (iii), (iv)

12. What is the carrier frequency, that a diode detector can detect if output circuit consists of R = 2 MΩ and C = 1 pF ? (a) fc >> 20 MHz (b) fc >> 0.2 MHz (c) fc = 0.5 MHz (d) fc >> 0.5 MHz 13. For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ, then the input signal voltage is (a) 10 mV (b) 20 mV (c) 30 mV (d) 15 mV. [NEET Phase II 2016] 14. The given circuit has two ideal diodes connected as shown in the figure. The current flowing through the resistance R1 will be   




(a) 2.5 A (c) 1.43 A



 





(b) 10.0 A (d) 3.13 A. [NEET Phase II 2016]

15. What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1 ?   

(a) 0, 1



(b) 0, 0





(c) 1, 0 (d) 1, 1 [NEET Phase II 2016]

16. A n-p-n transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Ω is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance 56

PHYSICS FOR YOU | JANUARY ‘17

(a) Simple diode, Zener diode, Solar cell, Light dependent resistance (b) Zener diode, Simple diode, Light dependent resistance, Solar cell (c) Solar cell, Light dependent resistance, Zener diode, Simple diode (d) Zener diode, Solar cell, Simple diode, Light dependent resistance. [JEE Main Offline 2016] 18. An experiment is performed to determine the I -V characteristics of a Zener diode, which has a protective resistance of R = 100 Ω, and a maximum power of dissipation rating of 1 W. The minimum voltage range of the dc source in the circuit is (a) 0 – 5 V (b) 0 – 24 V (c) 0 – 12 V (d) 0 – 8 V. [JEE Main Online 2016] 19. Choose the correct statement. (a) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal. (b) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(c) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal. (d) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal. [JEE Main Offline 2016] 20. A modulated signal Cm(t) has the form Cm(t) = 30 sin 300πt + 10 (cos 200πt – cos 400πt). The carrier frequency fc , the modulating frequency (message frequency) fω, and the modulation index μ are respectively given by 1 (a) fc = 200 Hz; fω = 50 Hz; μ = 2 2 (b) fc = 150 Hz; fω = 50 Hz; μ = 3 1 (c) fc = 150 Hz; fω = 30 Hz; μ = 3 1 (d) fc = 200 Hz; fω = 30 Hz; μ = . 2 [JEE Main Online 2016] SOLUTIONS 1. (a) : Here, forward current density, J = 105 A m–2 Saturation current density, J0 = 250 × 10–3 A m–2 As I = I0[exp(eV/kBT) – 1] or J = J0[exp(eV/kBT) – 1] 105 or exp(eV/kBT) −1 = J = = 4 × 105 −3 J 0 250 × 10 eV or ≈ ln(4 × 105 ) = 12.9 kBT or V = 0.33 V 2. (b) : Diode D1 is reverse biased as p-side is connected to negative potential and n-side is grounded. Diode D2 is forward biased as p-side is grounded and n-side is at negative potential. 3. (c) : Here, Pc = 10 kW ; Pt = 11.2 kW Depth of modulation, ⎛P ⎞ ⎛ 11.2 ⎞ m = 2 ⎜ t − 1⎟ = 2 ⎜ − 1⎟ ≈ 0.49 = 49% ⎝ 10 ⎠ ⎝ Pc ⎠ 4. (b) : Critical frequency, υC = 9 N max ∴ υCE : υCF : υCF 1

2

= 2 × 10 : 3 × 10 : 8 × 1011 = 2 : 3 : 8 11

11

= 1 : 3 / 2 : 8 / 2 = 1 : 1.22 : 2

5. (c) : For collector current, Ic = 10 mA Base current, Ib = 10% of Ic = (10/100) × 10 = 1 mA Emitter current, Ie = Ib + Ic = (1 + 10) mA = 11 mA. 6. (c) 7. (d) 8. (b) : Here, hR = 45 m, d = 40 km Since, d = 2RhT + 2RhR ∴ 40 × 1000 = 2 × 6.4 × 106 × h + 2 × 6.4 × 106 × 45

2 × 6.4 × 106 × h + 24 × 103

40 × 103 =

or

[(40 − 24)103 ]2 = 20 m 2 × 6.4 × 106 80 9. (c) : Here, It = 1.8 A ; m1 = = 0.80 ; m2 = 0.6 100 2 It = I c 1 + m1 / 2 h=

or

Ic =

It 1+ m / 2 2 1

It = I c 1 + =

1. 8 1.32

1. 8

=

1 + (0.8) / 2 2

=

1. 8 1.32

m12 m22 + 2 2

1+

(0.8)2 (0.6)2 + = 1.91 A 2 2

 10. (d) : VCC = RC (IC + IB) + IBRB + VBE    = RCIC + (RC + RB)IB + VBE   (V − I R ) − VBE  or I B = CC C C    (RC + RB )   VCE = VCC – (IC + IB) RC  ≈ (VCC – ICRC) (as IB RC υc

1 = 0.5 × 106 Hz = 0.5 MHz 2 × 10−6 s Therefore, υc >> 0.5 MHz i.e., υc >>

13. (b) : Here, RC = 2 kΩ = 2000 Ω, V0 = 4 V β = 100, RB = 1 kΩ = 1000 Ω, Vi = ? R 2000 Voltage gain, A = β C = 100 × = 200 1000 RB V V 2 4 Also, A = 0 or Vi = 0 = = V = 20 mV A 200 100 Vi 14. (a) : Diode D1 is reverse  biased so, it will   block the current and  
 diode D2 is forward    biased so, it will pass the current. Hence, equivalent circuit becomes as shown in the figure. Current in the circuit= Current flowing through the 10 resistance R1 = = 2.5 A 2+2 15. (c) 16. (c) : Here, load resistance, R0 = 800 Ω, input resistance, Ri = 192 Ω and current gain, β = 0.96 Voltage gain = Current gain × Resistance gain 800 =4 192 Power gain = (Current gain) × (Voltage gain) = 0.96 × 4 = 3.84 = 0.96 ×

17. (a) 18. (b) : Potential drop across Zener diode VZ = V – IR = V – 100 I   ∴ Power, P = VZIZ  = (V – 100 I) I  But P = 1 W (given) ∴ (V – 100 I) I = 1  or 100 I2 – VI + 1 = 0 58

PHYSICS FOR YOU | JANUARY ‘17

For I to be real, V2 – 4 × 100 × 1 ≥ 0 or V ≥ 20 V 19. (a) : Carrier wave : yc = Ac sin ωct Message signal : ym = Am sin ωmt Amplitude modulated carrier wave : y = (Ac + Am sin ωmt) sin ωct 20. (b) : Here, Cm(t) = 30 sin (300πt) + 10(cos(200πt) – cos(400πt)) Compare this equation with standard equation of amplitude modulated wave, μ Ac Cm (t ) = Ac sin 2 πfc t + 2 ( cos 2π( fc − f ω )t – cos 2π(fc + fω)t) 2πfc = 300π ⇒ fc = 150 Hz and fc – fω = 100 Hz ∴ fω = 150 – 100 = 50 Hz and

μ Ac

∴

μ=

2

= 10, Ac = 30 V 10 2 = 15 3

””

CLASS XII



Practice Paper 2017

Time Allowed : 3 hours Maximum Marks : 70

GENERAL INSTRUCTIONS (i) All questions are compulsory. There are 26 questions in all. (ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E. (iii) Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each. (iv) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to aĴempt only one of the choices in such questions.

SECTION-A

SECTION-B

1. Can there be a potential difference between two adjacent conductors carrying the same charge?

6. In a given sample two radioisotopes A and B are initially present in the ratio of 1:4. The half lives of A and B are respectively 100 years and 50 years. Find the time after which the amounts of A and B become equal.

2. Will the focal length of a lens for red light be more, same or less than that for blue light? 3. A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 Ω as shown in the figure. Find the value of the current in circuit. 


 




4. Relative permeability of a material is μr = 0.5. Identify the nature of the magnetic material and write its relation to magnetic susceptibility. 5. Explain the function of a repeater in a communication system.

7. A transmitting antenna at the top of a tower has a height of 36 m and the height of the receiving antenna is 49 m. What is maximum distance between them, for satisfactory communication in the LOS mode? (Radius of earth = 6400 km) 8. Two students X and Y preform an experiment on potentiometer separately using the given circuit. Keeping other parameters unchanged, how will the position of the null point be affected if

+ –






1



  + – 

 


 2

PHYSICS FOR YOU | JANUARY ‘17

59

(i) X increases the value of resistance R in the set-up by keeping the key K1 closed and the key K2 open ? (ii) Y decreases the value of resistance R′ in the set-up, while the key K2 remains open and the key K1 is closed ? 9. An object AB is kept in front of a concave mirror as 

shown in the figure. (i) Complete the ray diagram showing the image    formation of the object. (ii) How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black? OR Figure shows a ray of light passing through a prism. If the refracted ray QR is parallel to the base BC, show that (i) r1 = r2 = A/2, (ii) angle of minimum deviation, δm = 2i – A. 







 





 

10. Find the ratio of the de Broglie wavelengths, associated with (i) protons, accelerated through a potential of 128 V, and (ii) α-particles, accelerated through a potential of 64 V. SECTION-C



 

11. An electron moving horizontally with a velocity of 4 × 104 m s–1 enters a region of uniform magnetic field of 10–5 T acting vertically downward as  shown in the figure. Draw its   trajectory and find out the time it takes to come out of the region of magnetic field. 12. Define electric dipole moment. Is it a scalar or a vector quantity? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole. 60

PHYSICS FOR YOU | JANUARY ‘17

13. A cell of emf ε and internal resistance r is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) resistance R and (ii) the current I. It is found that when R = 4 Ω, the current is 1 A and when R is increased to 9 Ω, the current reduces to 0.5 A. Find the values of the emf ε and internal resistance r. 14. A rectangular loop of wire of size 2.5 cm × 4 cm carries a steady current of 1 A. A long straight wire carrying 2 A current is kept near the loop as shown in figure. If the loop and the wire are coplanar, find the (i) torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire. 15. Write any four characteristics of electromagnetic waves. Give two uses each of (ii) Microwaves (i) Radiowaves 16. A photon of energy 12.5 eV is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited? Calculate the wavelengths of the first member of Lyman and first member of Balmer series. 17. Draw V-I characteristics of a p-n junction diode. Answer the following questions, giving reasons. (i) Why is the current under reverse bias almost independent of the applied potential upto a critical voltage? (ii) Why does the reverse current show a sudden increase at the critical voltage? Name any semiconductor device which operates under the reverse bias in the breakdown region. 18. You are given a circuit below. Hence, identify the logic operation carried out by this circuit. Draw the logic symbol of the gate corresponds to it. Also, write its truth table.

19. Write two basic modes of communication. Explain the process of amplitude modulation. Draw a schematic sketch showing how amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave. 20. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if velocity of loop is 1 cm s–1 in a direction normal to the (i) longer side (ii) shorter side of the loop? For how long does the induced voltage last in each case? 21. In a plot of photoelectric current versus anode potential, how does (i) the saturation current vary with anode potential for incident radiations of different frequencies but same intensity? (ii) the stopping potential vary for incident radiations of different intensities but same frequency? (iii) photoelectric current vary for different intensities but same frequency of incident radiations? Justify your answer in each case. 22. State the necessary conditions for producing total internal reflection of light. Draw ray diagram to show how specially designed prisms make use of total internal reflection to obtain inverted image of the object by deviating rays (i) through 90° and (ii) through 180°. OR A parallel beam of monochromatic light of wavelength 500 nm falls normally on a narrow slit and the resulting diffraction pattern is obtained on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find (i) the width of the slit. (ii) the distance of the second maximum from the centre of the screen. (iii) the width of the central maximum. SECTION-D

23. Vishal purchased an induction stove and explained to his mother Kamla that due to shortage and heavy cost of LPG, she must utilize other sources that

are available to produce heat energy; Vishal also discussed with his younger brother that the oil and gas companies are trying their best to meet out the demand for LPG and that a good citizen must use other sources wherever feasible. The purchased induction stove have a value of 7 H inductor and the flow of current changes from 10 A to 7 A in a time of 9 × 10–2 s. Answer the following questions based on above information : (i) What qualities do you find in Vishal? (ii) What is the advantage of using induction stove over LPG? (iii) Calculate the induced emf in the induction stove when connected to the source? SECTION-E

24. (i) Explain briefly, with the help of a labelled diagram, the basic working principle of an ac generator. In an ac generator, coil of N turns and area A is rotated at υ revolutions per second in a uniform magnetic field B. Derive the expression for the emf produced. (ii) A 100-turn coil of area 0.1 m2 rotates at half a revolution per second. It is placed in a magnetic field 0.01 T perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil. OR (i) An alternating voltage ε = ε0 sin ωt applied to a series LCR circuit drives a current given by I =I0 sin (ωt + φ). Deduce an expression for the average power dissipated over a cycle. (ii) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (iii) Determine the current and quality factor at resonance for a series LCR circuit with L = 1.00 mH, C = 1.00 nF and R =100 Ω connected to an ac source having peak voltage of 100 V. 25. (i) A beam of unpolarised light is incident on the boundary between two transparent media. If the reflected light is completely plane PHYSICS FOR YOU | JANUARY ‘17

61

polarised, how is its direction related to the direction of the corresponding refracted light? Define Brewster’s angle. Obtain the relation between this angle and the refractive index for the given pair of media. (ii) Why does light from a clear blue portion of the sky show a rise and fall of intensity when viewed through a polaroid which is rotated? Explain by drawing the necessary diagram.

ε 190 V = = 5A R 38 Ω 4. The relative permeability is an intrinsic property of a magnetic material. A related quantity is the magnetic susceptibility, denoted by χm. μr = 1 + χm [& μr = 0.5]

OR Derive the expression for the total magnification of a compound microscope with labelled ray diagram.

5. A repeater is a combination of a receiver and a transmitter which picks up the signal from the transmitter, amplifies and retransmits it to the receiver sometimes with a change in carrier frequency.

Explain why both the objective and the eyepiece of a compound microscope must have short focal lengths. 26. State Gauss’s theorem in electrostatics and express it mathematically. Using it, derive an expression for electric field at a point near a thin infinite plane sheet of electric charge. How does this electric field change for a uniformly thick sheet of charge? OR Derive the expression for capacitance of a parallel plate capacitor with a dielectric medium of dielectric constant k between its plates, thickness of slab is less than the spacing between the plates. Obtain the expression for the energy stored in a charged capacitor. SOLUTIONS

Q , therefore potential difference between C two adjacent conductors can exist if their capacities are different or their sizes are different because capacitance depends on the size of conductor.

1. Since V =

2. According to lens maker’s formula 1 1 ⎞ ⎛1 = (μ − 1) ⎜ − ⎟ f R R ⎝ 1 2⎠ We know that μb > μr

∴ fb < fr

∴ Focal length of a lens for red light is more than that for blue light. 3. Since, the positive terminal of the batteries are connected together, so the equivalent emf of the batteries is given by ε = 200 V – 10 V = 190 V. Hence, the current in the circuit is given by 62

PHYSICS FOR YOU | JANUARY ‘17

I=

Here, μr < 1 (χm negative), so the material is termed diamagnetic.

6. We have, N = N0 e–λt For radioisotopes A and B, we can write N A = N 0e − λ At A , N B = 4 N 0e

− λB tB

Let t be the time after which NA = NB λ t −λ t

N 0e − λ At = 4 N 0e − λ Bt ⇒ 4 = e B A ⇒ loge4 = (λBt – λAt) loge e ⎛ log 2 log 2 ⎞ ⎛ log e 2 ⎞ e e − ⇒ 2 log e 2 = ⎜ ⎟ t ⎜∵ λ = TA ⎠ T1/2 ⎟⎠ ⎝ ⎝ TB1/2 1/ 2 1 ⎞ ⎛1 ⇒ 2=⎜ − t ∴ t = 200 years ⎝ 50 100 ⎟⎠ ∴

7. Here, hT = 36 m, hR = 49 m and R = 6400 km = 6400000 = 64 × 105 m ∴ Maximum line-of-sight (LOS) distance dM between the two antennae is dM = 2RhT + 2RhR dM = 2 × 64 × 105 × 36 + 2 × 64 × 105 × 49 = 48 × 102 × 20 + 56 × 102 × 20 = 208 × 2.236 × 100 = 46.51 km 8. (i) By increasing resistance R, the current through AB decreases, so potential gradient decreases. Hence a greater length of wire would be needed for balancing the same potential difference. So the null point would shift towards B. (ii) By decreasing resistance R′, the current through AB remains the same, potential gradient does not change. As K2 is open so there is no effect of R′ on the null point.

9. (i)



 









 



(ii) Position of image will remain same and intensity of image will decrease. OR (i) When QR is parallel to the base BC, we have i = e ⇒ r1 = r2 = r We know that, r1 + r2 = A ⇒ r + r = A ∴ r = A/2 (ii) Also, we have A+δ=i+e Substituting, δ = δm and i = e A + δm = i + i ∴ δm = 2i – A 10. The de Broglie wavelength λ = ∴

λp =

h 2mqV

and λ α =

12. Strength of an electric dipole is measured by its electric dipole moment, whose magnitude is equal to product of magnitude of either charge and separation between the two charges, i.e., p = q ⋅2a and is directed from negative to positive charge, along the line joining the two charges. Its SI unit is C m. Consider a dipole consisting of two electric charges +q and –q between a small distance AB = 2a with centre O.

h 2mqV h 2m′ q ′V ′

m ′ q ′V ′ mqV As per question, m′ = 4m, q = e, q′ = 2e, V = 128 V, V′ = 64 V λp 4m × 2e × 64 = λα m × e × 128 ∴ Required ratio = 2 : 1 λp : λα =

11. Let the time taken by the electron to come out of the region of magnetic field be t. Velocity of the electron, v = 4 × 104 m s–1 Magnetic field, B = 10–5 T Mass of the electron, m = 9 × 10–31 kg πr We know t = v mv where r = qB πm 3.14 × 9 × 10−31 = Now, t = Bq 10−5 × 1.6 × 10−19 ⇒ t = 17.66 × 10–7 s = 1.77 μs Thus, the time taken by the electron to come out of the region of magnetic field is 1.77 μs.

The magnitudes of the electric field at point P due to the two charges +q and –q are given by q 1 E+ q = , directed along BP 4 πε0 x 2 + a2 q 1 , directed along PA and E−q = 4 πε0 x 2 + a2 ∴ E+q = E–q The directions of E+q and E–q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up. Therefore, the resultant electric field at point P is given as; E = – (E+q + E–q) cosθ ...(i) Where the negative sign shows that the field is opposite to the dipole moment of the dipole. a From figure cos θ = 2 2 1/2 (x + a ) Putting the values of E+q, E–q and cosθ in eqn (i), we get E= =

−q ⎛ 1 1 ⎞ a + ⎟ ⎜ 2 2 2 2 4 πε0 ⎝ x + a x + a ⎠ ( x 2 + a2 ) − 2qa 4 πε0 (x 2 + a2 )3/2 PHYSICS FOR YOU | JANUARY ‘17

63

When point P lies at a large distance (x >> a) from the dipole, the above expression reduces to −p − 2qa −p ∴ E= E= = 4 πε0 x 3 4 πε x 3 4 πε x 3 0

0



13. Given situation is shown in figure ε I= r+R Terminal voltage, V = ε – Ir ε εR =ε– r= r+R r+R





(ii) V versus I, 



(i) V versus R,



∴

 

 

V = E – Ir When R = 4 Ω, then I1 = 1 A ε ;r+4=ε …(i) ∴1= r+4 1 When R = 9 Ω, then I = 0.5 A = A 2 1 ε r+4 [Using eqn. (i)] ∴ = = 2 r +9 r +9 r + 9 = 2r + 8; r = 1 Ω From eqn. (i) emf, ε = 1 + 4 = 5 V

14. Force between two current carrying wires, μ II l F= 0 1 2 2 πr Force on arm AB, μ × 2 × 1 × 4 × 10−2 FAB = 0 2 π × 2 × 10−2 2μ (Attractive, towards the wire) = 0N π Force on arm CD, μ × 2 × 1 × 4 × 10−2 FCD = 0 2 π × 4.5 × 10−2 8μ (Repulsive, away from the wire) = 0N 9π Force on arms BC and DA are equal and opposite. So, they cancel out each other. Net force on the loop is F = FAB – FCD 64

PHYSICS FOR YOU | JANUARY ‘17

μ ⎡ 8 ⎤ 10μ 0 10 × 4 π × 10 −7 = 0 ⎢2 − ⎥ = = 9⎦ 9π π ⎣ 9π

= 4.44 × 10–7 N

(Attractive, towards the wire)

15. Four characteristics of electromagnetic waves are : (a) Electromagnetic waves do not require any medium for their propagation. (b) These waves travel in free space with speed 3 × 108 m s–1. It is given by the relation 1 c= μ 0 ε0 (c) The energy in electromagnetic waves is divided equally between electric field and magnetic field. (d) Electromagnetic waves are produced by accelerated charged particles. (i) Uses of radiowaves : (a) In radio and television communication system. (b) In radio astronomy (ii) Uses of microwaves : (a) In radar system for aircraft navigation. (b) In microwave ovens. 16. Here, ΔE = 12.5 eV Energy of an electron in nth orbit of hydrogen atom is, 13.6 En = − eV n2 In ground state, n = 1; E1 = –13.6 eV Energy of an electron in the excited state after absorbing a photon of 12.5 eV energy will be En = –13.6 + 12.5 = –1.1 eV −13.6 −13.6 ∴ n2 = = = 12.36 ⇒ n = 3.5 En −1.1 Here, state of electron cannot be fraction. So, n = 3 (2nd exited state). The wavelength λ of the first member of Lyman series is given by ⎛1 1 ⎞ 3 1 = R⎜ − ⎟ = R ⎝ 12 22 ⎠ 4 λ 4 4 ⇒ λ= ⇒ λ = 1.215 × 10–7 m = 3R 3 × 1.097 × 107 ⇒ λ = 121 × 10–9 m ⇒ λ = 121 nm The wavelength λ′ of the first member of the Balmer series is given by ⎛1 1 1⎞ 5 = R⎜ − ⎟ = R 2 ⎝2 λ′ 32 ⎠ 36

⇒

λ′ =

36 36 = 5R 5 × 1.097 × 107

= 6.56 × 10–7 m = 656 × 10–9 m = 656 nm 17. V-I characteristics of a p – n junction diode



(ii) Broadcast communication Amplitude modulation : Amplitude modulation is produced by varying the amplitude of the carrier waves in accordance with the modulating wave. Let the carrier wave be c(t) = Ac sinωct and the modulating signal be m(t) = Am sinωmt, where ωm = 2πυm is the angular frequency of the message signal. Modulated signal cm(t) is cm(t) = (Ac + Am sin ωmt)sinωct ⎛ A ⎞ = Ac ⎜1 + m sin ωmt ⎟ sin ωc t Ac ⎝ ⎠



(i) The reverse current is mainly due to minority charge carriers and even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction. Here the current is not limited by the magnitude of the applied voltage but is limited due to the concentration of the minority carrier on either side of the junction. (ii) At critical voltage/breakdown voltage, a large number of covalent bonds break resulting in availability of large number of charge carriers. Zener diode operates under the reverse bias in the breakdown region. 18. Here, X = A, Y = B , Z = X + Y Z = A+B ∴ The output Z of the given combination of gates is, Z = A ⋅ B = A ⋅ B Hence, the equivalent gate is AND gate. Logic symbol :

∴ cm(t) = Ac sinωct + μAc sinωmt sinωct μA = Ac sin ωc t + c (cos(ωc − ωm )t − cos(ωc + ωm )t ) 2 Am where μ = is the modulation index. Ac ωc – ωm and ωc + ωm are the lower side band and upper side band, respectively. Production of amplitude modulated wave : Amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave as shown in the figure.    

 
   







Truth table : Input A 0 0 1 1

B 0 1 0 1

Output Z=A·B 0 0 0 1

19. The two basic modes of communication are (i) Point-to-point communication 66

PHYSICS FOR YOU | JANUARY ‘17




20. Here, A = 8 × 2 = 16 cm2 = 16 × 10–4 m2, B = 0.3 T, v = 1 cm s–1 = 10–2 m s–1, ε = ?

It is so because photoelectric current is directly proportional to the number of photoelectrons emitted per second. (i) When velocity is normal to longer side of the loop l = 8 cm = 8 × 10–2 m ε = Blv = 0.3 × 8 × 10–2 × 10–2 = 2.4 × 10–4 V Time, t =

distance 2 × 10−2 = =2 s velocity 10−2

(ii) When velocity is normal to shorter side of the top, l = 2 cm = 2 × 10–2 m ε = Blv = 0.3 × 2 × 10–2 × 10–2 = 0.6 × 10–4 V Time, t =

22. (i) Necessary conditions for total internal reflection : (a) Light should travel from a denser medium to a rarer medium. (b) Angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact. (ii) (a) To deviate a ray of light through 90° : 

distance 8 × 10−2 = =8 s velocity 10−2

21. (i) The saturation current remains same because the saturation current depends upon intensity of incident radiation.



 



















A totally reflecting prism is used to deviate the path of the ray of light through 90°. (b) To deviate a ray of light through 180° : When the ray of light comes to meet the hypotenuse face BC at right angles to it, it is refracted out of prism as such along the path RS. The path of the ray of light has been turned through 180° due to two total internal reflections. 

(ii) Stopping potential remains same. It depends upon the frequency of incident radiation.



  

 



 





OR Given, λ = 500 nm = 5 × 10–7 m, D = 1 m, y1 = 2.5 mm = 2.5 × 10–3 m (iii) At constant frequency and anode potential, photoelectric current is directly proportional to the intensity of light.

(i) If a is width of slit, then for first minimum λ sin θ1 = a y1 For small θ1, sin θ1 = D PHYSICS FOR YOU | JANUARY ‘17

67

y1 λ = D a λD 5 × 10−7 × 1 = = 2 × 10−4 m or a = y1 2.5 × 10−3 = 0.2 mm 1 ⎞ Dλ ⎛ (ii) Position of nth maximum, yn = ⎜ n + ⎟ ⎝ 2⎠ a For second maximum n = 2 ⎛ 1 ⎞ 1 × 5 × 10−7 ∴ ( y2 )max = ⎜ 2 + ⎟ ⎝ 2 ⎠ 2 × 10−4 = 6.25 × 10–3 m = 6.25 mm 2D λ ⎞ (iii) Width of central maximum, ⎛⎜ = ⎟ ⎝ a ⎠ = Separation between first minima on either side of centre of screen = 2.5 + 2.5 = 5 mm

∴

∴

23. (i) Vishal has the qualities of responsibility, concern for the nation, cost economic attitude, promptness to use latest technology, sharing attitude. (ii) LPG is costly and it is difficult to carry heavy gas cylinders. Also, availability of LPG is limited. On the other hand induction stove uses cheap and easily available electricity and is also easy to carry. 7 × (7 − 10) = 233.3 V (iii) ε = – L dI/dt = − 9 × 10−2 24. (i) Refer to point 4.8(2) page no. 275 (MTG Excel in Physics) (ii) Given, N = 100, A = 0.1 m2, B = 0.01 T 1 υ = revolution per sec = 0.5 rps 2 ∴ Maximum voltage generated ε0 = NBAω = NBA(2πυ) or ε0 = 100 × 0.01 × 0.1 × 2 × OR

22 × 0.5 = 0.314 V 7

(i) Refer to point 4.6(9) page no. 271 (MTG Excel in Physics) (ii) The power is P = εrms Irms cosφ. cosφ is small, then current considerably increases when voltage is constant. Power loss is I2R. Hence power loss increases. (iii) Given, L = 1.00 mH = 1 × 10–3 H, C = 1.00 nF = 1 × 10–9 F, R = 100 Ω, ε0 = 100 V At resonance Z = R 68

PHYSICS FOR YOU | JANUARY ‘17

ε 100 I0 = 0 = ⇒ I0 = 1 A R 100 I 1 Irms = 0 = = 0.707 A 2 2 Q=

1 L 1 1.0 × 10−3 = = 10 R C 100 1.0 × 10−9

25. (i) Refer to point 6.15(9) page no. 455(MTG Excel in Physics) (ii) Refer to point 6.15(10) page no. 455 (MTG Excel in Physics) OR Refer to point 6.9 (1(iv)) page no. 381 (MTG Excel in Physics) 26.

Refer to point 1.8 page no. 11 (MTG Excel in Physics) OR Refer to point 1.11(6, 9) page no. 16 (MTG Excel in Physics) ””

ON

R-C Circuit Er. Sandip Prasad Calculation of Instantaneous Charge on the Capacitor Carrying some Charge Before being Connected to the Battery :  Consider the basic RC  circuit shown in figure,  the capacitor carries a charge Q before the key K is closed.If now the key K is closed, charge on   the capacitor goes on increasing until it attains steady state. The circuit is similar to simple charging RC circuit but at t = 0 charge on the capacitor is Q instead of zero. By Kirchhoff ’s voltage law, q

t

dq 1 q dq Cε − q = dt ; IR + = ε; = ; Cε − q RC C dt RC Q O Cε − q t ln =− Cε − Q RC q =Cε(1 – e–t/RC) + Qe–t/RC q = q0 (1 – e–t/τ) + Qe–t/τ Here, q0 = Cε and τ = RC  If Cε > Q, then the  charging of capacitor  takes place and the capacitor will attain final charge  Cε and current   becomes finally zero.  If Cε < Q, then the  discharging of  capacitor takes place and the capacitor will attain final charge  q0 = Cε and current   becomes finally zero. Method-II: When the key K is closed at t = 0, the circuit can be treated as the combination of a charging and a discharging circuit.

∫

∫







 

 







 





Charge on the capacitor of circuit 1 q1 = q0 (1 – e–t/τ) and charge on the capacitor of circuit 2 q2 = Qe–t/τ Net charge on the capacitor q = q1 + q2 = q0 (1 – e–t/τ) + Qe–t/τ Method-III: (Short-Cut Method) When we deal with the R-C circuit in which a capacitor has initial charge, then we can use a very simple method to solve this type of problems quickly. Let us consider a resistor and a capacitor having some initial charges q0 is connected with a battery of e.m.f ε .  +0 –0 When key is closed two  situation arise, one in which capacitor is in charging condition and   another in which it is in discharging condition. To know whether the capacitor is in charging or in discharging condition , compare the voltage across the capacitor and e.m.f of the battery. q  If ε > 0 , then the charging C of capacitor takes place and the capacitor will attain a final charge q = Cε and current finally becomes zero.

Sandip Physics Classes, Girish Park-1/1 Shiv Krishna Daw Lane, Kolkata-700007 PHYSICS FOR YOU | JANUARY ‘17

69

From the graph, we can write the value of instantaneous value of charge i.e the value of charge on the capacitor at time t, q = initial charge + (amplitude of the curve )(1 – e–t/τ) Where, amplitude of the curve Q = Final charge – initial charge = Cε – q0 and τ= Req C ∴ charge on the capacitor at time t, q = q0 + (Cε – q0) (1 – e–t/τ) q  If ε < 0 , then the discharging C of capacitor takes place and the capacitor will attain a final charge q = Cε and current finally becomes zero. From the graph, we can write the value of instantaneous value of charge i.e. the value of charge on the capacitor at time t, q = Final charge + (amplitude of the curve) e–t/τ Where, amplitude of the curve Q = initial charge – Final charge = q0 – Cε and τ= ReqC ∴ charge on the capacitor at time t, q = q0 + (q0 – Cε) e–t/τ Example-1: In the circuit  shown capacitor is initially  connected to a 3 V cell. At    t =0 switch is thrown to B. Find the charge on the   capacitor at time t. Soln.: The capacitor is charged to 3CV initially i.e. ,at t=0 charge on the capacitor is +3CV . Now battery of emf V cannot pull out electrons from positive plate of capacitor as potential difference across the capacitor is 3 V . Instead of electrons move from negative plate of capacitor towards positive plate through battery. +3 –3  By Kirchhoff ’s laws, q  ∴ V − + 2iR = 0; electrons motion C q  V − = −2iR   C dq (i = − , here charge on the plate decreases with time) dt dq 1 q ⎛ dq ⎞ = dt V − = −2 ⎜ − ⎟ R or, V − q / C 2R ⎝ dt ⎠ C Integrating both side ; q

or,

∫

3CV

70

t

dq 1 = dt ⇒ q = CV + 2 CV e–t/2RC V − q /C 2R

∫ 0

PHYSICS FOR YOU | JANUARY ‘17

Method-II: When the key K is closed at t = 0, the circuit can be treated as the combination of a charging and a discharging circuit. 2 +3 –3

 2

 2 +3 –3



  1

2

Here, time constant of the given R-C circuit is τ = ReqC = 2RC Charge on the capacitor of the circuit 1 q1 = CV (1 – e–t/2RC ) and charge on the capacitor of the circuit 2 q2 = 3CV e–t/2RC) ∴ Net charge on the capacitor q = q1 + q2 = CV (1 – e–t/2RC) + 3CV e–t/2RC ⇒ q = CV + 2CV e–t/2RC Method-III: The capacitor is charged to 3 CV initially i.e., at t=0 charge on the capacitor is +3CV . Now battery of emf V cannot pull out electrons from positive plate of capacitor as potential difference across the capacitor is 3V. Instead of electrons move from negative plate of capacitor towards positive plate through battery. Step:I Firstly, check 2 +3 –3 the R-C circuit whether it is in charging condition  or discharging condition. Initial voltage across the capacitor i.e when switch is 3CV just closed i.e at t = 0, is VC = = 3V C And the voltage applied by the battery is V. Here initial voltage across capacitor i.e. voltage across capacitor at t = 0(when switch S is just closed) is greater than that of the voltage of the cell, hence after t > 0 capacitor gets discharged. So, in the given R-C circuit is in discharging condition.  Step:II Next draw a graph of  instantaneous charge versus time of the above circuit. Step:III From graph obtained from step II , we can write the  equation of the instantaneous   charge.

Hence, the charge on the capacitor at time t is , q = Final charge + (amplitude of the curve ) e–t/τ Here, initial charge on the capacitor q0 = 3CV Amplitude of the curve = initial charge - Final charge = 3CV – CV = 2CV Again, time constant , τ = ReqC = 2RC Hence, q =CV + 2CV e–t/2RC Example 2 : A capacitor hav i ng i n it i a l charge Cε is connected to a q0 = 2 cell of e.m.f ε through a resistor R as shown in the figure. Find the charge on the capacitor at time t .





  



Soln.: Step:I Firstly, check    the R-C circuit whether it  is in charging condition or discharging condition.  Initial voltage across the  capacitor i.e when switch is just closed i.e at t=0, is q Cε ε /C = VC = 0 = C 2 2 And the voltage applied by the battery is V = ε Here initial voltage across capacitor i.e. voltage across capacitor at t = 0(when switch S is just closed) is less than that of the voltage of the cell, hence after t >0 capacitor gets charged. So, in the given R-C circuit is in charging condition. Step:II Next draw a graph  of instantaneous charge  versus time of the circuit. Step:III From graph obtained from step II ,  we can write the equation    of the instantaneous charge. Hence, the charge on the capacitor at time t is , q = initial charge + (amplitude of the curve ) × (1 – e–t/τ) Cε Here, initial charge on the capacitor q0 = 2 Amplitude of the curve Q = Final charge – initial charge Cε Cε = Cε − = 2 2 Again , time constant , τ = ReqC = RC Cε Cε Hence, q = + (1 − e −t /RC ) 2 2

Simultaneous Charging and Discharging of Capacitors in a R - C circuit without battery  Consider a circuit in which a resistor of resistance R is  + 2 connected in series with two 1 – 1  capacitors of capacitance C1 and C2. Initially the capacitor C1 is charged to potential V1 and then at t = 0, the switch S is closed as shown. When switch S is closed at t = 0, the circuit is completed and discharging of C1 and charging of C2 takes place. The voltage across capacitor C1 does not rise instantaneously but builds up exponentially and not linearly. Charging current i is maximum at the start i.e. when capacitor C1 has voltage V1 and charge C1V1 and capacitor C2 is uncharged, then it decreases exponentially and finally ceases when potential difference across both the capacitors becomes equal. Hence as charge on capacitor C1 decreases and on capacitor C2 increases, current i in the circuit decreases. In other words, as the time passes, potential difference across capacitor C1 decreases and potential difference across capacitor C2 increases. Hence, whenever a R-C circuit having a charged capacitor connected with uncharged capacitor without a dc source, goes from initial state to final state (steady state condition), it passes through a transient state which is of short duration. In fact transient state lies in between initial state and final state(steady state condition).  Initial State: When switch    S is closed at t = 0, R-C  circuit is in its initial state.   In this state, current in the circuit i is maximum  
 because there is no potential difference across capacitor C2 to oppose the applied voltage by capacitor C1 i.e capacitor C2 behaves like being short circuited and capacitor C1 behaves as a dc source having voltage V1. If you wish to find initial current, short the terminals of the capacitor C2 and replace the capacitor C1 with a battery having voltage V1. Hence at t = 0 the circuit can be redraw as shown in the figure. V At this instant i.e at t = 0, i = 1 R Transient state: But as time  passes, i decreases gradually –   + 0 2 + – so does instantaneous voltage 1 – 0–   across capacitor C 1 also 

PHYSICS FOR YOU | JANUARY ‘17

71

decreases gradually, but instantaneous voltage cross capacitor C2 increases gradually till it reaches at its maximum value or steady state value i.e. when potential difference across both the capacitors becomes equal. Final state (Steady state):  
 As charging continues, charging current i decreases   gradually and it becomes  zero at final state or steady state, hence at condition, the circuit appears as an open circuit (i = 0). At this state capacitor C1 gets completely charged and the value of charge becomes maximum i.e. when switch is closed charge flows from capacitor C1 to capacitor C2 till both acquire the same potential. In steady state the circuit can be redrawn as shown in the figure. Calculation of charge and current in the R-C circuit in transient state  We need to find the charge on + + 0 –  2 1 the capacitor C1 – –  – 0 as a function of  time.  Initial charge on C1 is q0 = C1 V1 Let at any time t charge on C2 is, then charge of C1 at this instant is q0 – q. Hence current through the circuit at this time is, dq i= dt Applying KVL for circuit, q q(C1 + C2 ) dq q0 − q q or 0 − = R = iR + C1 C1C2 dt C1 C2 q

or

0 2

or q = or q =

t

dq

∫ C q − q(C

1 + C2 )

dt RC1C2 0

=∫

C2q0 CC −t / RCeq (1 − e ) where Ceq = 1 2 C1 + C2 (C1 + C2 )

C1C2V1 −t / RCeq (1 − e ) where q0 = C1V1 (C1 + C2 ) −t / RC

eq or q = CeqV1(1 − e ) Again, charge on capacitor C1 as a function of time t, q′ = q0 – q or q′ = C1V1 – CeqV1(1 – et/RCeq) We can also find the instantaneous value of charge in this case by another method. Method-II: Initially the capacitor C 1 is charged to potential V1 and then at t = 0, the switch S is closed

72

PHYSICS FOR YOU | JANUARY ‘17

as shown. When switch S is closed at t = 0, the circuit is completed and discharging of C1 and charging of C2 take place. Step-I : First of all we need to find time constant (τ) of the given R-C circuit. τ =Req Ceq = R

C1C2 C1 + C2

Step-II: Find the steady  = 0 state charge Q on the  1 + + 1 uncharged capacitor. 1 –  –  2 2 1 Let us consider in steady  state capacitor C1 and C2 has charge q1 and q2 respectively as shown in the figure. Also we know that in steady state both the capacitor has same potential difference, let V in this case. Charge on the capacitor C1 is, q1 = C1V Charge on the capacitor C2 is, q2 = C2V q C ∴ 1= 1 q2 C2 C1 C1 q , where (q1 + q2 ) = Hence, q1 = C1 + C2 0 C1 + C2 q0 = C1V1 is the initial charge on capacitor C1. C2 C2 C V =Q q0 = Similarly, q2 = C1 + C2 1 1 C1 + C2 where q0 = C1V1 is the initial charge on capacitor C1. Step-III: Putting the values of τ and Q in equation q = Q (1 – e–t/τ) . We can get the value of charge on the capacitor which is under charging condition, at time t. By differentiating the above equation w.r.t. time we can find current through the capacitor at time t. Then by using Kirchhoff ’s laws, calculate currents in different branches of the circuit. Also, charge on capacitor C1 having some initial charge q0, as a function of time t is given by, q′ = q0 – q where, initial charge on capacitor C1 is q0 = C1V1 and q is the value of charge on the capacitor which is under charging condition, at time t. Hence the charge on the capacitor at time t is given by following equation, As, q = Q(1 – e–t/τ) ∴ q =

C1C2V1 −t / RCeq (1 − e ) (C1 + C2 )

Again, charge on capacitor C1 as a function of time t, q = q0 – q or q′ = C1V1 − CeqV1(1 − e

−t / RCeq

)

””



Class XII

T

his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.

Dual Nature of Radiation and Matter Atoms and Nuclei Total Marks : 120 NEET / AIIMS / PMTs Only One Option Correct Type

1. Light described at a place by the equation E = (100 V m–1) [sin(5 × 1015 s–1)t + sin (8 × 1015 s–1)t] falls on a metal surface having work function 2.0 eV. The maximum kinetic energy of the photoelectrons is (a) 5.27 eV (b) 3.27 eV (c) 2.00 eV (d) 4.00 eV 2. An electron of mass m, when accelerated through a potential difference V, has de Broglie wavelength λ. The de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be M m M m (a) λ (b) λ (c) λ (d) λ m M m M 117 3. In a fission reaction, 236 X + 117Y + n + n, 92U → the binding energy per nucleon of X and Y is 8.5 MeV whereas that of 236 92U is 7.6 MeV. The total energy liberated will be about (a) 200 keV (b) 2 MeV (c) 200 MeV (d) 2000 MeV 4. A radioactive nucleus emits 3α-particles and 5β-particles. The ratio of number of neutrons to that of protons will be A − Z − 12 A−Z (b) (a) Z −6 Z −1 A − Z − 11 A − Z − 11 (c) (d) Z −6 Z −1 5. Let nr and nb be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time. Then, (b) nr < nb (a) nr = nb (c) nr > nb (d) None of these

Time Taken : 60 min

6. The ratio of the longest and shortest wavelengths in Brackett series of hydrogen spectra is 25 17 9 4 (b) (c) (d) (a) 9 6 5 3 7. According to Einstein’s photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is (a)

(b)

(c)

(d)

8. The radius of the hydrogen atom in its ground state is a0. The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged muon with mass 207 times that of an electron, is aμ equal to a a0 (a) 207a0 (b) 0 (c) (d) a0 207 207 207 9. A proton moving with u m s–1 strikes a stationary nucleus of mass A. The ratio of final to initial kinetic energy of proton is (1 − A)2 A2 (b) (a) ( A − 1)2 (1 + A)2 2 2 (1 − A) (1 − u ) A2 (1 + u2 ) (c) (d) ( A + 1)2 (1 − u2 ) (1 + A)2 (1 + u) 10. A nuclear reactor delivers a power of 10 W. Find fuel consumed by the reactor per hour, if its efficiency is 20%. (Given, c = 3 × 108 m s–1) PHYSICS FOR YOU | JANUARY ‘17

73

(a) 2 × 10–6 g h–1 (b) 9 × 10–12 g h–1 –9 –1 (c) 8 × 10 g h (d) 2 × 10–9 g h–1 11. Electrons with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is 2mcλ2 2h (b) λ0 = (a) λ0 = h mc 2m2c 2 λ3 (c) λ0 = (d) λ0 = λ h2 12. When the electron in hydrogen atom is excited from the 4th stationary orbit to the 5th stationary orbit, the change in the angular momentum of the electron is (Planck’s constant, h = 6.63 × 10–34 J s) (a) 4.16 × 10–34 J s (b) 3.32 × 10–34 J s –34 (c) 1.05 × 10 J s (d) 2.08 × 10–34 J s Assertion & Reason Type

irecti In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 13. Assertion : The de Broglie wavelength of a molecule varies inversely as the square root of temperature. Reason : The root mean square velocity of the molecule depends on the temperature. 14. Assertion : The binding energy per nucleon, for nuclei with atomic mass number A > 100, decreases with A. Reason : The nuclear force is weak for heavier nuclei. 15. Assertion : When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. Reason : If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change. JEE MAIN / JEE ADVANCED / PETs Only One Option Correct Type

16. A beam of 450 nm light is incident on a metal having work function 2.0 eV and placed in a magnetic field B. The most energetic electrons emitted perpendicular to the field are bent in circular arcs of radius 20 cm. Then value of B is (a) 1.46 × 10–5 T (b) 2.92 × 10–5 T –6 (c) 1.46 × 10 T (c) 2.92 × 10–6 T 17. An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of 74

PHYSICS FOR YOU | JANUARY ‘17

(a) 1 Å (b) 10–10 cm –12 (c) 10 cm (d) 10–15 cm 18. The electric potential between a proton and an ⎛r⎞ electron is given by V V0 ln ⎜ ⎟ where r0 is a ⎝r ⎠ 0

constant. Assuming Bohr’s model to be applicable, write variation of rn with n, where n being the principal quantum number. 1 1 (a) rn ∝ n (b) rn ∝ (c) rn ∝ n2 (d) rn ∝ 2 n n 19. A radioactive isotope A with a half life of 1.25 × 1010 years decays into B which is stable. A sample of rock from a planet is found to contain both A and B present in the ratio 1 : 15. The age of the rock is (a) 9.6 × 1010 years (b) 4.2 × 1010 years 10 (c) 5 × 10 years (d) 1.95 × 1010 years More than One Options Correct Type

20. Which of the following products in a hydrogen atom are independent of the principal quantum number n? The symbols have their usual meanings. (a) vn (b) Er (c) En (d) vr 21. Let λα , λβ and λ′α denote the wavelengths of the X-rays of the Kα , Kβ and Lα lines in the characteristic X-rays for a metal. Then (a) λ′α > λα > λβ (b) λ′α > λβ > λα 1 1 1 1 1 1 (c) = + (d) = + λβ λ α λ′α λ α λβ λ′α 22. When an α particle collides elastically with a nucleus, the nucleus recoils. A 5.00 MeV α particle has a head-on elastic collision with a gold nucleus, initially at rest. The mass of the α particle may be taken to be 4.00 u and that of the gold nucleus to be 197 u. Then (a) the kinetic energy of the recoiling nucleus is 0.390 MeV. (b) the kinetic energy of the rebounding α particle is 5.64 MeV. (c) the α particle will be stop. (d) the kinetic energy of rebounding α particle is 4.61 MeV. 23. A large electric generating station is powered by a pressurized water nuclear reactor. The thermal power in the reactor core is 3400 MW and 1100 MW of electricity is generated. The fuel consists of 86,000 kg of uranium, in the form of 110 tons of uranium oxide, distributed among 57,000 fuel rods. The uranium is enriched to 3.0% 235U. Then (The average energy released per fission event is 200 MeV.)

(a) the plant efficiency is 32% (b) fission events occur in the reactor core at the rate of 1.06 × 1020 fissions per second (c) the fuel supply will last after 580 days (d) the mass being lost in the reactor core at the rate of 4.5 kg per day. Integer Answer Type

24. The recoil speed (in m s–1) of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state is 25. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 970.6 Å. How many different wavelengths are possible in the resulting emission spectrum? 26. An α-particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de Broglie wavelengths are λα and λp respectively. λp , to the nearest integer, is The ratio λα Comprehension Type

The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition. 27. A diatomic molecule has moment of inertia I. By Bohr’s quantization condition, its rotational energy in the nth level (n = 0 is not allowed) is 1 ⎛ h2 ⎞ 1 ⎛ h2 ⎞ (a) 2 ⎜ 2 ⎟ (b) ⎜ 2 ⎟ n ⎝ 8π I ⎠ n ⎝ 8π I ⎠ ⎛ h2 ⎞ ⎛ h2 ⎞ (c) n ⎜ 2 ⎟ (d) n2 ⎜ 2 ⎟ ⎝ 8π I ⎠ ⎝ 8π I ⎠ 28. It is found that the excitation frequency from ground to the first excited state of rotation for the 4 CO molecule is close to × 1011 Hz. Then the π moment of inertia of CO molecule about its centre of mass is close to (Take h = 2π × 10–34 J s) (a) 2.76 × 10–46 kg m2 (b) 1.87 × 10–46 kg m2 (c) 4.67 × 10–47 kg m2 (d) 1.17 × 10–47 kg m2

Matrix Match Type

29. In performing photoelectric experiment to study photoelectric effect, intensity of radiation (I), frequency of radiation (υ), work function (φ0) of the photosensitive emitter, distance between emitter and collector (d) are changed or kept constant. Match the changes given in column I to their effects given in column II. Column I Column II (A) φ0 is decreased, (P) Saturation photoelectric keeping υ and I current increases constant (B) d is increased, (Q) Stopping potential (V0) keeping I, υ, φ0 increases constant (C) υ is increased, (R) Maximum kinetic keeping I, φ0, d energy (Kmax) of constant photoelectrons increases (D) I is increased, (S) Stopping potential keeping υ, φ0 remains the same and d constant A B C D (a) Q, R S Q, R P, S (b) Q, R, S S, P P, R S, Q (c) P S R, Q S, P (d) Q, R S P, S Q, R 30. Some laws/processes are given in column I. Match these with the physical phenomena given in column II. Column I Column II (A) Transition between (P) Characteristic X-rays two atomic energy levels (B) Electron emission (Q) Photoelectric effect from a material (C) Mosley’s law (R) Hydrogen spectrum (D) Change of photon (S) β-decay energy into kinetic energy of electrons A B C D (a) P, R P S Q, S (b) P, R Q, S P Q (c) Q, S P R, S Q (d) P S Q P, R ”” Keys are published in this issue. Search now! 

Check your score! If your score is > 90%

EXCELLENT WORK !

You are well prepared to take the challenge of final exam.

No. of questions attempted

……

90-75%

GOOD WORK !

You can score good in the final exam.

No. of questions correct

……

74-60%

SATISFACTORY !

You need to score more next time.

Marks scored in percentage

……

< 60%

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

PHYSICS FOR YOU | JANUARY ‘17

75

  

 P

hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / NEET / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.

 1. A particle moves in the x-y plane with velocity α 3 ^ ^ the magnitude of v = α i + βt j . At t = β tangential, normal and total accelerations respectively are 3 β β β (a) and β (b) , and β β, 2 2 2 2 (c)

2β,

2β and β (d) None of the above

2. A block of mass m slides down on a rough inclined plane of inclination θ with horizontal with zero intial velocity. The coefficient of friction between the block and the plane is μ with θ > tan–1(μ). The rate of work done by the force of friction at time t is (a) μmg2tsinθ (b) μmg2t(sinθ – μcosθ) (c) μmg2tcosθ(sinθ – μcosθ) (d) μmg2tcosθ 3. The minimum speed v with which a small ball should be pushed inside a smooth vertical circular tube of radius R from a height h such that it may reach the top of the tube is (a)

2 g (h + 2R)

μmg/α 2μmg/α 3μmg/α 6μmg/α

 

5 gR 2 (c) g (5R − 2h)

 

2 g (2R − h)



4. Two block A and B having mass ratio 2 : 1 are lying on a rough surface as shown in the figure. The friction coefficient between block B and ground is μ/2 and between blocks A and B is μ. Now a variable force F = αt is applied on A at t = 0. The mass of block B is m. Find the value of time till they move together. PHYSICS FOR YOU | JANUARY ‘17


 



5. An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is 2Mg 4Mg Mg Mg (a) (b) (c) (d) k 2k k k 6. A particle of mass m1 is fastened to one end of a massless string and another particle of mass m2 is fastened to the middle point of the same string. The other end of the string being fastened to a fixed point on a smooth horizontal table. The particles are then projected, so that the two particles and the string

Exam Alert!! Exam Date: 2nd April 2017 (Pen & Paper Based Examination) 8th & 9th April 2017 (Computer Based Examination) Last Date for applying online: 2nd January 2017 Last Date for fee payment: 3rd January 2017 Result Declaration: 27th April 2017

JEE Advanced Exam Date: 21st May 2017 Online registration begins: 28th April 2017 Online registration closes: 2nd May 2017 Result Declaration: 11th June 2017

By Akhil Tewari, Author Foundation of Physics for JEE Main & Advanced, Professor, IITians PACE, Mumbai.

76



JEE Main

(b)

(d)

(a) (b) (c) (d)

are always in the same straight line and describe horizontal circles. Then, the ratio of tensions in the two parts of the string is m1 m + m2 (a) (b) 1 (m1 + m2 ) m1 (c)

2m1 + m2 2m1

(d)

2m1 m1 + m2

The ratio (x) of magnitude of centripetal force and normal reaction on the particle at any point on the surface of bowl varies with respect to θ as 



(a)

7. In a circular motion of a particle, the tangential acceleration of the particle is given by at = 2t m s–2. The radius of the circle described is 4 m. The particle is initially at rest. Time after which total acceleration of the particle makes 45° with radial acceleration is (a) 1 s (b) 2 s (c) 3 s (d) 4 s 8. A small body of mass m can slide without friction along a trough bent which is in the form of a semi-circular arc of radius R. At what height h will the body be rest with respect to the trough, if the trough rotates with uniform angular velocity ω about a vertical axis.

(b) 







(c)

(d) 



”” SOLUTION OF DECEMBER 2016 CROSSWORD





(a) R 2g (c) R + 2 ω

2g (b) R − 2 ω g (d) R − 2 ω

9. If a particle starts from point A along the curved circular path as shown in the figure, with tangential acceleration a. Then acceleration of the particle at point B in magnitude is

Winner (December 2016) s Himani Bankoti, Kanpur

Solution Sender (November 2016)

(a) 2a 1 + π2

(b) a 1 + π2

(c) a π2 − 1

2 (d) aπ 1 + π

10. A small particle of mass  m is released from rest from point A inside a  smooth hemispherical bowl of radius R as shown in the figure.

s 6ANSH!GARWAL $ELHI

Solution Senders of Physics Musing 



s 2AJNI3INGH "IHAR

SET-41

1. Ravindar Negi, Jaipur  3URBHI!GARWAL 0UNE 3. Raman Reddy, Karnataka

PHYSICS FOR YOU | JANUARY ‘17

77

Exam Dates OFFLINE : 2nd April ONLINE : 8th & 9th April

1. In a vertical U-tube containing a liquid, the two arms are maintained at different  temperatures T1 and T2. The liquid columns in the two  arms have heights l1 and l2, respectively. The coefficient of volume expansion of the liquid is equal to l1 − l2 l1 − l2 (a) (b) l2T1 − l1T2 l1T1 − l2T2 (c)

l1 + l2 l2T1 + l1T2

(d)

 

l1 + l2 l1T1 + l2T2

2. A satellite is launched into a circular orbit of radius R around the Earth. A second satellite is launched into an orbit of radius 1.01 R. The period of second satellite is larger than that of the first one by approximately (a) 0.5 (b) 1.0% (c) 1.5% (d) 3.0% 3. An object is placed at f/2 away from first focus of a convex lens where f is the focal length of the lens. Its image is formed at a distance 3f/2 in a slab of refractive index 3/2, from the face of the slab facing the lens. Find the distance of this face of the slab from the second focus of the lens. (a) f/2 (b) 3f/2 (c) 2f (d) f 4. Pushing force making an angle θ with the horizontal is applied on a block of weight W placed on a horizontal table. If the angle of friction is φ, the magnitude of force required to move the body is equal to W cos φ cos(θ − φ) W tan φ (c) sin(θ − φ) (a)

78

W sin φ cos(θ + φ) W sin θ (d) tan(θ − φ) (b)

PHYSICS FOR YOU | JANUARY ‘17

5. Read the given statements and decide which is/are correct on the basis of kinetic theory of gases. (I) Energy of one molecule at absolute temperature is zero. (II) The rms speeds of different gases are same at same temperature. (III)For one gram of all ideal gases, kinetic energy is same at same temperature. (IV) For one mole of all ideal gases, kinetic energy is same at same temperature. (a) All are correct (b) I and IV are correct (c) IV is correct (d) None of these 6. A particle is moving on a straight line. Its acceleration ⎛ 100 ⎞ as a function of displacement is a = ⎜ 2 + 2 ⎟ m s −2. ⎝ s ⎠ If the velocity of particle is 5 ms–1 at s = 10 m, then the velocity of the particle at s = 25 m is (a) 9.8 m s–1 (b) 10 m s–1 –1 (c) 20 m s (d) 8 m s–1 7. The pressure and density of a diatomic gas 7⎞ ⎛ ⎜⎝ γ = ⎟⎠ change adiabatically from (P, d) to 5 d′ P′ (P′, d′). If = 32 then is d P 1 (a) (b) 32 (c) 128 (d) 256 128 8. Carbon monoxide is carried around a closed cyclic process abc, in which bc is an isothermal process, as shown in the given figure. The gas absorbs 7000 J of heat as its temperature is increased from 300 K to 1000 K in going from a to b. The quantity of heat ejected by the gas during the process ca is (a) 4200 J (b) 500 J (c) 9000 J (d) 9800 J

9. A magnetising field of 1600 A m–1 produces a magnetic flux of 2.4 × 10–5 Wb in an iron bar of cross-sectional area 0.2 cm2. The susceptibility of an iron bar is (a) 298 (b) 596 (c) 1192 (d) 1788 10. A projectile can have the same range R for two angles of projection. If t1 and t2 are the time of flights in the two cases, then the product of the two time of flights is proportional to 1 1 (a) R2 (b) 2 (c) (d) R R R 11. In the figure, the velocity v3 will be

above the surface of the earth is (here R is the radius of the earth) ⎛ n ⎞ ⎛ n ⎞ (b) ⎜ (a) ⎜ mgR mgR ⎟ ⎝ n + 1⎠ ⎝ n − 1 ⎟⎠ mgR (c) nmgR (d) n 16. A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is (a) 3.6 A (b) 2.8 A (c) 2.5 A (d) 5.0 A 17. Two solid spheres of same metal but of mass M and 8M fall simultaneously on a viscous liquid and their terminal velocities are v and nv, then value of n is (a) 16 (b) 8 (c) 4 (d) 2

(a) zero (c) 1 m s–1

(b) 4 m s–1 (d) 3 m s–1

12. A concave mirror of focal length 10 cm and a convex mirror of focal length 15 cm are placed facing each other 40 cm apart. A point object is placed between the mirrors, on their common axis and 15 cm from the concave mirror. Find the position and nature of the image produced by successive reflections, first at the concave mirror and then at the convex mirror. (a) 12 cm behind convex mirror, real (b) 9 cm behind convex mirror, real (c) 6 cm behind convex mirror, virtual (d) 3 cm behind convex mirror, virtual 13. The reading on the ammeter in the following figure will be (a) 0.8 A (b) 0.6 A (c) 0.4 A

(d) 0.2 A

14. A motorcycle starts from rest and accelerates along a straight path at 2 m s–2. At the starting point of the motorcycle, there is a stationary electric siren. How far has the motorcycle gone when the driver hears the frequency of the siren at 94% of its value when the motorcycle was at rest? (Speed of sound = 330 m s–1) (a) 49 m (b) 98 m (c) 147 m (d) 196 m 15. The change in the gravitational potential energy when a body of mass m is raised to a height nR

18. In Young's double slit experiment having slits of equal width, let β be the fringe width and I0 be the maximum intensity. At a distance x from the central bright fringe, the intensity will be 2πx ⎛x⎞ (a) I0 cos ⎜ ⎟ (b) I0 cos2 ⎝β⎠ β I πx πx (c) I0 cos2 (d) 0 cos2 β 4 β 19. When photons of energy 4.25 eV strike the surface of a metal A the ejected photoelectrons have a maximum kinetic energy EA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is EB = (EA – 1.50) eV. If the de Broglie wavelength of these photoelectrons is λB = 2λA, then (a) the work function of A is 2.25 eV (b) the work function of B is 4.20 eV (c) EA = 2.0 eV (d) All of these 20. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and the Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to m(YA + kL) m (a) 2π (b) 2π k YAk (c) 2π mYA kL

(d) 2π mL YA PHYSICS FOR YOU | JANUARY ‘17

79

21. In the circuit shown, which of the following statements is not correct?

(a) When S is open, charge on C1 is 36 μC. (b) When S is open, charge on C2 is 36 μC. (c) When S is closed, charges on both C1 and C2 change. (d) When S is closed, the charges on both C1 and C2 do not change. 22. The fundamental frequency of an open organ pipe is 300 Hz. The first overtone of this pipe has same frequency as first overtone of a closed organ pipe. If speed of sound is 330 m s–1, then the length of closed organ pipe is approximately (a) 41 cm (b) 37 cm (c) 31 cm (d) 80 cm 23. Two wires AO and OC carry  equal current i as shown in   figure. One end of both the wire  A extends to infinity. Angle AOC   is α. The magnitude of magnetic field at a point P on the bisector of these two wires at a distance r from point O is : μ0 i μ i α α cot ⎜⎛ ⎟⎞ (b) 0 cot ⎛⎜ ⎟⎞ ⎠ ⎝ ⎝ 2π r 4π r 2 2⎠ α⎞ ⎛ 1 + cos μ0 i ⎝ 2 ⎠ (d) μ0 i sin ⎛ α ⎞ (c) ⎜⎝ ⎟⎠ 2π r sin ⎛ α ⎞ 4π r 2 ⎝2⎠ 24 . A point mass oscillates along the X-axis according to the law x = x0 cos (ωt – π/4). If the acceleration of the particle is written as a = A (cos ωt + δ), then: (a) A = x0ω2, δ = 3π/4 (b) A = x0, δ = –π/4 (c) A = x0ω2, δ = π (d) A = x0ω2, δ = –π/4 (a)

25. A radioactive material of half-life T was produced in a nuclear reactor at different instants, the quantity produced second time was twice of that produced first time. If now their present activities are A1 and A2 respectively then their age difference equals :

80

(a)

2A T ln 1 A2 ln 2

A (b) T ln 1 A2

(c)

A T ln 2 ln 2 2A1

(d) T ln

PHYSICS FOR YOU | JANUARY ‘17

A2 2A1

26. A block of mass m is on an inclined  plane of angle θ. The coefficient of friction between the block and the plane is μ and tan θ > μ . The block is held stationary by applying a force F parallel to the plane. The direction of force pointing up the plane is taken to be positive. As F is varied from F1 = mg(sinθ – μcosθ) to F2 = mg(sinθ + μ cosθ), the frictional force f versus F graph will look like

(a)

(c)

 

 



(b)



(d)













27. Gravitational acceleration on the surface of a planet 6 g, where g is the gravitational acceleration on is 11 the surface of the earth. The average mass density of 2 the planet is times that of the earth. If the escape 3 speed on the surface of the earth is taken to be 11 km s–1, the escape speed on the surface of the planet will be (a) 1.5 km s–1 (b) 2 km s–1 (c) 3.0 km s–1 (d) 2.5 km s–1 28. If the time period (T) of vibration of a liquid drop depends on density (ρ) of the liquid, radius (r) of the drop and surface tension (S), then the expression of T is (where k is a dimensionless constant.) (a) T = k (c) T = k

ρr 3 S ρr 3 1/2

(b) T = k

ρ1/2r 3 S

(d) T = k

ρr S

S 29. Two identical metal plates are given positive charges Q1 and Q2 ( l and l2 < l. l1 l2 Also, l = = 1 + γT1 1 + γT2

T ⎛r ⎞ 2. (c): As 2 = ⎜ 2 ⎟ T1 ⎝ r1 ⎠ 3/2

= (1.01) ⇒

3/2

Here, μ = tan φ = 



⇒ l1 + γl1T2 = l2 + γl2T1 ⇒ γ =



3/2

 



l1 − l2 l2T1 − l1T2

⎛ 1.01 R ⎞ =⎜ ⎝ R ⎟⎠

1 ⎞ = ⎜⎛1 + ⎝ 100 ⎟⎠

3/2

3 1 ⎞ . = ⎛⎜1 + × ⎝ 2 100 ⎟⎠

T2 ⎛ 3 1 ⎞ = ⎜1 + × ...(i) ⎟ T1 ⎝ 2 100 ⎠ Percentage increase in period (from eqn (i) ⎛T −T ⎞ ⎛T ⎞ = ⎜ 2 1 ⎟ × 100 = ⎜ 2 − 1⎟ × 100 = 1.5% T T ⎝ 1 ⎠ ⎝ 1 ⎠

3f f 3. (d) : u = − ⎛⎜ f + ⎟⎞ = − ⎝ 2 2⎠ 3 f. 2 Distance of image formed due to convex lens from face 3⎞ (3/2) f ⎛ of the slab = =f ⎜⎝∵ μ = ⎟⎠ 2 μ Distance of final image from one face of the slab is

Let F2S = x v = 2f + x 1 1 1 From − = v u f 1 1 1 − = (x + 2 f ) (−3 f /2) f ⇒ x = f.

4. (b) : The various forces acting on the block are shown in figure.

f R

or f = R tanφ ...(i) The condition for the block just to move is Fcosθ = f = R tanφ ...(ii) and Fsinθ + W = R or R = W + Fsinθ ...(iii) From equations (ii) and (iii), we get Fcosθ = (W + Fsinθ) tanφ = Wtanφ + Fsinθ tanφ sin φ W sin φ = or F cos θ − F sin θ cos φ cos φ or F(cosθ cosφ – sinθ sinφ) = Wsinφ or Fcos(θ + φ) = Wsinφ W sin φ or F = . cos(θ + φ) 5. (c): If the gas is not ideal then its molecule will possess potential energy. Hence statement (I) is wrong. At same temperature rms speed of different gases 1 ⎞ ⎛ depends on its molecular weight ⎜ vrms ∝ ⎟ . Hence ⎝ M⎠ statement (II) is also wrong. Kinetic energy of one gram gas depends on the 1 molecular weight ⎛⎜ Egm ∝ ⎟⎞ . Hence statement (III) ⎝ M⎠ is also wrong. But kinetic energy of one mole of ideal gas does not 3 depend on the molecular weight ⎛⎜ E = RT ⎟⎞ . Hence ⎠ ⎝ 2 (IV) is correct. 100 6. (a) : ∵ a = 2 + 2 s dv 100 or v = 2 + 2 ds s v

25

25 100 ds 2

or

∫5 v dv = ∫10 2 ds + ∫10

or

⎡ −2+1 ⎤ 1 2v 25 [v ]5 = 2[s]10 + 100 ⎢ s ⎥ 2 ⎣ −2 + 1 ⎦10

or

v 2 25 5 − 2⎤ − = 30 + 100 ⎡⎢ 2 2 ⎣ 50 ⎥⎦

s

25

PHYSICS FOR YOU | JANUARY ‘17

81

100 × 3 = 36 50 or v2 = 72 + 25 ∴ v2 = 97

Magnetic susceptibility, χ =

= 30 +

v=

=

97 = 9.8 m s–1

(7.5 × 10−4 Wb A−1 m−1) (4π × 10−7 Wb A−1 m−1)

∴ P1V1γ = P2V2γ γ γ P2 ⎛ V1 ⎞ ⎛ d2 ⎞ =⎜ ⎟ =⎜ ⎟ P1 ⎝ V2 ⎠ ⎝ d1 ⎠ 7 /5

...(i)

t 1t 2 =

φ = = 1.2 Wb m−2 A 0.2 × 10− 4 m2 B Magnetic permeability, μ = H −2 1.2 Wb m = = 7.5 × 10−4 Wb A−1 m−1 1600 A m−1 ∴ B=

82

PHYSICS FOR YOU | JANUARY ‘17

...(i) ...(ii)

t 1t 2 = t 1t 2 =

4u2 sin θ cos θ g2 2u2 sin 2θ g2

=

2 2 u sin 2θ ⋅ g g

⎡ u2 sin 2θ ⎤ ⎢∵ R = ⎥ g ⎣ ⎦

2R g

t 1t 2 ∝ R

11. (c): According to steady flow, A 1v 1 = A 2v 2 + A 3v 3 or A3v3 = A1v1 – A2v2 1 or v3 = [A v − A2v2] A3 1 1 1 [0.2 × 4 − 0.2 × 2] = 1 m s −1 = 0. 4  = –10  = +15 12. (c) :  

Negative sign shows that heat is ejected.

2.4 × 10−5 Wb



From eqs. (i) and (ii), we get

= –μ(CV + R)700 ...(ii) For carbon monoxide: 7 γ= 5 R R 5R CV = ...(iii) = = γ − 1 7 −1 2 5 Hence, from eqn. (i) 20 5R 7000 = μ × 700 or μR = = 4 5 2 −7 ⎞ ⎛ 5R μR × 700 = –9800 J (ΔQ)ca = −μ ⎜⎝ + R ⎟⎠ 700 = 2 2 9. (b) : Here, magnetising field, H = 1600 A m–1 Magnetic flux, φ = 2.4 × 10–5 Wb Area, A = 0.2 cm2 = 0.2 × 10–4 m2

− 1 ≈ 596

10. (d) : A projectile can have  same range, if angles of  projection are complementary  of each other θ and (90° – θ). (90°– Q) Q Thus, in both cases, the time  of flights are 2u sin θ t1 = g 2u sin (90° − θ) 2u cos θ = and t2 = g g

7 7. (c): Here, γ = , P1 = P, d1 = d; P2 = P ′, d2 = d ′ 5 For an adiabatic process PV γ = constant

P ′ ⎛ d′ ⎞ = ⎜ ⎟ = (32)7/5 P ⎝d⎠ P′ = (25)7/5 = 27 = 128 P 8. (d) : (ΔQ)ab = +7000 = μCv(1000 – 300) For the process ca: Ta = 300 K, Tc = Tb = 1000 K (ΔQ)ca = μCp(300 – 1000) = –μCp × 700

μ −1 μ0

40 cm

2



1

15 cm a

a

According to given problem, for concave mirror, u = –15 cm and f = –10 cm 1 1 1 , i.e., v = –30 cm + = v 15 −10 i.e., concave mirror will form real, inverted, and enlarged image I 1 of object O at a distance of

30 cm from it, i.e., at a distance 40 – 30 = 10 cm from the convex mirror. For convex mirror, the image I1 will act as an object and so for it u = –10 cm and f = +15 cm. 1 1 1 i.e., v = +6 cm + = v −10 15 So, final image I2 is formed at a distance of 6 cm behind the convex mirror and is virtual as shown in figure. 13. (c): In the above circuit, the resistances 6 Ω, 3 Ω and 2 Ω are connected in parallel. Their effective resistance will be 1 1 1 1 1+ 2 + 3 6 = + + = = 6 6 Rp 6 3 2

⎛ v − v0 ⎞ 14. (b) : As υ′ = ⎜ υ, ⎝ v ⎟⎠ (source at rest, observer is moving away from the source) ⎛ υ′ ⎞ ⎜⎝ ⎟⎠ v = v − v0 υ or

330(0.94 – 1) = –v0

whence, v0 = 19.8 m s–1 and

v 2 − u2 (19.8)2 = m = 98 m s= 0 2a 2×2

(440 V) (2 A) (440 V) (2 A) (100) VI Ip = s s = = ⎛ 80 ⎞ (80) (220 V) ηV p ⎟⎠ (220 V) ⎜⎝ 100 =5A 17. (c) 18. (c) h 19. (d) : de Broglie wavelength, λ = 2mE where E is the kinetic energy h h λ EA and λ B = ∴ B= ∴ λA = λA EB 2mE A 2mEB or

or Rp = 1 Ω The equivalent circuit is as shown in the figure. The equivalent resistance of the circuit is Req = 1 Ω + 4 Ω = 5 Ω 2V Current in the circuit, I = = 0. 4 A 5Ω Hence, the reading of the ammeter is 0.4 A.

υ′ or v ⎛⎜ − 1⎟⎞ = − v0 ⎝υ ⎠

Change in gravitational potential energy is mgR ΔU = Uh – Us = − − (−mgR) (1 + n) mgR 1 ⎞ ⎛ ⎛ n ⎞ =− − (−mgR) = mgR ⎜1 − = mgR ⎜ ⎟ (1 + n) ⎝ 1+n⎠ ⎝ 1 + n ⎟⎠ 16. (d) : Here, Input voltage, Vp = 220 V Output voltage, Vs = 440 V, Input current, Ip = ? Output current, Is = 2 A Efficiency of the transformer, η = 80% Output power Efficiency of the transformer, η = Input power VI η= s s VpI p

(as u = 0)

15. (a) : Gravitational potential energy of mass m at any point at a distance r from the centre of earth is GMm U =− r At the surface of earth r = R, GM ⎞ ⎛ GMm ∴ Us = − = − mgR ⎜⎝∵ g = 2 ⎟⎠ R R At the height, h = nR from the surface of earth r = R + h = R + nR = R(1 + n) GMm mgR ∴ Uh = − =− R(1 + n) (1 + n)

EA EA or 4 = EA − 1.5 EA − 1.5 6 4E A − 6 = EA or E A = = 2 eV 3 ∴ EB = (EA – 1.5) eV = 2 eV – 1.5 eV = 0.5 eV 2=

According to Einstein’s photoelectric equation hυ = φ0 + KEmax where the symbols have their usual meaning or φ0 = hυ – KEmax φA = 4.25 eV – 2.00 eV = 2.25 eV φB = 4.70 eV – 0.50 eV = 4.20 eV 20. (b) : For two springs in series, the equivalent force kk YA constant is keq = 1 2 . Here, k1 = k and k2 = k1 + k2 L ∴

∴ 21. (d) 25. (c) 29. (d)

kYA kYA keq = L = YA kL + YA k+ L m(YA + kL) m T = 2π = 2π keq YAk 22. 26. 30.

(a) (a) (c)

23. (c) 27. (c)

24. (a) 28. (a)

PHYSICS FOR YOU | JANUARY ‘17

”” 83

PRESS RELEASE

VIT VARSITY, VELLORE GETS 4-STAR RATING FOR OVERALL PERFORMANCE

V

IT University, Vellore has become the first Indian university to get international recognition in terms of getting a 4-star rating for overall performance in a recently concluded audit of Quaquarelli Symonds (QS), UK. A certificate to this effect was handed over to the Chancellor of University, Dr G Viswanathan here by QS South Asia Director, Ashwani Fernandes in presence of Minister of State for Finance Santosh Kumar Gangwar. Speaking on the occasion, Mr Gangwar said that in deference to the wishes of Prime Minister Narendra Modi, the VIT university has gone ahead in promoting the meaningful education with wisdom and skill development. VIT University, India became the first Indian university to receive a 4 Star rating for overall performance in a recently concluded audit of Quacquarelli Symonds (QS), UK. They also became the first Indian university to receive 5 stars category rating in 5 different categories. The categories are Teaching, Innovation, Facilities, Employability and Inclusiveness. VIT is the only university in India now with 4 STAR overall rating of QS. Quacquarelli Symonds (QS) from the UK, founded in 1990, is the world’s leading network for top careers and education.

Selection Criteria The core criteria for QS Star evaluation are – Teaching, Employability, Research and Internationalization. The other criteria are – Innovation, culture, access, engagement, facilities, online learning, Discipline ranking and Accreditation. A university needs to score 550 points out of 1000 to get 4 star rating. The prerequisites of 4 star rating is at least 75 academic referees or must have at least 2 citations per faculty member, another prerequisite is the university should have at least 1% international students. However, these prerequisite are not applicable for 3 star rating. These have made QS 4 star rating very challenging and highly competitive. What this rating means for VIT University? This 4 star rating is valid for 3 years. In the meantime VIT has already started its work towards 5 star rating for subsequent evaluation by increasing number of foreign students and also the number of faculty citations. VIT has increased facilities in research and approaching the Government of India for more research funds. VIT University is also working to be one among the Top 200 universities of the world by next 5 years. This achievement of VIT University is a significant milestone towards our 84

PHYSICS FOR YOU | JANUARY ‘17

Honourable Prime Minister’s vision to upgrade 20 Indian Universities among the Top 100 universities of the World. About VIT University Chancellor of VIT, Dr.G.Viswanathan expressed his joy at a press conference held in New Delhi on November 16th saying that “in 1984, we began this institute as Vellore Engineering college with mere 180 students. With dedication and sincere efforts of faculties and students, we emerged as an autonomous university in the year of 2004. Apart from engineering, we offer various professional courses like catering technology, fashion designing, law and 24 different studies. In addition to this, we also provide post graduate courses like MCA, MBA, M.TECH and research courses like M.Phil and P.hd. We are proud to say that among 32000 students in our university, one third of them are female candidates. Gaining goodwill at Vellore, we thought of expanding our institution at Chennai and thus we established VIT campus at Chennai. Following that, we recently established two more VIT campuses, one at Bhopal district in Madhya Pradesh and other at Amravati district in Andhra Pradesh. With 1682 faculties and 1100 non teaching staffs, we offer international standard education for which we have been accredited A+ by NAAC and have been recognized by ABET and IET organisations from US and UK respectively. Last year almost 2.06 lakhs of students appeared for VIT’s Engineering Entrance Exam (VITEEE) which is a national level achievement and took a prominent place in Limca Book of Records and VIT holds this position for 3 years continuously. During the press meet Vice Presidents Mr. Sankar Viswnathan, Dr. Sekar Viswanathan, Mr. G.V. Selvam and Vice Chancellor Dr. Anand A Samuel were present. 

3. (a, c) : Let L1 and L3 are lengths of first and third air column respectively. Hence,

SOLUTION SET-41

1. (b) : Since, dV = − E . dr ⇒ ΔV = −

λ 2 πε0

2r

∫ r dr = 2πε0 [

r ]r = 2r

r

Now, ΔK + ΔU = 0 ΔK = –ΔU = – qΔV 1 2 qλ ln 2 or v = ⇒ mv = 2 2 πε0

−λ 2 πε0

2

qλ ln 2 πmε0

2. (a, d) : Electric field at any point P on the axis E=

1 4 πε0

⎡ ⎤ −Q 8Q + x ⎢ 2 2 3/ 2 2 2 3/ 2 ⎥ (9a + x ) ⎦ ⎣ (a + x )

For small x, E= =

8 ⎞ Q ⎛ −1 + ⎟x ⎜ 3 4 πε0 ⎝ a 27a3 ⎠ − Q 19 x 4 πε0 27a3

⇒ F ∝ – x, so motion of the particle is S.H.M. 19qQ F = mx 108 πa3ε0m

Time period, T =

5λ = L3 + e 4 By substracting eqn. (i) from eqn. (ii) λ = L3 – L1 = (82.8 – 15.6) cm = 67.2 cm = 0.672 m v = υλ = 500 Hz × 0.672 m = 336 m s–1 λ L2 = L3 − = 82.8 − 33.6 = 49.2 cm 2

...(ii)

4. (c) : Let a is acceleration and α is angular acceleration of sphere, then a = Rα. Using τnet = I α 2 4MgR cos 60° − fR = MR 2α 5 2 ...(i) ⇒ 2Mg − f = Ma 5 As f = Ma, we get 10 10 ⇒ a = g ⇒ f = Mg 7 7 For pure rolling,

Negative sign shows the net electric field is towards origin. So the force on the particle −19qQx F = qE = 108 πa3ε0

ω2 =

...(i)

and

−λ

1

λ = L1 + e 4

108 πa3ε0m 2π = 2π 19qQ ω = 12

π a mε 0 19qQ

Electric field will be zero for x given by, 5 a (9a2 + x2)3/2 = 8(a2 + x2)3/2 ⇒ x = 3 5 a , E will be in negative x direction. For x < 3 ∴ Both options (a) and (d) are correct.

f ≤ μ (Mg + 4Mg) 10 Mg ≤ 5μMg ⇒ 7 2 ⇒ μmin = 7 5. (a, c) : Just after cutting the string torque on the rod, 2

τ = I α ; Mg L = ML α 2 3 3g 3g α= ∴ aB = 2 2L Relative acceleration between block and end B of 3g g −g= rod = 2 2 When rod becomes vertical, ω will be given by, L 1 ML2 2 L 1 2 I ω = Mg ; ω = Mg 2 2 2 3 2 2 3g ω = L Tension in the rod at its mid point, M 3L M 3g 3L 9 T = ω2 = × × = Mg 2 4 2 L 4 8 PHYSICS FOR YOU | JANUARY ‘17

85

6. (a, c) : Total induced charge on the sphere is zero. kQ = Net potential at B and D Net potential at C,VC = 2R Potential at B due to induced charge, kQ kQ VB = − 2R 5R ⎛1 1 ⎞ ⎟ ⎜⎝ − 2 5⎠ Potential at D due to induced charge, kQ kQ −kQ VD = − = 2R R 2R 1 − 5 V 2 = Now, D = VB 5 −2 2− 5 2 5 kQ kQ and VB − VD = − R 5R If the sphere is earthed then potential at C is zero. −Q kQ kQ′ + = 0 ⇒ Q′ = 2R 2 R 7. (d) : Applying conservation of momentum, mv1 + 2mvc = mv v1 + 2vc = v ...(i) As collision is elastic so, ωl cos 45° − v1 vc + 2 e =1= v−0 ωl ...(ii) v = vc + − v1 2 =

⎡ v + v0 ⎛ v − v0 ⎞ ⎤ −⎜ υ 8. (5) : Beat frequency = ⎢ ⎝ v ⎟⎠ ⎥⎦ ⎣ v 2v 4 × 40 2×2 = 0υ= × 400 = =5 320 v 32 9. (6) :

kQ R

L and I show the position of lens and image respectively. From thin lens formula, 1 1 1 1 1 1 ...(i) − = ⇒ + = y (− x ) 9 y x 9 −1 1 1 1 1 1 − = ⇒ + = ...(ii) (− y ) (−(24 − x )) 9 y 24 − x 9 1 1 2 Using (i) and (ii), we get, + = x 24 − x 9 12 × 9 = 24x – x2 or 108 = 24x – x2 ⇒ x2 – 24x + 108 = 0 ⇒ x2 –18x – 6x + 108 = 0 ⇒ x(x – 18) –6 (x – 18) = 0 ⇒ (x – 6) (x – 18) = 0 ∴ x = 6, 18 Convex lens should be placed at 6 cm from the S1. 10. (6) : Charge in region R to r is r

Q ′ = ∫ 4πr 2 R

(r 2 − R2 ) 2 = 2πα(r2 – R2)

= 4πα

2παR 2 ⎤ 1 ⎡Q = + 2 − πα ⎢ ⎥ 4πε0r 2 4πε0 ⎣ r 2 r2 ⎦ Since, E is constant α 1 Q 2παR 2 and E = (2 πα) = ∴ 2= 2 4 πε0 2ε0 r r α 3α Comparing, E = and E = 2ε 0 mε0 We get, m = 6 E=

Applying conservation of angular momentum about C, 2 l l ⎛ l ⎞ ω = mv mv1 + 2m ⎜ ⎟ ⎝ 2⎠ 2 2 v1 + 2ωl = v ...(iii) Solving (i), (ii) and (iii) −v v1 = 7 v The speed of the striking mass after collision is . 7 86

PHYSICS FOR YOU | JANUARY ‘17

α dr r

(Q + Q′)

MPP-7 CLASS XII 1. 6. 11. 16. 21. 26.

(b) (a) (a) (a) (a, c) (3)

2. 7. 12. 17. 22. 27.

(d) (d) (c) (c) (a, d) (d)

3. 8. 13. 18. 23. 28.

””

ANSWER KEY (c) 4. (b) 9. (b) 14. (a) 19. (a, b, c) 24. (b) 29.

(d) (a) (c) (c) (4) (a)

5. 10. 15. 20. 25. 30.

(c) (d) (c) (a, b) (6) (b)

Y U ASK

WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. The best questions and their solutions will be printed in this column each month.

Q1. How does the gas lighter produce an electrical spark? – Harish Reddy Thalla, Hyderabad

Ans. A gas-stove lighter uses a phenomenon called piezo-electric effect to generate an electric spark which ignites the combustible gas from the stove burner. Piezo-electric effect is the ability of certain materials to generate an electric charge in response to applied mechanical stress.

Q2. How should a parachutist land so as to lessen the danger of injury? – Gauri Shelke, Pune

Ans. A parachutist is trained to collapse and roll by first making contact with the balls of the feet, then bending the knees and turning so as to come down on the side of the leg and finally the back side of the chest. The procedure has two advantages : it prolongs the collision (and so it reduces the force on the parachutist) and it spreads the force of the collision out over a large area. If the parachutist were to land standing up, the compression on the bones in the ankles would likely rupture the bones. Q3. When a spaceship is sent to the Moon, why is its path in the form of a distorted figure eight instead of an ellipse that encompasses Earth and the Moon? – Preet Gurung, Shilong

Ans. The figure eight path requires less energy by the spaceship because for much of the trip it stays close to the line between the centers of Earth and the Moon on the spaceship. Since along that line the gravitational pulls from the Earth and the Moon compete, the net force on the spaceship is smaller than if the spaceship is in an elliptical orbit. So, less energy is required to overcome the net force. Q4. What is the difference between neutral and ground in an electrical connection? – Mr. Sandeep Kumar Singh, Ludhiana

The gas lighter consists of a piezo-electric crystal over which a spring loaded hammer is placed. The hammer and spring set up is attached to a button. When this button is pressed, the hammer is moved away from the piezo-electric crystal. The spring releases the hammer which hits the piezo-electric crystal. Due to piezo-electric effect, a high voltage is generated in the range of 700 V- 800 V. The lighter is wired in such way that this voltage is applied in a small air gap between two metallic points. Due to high voltage generated, the air is ionized and acts as a path for the discharge which creates a spark, which when exposed to the combustible gas (LPG and air mix) it produces flame. In home gas lighters, piezo-electric ceramics like lead zirconate titanate (PZT) are used due to low cost and high sensitivity.

Ans. As its name suggests, the ground wire is connected to the ground; the neutral is not. The ground wire ensures that in case of a leakage, an appliance does not acquire a voltage that might cause injury or malfunction. It appears that there is no need or separate identification of the live and neutral terminals when you are dealing with a single phase of alternating current. You just need an incoming wire and an outgoing wire to complete the circuit. But many homes and establishments are served by three phases. In that case you would need three separate pairs of electric cables, in other words a total of six cables. On the other hand, if you make one of the cables as a common neutral you would need only four cables, resulting in great saving. You would have noticed therefore that in your switchboard serving a three-phase supply, there are four cables coming in and one of them, called the neutral, is connected to all your appliances and light fixtures. ”” PHYSICS FOR YOU | JANUARY ‘17

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PHYSICS FOR YOU 2016 MONTHS SOLVED PAPERS (2016)

PRACTICE PAPERS (2016-17)

BRAIN MAP

XI

MARCH

FEBRUARY

JANUARY

PMT 2016, ACE Your Way CBSE XII 2016, CBSE XI 2016

APRIL

CBSE Board 2016

DECEMBER

NOVEMBER

OCTOBER

SEPTEMBER

AUGUST

JULY

JUNE

MAY

JEE Main 2016

88

CORE JEE ACCELERATED CONCEPT LEARNING SERIES

XII Dual Nature of Radiation and Matter

XI Superposition of Waves

JEE Main 2016, AIPMT 2016, ACE Your Way CBSE XII 2016

Atoms Nuclei

AIPMT 2016, JEE Main 2016, JEE Advanced 2016, AIPMT Model Test Paper 2016 AIPMT 2016, JEE Advanced 2016, AIPMT Model Test Paper 2016, BITSAT 2016

Semiconductor Torque Electronics

Review I

Diffraction

Review II

Electric Flux and Gauss’s Law

Units and Measurements

Electric Charges and Fields

Gravitation

Units and Measurements

Motion in a Straight Line

Electrostatic Potential and Capacitance

Electro

Kinematics

NEET Phase I 2016, PMT 2016, Karnataka CET 2016, NEET Phase II 2016, ACE Your Way Kerala PET 2016 CBSE XII Series-1

ACE Your Way CBSE XII Series-3, CBSE XI Series-2

NEET Phase II 2016 ACE Your Way CBSE XII Series-4, CBSE XI Series-3

and Capacitive Circuits

Communication Alternating Systems Current

JEE Advanced 2016, PMT 2016, BITSAT 2016, AIIMS 2016, JIPMER 2016

JEE Advanced 2016 NEET Phase II,2016, ACE Your Way CBSE XII Series-2, CBSE XI Series - 1

AT A GLANCE

Motion in a Plane

Current Electricity

ing of Conductors Relative Motion

Laws of Motion, Work Energy and Power

OTHERS NEET |JEE ESSENTIALS

XII Electromagnetic Physics Musing Problem Waves | Optics Set 30, Solution Set 29, JEE Workouts XI, Exam Prep 2016, Live Physics, You Ask We Answer, Crossword, Thought Provoking Problems Physics Musing Problem Set 31, Dual Nature Solution Set 30, JEE Workouts of Radiation XII, Exam Prep 2016, You Ask and Matter| We Answer, Live Physics, At a Atoms and Glance 2015, Crossword Nuclei Physics Musing Problem Set 32, Electronic Solution Set 31, Exam Prep 2016, Devices| Communication Live Physics, You Ask We Answer, Crossword Systems Physics Musing Problem Set 33, Solution Set 32, AIIMS Special : Assertion and Reason, Thought Provoking Problems, Exam Prep 2016, Olympiad Problems, Live Physics, You Ask We Answer, Crossword Physics Musing Problem Set 34, Solution Set 33, Exam Prep 2016, Olympiad Problems, Live Physics, You Ask We Answer, Crossword Physics Musing Problem Set 35, Solution Set 34, Exam Prep 2016, Olympiad Problems, Live Physics, You Ask We Answer, Crossword Electrostatics Physics Musing Problem Set 36, Solution Set 35, Cracking the JEE Advanced Exam, You Ask We Answer,Live Physics, Crossword, MPP-1 (XI-XII) Current Physics Musing Problem Set Electricity 37, Solution Set 36, You Ask We Answer, Live Physics, Crossword, Exam Prep, AIIMS Topper Interview, MPP-2 (XI - XII) Musing Problem Magnetic Effect Physics of Current and Set 38, Solution Set 37, Crossword, MPP-3 (XI - XII) Magnetism Electromagnetic Induction, Alternating Current and Electromagnetic Waves Optics

ACE Your Way CBSE XII Series-5, CBSE XI Series - 4

Induced Moving Laws of Electric Charges Motion Electromagnetic and Magnetism Field Wave

System of Particles and Rotational Motion

ACE Your Way CBSE XII Series-6, CBSE XI Series - 5

Work, Energy and Power

Magnetism and Matter

Gravitation

ACE Your Way CBSE XII Series-7, CBSE XI Series - 6

System of Particles and Rotational Motion

Electromagnetic Induction

Dual Nature of Mechanical Properties of Solids Radiation and Matter, and Fluids Atoms and Nuclei

PHYSICS FOR YOU | JANUARY ‘17

Physics Musing Problem Set 39, Solution Set 38, You Ask We Answer, Crossword, MPP-4 (XI - XII), Exam Prep Physics Musing Problem Set 40, Solution Set 39, You Ask We Answer, Crossword, MPP-5 (XI XII), Exam Prep, JEE Workouts (XI - XII) Physics Musing Problem Set 41, Solution Set 40, Key Concept, Olympiad Problems, You Ask We Answer, Crossword, MPP-6 (XI - XII), JEE Workouts (XI - XII)

Readers can send their responses at [email protected] or post us with complete address by 25th of every month to win exciting prizes. Winners' name with their valuable feedback will be published in next issue. ACROSS 3. An imaginary line connecting points on the earth’s surface where the magnetic declination is the same. [8] 5. A thermionic valve having five grids between the cathode and anode. [7] 7. The point in a solar orbit that is nearest to the sun. [10] 10. A device used for measuring the velocity of a fluid. [10] 11. The derived SI unit of absorbed dose of ionizing radiation and of specific energy. [4] 12. Prefix denoting 10–15. [5] 15. A network of fine lines dots, cross wires in the focal plane of the eyepiece of an optical instrument. [9] 16. A device, use to receive visual and aural information using broadcast mode of communication. [10] 19. An interstellar cloud of dust, hydrogen, helium and other ionised gases that can be observed either as a luminous patch or as a clark region against a brighten background. [6] 22. A dimensionless unit used for comparing two currents entering and leaving a transmission line, almost exclusively in telecommunication engineering. [5] 23. Frequencies of integral multiple of fundamental or lowest tone. [8] 24. Alternate rise and fall arising by superimposing two sound waves of slightly different frequencies travelling in same direction. [5] 25. A p-n junction between two regions of opposite polarity within a semiconductor. [12] 26. The most common method is to measure the ionization caused by the radiation, as in ionization chamber. [9] 27. Detection system which works on the principle of radar like technique using continuous wave laser beams for remote sensing. [5] 28. The outermost layer of the Earth’s atomosphere. (9). 29. The word used for the progressive decrease in the amplitude of oscillation. [7] 30. The amount of potential energy stored in an elastic substance by means of elastic deformation. [10] 31. Patterns demonstrating the presence of domains in ferromagnetic crystal. [6, 8] DOWN 1. A device that converts a signal in the form of one type of energy into a signal of another form. [10] 2. The apparent motion of celestial bodies across the sky from east to west. [7, 6]

4.

6. 8. 9. 12. 13. 14.

17. 18. 20. 21.

Component, usually used in microwave frequencies, that reverses the phase of signals transmitted in one direction without affecting the phase of signals transmitted in opposite direction. [7] An instrument designed to measure brief flow of charge. [9, 12] An instrument for electron amplifying and directing sound. [9] A high-powered electron vacuum tube, that works as self-excited microwave oscillator. [9] A nuclide capable of undergoing fission by interaction with slow neutrons. [7] A substance that can sustain an electric field and acts as an insulator. [10] A measure of the ability of a ferromagnetic material to with stand an external magnetic field without becoming demagnetised. [8, 5] A device for measuring a fluid pressure. [9] An aura of plasma that surrounds the Sun and other stars. [6] A technique enabling a system to bring itself into some desired state. [9] The fraction of incident light diffusely reflected from a surface. [6] ”” PHYSICS FOR YOU | JANUARY ‘17

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PHYSICS FOR YOU | JANUARY ‘17