Physics for You - January 2016

Volume 24 Managing Editor Mahabir Singh Editor Anil Ahlawat (BE, MBA)

No. 1

January 2016

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CONTENTS

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Physics Musing Problem Set 30

8

Core Concept

12

Thought Provoking Problems

22

PMT Practice Paper

25

JEE Accelerated Learning Series Brain Map

31 46

Ace Your Way CBSE XI

57

JEE Workouts

64

Ace Your Way CBSE XII

68

Exam Prep 2016

75

Physics Musing Solution Set 29

81

Live Physics You Ask We Answer

83 84

Crossword

85

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Physics for you | january ‘16

7

P

PHYSICS

MUSING

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 / 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.

30 single oPtion correct tyPe

1. Two batteries of emf e1 and e2 having internal resistance r1 and r2 respectively are connected in series to an external resistance R. Both the batteries are getting discharged. The above described combination of these two batteries has to produce a weaker current than when any one of the batteries is connected to the same resistor. For this requirement to be fulfilled e2 r r (a) e must not lie between 2 and 1 r2 + R 1 r1 + R r r +R e (b) 2 must not lie between 2 and 2 r1 + R r1 e1 r e r (c) 2 must lie between 2 and 1 r2 + R e1 r1 + R e2 r +R r (d) must lie between 2 and 2 . e1 r1 r1 + R 2. A particle of charge per unit mass a is released from  origin with velocity v = v0i in a magnetic field  3 v0 B = − B0  k for x ≤ 2 B0a  3 v0 and B = 0 for x > 2 B0a π   The x-coordinate of the particle at time t  >  3B0a  would be

8

(a)

π  3 v0 3  + v t− 2 B0a 2 0  B0a 

(b)

π  3 v0  +v t − 2 B0a 0  3B0a 

(c)

3 v0 v0  π  + t − 2 B0a 2  3B0a 

(d)

3 v0 v0t + 2 B0a 2 Physics for you | January ‘16

3. An ideal choke takes a current of 10 A when connected to an ac supply of 125 V and 50 Hz. A pure resistor under the same conditions takes a current of 12.5 A. If the two are connected to an ac supply of 100 V and 40 Hz, then the current in series combination of above resistor and inductor is (a) 10/ 2 A (c) 10 2 A

(b) 12.5 A (d) 10 A

4. When an athlete runs with some acceleration, he leans forward. The line joining his centre of mass to his foot which is in contact with the ground makes an angle q with the vertical. The coefficient of friction between his foot and the ground is m. His foot does not slip on the ground. His acceleration is (a) mg (b) mgtanq (c) gtanq (d) (gtanq)/m 5. In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges are placed at (n – 1) corners. At the centre, the magnitude of intensity is E and the potential is V. The ratio V/E is (a) rn (b) r(n – 1) (c) (n – 1)/r (d) r(n – 1)/n Solution Senders of Physics Musing 1. 2. 3. 4.

set-29 Amatra Sen (WB) Manmohan Krishna (Bihar) Meena Chaturvedi (New Delhi) Naresh Chockalingam (Tamil Nadu) set-28

1. Azhar Qureshi (Bihar) 2. Amatra Sen (WB) 3. Preeti Puri (Haryana)

subjective tyPe

6. In figure, a long thin wire carrying a varying current I = I0 sin wt lies at a distance y above one edge of a rectangular wire loop of length L and width W lying in the X-Z plane. What emf is induced in the loop? X

I

W y

L

Z

Y

7. A wire is wrapped N times over a solid sphere of mass m near its centre, which is placed on a smooth horizontal surface. A horizontal magnetic field of  induction B is present. Find the m R angular acceleration experienced  B by the sphere. Assume that the I mass of the wire is negligible compared to the mass of the sphere.

8. A stone is dropped from a balloon going up with a uniform velocity of 5.0 m s–1. If the balloon was 50 m high when the stone was dropped, find its height when the stone hits the ground. Take g = 10 m s–2. 9. A body weighs 98 N on a spring balance at the north pole. What will be its weight recorded on the same scale if it is shifted to the equator? Use g = GM/R2 = 9.8 m s–2 and the radius of the earth R = 6400 km. 10. A wheel of radius r and moment of inertia I about its axis is fixed at the top of an inclined plane of inclination q as shown in figure. A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane. Initially, the wheel is rotating at a speed w in a direction such that the block slides up the plane. After some time, the wheel stops rotating. How far will the block move before stopping? 

M 

nn

10

Physics for you | January ‘16

Superposition of Waves When two or more waves of similar kind, simultaneously arrive at a point then the resultant disturbance (or displacement in case of particles) at the point of meeting is given by a vector sum of the disturbances produced by each of the arriving waves. After superposition, the waves pass through as if they did not encounter each other, hence there is no change in the properties of either of the arriving waves after superposition. If the disturbances are produced along same line, vector sum becomes algebraic sum. Let us see what we understand from this. Consider two disturbances travelling in opposite directions meet each other as shown. 1 cm s–1

1 cm s–1

1cm 1 cm 1 cm 1 cm 1 cm 1 cm 1 cm

We are going to draw the shape of the resultant waveform at (i) t = 2 s (ii) t = 2.5 s (iii) t = 4 s. To do this, imagine individual waves travelling as if they are not meeting each other and then we apply superposition. t=2s 2 cm

1 cm 1 cm

t = 2.5 s

t=4s

1 cm s–1

1 cm 1 cm

1 cm s–1

1 cm 1 cm

Now, you see, that during superposition, the waves individually appear to loose their identity, but after superposition we can clearly see, they were always there! Applications of Superposition of waves

• Interference • Standing waves • Beats

Interference

Before we begin with the topic, we need to understand what are coherent sources. These are such sources for which the phase difference is independent of time. This clearly is possible only if the frequency of both the waves are identical else (w1 – w2)t will come out to be time dependent expression. Coming back to interference; when waves of similar kind from two or more coherent sources simultaneously arrive at a point then the resultant intensity at the point of superposition is different from the sum of intensity of the arriving waves and is dependent on the phase difference between the arriving waves which is directly dependent on path difference between the arriving waves. To put this in simple words, let me put a simple statement, imagine two coherent sources of light made to interfere. There might be a situation where light + light = darkness Isn't it opposite to our common sense? Yes, it is since we expect more bright light when two light sources are made to superimpose. But this logic is true only for non-coherent sources. Let us see this through an example. Consider waves S1 from two coherent sources S 1 and S 2 meeting at a point P travelling different S2 distances.

Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata

12

physics for you | January ‘16

x1 x2

P

The displacement equations for the two sources are say S1 : y1(x1, t) = A1sin(wt – kx1) S2 : y2(x2, t) = A2sin(wt – kx2) Clearly, the phase difference between them is Df = (wt – kx1) – (wt – kx2) = k(x2 – x1) 2p \ Df = (Dx ) l where Dx = x2 – x1 = path difference between the arriving waves. Hence from phasor analysis of SHM, we now understand A2 that the resultant amplitude of  oscillation can be found out.

\ AR = A12 + A22 + 2 A1 A2 cos(Df) where AR = resultant amplitude of oscillation. Now we understand that A1 and A2 are fixed for wave but if we change the location of sources from the point P, x1 and x2 changes, hence Dx changes, hence Df changes. Therefore AR becomes dependent on Dx. Now, considering intensity of waves, as we know, I ∝ A2 2 2 2 \ from, AR = A1 + A2 + 2A1A2cos(Df) we have , I R = I1 + I2 + 2 I1I2 cos(Df) where IR = intensity of the resultant wave due to arriving waves of intensities I1 and I2. Now extreme cases of interference may arise Constructive interference: When the waves arrive such that crest of one wave coincides with the crest of the other, the waves are said to be in phase and in such case the amplitude and hence intensity of resultant wave is maximum. A1 

A1 + A2 = AR

A2

This clearly is possible only if one wave is shifted with respect to the other wave by an integral multiple of complete wavelength (l).  2p  \ Dx = nl ⇒ Df =   (nl) = n(2p)  l  \ AR = A1 + A2 max

IR

max

14

=

(

I1 + I2

)2


A1 +

AR = A1 – A2



AR

A1

+

Destructive interference: If the waves arrive such that crest of one wave coincides with the trough of the other, the waves are said to be out of phase and in such case the resultant amplitude and hence intensity of wave is minimum.

A2

This clearly is possible only if one wave is shifted with respect to the other by an odd multiple of half wavelengths. l \ Dx = (2n + 1) ⇒ Df = (2n + 1)p 2 \ AR = A1 – A2 min

I Rmin =

(

I1 − I2

)2

Hence, generalising, A1 – A2 ≤ AR ≤ A1 + A2

(

)2

(

)2

I1 − I2 ≤ I R ≤ I1 + I2 Special case: If the arriving waves are of identical amplitude and hence intensity, then A1 = A2 = A0 (say) I1 = I2 = I0 (say) \ AR = 2A0, IR = 4I0 max

and AR

min

max

= 0, IR

min

= 0.

Now, this is what I was talking about when I said light + light = darkness! In general at any point in such case,  Df   Df  AR = 2 A0 cos  , I R = 4 I0 cos2    2   2  Let us now see an application of interference. Q1. Two coherent sources S1 and S2 are symmetriR cally placed with S1 S2 respect to centre at D 3 a separation 3l as R >> 3 shown. D is a detector which can measure the resultant intensity at all points on circumference. Find

(i) the number of maximas and minimas detected in one complete rotation. (ii) the angular position of the 1st maxima after starting. A(x = 0) Soln.: (i) In such x =  question analyse one x = 2 quadrant, since all quadrants are identical. S1 S2 B (x = 3) Let us see how. 3 At point A, S1A – S2A = 0 \ Dx = 0, since the same path length for waves arriving from S1 and S2. At point B, S1B – S2B = 3l \ Dx = 3l Hence A and B are locations of maximas but in between A and B, there must be other locations where Dx would gradually increase from Dx = 0 to Dx = 3l. \ In between 0 and 3l, there would be locations of Dx = l and Dx = 2l too. Hence total maximas = 2 × 4 (each quadrant) + 1 × 4(each point on x and y axis) = 12 l For minimas, Dx = (2n + 1) 2 Hence between each successive maxima, there would be minima too, so, l 3l 5l Dx = , , in each quadrant. 2 2 2 \ Total minimas = 3 × 4 = 12 (ii) Given : R >> 3l hence, from all points on circumference S1 and S2 will appear to be close by points. \ Dx = 2l ∆x = 2λ sθ o ⇒ 3lcosq = 2l c 3λ θ θ 2 S1 θ S2 ⇒ cos q = 3 ∆ x = 3λ 3λ 2 ⇒ q = cos −1   3

Standing Waves/Stationary Waves

When two waves of similar kind, same frequency and amplitude travelling from opposite directions superimpose we get standing waves. Now to understand better what this actually means, let us start by considering two such waves. 16


y1(x, t) = Asin(wt – kx) y2(x, t) = Asin(wt + kx) Hence, considering superposition, yR(x, t) = y1(x, t) + y2(x, t) = A[sin(wt – kx) + sin(wt + kx)] = A[2sin(wt) cos(kx)] = 2Acos(kx) sin(wt) = Axsin(wt) where Ax = 2Acos(kx) This resultant wave clearly shows that it is not a travelling wave atleast.

The quantity, Ax = 2Acos(kx) is constant for a location and the value is position dependent whereas y(x, t) = Axsin(wt) indicates that the displacement along y-axis of a particle located at x varies simple 2p harmonically with time period, T = . w Hence, sin(wt) being same for all particles, each executes SHM with same time period. Hence, we conclude that Ax indicates the amplitude of oscillation at a location x. Ax = 2Acos(kx) \ –2A ≤ Ax ≤ 2A But amplitude indicates magnitude of maximum particle displacement. Hence –2A is same as 2A when we talk of amplitude. \ 0 ≤ |Ax| ≤ 2A Hence at all locations the amplitude varies between 0 to 2A. Imagine what do we mean when we say amplitude of oscillation is zero? Amplitude indicates maximum displacement and maximum displacement itself is zero indicates particles are not oscillating at all, i.e. they are stationary at their respective points. It is due to these particles that the name stationary wave is coined. The locations where amplitude of oscillation is (i) maximum (Ax = ±2A) are said to be antinodes. Ax = ±2A = 2Acos(kx) cos(kx) = ±1 ⇒ (kx) = np p  p  ⇒ x = n  = n k  2 p / l  l 3l , ... l ⇒ x = n   = 0, , l, 2 2 2 (ii) minimum (Ax = 0) are said to be nodes. Ax = 0 = 2Acos(kx)

⇒ cos(kx) = 0

p ⇒ kx = (2n + 1) 2 l l 3l 5l ⇒ x = (2n + 1) = , , , ... 4 4 4 4 Let us try to draw the envelope of the particle's displacement (for standing waves in a stretched string) with these locations of nodes and antinodes. Antinodes

Nodes

2A 2A x=0 x= 4

x= 2

x= x =  4

x =  4

x =  2

The separation between two consecutive nodes or l antinodes is said to be loop length which is . 2 The bold lines drawn above indicate the boundary within which the particles are confined to move. How would standing waves in stretched string appear? As below,

\ f1 = 2f0

All the particles located between two consecutive nodes oscillate in same phase, i.e., they reach their extreme ends and mean position together. Particles located on opposite sides of a node always move opposite to each other, hence are 180° out of phase with respect to each other. Hence, unlike travelling wave, the energy does not propagate from one end to the other, it gets confined between a loop. Application of standing waves Standing waves in stretched string fixed at both ends: The fixed ends will always be a displacement node. The smallest frequency l with which standing waves can be set up is said to be fundamental frequency (f0). 18


T m where, T = tension in string m = mass per unit length. Since velocity is only medium dependent, v = f l suggests that for minimum frequency, wavelength l should be maximum. Hence keeping the two extreme ends fixed (nodes) the largest wavelength that can be drawn by inserting an antinode in between is shown here. \ v = f0 l0 v v ⇒ f0 = = l0 2l v  ⇒ f0 = l= 0 2l 2 The frequencies higher than the fundamental frequency with which standing waves can be set up in the system are said to be overtones, whereas all integral multiples of fundamental frequency are said to be harmonics. 1st overtone The wavelength decreases hence frequency increases. v v v f1 = = = 2   2l  l1 l Velocity of wave, v =

1st

2nd

overtone = harmonic Similarly, for nth overtone,

1 = l

fn = (n + 1)f0 = (n + 1)th harmonic v = (n + 1)    2l  Standing waves in organ pipe: The open ends of the organ pipe behave as displacement antinode whereas the closed end behaves as displacement node. Open organ pipe Closed organ pipe It has both ends open. It has one end closed. Fundamental mode

l=

0 2

v \ f0 = l0 v ⇒ f0 = 2l

l=

0 4

v \ f0 = l0 v ⇒ f0 = 4l

1 w2 − w1 \ fb = = = f2 − f1 Tb 2p

1st overtone

\ fb = f2 – f1 1 = l

31

=l

4 v \ f1 = × 2 = 2 f0 v 2l \ f1 =   × 3 = 3 f0  4l  \ 1st overtone = 2nd st \1 overtone = 3rd harmonic harmonic Generalising th nth overtone n overtone th = (n + 1) harmonic = (2n + 1)th harmonic v v = (n + 1)   = (2n + 1)    2l   4l 

Beats

When two sound waves of frequency close to each other (but not equal) superpose at a point, then at the point of superposition the phase difference keeps on changing with respect to time. Df = (w1 – w2)t Hence the amplitude of resultant wave will also keep on changing at a location as time passes by. But everytime Df = n(2p) we get to hear a maxima since waves become in phase and everytime Df = (n + 1)p we get to hear a minima. So, the appearance is a sound of alternating intensity, alternating between a certain maximum and minimum intensity. This phenomenon is beats and the number of maximas heard per second is beat frequency. To find beat frequency (fb) Df = n(2p) ⇒ (w2 – w1)t = n2p  2p  ⇒ t = n  w2 − w1  2p 2(2 p) 3(2 p) = , , , .... w2 − w1 w2 − w1 w2 − w1 Tb Tb The time gap between two successive maximas is beat period (Tb). 2p \ Tb = w2 − w1

Therefore beat frequency is just the difference of two frequencies of the source. Applications: Q2. An aluminium rod having a length 100 cm is clamped at its mid point and set into longitudinal vibrations of fundamental mode. Given r = 2600 kg m–3 and Y = 7.8 × 1010 N m–2 for aluminium. Find the frequency of sound emitted. Soln.: Velocity of wave v=

7.8 × 1010 Y = = 30 × 103 m s −1 r 2600

30 × 103 v \ f0 = = ≈ 2740 Hz. 2l 2 ×1 Q3. A tuning fork of unknown frequency produced 4 beats per second when sounded with another of frequency 254 Hz. It gives the same number of beats per second whenever unknown tuning fork is loaded with wax. Find its frequency before waxing. Soln.: Remember, two terms with tuning forks (i) Waxing : Here wax is added to the prongs due to which frequency of wave decreases. (ii) Waning : Here the prongs are filed due to which frequency of wave increases. Coming back to the question, having 4 beats/second gives two possible situations 254 +4 258

–4 250

waxed

waxed f < 250

f < 258 Here the difference with 254 will initially decrease as long as f > 254 but if f < 254 then it might happen that it reaches f = 250 Hz and again we get fb = 4 Hz, hence 258 Hz is the required frequency.

Here difference with 254 will clearly be greater than 4. Hence fb > 4 which is not possible.


19

Q4. For a certain organ pipe three successive frequencies are observed at 425, 595, 765 Hz respectively. Taking speed of sound in air to be 340 m s–1, find the fundamental frequency. Soln.: Note the ratio, 425 : 595 : 765 = 5 : 7 : 9 \ Odd multiples ⇒ closed organ pipe \ 5f0 = 425 ⇒ f0 = 85 Hz

Q7. vw A

fa T 324 18 = = = fa′ T′ 289 17 4 + ft 18 ⇒ = ⇒ ft = 50 Hz 1 + ft 17 \

20


B

A vehicle is moving towards a large building with a speed v0 and blows a horn of frequency f0 which travels with a speed vw. Find the number of beats detected by A who is moving with the vehicle and B who stands in between vehicle and building. Soln.: Using method of images, we create a virtual source behind the building. vw

vw A

f0

f0

B

v0

v0

S

⇒ ft = 205 Hz

Soln.: v ∝ T \ Decrease in temperature ⇒ decrease in velocity of wave ⇒ decrease in frequency of air column (fa) Two situations were possible. fa – ft = 4 or ft – fa = 4 The second equation suggests that if fa decreases, fb > 4 (increases) which is a contradiction. Hence, fa – ft = 4 ... (i) Again on decreasing temperature, fa′ – ft = 1 ... (ii)

f0 v0

Q5. A tuning fork vibrating with a sonometer having a wire of length 20 cm produces 5 beats per second. The beat frequency does not change if the length of wire is changed to 21 cm. Find frequency of tuning fork. 1 Soln.: For a sonometer wire, f s ∝ l Hence if length increases, fs decreases. Hence we conclude that initially fs was greater than ft (frequency of tuning fork) and later it became smaller by same amount. \ fs – ft = 5 ft – fs′ = 5 f l ′ 21 5 + ft 21 \ s = = ⇒ = f s′ l 20 ft − 5 20 Q6. A tuning fork and an air column whose temperature is 51°C produce 4 beats in one second when sounded together. When the temperature of air column is 16°C only one beat is detected per second. Find frequency of tuning fork.

Building

S

For both A and B we have two sources S and S′, but for B, the frequency of waves detected from both of them would be identical and equal to  vw  fappB =  f0  vw − v0  hence no beat is heard by him. Speaking of A, he hears two frequencies (i) f1 = f0 (since there is no relative motion between him and S) v +v  (ii) f2 =  w 0  f0  vw − v0  \ fb = f2 – f1 2v0 f0 v +v  =  w 0 − 1 f 0 = vw − v0  vw − v0  nn

EXAM DATES 2016 jee main

:

vIteee mgImS kerala pet kerala pmt AIpmt WBjee jee Advanced AIImS

: : : : : : : :

3rd april (offline), 9th & 10th april (online) 6th to 17th april 17th april 25th & 26th april 27th & 28th april 1st May 17th May 22nd May 29th May

PRESS RELEASE

VITEEE-2016

B.Tech Engineering Entrance Exam

Application forms sales begins

VIT University Chancellor Dr. G. Viswanathan is seen inaugurating the sale of the VITEEE-2016 application forms, to some of the aspiring students for the tests, in April 2016 at the Vellore Head Post Office on Friday. Seen others in the picture are Superintendent of Posts A. Natarajan, Deputy Superintendent R. Mahendran, VIT University’s VC Dr. Anand A Samuel, Pro-VC Dr. S. Narayan, VPs Sankar Viswanathan, Sekar Viswanathan and G.V. Selvam, Director UG Admissions, K. Manivannan, PRO- (Posts) Kathir Ahmed and Marketing Executive S. Selvakumar.

T

he sale of application forms for the VIT university Entrance Examinations (VITEEE-2016) to be held in april 2016, for B.Tech courses various streams, began in all the 92 Head Post Offices, with the VIT university Chancellor Dr. G. Viswanathan inaugurating it at the Head Post Office, here, on Friday. It is scheduled that the entrance examinations for the B.Tech offered in the VIT university, Vellore and Chennai Campuses, will be held from April 6th to April 17th 2016. This Computer Based Test (CBT) is held in 118 cities including Dubai, Kuwait and Muscat. The university offers courses in Vellore Campus - Bio-Medical Engineering, Biotechnology, Computer Science and Engineering (Specialisation in Bioinformatics), Civil Engineering, Chemical Engineering, Computer Science and Engineering, Electronics and Communication Engineering, Electrical and Electronics Engineering, Electronics and Instrumentation Engineering, Information Technology, Mechanical Engineering, Mechanical Engineering (Specialisation in automotive Engineering),

Mechanical Engineering (Specialisation in Energy Engineering), Production and Industrial Engineering and in Chennai Campus – B.Tech in Civil Engineering, Computer Science and Engineering, Electronics and Communication Engineering, Electrical and Electronics Engineering and Mechanical Engineering. The entrance examination application forms from the Head Post Offices, it can be obtained by sending a Demand Draft for Rs.990/- drawn in favour of VIT university, payable at Vellore to the Director – uG admissions or by cash payment at selected post offices across the country. Issuing of online and offline application has commenced from November 27th 2015. Candidates can also apply online at www.vit.ac.in (online applicants need to pay Rs.960/- only). The last date for applying is 29th February, 2016. VIT university has been consistently been ranked among the premier engineering institutions of the country by India Today. VIT has an impressive track record of placement. For the 8th year in a row, VIT has achieved the top slot in placements. Visit: www.vit.ac.in for further details. physics for you | january ‘16

nn 21

By : Prof. Rajinder Singh Randhawa*

THERMODYNAMICS 1.

A large tank A of water is kept at a constant temperature q0. It is connected to another body B of specific heat c and mass m by a conductor of length l, area of cross-section A and thermal conductivity K. Initial temperature of body B is q1. Find the variation of temperature q of the body B at time t.

K l

A

2.

1

t=0

B

One end of a rod of length L and cross-section area A is kept in a furnace at temperature T1. The other end of rod is kept at a temperature T2. The thermal conductivity of the material of the rod K and emissivity of the rod is e. It is given that T2 = TS + DT, where DT < < TS, TS is the temperature of surroundings. If DT ∝ (T1 – TS), find the proportional constant. Furnace T1

3.

by w2. What is the ratio of 4.

Tank of water 0

in the vessel which is maintained at a temperature T. The mean square of velocity of the molecules of B type is denoted by v2 and the mean square of the x-component of the velocity of A type is denoted

Insulated L Insulated

N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained

v2

?

An ideal monoatomic gas is confined in a cylinder by a spring-loaded piston of area of cross-section 8 × 10–3 m2. Initially, the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is in its relaxed state as shown in figure. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied (in J) by the heater. The force constant of the spring is 8000 N m–1, atmospheric pressure is 1 × 105 N m–2. The piston is massless and no friction exists between the piston and the cylinder.

Heater

TS T2

w2

Piston

5.

Rigid support

Figure shows three isotherms at temperatures T1 = 4000 K, T2 = 2000 K, and T3 = 1000 K. When 1 mole of an ideal monoatomic gas is taken through paths AB, BC, CD and DA, (a) find the change in

*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699

22


internal energy DU, (b) find the work done by the gas W. A

P

B

VA

1.

V

C T = 2000 K 2 T3 = 1000 K VB

B

K (T1 − TS − DT ) = 4σeTS3 DT L K (T1 − TS )  K =  4σeTS3 +  DT  L L \ DT =

The rate of increase in thermal energy of the tank B is dU B dq = mc dt dt dU dQ B because no work is done in change Since = dt dt in volume, we have

= 3.

− KAt

The mean square velocity of gas molecules is given 3kT . by v 2 = m 3kT ... (i) For gas A, v 2A = m For a gas molecule, ( v 2x = v 2y = v z2 )

2 2 3kT For gas B, v B = v = 2m Dividing eqn. (ii) by eqn. (iii), we get kT w2 2 = m = 2 3kT 3 v 2m

or (q0 − q) = (q0 − q1 )e mlc

Given that T2 = TS + DT

.

 3kT    kT w2 = v 2x =  m  =  3  m

q − q −KA dq KA = t or ln 0 = t ∫ q0 − q1 mlc q1 q0 − q mlc

t

Rate of heat conduction through rod = rate of heat lost from right end of the rod. KA(T1 − T2 ) ... (i) = σAe(T24 − TS4 ) L

(4σeLTS3 + K )

or v 2x =

q

2.

K

v3 3 From eqn. (i), we get

 q − q  or dq = KA dt dq =K 0 A  l  q0 − q mlc dt Integrating both sides, we get

− KA

(4eσLTS3 + K )

v 2 = v 2x + v 2y + v z2 = 3v 2x

mc

or q = q0 − (q0 − q1 )e mlc

K (T1 − TS )

Comparing with the given relation DT ∝ (T1 – TS), Proportionality constant

At t



0

(as DT << TS)

Substituting in eqn. (i), we have

m c

= TS4

\ T24 − TS4 = 4(DT )(TS3 )

solutions Let the temperature of the body B be q, in time dt, let a heat dQ flow through the conductor rod. dQ q0 − q Then = KA dt t A

4

DT   T24 = TS4 1 + 4 TS  

T1 = 4000 K D

4

 DT  \ = (TS + DT ) 1 + T  ,  S  Using binomial expansion, we have T24

4.

... (ii) ... (iii)

Since F = kx = 8000 × 0.1 = 800 N Pressure exerted on the piston by the spring is physics for you | January ‘16

23

F 800 DP = = = 1 × 105 N m −2 − 3 A 8 × 10 The total pressure of the gas inside the cylinder is P2 = P1 + DP = Patm + DP P2 = 1 × 105 + 1 × 105 = 2 × 105 N m–2 Since the piston has moved outwards, then increase in volume of the gas is DV = A × x = 8 × 10–3 × 0.1 = 8 × 10–4 m3 The final volume of the gas, V2 = V1 + DV = 2.4 × 10–3 + 8 × 10–4 = 3.2 × 10–3 m3 Let T2 be the final temperature of the gas, then P1V1 P2V2 PV = ⇒ T2 = 2 2 T1 T1 T2 P1V1

T2 =

2 × 105 1 × 105

×

3.2 × 10 −3

2.4 × 10 −3

× 300

= 800 K Let the heat supplied by the heater be Q, then Q = W + DU

= 3000 × 8.314 = 2.49 × 104 J DUBC = CV (TC – TB) 3 = × R(2000 − 4000) = –2.49 × 104 J 2 DUCD = CV(TD – TC) 3 = × R(1000 − 2000) = –1.25 × 104 J 2 DUDA = CV(TA – TD) 3 = × R(2000 − 1000) = 1.25 × 104 J 2 \ Total change in internal energy DU = 0 (b) Work done by the gas during path AB under constant pressure is P(VB – VA) = R(TB – TA) = 8.314 × 2000 = 16.628 × 103 J No work is done during BC and DA. Work done on the gas during CD is R(TC – TD) = 8.314 × 1000 = 8.314 × 103 J. .. . Net work done by the gas during the cycle is (16.628 × 103 – 8.314 × 103) J = 8.314 × 103 J.

V2 x kx  Now, W = ∫ PdV = ∫  Patm +  Adx  A V 0

nn

1

kx 2 = Patm Ax + 2 kx    P = Patm + A and dV = Adx  Substituting the values, we get W = 120 J. Also, U = nCV DT Number of moles of gas, PV n= RT 3 For monoatomic gas, Cv = R 2 PV  3  3 PV \ DU = DT  R  DT = × RT  2  2 T 3 1 × 105 × 2.4 × 10−3 = × × 500 2 300 = 600 J \ Heat supplied by the heater Q = 120 + 600 = 720 J 5.

24

(a) The change in internal energy is given by DUAB = CV(TB – TA) (for n = 1 mole) 3 = × R(4000 − 2000) = 3000R 2 physics for you | January ‘16

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PRACTICE PAPER

CLASS XI

single oPtion correct This paper contains 45 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which only one is correct. (Mark only One Choice). Marks : 45 × 4 = 180

Negative Marking (–1)

1. A body of mass 7m initially at rest explodes into two fragments of masses 4m and 3m. If the momentum of the lighter fragment is p then the kinetic energy released in the explosion will be 11 p2 5 p2 7 p2 9 p2 (a) (b) (c) (d) 24 m 14 m 16 m 24 m 2. A particle of mass m describes a circle of radius r. 4 The centripetal acceleration of the particle is . r2 The momentum of the particle is 4m 2m 4m 2m (c) (d) (a) (b) r r r r 3. Figure shows a block of mass m1 on a smooth horizontal surface pulled by a m string which is attached to a block of mass m2 hanging over a frictionless pulley which has no m mass. The blocks will move with an acceleration  m + m2  (a) g (b)  1 g  m2  1

2

 m2  (c)  g  m1 + m2 

 m1  (d)  g  m1 + m2 

4. Two blocks m1 and m2 are in contact over a frictionless table; m1 = 2.0 kg, m2 = 1.0 kg. In the first case a horizontal F = 3 N m1 m2 force of magnitude 3 N is applied to block m1. In Case 1 the second case this force F=3N is applied to block m2. The m1 m2 forces of contact between Case 2

the blocks in the first and the second cases, respectively are (a) 3 N, 3 N (b) 2 N, 2 N (c) 1 N, 2 N (d) 2 N, 1 N f1 5. A uniform ladder is in equilibrium against a rough A N1 wall as shown. Points A and B respectively are the point N2 C of contact of ladder with wall B  and with the ground. Point C is the CM of the ladder. f2 W 1. Torque due to friction f2 about point A is not zero. 2. Torque due to friction f2 about point B is zero. 3. Torque due to weight is zero about A, B and C. 4. Torque due to f1 and N1 is not zero about A. (a) Only 1 and 3 are correct (b) Only 2 and 3 are correct (c) Only 3 and 4 are not correct (d) Only 1 and 4 are correct 6. The moment of inertia of an annular disc (a disc with concentric cavity) of mass M radius R and cavity radius r about an axis passing through its CM and normal to its plane will be 1 1 2 2 (b) M R2 − r 2 (a) M R + r 2 2 1 1 2 2 2 2 (c) M 2R + r (d) M R + r 8 4 7. A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to (a) t1/2 (b) t3/4 (c) t3/2 (d) t2

( (

)

)

( (

) )

Contributed by : K. P. Singh, KP Institute of Physics, Chandigarh, 09872662552 Physics for you | January ‘16

25

8. A body of mass 3 kg is under a force which causes t2 in metre, with displacement in it, given by s = 3 time t in seconds. What is the work done by the force between time t = 0 and t = 2 s ? (a) 8 J (b) 5.2 J (c) 3.9 J (d) 2.6 J 9. The linear momentum p of a body varies with time as p = 5a + 7bt2 where a and b are constants. The net force acting on the body for one dimensional motion varies as (a) t2 (b) t–1 (c) t–2 (d) t 10. A body is acted upon by a force proportional to square of distance covered. If the distance covered is denoted by x, then work done by the force will be proportional to (a) x (b) x2 (c) x3 (d) x–2 11. The potential energy function along the positive x-axis is given by U (x ) = −ax + b , a and b are x constants. If it is known that the system has only one stable equilibrium configuration, the possible values of a and b are (a) a = – 1, b = 2 (b) a = – 5, b = 1 (c) a = 1, b = – 2 (d) a = 5, b = – 3 12. Neglecting the friction and weights of the pulley, which one of the T following is the force F required to T F B lift a 100 N load in the system of T T pulleys as shown in the figure? A (a) 20 N (b) 25 N (c) 30 N (d) 35 N W 2

2

1

1

13. A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall? (a) Kinetic energy (b) Potential energy (c) Total mechanical energy (d) Total linear energy 14. A motor drives a body along a straight line with a constant force. The power P developed by the motor must vary with time t as shown in figure (a)

P

(b)

(c)

P

t

26

17. A block of mass 5 kg is resting on a smooth surface. At what angle a force of 20 N be acted on the body so that it will acquire a kinetic energy of 40 J after moving 4 m ? (a) 30° (b) 45° (c) 60° (d) 120° 18. The moment of inertia of a body about a given axis is 1.2 kg m2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad s–2 must be applied about that axis for a duration of (a) 4 s (b) 2 s (c) 8 s (d) 10 s 19. A particle performs uniform circular motion with an angular momentum L. If the frequency of particle’s motion is halved and its kinetic energy doubled, the angular momentum becomes (a) 2 L (b) 4 L (c) L/2 (d) L/4 20. The moment of inertia of two spheres of equal masses about their diameters are equal. If one of them is solid and other is hollow, the ratio of their radii is (a) 3 : 5 (b) 3 : 5 (c) 5 : 3 (d) 5 : 3 21. The curve between log L and log p is (L is angular momentum and p is linear momentum) log L


log L

(b)

(a) log p

log p

log L t

(d)

–1

P

t

P

V(m s ) 15. Velocity-time graph of a particle 20 of mass 4 kg moving in a straight line is as shown in figure. Work done by all t(s) 2 forces on the particle is (a) 400 J (b) – 800 J (c) – 400 J (d) 200 J 16. 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 4 Mg 2 Mg Mg (d) Mg (a) (b) (c) 2k k k k

t

log L

(c)

(d) log p

log p

22. A uniform rod of mass m and length L is suspended by means of two light inextensible strings as shown in figure. B Tension in one string immediately A after the other string is cut is mg mg (a) (b) mg (c) 2 mg (d) 4 2 23. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now the platform is given an angular velocity w0. When the tortoise moves along a chord of the platform with a constant velocity with respect to the platform, the angular velocity of the platform will vary with the time t as (t)

(a)

(b)

0

(t)

t

t

(t)

(c)

(d) 0

0

t

t

24. A wheel has angular acceleration of 3.0 rad s–2 and an initial angular speed of 2.00 rad s–1. In a time of 2 s it has rotated through an angle (in radian) of (a) 6 (b) 10 (c) 12 (d) 4 25. The moment of inertia of a rod about an axis through 1 its centre and perpendicular to it is ML2 where, 12 M is the mass and L the length of the rod. The rod is bent in the middle so that the two halves make an angle of 60°. The moment of inertia of the bent rod about the same axis would be ML2 ML2 ML2 ML2 (a) (b) (c) (d) 48 24 8 3 12 26. A body of mass 15 kg is suspended by the strings making angles 60° and 30° with the horizontal as shown in figure. Then (take g = 10 m s–2) T2 60°

T1 15 kg

29. For shown atwood machine m1 = 8 kg, m2 = 2 kg. The string and the pulley are assumed to be smooth and massless. Take g = 10 m s–2. The acceleration of center of mass of the system is. (a) 3.6 m s–2 (b) 6 m s–2 –2 (c) 5 m s (d) 0

m2 m1

0

(t)

28. A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 m s–1. To give an initial upward acceleration of 20 m s–2, the amount of gas ejected per second to supply the needed thrust will be (g = 10 m s–2) (a) 127.5 kg s–1 (b) 187.5 kg s–1 –1 (c) 185.5 kg s (d) 137.5 kg s–1

30°

(a) T2 = 75 N (b) T2 = 75 3 N (c) T1 = 25 3 N (d) T1 = 12 N 27. A car of mass 1000 kg negotiates a banked curve of radius 40 m on a frictionless road. If the banking angle is 45°, the speed of the car is (a) 20 m s–1 (b) 30 m s–1 –1 (c) 5 m s (d) 10 m s–1

30. A wide hose pipe is held horizontally by a fireman. It delivers water through a nozzle at two litre per second. On increasing the pressure, this increases to four litres per second. The fireman has now to (a) push forward twice as hard (b) push forward four times as hard (c) push backward four times as hard (d) push backward twice as hard. 31. A block of mass m is placed on a smooth wedge of inclination q. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude (a) mg tan q (b) mg cos q (c) mg sec q (d) mg 32. A system consists of 3 particles each of same mass and located at points (1, 2), (2, 4) and (3, 6). The co-ordinates of the center of mass are (a) (1, 2) (b) (2, 4) (c) (4, 2) (d) (3, 6) 33. Out of the following bodies of same mass, which one will have maximum moment of inertia about an axis passing through its center of gravity and perpendicular to its plane ? (a) ring of radius r (b) disc of radius r (c) square frame of sides 2r (d) square lamina of sides 2r 34. Consider a two particle system with particles having masses m1 and m2. If the first particle is pushed towards the center of mass through a distance d, by what distance should the second particle be moved so as to keep the center of mass at the same position ? Physics for you | January ‘16

27

m1 d m2 m2 (c) m d 1

(b) d

(a)

(d)

m1 d m1 + m2

35. Two particles of mass m1 and m2 (m1 > m2) attract each other with a force inversely proportional to the square of the distance between them. The particles are initially held at rest and then released. Then the CM (a) moves towards m1 (b) moves towards m2 (c) remains at rest (d) moves at right to the line joining m1 and m2 36. Two skaters A and B of masses 50 kg and 70 kg respectively stand facing each other 6 m apart. Then they pull on a rope stretched between them. How far has each moved when they meet? (a) both have moved 3 m (b) A moves 2.5 m and B 2.5 m (c) A moves 3.5 m and B 2.5 m (d) A moves 2 m and B 4 m 37. A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with an angular velocity w. Another disc of 1 same dimensions but of mass M is placed gently 4 on the first disc co-axially. The angular velocity of the system is 3 4 1 (a) 2 w (b) w (c) w (d) w 4 5 3 38. A wheel is rolling uniformly along a level road (see figure). The speed of translational motion of the wheel axis is v. What are the speeds of the points A and B on the wheel rim relative to the road at the instant shown in the figure? B (a) vA = v ; vB = 0 (b) vA = 0 ; vB = v v (c) vA = 0 ; vB = 0 (d) vA = 0 ; vB = 2v A ^

^

^

39. If a force 10 i + 15 j − 25 k acts on a system and gives ^

^

^

an acceleration 2 i + 3 j − 5 k to the centre of mass of the system, the mass of the system is (a) 5 units (b) 38 units (c) 5 38 units (d) None of these u = 5 m s–1 40. Two particles A and B are situated at a distance v d = 3 m apart. Particle A –1 has a velocity of 5 m s at 30° 60° an angle of 60° and particle A B has a velocity v at an angle of 30° as shown in

28


figure. The distance d between A and B is constant. The angular velocity of B with respect to A is 5 (a) 5 3 rad s −1 (b) rad s −1 3 5 10 −1 rad s −1 rad s (c) (d) 3 3 41. The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of (a) Frequency (b) Velocity (c) Angular momentum (d) Time 42. A ball is dropped to the ground from a height of 8 m. The coefficient of restitution is 0.5. To what height will the ball rebound? (a) 2 m (b) 1.42 m (c) 4 m (d) 0.5 m 43. A mass m moving horizontally with velocity vo strikes a pendulum of mass 2m. If the two masses stick together after the collision, then the maximum height reached by the pendulum is v2 v02 v2 v2 (c) 0 (d) 0 (a) 0 (b) 18 g 2g 6g 12 g 44. A ball is projected in vacuum as shown. Average power delivered by gravitational force B (a) for A to C is positive (b) for B to C is zero. u (c) for A to B is negative. C  (d) for A to B is zero. A 45. 1. During any collision, velocity along common tangent doesn’t change. 2. In an elastic collision with equal masses, the velocity along common normal is interchanged. 3. When a ball makes an oblique inelastic collision with a fixed target the reflection angle is less than incidence. 4. In a one dimensional elastic collision the fraction of kinetic energy transferred by a projectile to a 4 m1 m2 stationary target is . (m1 + m2 )2 (a) Only 3 is wrong (c) Only 4 is wrong

(b) 2 and 3 are wrong (d) 3 and 4 are wrong

solutions

1. (a) : 7m . 0 = 4mv1 + 3mv2 = p1 + p2 ⇒ p2 = – p1 = p (given) K= =

p12 p2 + 2 2m1 2m2

p2  1 1  7 p2 p2 p2 = +  + = 2(4m) 2(3m) 2m  4 3  24m

2. (d) : acp = ⇒v =

2

v2 4 = r r2

2m . . . p = mv = r

r 3. (c) : For m2 : m2 a = mg – T For m1 : T = m1a m2 g On adding, (m1 + m2) a = mg or a = m1 + m2 Fnet 3 −2 4. (c) : a = = =1m s mtotal 2 + 1 Case 1: for m1 : 3 – N = 2 (1) ⇒ N = 1 N Case 2: for m2 : 3 – N = (1) (1) ⇒ N = 2 N 5. (c) : Torque of a force about a point is zero only when the line of action of the force passes through that point. 6. (a) : Check: When r 0, it becomes disc and when r R, it becomes ring. 7. (c) : P = M1 L2 T –3 = constant ⇒ x2 t–3 = constant or x2 ∝ t3 or x ∝ t3/2 2t 2 t2 ⇒v = v(t = 0) = (0) = 0 3 , 3 3 2 4 v(t = 2) = (2) = 3 3 2   1 4 4 8 1 4  W = DK = .3.   − 0  = .3. . = = 2.6 J  2 3 3 3 2  3   

8. (d) : s =

9. (d) : p = 5a + 7bt2 dp F= = 0 + 7 × 2bt ⇒ F ∝ t dt 10. (c) : F ∝ x2 , W = ∫F dx ∝ x3 dU b b = −a − 2 11. (a) :U = −ax + ⇒ −F = dx x x 2b b d 2U or F = a + =0+ 3 x 2 dx 2 x At stable equilibrium, F = 0,

d 2U

dx 2 2b b −b a + 2 = 0 ⇒ a = 2 Also, >0 x x x3

12. (b) : For rope: F = T2 For pulley B: 2 T2 = T1 For pulley A: 2 T1 = W = 100 Using above eqns. 100 F= = 25 N 4

>0

13. (c) 14. (a) : P = Fv = (constant) (u + at) ⇒ P ∝ t, so linear variation. 15. (b) : Speed becomes 0 from 20 m s–1. 1 ... W = DK = .4. 0 − (20)2  = −800 J  2  16. (b) : Let the maximum extension in the spring be x. By work energy theorem, W = DK 2 Mg 1 Mgx − kx 2 = 0 − 0 ⇒ x = k 2 17. (c) W = DK

1 20 × 4 × cos q = 40 ⇒ cos q = or q = 60° 2 1 18. (b) : K = I (w0 + at )2 2 1 1500 = ×1.2 × (0 + 25t )2 ⇒ t = 2 s 2 1 K ′ L′ w′ 19. (b) : K = Lw ⇒ = 2 K L w ⇒ 2=

L′  1    ⇒ L′ = 4 L L 2

5 2 2 2 ⇒ RS 20. (c): MRS2 = MRH = RH 3 5 3 21. (b): L = pr ⇒ log L = log (pr) log L = log p + log r; y = mx + C So, straight line with positive intercept. 22. (a): Before cutting, TA + TB = mg For rotational equilibrium about center L L TA = TB ⇒ TA = TB 2 2 mg So, 2TA = mg or TA = 2 If string B is cut, just after cutting tension in mg A remains same i.e., . 2 23. (c): Moment of inertia first decreases and then increases, thus by law of conservation of angular momentum, L = Iw = constant w first increases and then decreases. 1 1 2 24. (b) : q = w0t + at 2 = 2(2) + (3)(2) = 10 radian 2 2 2

1 M L 1 M L 25. (b) : I =     +     3  2  2  3  2  2 

2


29

35. (c) : Internal forces cannot change velocity of CM of a system. As CM was initially at rest, it will remain at rest. 36. (c) : Two bodies under mutual internal forces always meet at their CM. From theorem of moment of masses, we can write 50 x = 70 (6 – x) ⇒ 120 x = 420 ⇒ x = 3.5 m 37. (b) : From law of conservation of angular momentum,

2

2 1  M  L  ML =    ×2 = 3  2  2  12

26. (b) : Horizontally: T2 cos 60° = T1 cos 30° ⇒ T2 = T1 3 Vertically : T2 sin 60° + T1 sin 30° = 150 Thus T1 = 75 N and T2 = 75 3 N 27. (a) : As tan q =

v2 gr

M  M 2 1 M  4 R w + 0 =  R2 +   R2  w′ ⇒ w′ = w 2 2 2 4 5     5 As total moment of inertia increases to times, 4 4 then w becomes th. 5 38. (d) : For pure rolling, point of contact is at rest and topmost point has double speed of that of CM.

v2 tan 45° = ⇒ v = 20 m s–1 10(40) dm 28. (b) : vgr = m( g + a) dt dm dm = 187.5 kg s −1 800 = 5000(10 + 20) ⇒ dt dt 2

29. (a) : acm

→

2  m − m2   8 − 2 = 1 10 = 3.6 m s −2  g=  8 + 2   m1 + m2 

^

dm dm , = ρAv dt dt dm On doubling , v also doubles, this increases dt force to 4 times.

30. (b) : F = v

N

31. (c) : N = m g 2 + a2 ...(i) For no slip of block with respect to wedge ma cos q = mg sin q or a = g tan q ...(ii) Using (ii) in (i), we get

a

ma mg



N = m g 2 + g 2 tan2 q ⇒ N = mg sec q →

→

32. (b) : r = CM =

→

→

m1 r1 + m2 r2 + m3 r3 m1 + m2 + m3

m(1, 2) + m(2, 4) + m(3, 6) (6, 12) = = (2, 4) m+m+m 3

33. (c) : Iring = mr2 , Idisc = 0.5 mr2 4 2 1 m 2 m 2 Isq.frame 12 4 (2 r ) + 4 (r )  = 3 mr   1 2 2 2 Isq. lamina = m (2r ) + (2r )  = mr 2   12 3 34. (a) : (m1 + m2) DxCM = m1 Dx1 + m2 Dx2 (m1+ m2) (0) = m1 (–d) + m2 Dx2 md ⇒ Dx2 = 1 m2 = 4×

30


→

39. (a) : | F | = m | a | ^

^

^

^

^

10 i + 15 j − 25 k = m 2 i + 3 j − 5 k ⇒ m = 5 units 40. (b) : v cos 30° = 5 cos 60° ⇒ v = 5 m s −1 3 v v AB 5 sin 600 − v sin 300 w= = = r AB 3 3 5 1 − . 2 2 5 3 1 = = rad s– 3 3 41. (a) : Planck’s constant and angular momentum have same dimensions. [ h] [ L ] = =w [I ] [I ] 5

42. (a) : hr = e 2 hs = (0.5)2 (8) = 2 m 43. (a) : Using law of conservation of linear momentum: mv0 + 0 = (m + 2m) vs ⇒ ...

vs = v0 / 3 v 2 (v / 3)2 v02 h= s = 0 = 2g 2g 18 g

44. (c) : During upward motion gravity does negative work and thus negative power is delivered. 45. (a) : When a ball makes an oblique inelastic collision with a fixed target, the reflection angle is more than tan i incidence, as tan r = . e nn

7

Electromagnetic Waves | Optics ElEctromagnEtic wavEs

Maxwell predicted the existence of electromagnetic waves on the basis of his equations in 1865. According to him, an accelerated charge produces a sinusoidal time varying magnetic field, which in turn produces a sinusoidal time varying electric field. The two fields so produced are mutually perpendicular and are sources to each other. Such a combination of electric and magnetic fields constitute electromagnetic waves which can propagate through empty space in the direction perpendicular to both fields. Displacement current It is that current which comes into play in the region, whenever the electric field and hence, the electric flux is df changing with it. It is given by I D = e 0 E dt where e0 = absolute permittivity of free space, df E = rate of change of electric flux. dt In case of a steady electric flux linked with a region, the displacement current is zero. maxwell’s Equations •





q

∫s E ⋅ dS = e 0 , Gauss’s law for electrostatics.

This law gives the total electric flux in terms of charge enclosed by the closed surface. This law states that the electric lines of force start from positive charge and end at negative charge i.e., the electric lines of force do not form a continuous closed path.

•





∫s B ⋅ dS = 0, Gauss’s law for magnetism.

This law shows that the number of magnetic lines of force entering a closed surface is equal to number of magnetic lines of force leaving that closed surface. This law tells that the magnetic lines of force form a continuous closed path. This law also predicts that the isolated magnetic monopole does not exist.   df • ∫ E ⋅ dl = − dtB , Faraday’s law of electromagnetic induction. This law gives a relation between electric field and a changing magnetic flux. This law tells that the changing magnetic field is the source of electric field.   df • ∫ B ⋅ dl = m 0I + m 0e 0 dtE , Ampere-Maxwell law. This law states that the magnetic field can be produced by a conduction current as well as by displacement current. This law also states that the conduction current and displacement current together have a property of continuity. At an instant, in a circuit, the conduction current is equal to displacement current. Production of Electromagnetic waves

The electromagnetic wave is emitted when an electron orbiting in higher stationary orbit of atom jumps to one of the lower stationary orbit of that atom. The electromagnetic waves are also produced when fast moving electrons are suddenly stopped by the metal of high atomic number. Physics for you | january ‘16

31

SELF CHECK

SELF CHECK

1. For plane electromagnetic waves propagating in the z direction, which one of the following combination gives the correct possible direction   for E and B field respectively? ^

^

^

^

(a)

(i + 2 j) and (2 i − j)

(b) (c)

(−2 i − 3 j) and (3 i − 2 j)

(d)

(3 i + 4 j) and (4 i − 3 j)

^

^

^

^

^

^

^

^

^

momentum of Electromagnetic wave

^

(2 i + 3 j) and (i + 2 j) ^

^

(JEE Main 2015)

Energy Density and intensity

The electric field and magnetic field in a plane electromagnetic wave propagating along x-axis are given by x x   E = E0 sin w  t −  , B = B0 sin w  t −  c c Here, E0 and B0 are the amplitudes of the fields and c is the speed of light. 1 The energy of electric field is given by U E = e 0 E 2dV 2 and the energy of the magnetic field is given by 1 2 UB = B dV 2m 0 The average energy density of electric field is 2

1 1 E  1 uE = e 0 < E 2 > = e 0  0  = e 0 E02 2 2  2 4 The average energy density of magnetic field is 2

  B0   2 B02 < B >   2   uB = = =  2m 0 4m 0  2m 0  E 1 Using the relation, c = 0 = B0 m 0e 0 Total average energy density is given by 2 1 uav = u E + u B = 2uE = 2uB = e 0 E02 = 1 B0 2 2 m0 The unit of uE and uB are J m–3. Intensity of electromagnetic wave is defined as energy crossing per unit area per unit time perpendicular to the directions of propagation of electromagnetic wave. The intensity I is given by the relation I = uav c = 32

2. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is (a) 5.48 V/m (b) 7.75 V/m (c) 1.73 V/m (d) 2.45 V/m (JEE Main 2015)

B02c

1 1 e E 2c = 2 0 0 2 m0


The electromagnetic wave also carries linear momentum with it. The linear momentum carried by the portion of wave having energy U is given by p = U/c. If the electromagnetic wave incident on a material surface is completely absorbed, it delivers energy U and momentum p = U/c to the surface. If the incident wave is totally reflected from the surface, the momentum delivered to the surface is U/c – (–U/c) = 2U/c. It follows that the electromagnetic wave incident on a surface exert a force on the surface. radiant flux of electromagnetic wave According to Maxwell, when a charged particle is accelerated it produces electromagnetic wave. The total radiant flux at any instant is given by q 2a 2 P= (6pe 0c 2 ) where q is the charge on the particle, and a is its instantaneous acceleration. Poynting vector

The total energy flowing perpendicularly per second per unit area into the surface in free space is called a poynting vector S.    2   E×B . S = c e 0 (E × B) = m0 The S.I. unit of S is watt m–2. The rate of energy transfer for electromagnetic wave is proportional to the product of the electric and magnetic field strength, i.e. to the surface integral of the poynting vector formed by the component of the field in the plane of the surface.  The average value of poynting vector ( S ) over a convenient time interval in the propagation of electromagnetic wave is known as radiant flux density. When energy of electromagnetic wave is incident on a surface, the flux density is called intensity of wave (denoted by I). Thus I = S. Electromagnetic spectrum

The orderly distributions of electromagnetic radiations according to their wavelength or frequency is called the electromagnetic spectrum.

•

•

•

•

Radiowaves have wavelength longer than 0.1 m. They can be produced by electrons oscillating in wires of electric circuits and antennas can be used to transmit or receive radiowaves that carry AM or FM radio and TV signals. Microwaves can be regarded as short radio waves, with typical wavelength in the range 1 mm to 0.1 m. They are commonly produced by oscillating electric circuits, as in the case of microwave oven. Infrared radiation, which has wavelengths longer than the visible (from 7 × 10–7 m to about 1 mm), is commonly emitted by atoms or molecules when they change their vibrational motion or vibration with rotation. Infrared radiation is sometimes called heat radiation. Visible light, the most familiar form of electromagnetic wave, is that part of the spectrum the human eye can detect. The limits of wavelength

•

•

•

of the visible region are from 430 nm (violet) to 740 nm (red). Ultraviolet light covers wavelengths ranging from 4.3 × 10–7 m (430 nm) down to 6 × 10–10 m (0.6 nm). Ultraviolet rays can be produced by electrons in atoms as well as by thermal sources such as the sun. X-rays typical wavelength in the range of about 10–9 m (1 nm) to 10–13 m (10–4 nm) can be produced with discrete wavelength in atoms by transitions involving the most tightly bound electrons and they can also be produced in continuous range of wavelengths when charged particles such as electrons are decelerated. Gamma rays which have shortest wavelengths in the electromagnetic spectrum (less than 10 pm), are emitted in the decays of many radioactive nuclei and certain elementary particles.

Electromagnetic Spectrum Name

Frequency range (Hz)

Wavelength range

Production

Main properties and uses

Radiowaves 104 to 109

> 0.1 m

Rapid acceleration and decelerations of electrons in aerials.

Different specialised uses in radio communication.

Microwaves 109 to 1011

0.1 m to 1 mm

Klystron valve or magnetron valve.

Infrared 1011 to 5 × 1014 radiation Visible light 4 × 1014 to 7 × 1014

1 mm to 700 nm 700 nm to 400 nm

Vibration of atoms and molecules. Electrons in atoms emit light when they move from one energy level to a lower energy level.

(a) Radar communication. (b) Analysis of fine details of molecular and atomic structure. (a) Useful for molecular structure. (b) Useful for haze photography. Detected by stimulating nerve endings of human retina.

Ultraviolet radiation

1014 to 1017 400 nm to 1 nm

Inner shell electrons in atoms (a) Absorbed by glass. moving from one energy level to (b) Can cause the tanning of the a lower level. human skin. (c) Ionize atoms in atmosphere, resulting in the ionosphere.

X-rays

1017 to 1019 1 nm to 10–3 nm

X-ray tubes or inner shell electrons.

(a) Penetrate matter (b) Ionize gases (c) Cause fluorescence (d) Cause photoelectric emission from metals.

Gamma rays

1018 to 1022 < 10–3 nm

Radioactive decay of the nucleus.

Similar to X-rays.


33

oPtics

Light is a form of energy which makes objects visible to our eyes. The branch of physics, which deals with nature of light, its sources, properties, measurement, effects and version is called optics. It is divided in to two parts, ray optics and wave optics. reflection

A ray of light is composed of packets of energy known as photons. The photons have the ability of colliding with any surface. Thus, the photons transfer momentum and energy in the same way as the transfer of momentum and energy take place between any two particles during collision. When a photon is incident on a plane surface, it gets rebounded. This is known as reflection. Laws of Reflection : A light ray is reflected by a smooth surface in accordance with the two laws of reflection.

• •

The angle of incidence is equal to the angle of reflection i.e., ∠i = ∠r. The incident ray, the reflected ray and the normal to the reflecting surface are coplanar.

reflection by Plane mirror

If the object is real, the image formed by a plane mirror is virtual, erect, of same size and at the same distance from the mirror. characteristics If keeping the incident ray fixed, the mirror is rotated by an angle q, about an axis in the plane of the mirror, the reflected ray is rotated through an angle 2q. As every part of a mirror forms a complete image of an extended object and due to superposition of images brightness will depend on its light reflecting area, a large mirror gives more bright image than a small one. This in turn also implies that if a portion of a mirror is obstructed, complete image will be formed but of reduced brightness. If two plane mirrors are kept facing each other at an angle q and an object is placed between them, multiple images of the object are formed as a result of successive reflections. The number of images, n is given by • 34

Case-I : If m =

360 is even integer, number of q


images formed n = (m – 1) for all positions of object. 360° • Case-II : If m = is odd integer, number of q images formed n = m – 1, if the object is on the bisector of mirrors and n = m, if the object is not on the bisector of mirrors. 360° • Case-III : If m = is a fraction, number of q images formed will be equal to its integral part. When two plane mirrors are placed parallel to each other, then q = 0° ⇒ n = ∞. Therefore, infinite number of images are formed. If the mirror moves away or towards an object by a distance d, then the image moves away or towards the object by a distance 2d. If the mirror moves with speed v towards or away from fixed object then image appears to move towards or away from the object with speed 2v. If the object moves with speed v towards a fixed mirror, the image also moves towards the mirror with speed v. The speed of the image relative to the object in this case is 2v. KEY POINT •

The image formed by a plane mirror is laterally inverted and the linear magnification is unity.

reflection from a spherical surface

There are two types of spherical mirrors, concave and convex mirror as shown in figure.

For a concave mirror, the centre C of the sphere of which the mirror is a part is in front of reflecting surface, for a convex mirror it is behind. C is centre of curvature of the mirror, and P, the centre of the mirror surface is called pole. The line CP produced is the principal axis. AB is the aperture of the mirror. A narrow beam of rays, parallel and near to the principal

axis, is reflected from a concave mirror so that all rays converge to a point F on the principal axis. In figure, F is called the principal focus of the mirror and it is a real focus since light actually passes through it. Concave mirrors are also known as converging mirrors because of their action on a parallel beam of light.

sign conventions

All distances have to be measured from the pole of the mirror. Distances measured in the direction of incident A narrow beam of rays, parallel and near to the principal light are positive, and those measured in opposite axis, falling on a convex mirror is reflected to form a direction are taken as negative. diverging beam which appears to come from a point Heights measured upwards and normal to the principal F behind the mirror. A convex mirror has a virtual axis of the mirror are taken as positive, while those measured downwards are taken as negative. principal focus and it is also a diverging mirror. Image formation by concave mirror S.

Position of object

No 1. At infinity

Ray diagram

Position of image At focus F

Nature of image

Size of image

Real and inverted

Diminished

2.

Between infinity and centre of curvature C

Between centre of Real and inverted curvature C and focus F

Diminished

3.

At the centre of curvature C

At the centre curvature C

of Real and inverted

Same size

4.

Between the centre of curvature C and focus F

Between infinity and Real and inverted the centre of curvature C

Magnified

5.

At the focus F

At infinity

Real and inverted

Magnified

6.

Between focus F and pole P

Behind the mirror

Virtual and erect

Magnified


35

Image formation by convex mirror S.

Position of object

Ray diagram

No 1. At infinity

2.

Between pole P and infinity

mirror formula

If an object is placed at a distance u from the pole of a mirror and its image is formed at a distance v (from the 1 1 1 pole) then, = + f v u In this formula, to calculate unknown, known quantities are substituted with proper sign.

Position of image

Nature of image

Size of image

At focus F

Virtual and erect

Diminished

Between focus F and pole P

Virtual and erect

Diminished

4. A thin convex lens of focal length f is put on a plane mirror as shown in the figure. When an object is kept at a distance a from the lens - mirror combination, a in front of the its image is formed at a distance 3 combination. The value of a is

KEY POINT •

Mirror formula is independent of the angle the incident ray makes with the axis. Therefore all paraxial rays from point object must, after reflection, pass through image to give point image.

Power of mirror

1 100 where f should = f (in m) f (in cm) be taken with proper sign i.e., –ve for concave and +ve for convex mirror. The unit of power is dioptre (D) or m–1.

Power of mirror, P =

lateral magnification

A real image is always inverted one and a virtual one is always erect. Keeping these points in mind and that the real object and its real image would lie on the same sides in case of mirrors, we define magnification (m) in v case of reflection by spherical mirror as m = − . u

SELF CHECK

3. You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face and views the magnified image of the face at the closest comfortable distance of 25 cm. The radius of curvature of the mirror would then be (a) 30 cm (b) 24 cm (c) 60 cm (d) –24 cm (JEE Main 2015) 36


(a) f refraction

(b) 2f

(c) 3f

3 f 2 (JEE Main 2015) (d)

Light travels through vacuum at a speed of c = 3 × 108 m s–1. It can also travel through many materials, such as air, water and glass. However, atoms in the material absorb, re-emit, and scatter the light. Therefore, light travels through the material at a speed that is less than c, the actual speed, depending on the nature of the material. The change in speed as a ray of light goes from one material to another causes the ray to deviate from its incident direction. The index of refraction or refractive index m of a material is the ratio of the speed of light (c) in vacuum to the speed of light in the material (v). Mathematically, refractive index is given by the relation Speed of light in vacuum c m= = . Speed of light in the material v The absolute refractive index of a substance is the relative refractive index for light travelling from vacuum into the substance. It is commonly referred to as the refractive index.

Relative refractive index is a measure of how much light bends when it travels from any one substance into any other substance. Laws of refraction (snell’s Law)

When light strikes the interface between two transpanent materials, such as air and water, the light generally divides into two parts. Part of the light is reflected, with the angle of reflection being equal to the angle of incidence. The remainder is transmitted across the interface. If the incident ray does not strike the interface at normal incidence, the transmitted ray has a different direction than the incident ray. The ray that enters the second material is said to be refracted and behaves in one of the following two ways : • When light travels from a medium with smaller refractive index into a medium with larger refractive index, the refracted ray is bent toward the normal, as shown in figure (i). • When light travels from a medium with larger refractive index into a medium with smaller refractive index, the refracted ray is bent away from the normal, as shown in figure (ii). These two possibilities illustrate that both the incident and refracted rays obey the principle of reversibility. Thus, the directions of the rays in part (i) of the drawing can be reversed to give the situation depicted in part (ii). In part (ii) the reflected ray lies in the water rather than in the air. In both parts of figure, the angles of incidence, refraction, and reflection are measured relative to the normal. The angle of refraction q2 depends on the angle of incidence q1 and on the indices of refraction, m2 and m1, of the two media. The relation between these quantities is known as Snell’s law of refraction. According to Snell’s law, when light travels from a material with refractive index m1 into a material with refractive index m2, the refracted ray, the incident ray, and the normal to the interface between the materials all lie in the same plane. The angle of refraction q2 is related to the angle of incidence q1 by m1sinq1 = m2sinq2. Normal Incident ray Air(1) Water(2)

ray 1 1 2 (i)

Refracted ray

Normal Refracted ray Air(1)

2

Water(2)

1 1

ray

Incident ray

(ii)

Refractive Index of a Material Depends Upon Four Factors as Follows Material : Refractive index is maximum for diamond. Its value is 2.42. It is rated to be one for vacuum or almost unity for air. This value of refracting index is minimum. Colour of light : Refractive index of violet colour is maximum while that for red light is minimum. mV > mR. VIBGYOR denotes seven colours in descending order of refractive index. Thus mV > mI > mB > mG > mY > mO > mR. constant generally. m= λ2 B C Specifically, m = A + 2 + 4 . λ λ This is Cauchy’s formula. m −1 = constant r where r denotes density of material. Since density decreases with rise of temperature, refractive index also decreases with rise of temperature. i.e. m is inversely proportional to temperature, 1   m ∝ T  . Surrounding media : Refractive index of medium 2 with respect to medium 1 is represented as 1m2 or m21 and here the ray of light travels from 1 to 2. Refractive index of medium 1 with respect to medium 2 is denoted by 2m1. 1 = 1m 2 or 1m 2 × 2m1 = 1. 2 m1 If glass is surrounded by water, refractive index = wmg. velocity of light in water w mg = velocity of light in glass a mg 3/2 9 w mg = a = = . mw 4 / 3 8 Temperature and density :


37

KEY POINT • Quantities

which remain constant during refraction are :  Frequency of light as frequency depends upon source of light.  Phase angle. • Quantities which change during refraction are :  Speed of light, denser the medium, less the velocity.  Wavelength of light, denser the medium, less the wavelength.  Amplitude of light.  Intensity of light. This is due to the partial reflection and partial absorption of light. apparent Depth

If an object is placed below the surface of water or under a glass slab, it appears to be raised, i.e. the apparent depth is less than the real depth. This is due to refraction. Refractive index (m) can also be given as Real depth m= Apparent depth

For any angle of incidence q1 in medium 1, corresponding angle of refraction in medium 2 is q2. m sin q2 = 1 sin q1 m2 As m1 > m2, sinq2 > sinq1 and hence q2 > q1. In this way for a certain value of q1, q2 will be equal to 90°. This value of q1 is known as critical angle (qc). If q1 > critical angle, q2 should be greater than 90°. This implies that the ray will return to the previous medium. For the angle of incidence greater than critical angle, light does not transmit into medium 2 rather it reflects back to the medium 1. This phenomenon is called total internal reflection. Applications of Total Internal Reflection • Totally reflecting prisms • Optical fibres • Brilliance of diamond • Mirages refraction at curved surface

Lateral shift due to slab

For a spherical surface, m 2 − m1 = m 2 − m1 v u R The symbols should be carefully remembered as, m2 is refractive index of the medium in which light rays are entering, m1 is refractive index of the medium from which light rays are coming. Care should also be taken while applying the sign convention to R. lens

The perpendicular distance between the incident ray and the emergent ray, when the light is incident obliquely on a parallel sided refracting slab, is called lateral shift. In figure, distance d is lateral shift and is given as t d= sin(i − r), t = thickness of slab. cos r total internal reflection The medium whose refractive index is greater is known as optically denser medium whereas one which has smaller value is known as optically rarer medium. In the figure shown, light travels from medium 1 to 2, of corresponding refractive indices m1 and m2 and m1 > m2. 38


Lens is a transparent medium bounded by two surfaces, at least one of them must be curved. Thin Lens : A lens is said to be thin if the gap between two surfaces is very small. lens maker’s formula

If the refractive index of the material of the lens is m, R1 is the radius of curvature of the surface facing the object and R2 that of the other surface then focal length is 1 1  1 = (m − 1)  −  f  R1 R2 

lens formula

If any of the limitation is violated, then we have to use the refraction at the curved surface formula for both the surfaces.

1 1 1 − = v u f Here, u is the distance of the object from the lens, v is the distance of the image from the lens and f is the focal length. Values to be introduced with proper sign.

laws of formation of images by lens

limitation of the lens maker’s formula

•

• •

•

The rays coming parallel to principal axis of lens pass through the focus after refraction. The rays coming from the focus of lens go parallel to the principal axis of lens after refraction. The rays of light passing through optical centre go straight after refraction without changing their path.

The lens should be thin so that the separation • between the two refracting surfaces is small. The medium on either side of the lens should be same. Image formation by Convex Lens

S. No. Position of object 1. At infinity

Ray diagram

Position of image Nature and size of image At the principal Real, inverted and focus (F2) extremely diminished

2.

Beyond 2F1

Between F2 and Real, inverted and 2F2 diminished

3.

At 2F1

At 2F2

Real, inverted and of same size as the object

4.

Between F1 and 2F1

Beyond 2F2

Real, inverted and magnified

5.

At F1

At infinity

Real, inverted and highly magnified

6.

Between F1 optical centre

and

Beyond F1 on Virtual, erect the same side as magnified. the object


and

39

Power of the lens

1 Power of lens is, P = , where f should be in meters with f proper sign i.e., +ve for convex and –ve for concave, P is in dioptre. combination of the lenses

For a system of lenses, the net power of the system is P = P1 + P2 + P3 + ...... provided all the thin lenses are in close contact. The focal length of the net system can be written as 1 1 1 1 = + + + ........... f f1 f 2 f 3 f1, f2, f3 ... should be taken with proper sign. When a convex and a concave lens of equal focal length are placed in contact, the equivalent focal length is equal to infinite. Therefore, the power becomes zero. If the lenses are kept at a separation d, then the effective focal length f is d 1 1 1 = + − f f1 f 2 f1 ⋅ f 2 P = P1 + P2 – dP1P2 If the lens is converging, put +ve for its focal length and –ve for the diverging lens. silvering of lenses

If any surface of a lens is silvered, it will ultimately behave as a mirror and the power of mirror thus formed will be equal to the sum of powers of the optical lenses and the mirrors in between. change of focal length of a lens When a lens (convex lens) of refractive index m, is kept in any medium other than air, the focal length is given 1  m 1  1 as =  − 1  −  f  m m   R1 R2 

Case-II : When the refractive index of the medium is equal to that of the lens 1 i.e., mm = m ⇒ f = ∞ and P = = 0 f Therefore, the focal length of the lens becomes infinity and the power becomes zero. Consequently, the lens behaves like a glass slab (plate). • Case-III : When the refractive index of the medium is greater that of the lens, •

i.e., mm > m ⇒ f < f0. Therefore, the focal length of the lens decreases. Consequently, the power of the lens increases. Numerically, since mm > m ⇒ m – mm < 0 f is negative ⇒ f is negative. ⇒ f0 The negative focal length indicates that the nature of the lens changes from converging to diverging and vice versa. When a bi-(equi) convex lens is cut transversely into two equal parts, the radius of curvature of the lens in the cutting side increases to infinity. Now the focal length of each part increases to twice the previous value. When a lens is cut into two equal halves parallel to principal axis the focal length of each part remains constant. Prism

A prism has two plane surfaces inclined to each other.

Angle of prism, A = r1 + r2 Deviation of incident ray, d = i + e – A Minimum Deviation : From (i – d) graph, we conclude that deviation d is minimum, when ray of light passes symmetrically through the prism. Case-I : When the refractive index of the medium is less than that of the lens mm < m ⇒ f > f0 and P < P0 Therefore, the focal length increases and consequently the power of the lens decreases. The nature of the lens remains unchanged, since f is still positive, it remains converging. •

40


i = e ⇒ dm = 2i – A ⇒ i =

A + dm 2

i = e ⇒ r1 = r2 = r A ∴ A = 2r ⇒ r = 2 Putting i and r in terms of A and dm in Snell’s law sin i m= sin r  A + dm  sin  2  m= A sin 2 If the angle of prism is very small, sin A ≈ A A + dm 2 ⇒ m= ⇒ d m = A(m − 1) A 2 Since,

Dispersion

White light is actually a mixture of light of different colours. Therefore, when white light gets refracted under certain conditions, its components bend by different amounts and separate out. This phenomenon of splitting of light into its component colours, due to different refractive indices for different colours, is called dispersion of light.

where, w and w′ are the dispersive powers of the two prisms and d and d′ their mean deviations. Total deviation is given by w  d net = (mY − 1)A  1 −  w′ Dispersion without Deviation

A combination of two prisms in which deviation produced for the mean ray by the first prism is equal and opposite to that produced by the second prism is called a direct vision prism. This combination produces dispersion without deviation. For deviation to be zero, (d + d′) = 0 ⇒ (m – 1)A + (m′ – 1)A′ = 0 (m − 1) ⇒ A′ = − A (m′ − 1) (–ve sign implies prism A′ has to be kept inverted). Total angular dispersion q = (mY – 1)A(w – w′)

SELF CHECK 5. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is m, a ray, incident at an angle q, on the face AB would get transmitted through the face AC of the prism provided A θ

Angular dispersion, q = dV – dR Dispersive power : Ratio of angular dispersion to mean deviation d − dR w= V d where d is deviation of mean ray (yellow). As, dV = (mV – 1)A, dR = (mR – 1)A, m − mR ∴ w= V mY − 1 m + mR Here, mY = V 2 Deviation without Dispersion This means an achromatic combination of two prisms in which net or resultant dispersion is zero and deviation is produced. For two prisms, (mV − m R )A + ( mV′ − m′R ) A′ = 0 ⇒

(m − m R )A and wd + w′d ′ = 0 A′ = − V (mV′ − m′R )

B

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

C

   1 q > cos −1 m sin  A + sin −1      m       1 q < cos −1 m sin  A + sin −1      m       1 q > sin −1 m sin  A − sin −1      m       1 q < sin −1 m sin  A − sin −1      m    (JEE Main 2015)

optical instruments simple microscope

•

Distinct vision : Image is formed at least distance of distinct vision (D). D is about 25 cm to 30 cm. Physics for you | january ‘16

41

 D Magnifying power =  1+  . f  Normal vision : Image is formed at infinity during normal vision. D Magnifying power = . f • A convex lens of short focal length is used to form an image of object situated between focus and optical centre of lens. A magnified, erect and virtual image is formed. •

compound microscope •

Magnifying power for distinct vision v  D m = o 1 +  uo  fe 

vo and uo refer to objective lens, fe refers to eye-lens. • Magnifying power for normal vision v D m= o uo f e l 2m sin q

•

Limit of resolution (Dx) =

•

 1  2m sin q Resolving power =   = l  Dx 

m = Refractive index of medium q = Angle subtended by the ray with the axis of the microscope.

•

Compound microscope is mainly used in biological laboratories.

astronomical telescope •

42

Magnifying power for distinct vision f  f  m = o 1 + e  fe  D  Physics for you | january ‘16

fo and fe respectively refer to focal lengths of objective and eyepiece. D = 25 cm to 30 cm for normal eye. f • Magnifying power for normal vision m = o fe • Maximum length of tube = (fo + fe) 1.22 l • Limit of resolution = dq = where a l = Wavelength of light, a = Diameter of objective. a  1 • Resolving power =   =  dq  1.22 l

•

Telescope is used to observe astronomical bodies like sun, star, moon and planets.

SELF CHECK

6. A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is q, then q is close to (a) 1° (b) 15° (c) 30° (d) 60° (JEE Main 2015) wavE oPtics In case of mirrors and optical instruments, we use the model of geometrical optics in which we represent light as rays which are straight lines that are bent at a reflecting or refracting surface. Many aspects of the behaviour of light can’t be understood on the basis of propagation of rays. Light is fundamentally a wave and in some situations we have to consider its wave properties explicitly. If two or more light waves of the same frequency overlap at a point, the total effect depends on the phases of the waves as well as their amplitudes. The resulting patterns are as a result of the wave nature of light and can’t be understood on the basis of rays. Wave front : Particles of light wave which are equidistant from the light source and vibrate in the same phase constitute a wavefront. Depending on the type of light source, we generally have three types of wavefronts spherical, cylindrical, and plane.

•

Spherical wavefront : It is formed by a point source of light. As we move away from the point source of light, the radius of the spherical wavefront increases.

•

Cylindrical wavefront : When the source of light is linear, such as a slit, the light propagates in the medium by forming a cylindrical wavefront. The radius and height of a cylindrical wavefront increases as we move away from the linear source of light. Plane wavefront : For a point source or linear source of light, if the point of observation lies far away from the source, a plane wavefront of light is obtained.

•

huygens Principle

According to Huygens principle : • Each and every point on the given wavefront, called primary wavefront, acts as a source of new disturbances, called secondary wavelets, that travel in all directions with the velocity of light in the medium. • A surface touching these secondary wavelets tangentially in the forward direction at any instant gives a new wavefront at the instant, which is known as the secondary wavefront. Let us consider two wavefronts AB, shown in figure, for understanding Huygens principle. A2

B2

A1

A

A2

t a c

A a ct

b c

c

d

d

e

e B1

change in wavelength and velocity

The frequency of a wave is the characteristic of its source. Thus it does not change due to medium. Thus the velocity of a wave changes from one medium to another medium due to its change in wavelength. In a medium of absolute refractive index m, the velocity of wave is v = ul′ where l is wavelength in this medium. In free space, c = ul c ul l l ∴ = ⇒ = m ⇒ l′ = v ul ′ l′ m

SELF CHECK 7. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens’ principle leads us to conclude that as it travels, the light beam (a) bends downwards (b) bends upwards (c) becomes narrower (d) goes horizontally without any deflection (JEE Main 2015) coherent sources

b

B

A1

Let us assume that AB is the primary wavefront at any given instant of time. One of them is a plane wavefront and the other is a curved wavefront. Let us take points a, b, c, d and e on the two primary wavefronts AB. These points act as secondary wavelets. The distance travelled by light in Dt seconds is cDt, where c is the speed of light. Now with a, b, c, d, and e as centre and radius equal to cDt, draw circles. Then draw surface A1B1 and A2B2 touching these dotted circles tangentially as shown in figure. Since light travels in forward direction, wavefront A1B1 is the newly formed secondary wavefront. Thus by using Huygens principle, if we know the primary wavefront at any instant, we can calculate the secondary wavefront after instant t.

B2

B

B1

Two sources are said to be coherent if they emit light waves of the same frequency, nearly the same amplitude and are always in phase with each other. It means that the two sources must emit radiations of the same colour (wavelength). Presence of coherent sources is necessary to observe the phenomenon like interference. Phase difference and path difference : If the path difference between the two waves coming from two coherent sources is l (i.e. equal to the wavelength of light coming from the sources) the phase difference is Physics for you | january ‘16

43

2p. Therefore for a path difference x, the phase difference d is given as d = 2p x. l interference The term interference refers to any situation in which two or more waves overlap in space. When this occur, the total wave at any point at any instant of time is governed by the principle of superposition, which states that, when two or more waves overlap, the resultant displacement at any point and at any instant may be found by adding the instantaneous displacements that would be produced at the point by the individual waves if each were present alone. Interference effects are most easily seen when we combine sinusoidal waves with a single frequency u and wavelength l. In optics, sinusoidal waves are characteristic of monochromatic light (light of a single colour) as we analyze characteristic of interference. constructive and Destructive interference

When waves from two or more sources arrive at a point in phase, the amplitude of resultant wave is the sum of the amplitudes of the individual wave, the individual waves reinforce each other. This is called constructive interference. A crest of one wave arrives at the same time a trough of the other wave. The resultant amplitude is the difference between the two individual amplitudes. If the individual amplitudes are equal, then the total amplitude is zero. This cancellation or partial cancellation of the individual waves is called destructive interference. young’s Double slit Experiment

Here two coherent sources are obtained by allowing light from a source to pass through two narrow slits S1 and S2 separated by a small Screen distance d, and equidistant from a screen. For constructive interference (i.e. for the formation of bright fringes). • Path difference for nth bright fringe is given by d xn = nl = y n D where n = 1, 2, 3, ....., l = wavelength of light, yn = distance of nth bright fringe from central maxima on the screen. d = separation between the two coherent sources. D = perpendicular distance between screen and the two coherent sources. 44


• •

Phase difference dn = 2pn For central maximum, n = 0 and hence path difference and phase difference are zero.

Distance between two consecutive bright fringes lD . = d For destructive interference (i.e. for the formation of dark fringes) • Path difference for nth dark fringe is given by l d x n = (2n − 1) = y n (where n = 1, 2,...) 2 D Here yn is the distance of nth dark fringe on the screen from centre of the interference pattern. xn is an odd number multiple of half wavelength. 2p 1  • Phase difference d = x = 2p  n −  . n l n 2 th • Distance of n dark fringe from the centre of the pattern is given as (2n − 1)lD , where n = 1, 2, 3..... yn = 2d •

Alternately formed bright and dark fringes are parallel lD and are of equal width b = . d Thus b is proportional to l and D and inversely proportional to d. If the light waves coming from the two coherent sources be of equal amplitude a then If phase difference d = 0, 2p, 4p, ....... and x = 0, l, 2l ..... . Then, Intensity I = 4a2 Intensity is maximum for bright fringes. When phase difference is d = p, 3p, ...... and path difference is x = l/2, 3l/2, ........ I = 0 (i.e. Intensity is minimum.) Formation of interference fringes is in accordance with law of conservation of energy. intensity Distribution

If a, b are the amplitudes of interfering waves due to two coherent sources and f is constant phase difference between the two waves at any point P, then the resultant amplitude at P will be R = a 2 + b 2 + 2ab cos f If a2 = I1, b2 = I2, then Resultant intensity I = R2 = a2 + b2 + 2 ab cosf I = I1 + I 2 + 2 I1I 2 cos f When cos f = 1; I max = I1 + I 2 + 2 I1I 2 = When cos f = − 1, I min =

(

I1 − I 2

)

2

(

I1 + I 2

)

2

I max = I min

( (

I1 + I 2 I1 − I 2

) )

2 2

f If I1 = I2 = I0, then Imax = 4I0; Imin = 0. I = 4I0 cos2 2 If the sources are incoherent, I = I1 + I2 conditions for sustained interference Pattern • • • • • •

Superimposing light waves should be from coherent sources. The amplitude of the interfering waves should be same. Fringe width, b = lD should be of suitable value. d The angle at which superimposing waves are inclined should not be very large. Interfering waves should be in same state of polarisation. Two sources should be as narrow as possible. KEY POINT

• Fresnel used a biprism called Fresnel’s Biprism to

show the interference phenomenon.

SELF CHECK 8. In a Young’s double slit experiment with light of wavelength l the separation of slits is d and distance of screen is D such that D >> d >> l. If the fringe width is b, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is b b b b (a) (b) (c) (d) 2 4 3 6 (JEE Main 2015) interference in thin films

The coloured pattern formed by a thin film on water surface is due to the interference pattern between two beams. One is reflected from the top surface of the film and the other from the bottom surface. • For reflected light : l The correct path difference x = 2mt cos r − 2 t = thickness of film; m = refractive index of the film, r = angle at which light is refracted in film. For bright fringes : 2mt cos r = (2n − 1) l 2 n = 1, 2, .... For dark fringes : 2mtcosr = nl n = 0, 1, 2, ......

•

For transmitted light : Actual path difference x = 2mtcosr For bright fringes : 2mtcosr = nl n = 0, 1, 2, 3....

l For dark fringes: 2mt cos r = (2n − 1) 2 n = 1, 2, 3.... Diffraction Diffraction of light is the phenomenon of bending of light around corners of an obstacle or aperture in the path of light. The bending light penetrates into the geometrical shadow of the obstacle. The light thus deviates from its linear path. This deviation is more effective when the dimensions of the aperture or the obstacle are comparable to the wavelength of light. The phenomenon of diffraction is divided mainly in the following two classes • Fresnel diffraction • Fraunhoffer diffraction Fresnel Diffraction Fraunhoffer Diffraction The source is at a The source is at infinite finite distance. distance. No opticals are Opticals are in the form required. of collimating lens and focusing lens are required. Fringes are not sharp Fringes are sharp and well and well defined. defined. Greater the wavelength of a wave, higher will be its degree of diffraction. All types of waves exhibit diffraction . Appearance of a shining circle around the section of sun just before sun rise is due to diffraction of light. When an intense source of light is viewed with the partially opened eye, colours are observed in the light. It is due to diffraction. A single slit of width a gives a diffraction pattern with a central maximum. The intensity falls to zero at angles of ± l , ± 2l etc., with successively weaker secondary a a maxima in between .


45

Albert Einstein (1879-1955) One of the greatest physicists of all time, was born in Ulm, Germany. In 1905, he published three path breaking papers. In the first paper, he introduced the notion of light quanta (now called photons) and used it to explain the features of photoelectric effect. In the second paper, he developed a theory of Brownian motion, confirmed experimentally a few years later and provided a convincing evidence of the atomic picture of matter. The third paper gave birth to the special theory of relativity. In 1921, he was awarded the Nobel Prize in physics for his contribution to theoretical physics and the photoelectric effect.

BRAIN

Radiant energy

Evacuated chamber

Radiant energy

Emitted electrons

Metal surface

Metal surface

Photoelectric Cell l

Positive terminal Current indicator

+

Voltage source

Electron Microscope l

l

An electrical device which converts light energy into electrical energy, is called as photocell or photoelectric cell. It works on the principle of photoelectric emission of electrons

l

Transmission electron microscope

Electron microscope is a device designed to study very minute objects like viruses microbes and the crystal structure of solids It is based on principle of de Broglie wave and the fast moving electrons can be focussed by E or B field in a same way as beam of light is focussed by glass lenses.

Electron source Condenser lens Objective lens Specimen Diffraction lens

Fluorescent screen

Applicatio n of

es

DUAL NATURE OF RADIATION AND MATTER

Particle Nature of Radiation

de Bro gli eW av

Wave Nature of Matter

Nature's Love with symmetry arises the matter-wave duality

Photoelectric Effect The phenomena of emission of electrons from a metal surface when an electromagnetic wave of suitable frequency is incident on it is called photoelectric effect.

l

Davisson-Germer Experiment

Photoelectric Effect Electrons ejected Light from the shining surface Showed on clean sodium particle properties metal in a of light vacuum

de Broglie Hypothesis l

50°

Nickel crystal

Sodium metal

Photoelectric Equation

de Broglie Wavelength

E = Kmax + f₀ ; where f₀ = work function E = energy of incident light Kmax = maximum kinetic energy of electrons

l

Due to symmetry in nature, as radiation behaves like a particle, the material particle in motion also possesess wave-like properties. And the waves are called matter waves.

l d

Þ

Scattering angle

Lattice spacing

Bragg condition for constructive interference nλ = 2dsin

From quantum theory of radiation, E = hu and from Einsteinenergy of relativistic particles, E = pc Combining both, we have de Broglie wavelength,

l

de Broglie wavelength in terms of K.E. of electrons,

l

de Broglie wavelength of a charged particle having charge q and accelerated by potential V is

Accelerated electrons

Þ 54V

Electrons Light

+

Scattered electrons

Intensity

Photocathode

and momentum of charged particle,

Evacuated quartz tube 0

V

5

10 15 20 Accelerating voltage

25

A

Experimental Study and Conclusion of Photoelectric Effect

I3 > I2 > I1 u = constant

ip

I3

l

I2 I1 V0

u3 u2 u1 V03 V02 V01

V l

ip

I Saturation current

u3 > u2 > u1 V03 > V02 > V01

l

l

l

V

Photo-current is directly proportional to the intensity of incident light i.e., ip µ I. (At constant frequency u and potential V ) At constant frequency and intensity, the minimum negative potential at which the photocurrent becomes zero is called stopping potential (V0). At stopping potential V0 , Kmax of e– = eV0 For a given frequency of the incident radiation, the stopping potential is independent of its intensity. The stopping potential varies linearly with the frequency of incident radiation but saturation current value remains constant for fixed intensity.

Davisson Germer Experiment l l

Davisson Germer Experiment: Study of wave nature of particle. At a suitable potential V, the fine beam of electrons from electron gun is allowed to strike on the nickel crystal. The electrons are scattered in all directions and following assumptions were made : Ø Intensity of scattered electrons depends over scattering angle f. Ø Always a kink occurs in curve at f = 50°. Ø The peak is maximum at accelerating voltage 54 V. After this voltage, peak starts decreasing. Ø Here, Þ q = 65° at f = 50° From Bragg’s law (particle nature), l = 2d sin q Þ l = 1.65 Å. Also, from wave nature at V = 54 volt,

Central maximum is of maximum intensity and twice as wide as a secondary maximum. Path difference is odd multiple of l/2 for secondary maxima. For nth secondary maximum l xn = (2n + 1) ; n = 1, 2, 3, ...... 2 Phase difference dn = (2n + 1)p; n = 1, 2, 3, .... fl Width of such secondary maxima = a a = width of the slit f = focal length of the converging lens Path difference of nth secondary minimum xn = nl n = 1, 2, 3, ...... Therefore phase difference, dn = n2p fl Width of any secondary minimum is given as = a Intensity of central maximum is largest and its width = 2 f l . a Diffraction limits the angular resolution of a telescope to l/D where D is the diameter of objective lens. Two stars closer than this give strongly overlapping images. A beam of width a travels a distance a2/l, called the Fresnel distance, before it starts to spread out due to diffraction. KEY POINT • When the source of light is monochromatic, the diffraction patterns consists of alternate bright and dark bands of unequal widths. • When we use white light, the diffraction pattern is colored. Central maximum is white and the other bands with higher wavelength (red) is wider than the band with smaller wavelength (violet). Polarisation of light

Polarisation of light confirms the transverse nature of light. Polarisation of light is the phenomenon of restricting the vibrations of light (electric vector) in a particular direction, on passing ordinary light (unpolarised) through certain crystals like tourmaline crystal. The crystal acts as a polariser. The plane in which vibrations of polarised light are confined is called plane of vibration. (Plane ABCD) A plane perpendicular to the plane of vibration is called plane of polarisation. (Plane EFGH)

Polarisation can be obtained by three ways 48


Scattering Reflection Refraction Polariser like Nicol prism are used to analyse plane polarised light. • • •

KEY POINT • Light scattered in direction perpendicular to

incident light is always plane polarized.

Brewster's law and malus law

Angle of Polarisation : The angle of incidence for which an ordinary light is completely polarised in the plane of incidence when it gets reflected from a transparent medium. Brewster’s Law : It states that tangent of the angle of polarisation (i) is numerically equal to the refractive index (m) of the medium. sin i ⇒ m = tan i = cos i

The reflected and refracted rays are perpendicular to each other in this case. Law of Malus : It states that when a beam of completely plane polarised light is incident on an analyser, the resultant intensity of light (I) transmitted from the analyser varies directly as the square of the cosine of the angle (q) between plane of transmission of analyser and polariser, i.e., I ∝ cos2q.

SELF CHECK

9. Unpolarized light of intensity I0 is incident on surface of a block of glass at Brewster’s angle. In that case, which one of the following statements is true? (a) transmitted light is partially polarized with intensity I0/2. (b) transmitted light is completely polarized with intensity less than I0/2. (c) reflected light is completely polarized with intensity less than I0/2. (d) reflected light is partially polarized with intensity I0/2. (JEE Main 2015)

optical activity

When a plane polarised light enters the quartz plate, its plane of vibration is gradually rotated. The amount of rotation through which the plane of vibration is turned depends upon the thickness of the quartz plate and the wavelength of light. The property of rotating the plane of vibration by certain crystals or substances is known as optical activity and the substance is known as an optically active substance. Dextro-rotatory (right handed) substance rotates the plane of vibration to the right i.e. in clockwise direction if observer is looking towards the light approaching him. Laevo-rotatory (left handed) substance rotates the plane of vibration to the left (anticlockwise direction). specific rotation

Specific rotation for solids, is given as q = Sl where q is the rotation measured in degrees, l is the path length in mm and S is the specific rotation in degree mm–1. For solutions, S = 100 lc

where c is concentration of the solute in g cm–3 of the solution and l is the path length in centimetre. Hence units of S are degree cm–1. From above relation, specific rotation is defined as the rotation produced by a decimetre long column of the solution containing 1 g of the active substance in one cc of the solution.

KEY POINT • Polarimeter is the optical device used to determine

specific rotation of optically active solutions.

Polaroid

A synthetic substance in which one of the rays, with a particular direction of the electric vector along some dye molecules, is absorbed much more than the other is called polaroid. For practical purposes, a polaroid can be said to transmit waves having only one specific direction of the electric vector.

SELF CHECK 10. Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB respectively, I then A equals IB 1 3 (a) (b) 3 (c) (d) 1 3 2 (JEE Main 2014) 1. (b) 6. (d)

answEr kEys (sElf chEck) 2. (d) 3. (c) 4. (b) 5. (c) 7. (b) 8. (b) 9. (c) 10. (a) nn

POLAROID SUNGLASSES We use polaroid sunglasses to minimise glare. When light is reflected from a bright non-metallic surface, it suffers partial plane polarisation. This means that more of the reflected light is vibrating in one plane than the other. For example, if you look at the morning sun reflected from a lake’s surface the light waves that vibrate parallel to lake’s surface have larger amplitude than the reflected wave that vibrate perpendicular to it. Therefore if you look at the surface of the water from an angle close to the horizontal you get more light that is reflected off the surface than light that comes out of the water by refraction. You therefore cannot see what is under the surface, just a bright flash. To reduce the glare of this light and to view objects under the surface more clearly we can wear polaroid sunglasses. These contain polaroid cores with transmission axes that are oriented vertically so the glasses absorb the horizontally polarised light and as the glare is composed of that orientation of light they therefore reduce the glare. But they are not recommended for safe viewing of sun during a solar eclipse because they do not block out the damaging rays. They just cut down the glare. Fishermen wear polaroid glasses to eliminate reflected glare from the surface of a lake or stream and they can therefore see beneath the water more clearly. Photographers sometimes use polaroid sheets in front of lens to reduce the glare of light reflected from a surface. Physics for you | january ‘16

49

1. A spherical convex surface separates object and image space of refractive indices 1.0 and 4/3 respectively. If radius of curvature of the surface is 10 cm, then its power is (a) + 2.5 D (b) – 2.5 D (c) + 1.0 D (d) – 1.0 D 2. A combination is made of two lenses with focal lengths f1 and f2 and dispersive powers w1 and w2 respectively. The combination will be achromatic, if (a) w1 = 2w2 and f1 = 2f2 (b) 2w1 = w2 and f1 = –2f2 (c) w1 = 2w2 and f1 = –2f2 (d) 2w1 = w2 and 2f1 = f2 3. A lamp rated at 100 cd hangs over the middle of a round table with diameter 3 m at a height of 2 m. It is replaced by a lamp of 25 cd and height of the lamp is changed to 1 m so that the illumination at the centre of the table remains as before. The illumination at edge of the table becomes X times the original. Then X is 16 1 (a) (b) 27 3 1 1 (c) (d) 9 4 4. In a single slit diffraction experiment, first minimum for l1 = 660 nm coincides with first maxima for wavelength l2. Then l2 is (a) 220 nm (b) 330 nm (c) 440 nm (d) 660 nm 5. Interference fringes were produced in Young’s double slit experiment using light of wavelength 5000 Å. When a film of thickness 2.5 × 10–3 cm was placed in front of one of the slits, the fringe pattern shifted by a distance equal to 20 fringe widths. The refractive index of the material of the film is (a) 1.25 (b) 1.35 (c) 1.4 (d) 1.5 50


6. A parallel beam of light of intensity I is incident on a glass plate. 25% of light is reflected in any reflection by upper surface and 50% of light is reflected by any reflection from lower surface. Rest is refracted. The ratio of maximum to minimum intensity in interference region of reflected rays is 1  − (a)  2 1+ 2 1−  (c)  1+ 

3 8  3 8 3 8 3 8

    

2

2

1 2+ (b)  1− 2 1+  (d)  1− 

3 8  3 8 3 8 3 8

    

2

2

7. The electric field of an electromagnetic wave in free  space is given by E = 10 cos(107 t + kx )j V m −1 , where t and x are in seconds and metres respectively. It can be inferred that (i) the wavelength l is 188.4 m (ii) the wave number k is 0.33 rad m–1 (iii) the wave amplitude is 10 V m–1 (iv) the wave is propagating along +X-direction Which of the following pairs of statements is correct? (a) (iii) and (iv) (b) (i) and (ii) (c) (ii) and (iii) (d) (i) and (iii) 8. Consider the radiation emitted by the human body. Which one of the following statements is true? (a) The radiation emitted is in the infrared region. (b) The radiation is emitted only during the day. (c) The radiation is emitted during the summers and absorbed during the winters. (d) The radiation emitted lies in the ultraviolet region and hence is not visible. 9. A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0 cm from the center. An eye is placed

at a position such that the edge of the bottom is just visible. The particle P is in the plane of drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible? 4 (Take, refractive index of water = ) 3

(a) 3.5 cm (c) 18.9 cm

(b) 12.3 cm (d) 26.7 cm

10. The speed of electromagnetic wave in a medium of dielectric constant 2.25 and relative permeability 4 is (a) 1 × 108 m s–1 (b) 2.5 × 108 m s–1 8 –1 (c) 2 × 10 m s (d) 3 × 108 m s–1 11. A glass rod having square cross-section is bent into the shape as shown in the figure. The radius of the inner semi-circle is R and width of the rod is d. Find the maximum value of d/R so that the light that enters at A will emerge at B. (Take, refractive index of glass = 1.5) (a) 0.2 (b) 0.5 (c) 0.8 (d) 1.0 12. Find the focal length of equivalent system shown in figure. Radius of curvatures of different surfaces and refractive indices are given in the figure.

13. An object ABED is placed in front of a concave mirror beyond the centre of curvature C as shown in figure. Then

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

|mAB| < 1 and |mED| < 1 |mAB| > 1 and |mED| < 1 |mAB| < 1 and |mED| > 1 |mAB| > 1 and |mED| > 1

14. Light from a source emitting two wavelengths l1 and l2 is allowed to fall on Young’s double-slit apparatus after filtering one of the wavelengths. The position of interference maxima is noted. When the filter is removed both the wavelengths are incident and it is found that maximum intensity is produced where the fourth maxima occurred previously. If the other wavelength is filtered, at the same location the third maxima is found. Find the ratio of wavelengths. 3 2 3 4 (a) (b) (c) (d) 5 3 4 5 15. An electromagnetic wave in vacuum has the   electric and magnetic fields E and B, which are always perpendicular to each other. The direction  of polarization is given by X and that of wave propagation by k. Then      (a) X || B and k ||(B × E )      (b) X || E and k ||(E × B)      (c) X || B and k ||(E × B)      (d) X || E and k ||(B × E ) 16. The electric field associated with an electromagnetic  waveinvacuumisgivenby E = i 40 cos(kz − 6 × 108 t ), where E, z and t are in volt m–1, metre and second respectively. The value of wave vector k is (a) 2 rad m –1 (b) 5 rad m –1 (c) 6 rad m –1 (d) 3 rad m –1

50 cm 9 100 cm (c) + 13 (a) +

(b) −

100 cm 7

(d) −

50 cm 3

17. On a horizontal plane mirror, a thin equiconvex lens of glass is placed and when the space between the lens and mirror is filled with a liquid, an object held at a distance D which is 30 cm vertically above the lens Physics for you | january ‘16

51

is found to coincide with its own image as shown in figure. If equiconvex lens of glass has refractive index m = 1.5 and radius of curvature R, then find refractive index of the liquid. R R +1 (a) (b) D D R 2R (c) 2 − (d) 1 − D D 18. Choose the correct ray diagram of a thin equi-convex lens which is cut as shown in the figure. (a)

(b)

(c)

(d)

19. A man is standing on a straight road in a summer noon. The refractive index of the air changes with y  due to variation of height as m = m0 1 +  3(metre)  the vertical temperature where m0 is the refractive index of the air at road surface and y height. Eyes of the man are at height of 1.5 m above the road. What is the apparently visible length of the road? 3 m (b) 2 m (c) 4 m (d) 6 m (a) 5 5 5 5 20. An object AB of length 2 cm is kept in front of concave mirror of radius of curvature 20 cm as shown in figure. The upper half of mirror is exposed to air and lower half is exposed to a medium having refractive index 1.5. For this situation mark out the correct statement about the image of AB formed by mirror.

(a) The difference between x-coordinates of image of ends A and B is zero. (b) The difference between x-coordinates of image of ends A and B is 2 cm. (c) The difference between x-coordinates of image 2 of ends A and B is cm. 3 (d) None of these. 52


21. A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength l travelling in air is incident normally on the layer. It is partly reflected at the upper and lower surfaces of the layer and the two reflected rays interfere. If l = 648 nm, obtain the least value of t for which rays interfere constructively. (a) 45 mm (b) 90 mm (c) 45 nm (d) 90 nm 22. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young’s double-slit experiment is (a) infinite (b) five (c) three (d) zero 23. When an unpolarised light of intensity I0 is incident on a polarising sheet, the intensity of the light which does not get transmitted is I0 I (a) 0 (b) 4 2 (c) zero (d) I0 24. A telescope with objective of focal length 60 cm and eyepiece of focal length 5 cm is focussed on a far off distant object such that parallel rays emerge from the eyepiece. If object subtends an angle of 2° on the objective, angular width of the image will be (a) 10° (b) 30° (c) 24° (d) 60° 25. An astronaut is looking down on Earth’s surface from a space shuttle at an altitude of 400 km. Assuming that the astronaut’s pupil diameter is 5 mm and the wavelength of visible light is 500 nm, the astronaut will be able to resolve linear objects of the size of about (a) 0.5 m (b) 5 m (c) 50 m (d) 500 m 26. The frequency of electromagnetic wave which is best suited to observe a particle of radius 3 × 10–6 m is of the order of (a) 1015 Hz (b) 1013 Hz 14 (c) 10 Hz (d) 1012 Hz 27. A ray of light falls on a transparent sphere with centre at C as shown in figure.

The ray emerges from the sphere parallel to line AB. The refractive index of the sphere is 3 1 (a) 2 (b) 3 (c) (d) 2 2 28. Light of wavelength 500 nm travelling with a speed of 2.0 × 108 m s–1 in a certain medium enters another medium of refractive index 5/4 times that of the first medium. What are the wavelength and speed in the second medium? Wavelength (nm) speed (m s–1) (a) 400 1.6 × 108 (b) 400 2.5 × 108 (c) 500 2.5 × 108 (d) 625 1.6 × 108 29. A cubic container is filled with a liquid whose refractive index increases linearly from top to bottom. Which of the following represents the path of a ray of light inside the liquid? (a)

(b)

4 / 3 1 . 0 4 / 3 − 1. 0 + = f +10 ∞ f = 40 cm = 0.4 m

\ or

Since, the rays are converging, its power should be positive. Hence, +1 +1 P(in dioptre) = or P = +2.5 D = f (metre) 0.4 2. (c) : Condition of achromatic combination is ω1 ω2 + =0 satisfied if f1 f2 3. (a) : For Ist case : 100 IP = 2 = 25 lux 2 IQ =

(c)

(d)

=

30. Match the electromagnetic waves given in column-I with their detection techniques given in column-II and select the correct option from the given codes. Column-I

Column-II

P.

Microwaves

1.

Photo cells

Q.

IR waves

2.

GM counter

R.

UV waves

3.

Point contact diodes

S.

g-rays

4.

Bolometer

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

P - 3, Q - 4, R - 2, S - 1 P - 3, Q - 4, R - 1, S - 2 P - 4, Q - 3, R - 1, S - 2 P - 4, Q - 3, R - 2, S - 1 SolutionS

1. (a) : The point where the parallel rays converge (or diverge) on the principal axis is its focus and the corresponding length is the focal length f. µ µ µ − µ1 Using 2 − 1 = 2 v u R Here, u = –∞, v = f, R = 10 cm 4 µ1 = 1.0, µ2 = 3

100 cos q (2.5)2 100 2 200 lux × = 2 2. 5 (2.5) (2.5)3

For IInd case : IQ′ = XIQ IQ′ =

25 lux (3.25)3/2 IQ′

(2.5)3 25 1 × = IQ 200 (3.25)3/2 3 1 \ X= 3 4. (c) : Position of minima in diffraction pattern is given by, dsinq = nl For first minima of l1, we have dsinq1 =(1)l1 l or sin q1 = 1 d Position of maxima in diffraction pattern is given (2n + 1) by, d sin q = l 2 \ For wavelength l2, 3l 3 d sin q2 = l2 or sin q2 = 2 2 2d \

=


53

The two will coincide if, q1 = q2 or sinq1 = sinq2 l1 3l2 \ = d 2d 2 2 or l2 = l1 = × 660 nm = 440 nm 3 3

\

 lD  (m − 1)tD = 20   d  d m = 1+

\

 1  4 3 1 = sin q or sin q =   2 3 4 2 3 3 tan q = = 16 × 2 − 9 23

⇒ In DAPM, PM 5 + x 5 + h − 15 tan q = = = AM h h 3 h − 10 = ⇒ 3h = h 23 − 10 23 h 23

(m − 1)tD d

5. (c) : Shift in fringe pattern = or

⇒

 10 23  h= cm = 26.7 cm  23 − 3 

20(5.0 × 10−7 ) 20 l = 1+ = 1. 4 t 2.5 × 10−5

Hence, water should be poured upto height 26.7 cm to make the particle P visible.

6. (d) :

10. (a) : n = εr mr = 4 × 2.25 = 3 v= I 9I I1 = and I2 = 4 32 \

\

I1 8 = or I2 9 I max I min

  =   

I1 2 2 = 3 I2 2

 I1 1+ +1 2 I2    + 2 2 3  = =     2 2 − 3  1− I1 −1  I2 

3 8 3 8

    

2

7. (d) 8. (a) 9. (d) : If we pour water in vessel, refraction will take place at air and water interface. Applying Snell’s law at A, we get 4 1 × sin 45° = sin q 3

A  = 4/3 5 cm P

54

45°



h

x X M 30 cm


30 cm

c 3 × 108 8 –1 = m s −1 = 1 × 10 m s n 3

11. (b) : If smallest angle of incidence q is greater than critical angle, then all light will emerge out at B. 1 1 ⇒ q ≥ sin −1   ⇒ sin q ≥ m m From figure, sin q = R R+d R 1  d ≥ ⇒ 1 +  ≤ m R+d m  R d ⇒ ≤ m −1 R d  = 1.5 − 1 = 0 .5 ⇒   R  max ⇒

12. (c) : Using lens makers formula 1 1 1  = (m − 1)  −  , f R R  1

2

1 1 3  1 1  1 2 =  − 1  +  = × =     10 10 2 10 10 2 f1 1  6   −1 1  1  −30  −3 =  − 1  −  = ×  = f2  5   10 20  5  10 × 20  100 1 8  1 1  3 =  − 1  +  = f3  5   20 20  50 Hence, equivalent focal length, 1 1 1 1 1 −3 3 = + + = + + f f1 f2 f3 10 100 50 ⇒ f = 100/13 cm

13. (a) : Object is placed beyond C. Hence, the image will be real and it will lie between C and F. Further, u and v are negative, hence the mirror formula will become 1 1 1 + = v u f f 1 1 1 u− f = − = or v = v f u uf 1 − ( f / u)

1 1 1 = + F f g fl 1 2 1 = (1.5 − 1) = fg R R (m − 1) 1  1 1 = (m − 1)  − −  = −  R ∞ fl R 1 1  m −1  2 − m  = −  = F R  R   R  Since, F = D R \ m=2− D 18. (c) E

19. (a) : Road

Now, uAB > uED \ vAB < vED

A

v  \ | mAB | < | mED |  as m = −   u Therefore, shape of the image will be as shown in figure. Also note that, vAB < uAB and vED < uED. So, |mAB| < 1 and |mED| < 1. 14. (c) : The maxima is obtained whenever nlD xcm = . d For first wavelength, ( x 4 )l = 1

4 l1D d

For second wavelength, (x3) l2 = 

( x 4 )l = ( x 3 )l

\

l 4 l1D 3l2 D 3 = ⇒ 1= d d l2 4

1

3l2 D d

2

15. (b) 16. (a) : As w = 6 × 108 rad s–1, v = c = 3 × 108 m s–1 8 −1 w 6 × 10 rad s = 2 rad m −1 k= = 8 − 1 v 3 × 10 m s

90°



h = 1.5 m

B l C Virtually visible length Actual length

As per Snell’s law,  1.5  m0 sin 90° = m0 1 +  sin q  3  2 or sin q = 3 5 \ cot q = 2 h Now, cot q = l h 2 3 \ l= m = 1.5 × = cot q 5 5 20. (c) : As focal length of mirror is independent of surrounding medium, so the image of ends A and B formed by concave mirror is having the same x-coordinates. 21. (d) : R 1 and R 2 are the two rays considered for interference. R 1 is the result of reflection at denser medium. Hence, it suffers an additional path difference l/2 and phase difference p. Ray R2 originates after reflection at lower surface, this reflection takes place at rarer medium.

17. (c) : If an object has to coincide with its image, then the ray have to retrace its path; hence, it is concluded that object is at the focus of the combination. Focus of equivalent combination Physics for you | january ‘16

55

Net path difference, l 2mt + = nl [for constructive interference] 2 or 2mt = (2n − 1)l 2 t will be minimum for n = 1. l 648 nm = 90 nm = tmin = 4 m 4 × 1. 8 22. (b) : As d sinq = nl and d = 2l, 2 sinq = n and as such maximum value of n = 2. Thus, interference maxima are formed when n = 0, n = ±1, n = ±2, i.e., maximum number of possible interference maxima is 5. 23. (a) : As intensity of transmitted light, I = I0cos2q I 1 and the average value of cos2 q = , \ I = 0 . 2 2 Intensity of light which is not transmitted, I I I0 − 0 = 0 2 2 24. (c) : M =

b f0 60 cm = = = 12, a fe 5 cm

b = 12a = 12 × 2° = 24° 25. (c) : As dq = =

1.22 ld x 1.22 l = ,\ x= D d D

1.22(500 × 10−9 m)(400 × 103 m) = 48.8 m ≈ 50 m (5 × 10−3 m)

26. (c) : As the particle can be best observed if l is of the order of size of the particle, i.e., 3 × 10–6 m υ=

8 −1 c 3 × 10 m s = 1014 Hz = l 3 × 10−6 m

27. (b) : Deviation by a sphere is 2(i – r). Here, deviation d = 60° = 2(i – r) or i – r = 30° \ r = i – 30° = 60° – 30° = 30° sin i sin 60° = 3 \ m= = sin r sin 30° 28. (a) : The refractive index, 1n2 (from medium 1 to medium 2) is speed of light in medium 1(v1 ) 1 n2 = speed of light in medium 2(v2 ) 56


=

wavelength of light in medium 1(l1 )

wavelength of light in medium 2(l2 ) n2 5 Now, 1 n2 = = n1 4 v1 = 2.0 × 10–8 m s–1 l1 = 500 nm \

8 −1 500 nm 5 2.0 × 10 m s = = 4 v2 l2

Hence, l2 = 400 nm v2 = 1.6 × 108 m s–1 29. (a) : Since the refractive index is changing, the light cannot travel in a straight line in the liquid. Initially, it will bend towards normal and after reflecting from the bottom, it will bend away from the normal as shown here.

30. (b)

nn

PRACTICE QUESTIONS 2016 GENERAL INSTRUCTIONS (i) (ii) (iii) (iv) (v) (vi) (vii)

All questions are compulsory. Q. no. 1 to 5 are very short answer questions and carry 1 mark each. Q. no. 6 to 10 are short answer questions and carry 2 marks each. Q. no. 11 to 17 are also short answer questions and carry 3 marks each. Q. no. 18 is a value based question and carries 4 marks. Q. no. 19 and 20 are long answer questions and carry 5 marks each. Use log tables if necessary, use of calculators is not allowed.

1. What is the advantage in choosing the wavelength 2. 3. 4.

5. 6. 7.

8.

of a light radiation as a standard of length? Why does a tennis ball bounce higher on hills than on plains? If the kinetic energy of a satellite revolving around the Earth in any orbit is doubled, what will happen to it? How much height can a 60 kg man climb by using the energy from a slice of bread which produces 100,000 cal? Assume that the efficiency of human body is 28%? The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe 2 m long. Calculate the length of the open pipe. A block of mass m moving at a speed v0 compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring. The angular speed of motor wheel is increased from 1200 rpm to 3120 rpm in 16 second. (i) What is its angular acceleration, assuming the acceleration to be uniform ? (ii) How many revolutions does the engine make during this time? Obtain an expression for the height to which a liquid of density r and surface tension T, will rise 1 in a capillary tube of radius r. Given that h ∝ . r

OR Two straight lines drawn on the same displacementtime graph make angles 30° and 60° with time axis, as shown in figure. Which line represents greater velocity? What is the ratio of the two velocities?

9. Determine the radius of a drop of water falling

through air if it covers 4.8 cm in 4 s with a uniform velocity. Assume density of air as 0.0012 g cm–3 and h for air as 1.8 × 10–4 poise.

10. Calculate the temperature at which rms speed of

gas molecules is double of its rms speed at 27°C, pressure remaining constant. 11. A ball of mass 1 kg hangs in equilibrium from two strings OA and OB as shown in figure. What are the tensions in strings OA and OB? (Take g = 10 m s–2) Physics for you | january ‘16

57

12. How will the value of g be affected if

(i) the rotation of the earth stops (ii) the rotational speed of the earth is doubled (iii) the rotational speed of the earth is increased to seventeen times its present value? 13. A truck can move up a road rising 1 in 25 with a speed of 24 km h–1 with frictional force (1/50)th of the weight of truck. What will be the speed of the truck moving down the same road with the same power? 14. A bug shown in figure has just lost its footing near

the top of the stationary bowling ball. It slides down the ball without appreciable friction. Show that it will leave the surface of the ball at the angle q = cos–1(2/3). Bug



O

15. A rocket is fired vertically from the surface of Mars

with a speed of 2 km s–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of Mars before returning to it? (Take mass of Mars = 6.4 × 1023 kg; radius of Mars = 3395 km; G = 6.67 × 10–11 N m2 kg–2) OR

For the travelling harmonic wave, y (x, t) = 2.0 cos 2p[10t – 0.0080 x + 0.35], where x and y are in cm and t in s, what is the phase difference between oscillatory motion at two points separated by distance of (i) 4 m (ii) 0.5 m ? 16. A metallic sphere of radius 1.0 ×10

–3

m and density 1.0 × 104 kg m–3 enters a tank of water, after a free fall through a distance of h in the earth's gravitational field. If its velocity remains unchanged after entering water, determine the value of h. Given coefficient of viscosity of water = 10 × 10–3 N s m–2, g = 10 m s–2 and density of water = 1.0 × 103 kg m–3.

17. A certain volume of dry air at NTP is allowed

to expand four times its original volume under (i) isothermal conditions (ii) adiabatic conditions. Calculate the final pressure and temperature in each case. Given g = 1.4.

58


18. Read the given passage and answer the following

questions. Surface tension is the property of a liquid by virtue of which free surface of liquid at rest tries to have minimum surface area. In doing so, the free surface of liquid at rest behaves as if it is covered with a stretched membrane. Surface tension (S) of a liquid is measured by the force (F) acting on unit length of a line (l) imagined to be drawn tangentially anywhere on the free surface, i.e. S = F . S is measured in N m–1. l (i) What is the cause of surface tension? (ii) A wire ring of 30 mm diameter resting flat on the surface of a liquid is raised. The pull required is 1.5 gf more before the film breaks than it is after. What is surface tension of the liquid? (iii) What are the implications of this phenomenon in day to day life? 19. A stone is thrown from the ground towards a wall of height H with a speed v0 at an angle q with the horizontal. Show that the stone must be thrown from a point at a distance 1  2 2 2  2 g v 0 sin 2q ± 2v 0 cos q v 0 sin q − 2 gH  from the foot of the wall in order that it may just clear it. OR A body of weight W can just be supported on a rough inclined plane by a force P parallel to the floor. It can also be supported by a force Q parallel to the plane. If q is the angle of friction, find cosq in terms of P, Q and W only. 20. Show that for a particle in linear S.H.M., the average kinetic energy over a period of oscillation equals the average potential energy over the same period. OR

An air chamber of volume V has a neck area of cross section A into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes S.H.M. Obtain an expression for the time period of oscillations assuming pressure volume variations of air to be isothermal. solutions 1. (i) The wavelength of light is independent of both

environment and time. (ii) The length-standard does not undergo any change with place.

2. The maximum height of a projectile is inversely

proportional to the value of acceleration due to gravity. So smaller the acceleration due to gravity, greater is the maximum height. Since the value of acceleration due to gravity is less on the hills than on the plains, therefore, a tennis ball will bounce higher on hills than on plains.

1 = (40 p × 16 + × 4p × 162 ) rad = 1152 p rad 2 1152 p Number of revolutions = = 576 2p 8.

h = k ra T bg cr–1 [M0L1T0]=[ML–3]a[MT–2]b[LT–2]c[L–1] =[Ma+b L–3a+c–1T–2b–2c] On comparing, we get a + b = 0, – 3a + c – 1 = 1 and –2b – 2c = 0 Hence, a = – 1, b = 1, c = –1 kT T Thus, h = kr −1Tg −1r −1 = or h ∝ rrg rrg or \

The total energy of satellite in any orbit, E = – K, where K is its kinetic energy in that orbit. If its kinetic energy is doubled, i.e., an additional kinetic energy (K) is given to it, E = – K + K = 0 and the satellite will leave its orbit and go to infinity. 4. Effective work = mgh 28 × 100, 000 × 4.2 = 60 × 9.8 × h 100 28 × 100, 000 × 4.2 or h = m = 200 m 100 × 60 × 9.8 5. Let lo and lc represent the lengths of open organ pipe and closed organ pipe respectively. 3v Frequency of second overtone of open pipe = 2lo 3.

3v Frequency of first overtone of closed pipe = 4lc 3v 3v Equating, = or 2lo = 4lc 2lo 4lc or lo = 2lc = 2 × 2 m = 4 m 6.

Loss in kinetic energy of the block

2 1 1 3 v K = mv 02 − m  0  = mv 02 2 2 2 8 Gain in elastic potential energy of the spring, 1 U = kx 2 2 1 2 3 2 As U = K , \ kx = mv 0 2 8

or 7.

k=

3 mv 02 4x 2

(i) ω 0 = 2p × 1200 rad s −1 = 40p rad s −1 60 2p × 3120 ω= rad s −1 = 104p rad s −1 60 ω − ω0 \ Angular acceleration, a = t 104 p − 40 p = = 4 p rad s −2 16 1 2 (ii) q = ω 0t + at 2

1   as h ∝ r 

h ∝ ra T bg cr–1

OR

Slope of displacement-time graph = Velocity of the object As slope of line OB > slope of line OA \ The line making angle of 60° with time-axis represents greater velocity. v tan 30° Ratio of the two velocities = A = v B tan 60° 1/ 3 1 = = = 1: 3 3 3 9.

Since the drop covers 4.8 cm in 4 s with uniform velocity, i.e., the terminal velocity vt = 4.8 cm/4 s = 1.2 cm s–1 Density of drop (water), r = 1 g cm–3 h = 1.8 × 10–4 poise As vt =

2r 2(r − s)g 9h

9hvt 9(1.8 × 10 −4 )1.2 = 2(r − s)g 2(1 − 0.0012)980 –4 or r = 9.97 × 10 cm r=

10. Let t°C be the temperature at which the rms speed

of the gas molecules (vrms)t is double of its value at 27°C [i.e., (vrms)27]. (v rms )t 273 + t 273 + t = = 273 + 27 300 (v rms )27 According to the problem, (v rms )t (v rms )t = 2(v rms )27 or =2 (v rms )27

Now,

or

273 + t 273 + t = 2 or = 4 or t = 927 ° C 300 300 Physics for you | january ‘16

59

11.

Various forces acting at the point O are as shown in figure. The three forces are in equilibrium. Using Lami's theorem, T1 T2 10 = = sin 150° sin 120° sin 90° or \

T1 T2 10 = = sin 30° sin 60° 1 T1 = 10 sin 30° = 10 × 0.5 = 5 N

and T2 = 10 sin 60° = 10 ×

3 =5 3N 2

12. (i) If the rotation of the earth stops, no centrifugal

force will act on the bodies lying on it. The value of g increases maximum at the equator. As no centrifugal force acts on body at poles, so the value of g is not affected there. (ii) If the rotational speed of the earth is doubled, the centrifugal force on the bodies increases. The value of g decreases maximum at the equator and is not affected at poles. (iii) If the rotational speed of the earth is increased to seventeen times its present value, the value of g at the equator will become zero. At the poles, the value of g remains unchanged. 1 25 Total upward force required, F = mg sinq + f or F = mg  1  +  1  mg =  3  mg  25   50   50 

13. Here, sin q =

3 Power, P = Fv =   mgv  50  Net downward force, F′ = mgsinq – f or F ′ = mg  1  −  1  mg =  1  mg  25   50   50  As P = F ′v′, \  3  mgv =  1  mgv′  50   50  or v′ = 3v = (3 × 24) km h–1 = 72 km h–1 14. Let the bug lose its footing at point near A and then

leave the ball at B where its velocity is v.

60


2 At B, mv = mg cos q − R ... (i) r Decrease in potential energy of the bug as it slides down from A to B, i.e., mgh = mgr(1 – cosq) ... (ii) [as h = AC = OA – OC = r – rcosq = r(1– cos q)] 1 Increase in kinetic energy of the bug = mv 2 ... (iii) 2 From eqns. (ii) and (iii), 1 2 mv = mgr(1 − cos q) 2 mv 2 or ...(iv) = 2mg (1 − cos q) r From eqns. (i) and (iv), 2mg(1 – cosq) = mg cos q – R Since at B, bug loses its contact with the ball, R = 0. Thus, 2mg(1– cosq) = mgcosq 2 2 or cos q = or q = cos −1   3 3

15. Let h be the maximum height attained by the

rocket. Change in potential energy of the rocket, = final potential energy – initial potential energy Mm  1  Mm  = GMm  1 = −G − −G −   (R + h)   R R + h  R h ... (i) = GMm R(R + h)

Since 20% of the kinetic energy of the rocket is lost due to martian atmosphere, Kinetic energy of the rocket which is converted into its potential energy 80 1 2 = × mv = 0.4 mv 2 ... (ii) 100 2 Applying the law of conservation of energy, h GMm = 0.4 mv 2 R(R + h) h or GM = 0. 4 v 2 2 R + Rh

or

h=

R2 11.526 × 1012 = m 6 6  GM  × − × 26 . 68 10 3 . 395 10  −R 0.4v 2 

= 495 × 103 m = 495 km OR

The given equation can be rewritten as y = 2.0 cos [2p(10 t – 0.0080 x) + 2p × 0.35]   10t  y = 2.0 cos 2p × 0.0080  − x  + 0.7 p   0 . 0080   Comparing it with the standard equation of a travelling harmonic wave, 2p y = r cos  (vt − x) + φ 0   λ  2p we get, = 2p × 0.0080 λ Further, we know that for the path difference x, the 2p phase difference φ = x λ (i) when, x = 4 m = 400 cm 2p φ= x = 2p × 0.0080 × 400 = 6.4p rad λ (ii) when, x = 0.5 m = 50 cm 2p φ= x = 2p × 0.0080 × 50 = 0.8 p rad λ 16. The velocity attained by the sphere after falling freely from height h is v = 2 gh ...(i) After entering water, the velocity of the sphere does not change. So v is also the terminal velocity of the sphere. Hence v=

2 r2 (r − s)g 9 h

But r = 104 kg m–3, s = 103 kg m–3, r = 10–3 m, g = 10 m s–2, h = 10–3 N s m–2

2 (10 −3 )2 × (10 4 − 103 ) × 10 \ v= × = 20 m s −1 −3 9 10 v 2 20 × 20 = = 20 m From (i), h = 2 g 2 × 10 17. (i) Isothermal expansion

V1 = V, V2 = 4V P1 = 76 cm of mercury, P2 = ? P2V2 = P1V1 V P 76 or P2 = P1 1 = 1 = cm of Hg = 19 cm of Hg 4 V2 4

Since the process is isothermal, therefore the final temperature will be the same as the initial temperature, i.e. 273 K. (ii) Adiabatic expansion V1 = V, V2 = 4 V P1 = 76 cm of Hg, P2 = ? T1 = 273 K , T2 = ? g = 1.4 g

g

Now, P2V2 = P1V1 g

1. 4 V  V  or P2 = P1  1  = 76   4V   V2 

or

P2 = 76 × (0.25)1.4 = 10.91 cm of Hg

Again, T2V2g–1 = T1V1g–1 V  or T2 = T1  1   V2 

g −1

1.4 −1

V  or T2 = 273   4 V  = 273 × (0.25)0.4 K = 156.8 K 18. (i) Every individual molecule in the free surface

of a liquid has maximum potential energy. And the free surface on the whole tends to have minimum potential energy for stability. This is possible when free surface of liquid contains minimum number of molecules. As area of each molecule is fixed, the free surface of liquid at rest tries to have minimum area. This is the cause of surface tension. 30 (ii) Here, r = = 15 mm = 1.5 cm. 2 F = 1.5 gf = 1.5 × 980 dyne As liquid is touching the ring inside as well as outside, therefore, F = S × 2l = S × 2(2 p r) F 1.5 × 980 S= = = 78.0 dyne cm–1 4pr 4 × 3.14 × 1.5 (iii) Suppose you are the owner of a factory where 20 people are working. Each of them wants to get maximum wages and you want to save maximum. The only practical solution is to pay them the desired wages with the promise that each one will put in the best of efforts, so that you will not engage extra labour. This is the practical implication of the concept of surface tension, where free surface tries to contain minimum number of molecules, with each molecule having maximum energy. Physics for you | january ‘16

61

19. Let the stone be thrown from a point O, distant

x from the foot of the wall.

From figure (a) R = W cosa + P sina ... (i) and P cosa + f = W sina ...(ii) From eqns. (i) and (ii) Pcosa + m(Wcosa + Psina) = Wsina (as f = mR) or P[cosa + msina] = W(sina – mcosa) W cos a + m sin a or = P sin a − m cos a cos a + tan q sin a sin a − tan q cos a cos a cos q + sin a sin q = sin a cos q − sin q cos a =

For the stone to just clear the wall, time taken to cover a horizontal distance x = time taken to rise vertically to a height H For motion along X-axis, x = vx0t x x or t = = ... (i) v x0 v 0 cos q For motion along Y-axis, 1 y = v y 0t + a yt 2 2 1 ... (ii) or H = (v 0 sin q)t + (− g )t 2 2 From eqns. (i) and (ii),

or or

 x  1  x  − g H = (v 0 sin q)   v 0 cos q  2  v 0 cos q  gx 2 = x tan q − 2 2v 0 cos 2 q

2

2

gx − (2v 02 cos 2 q tan q)x + 2Hv 02 cos 2 q = 0 gx 2 − (v 02 sin 2q)x + 2Hv 02 cos 2 q = 0 2 2

or or

W cos(a − q) = P sin(a − q) W2 P2

or 1 +

=

cos 2(a − q)

W2 P2

sin2(a − q)

cos 2(a − q)

=

sin2(a − q)

= or

(as m = tanq)

P2 +W 2 P

2

+1

cos 2(a − q) + sin2(a − q)

=

or sin(a − q) = From figure (b),

sin 2(a − q)

2

1

sin (a − q) P 2

P +W2

... (iii)

[as 2cos q tanq = 2 cos q(sinq/cosq) = 2 sinq cos q = sin 2q]

v 02 sin 2q ± v 04 sin 2 2q − 8 gHv 02 cos 2 q or x = 2g 1  2 4 2 2 2 2  = 2 g v 0 sin 2q ± 4v 0 sin q cos q − 8 gHv 0 cos q  1  2 2 2  = 2 g v 0 sin 2q ± 2v 0 cos q v 0 sin q − 2 gH  OR R Wsin 

62

Pcos P f  Psin W Wcos

Figure (a)


Q + f = Wsina ... (iv) and R = Wcosa ... (v) From eqns. (iv) and (v), Q + mW cosa = Wsina (as f = mR) or Q = W(sina – mcosa) = W(sina – tanqcosa) W sin(a − q) or Q = cos q W or cos q = sin(a − q) ... (vi) Q From eqns. (iii) and (vi), P  W cos q =   2 2 Q  P +W 

20. Consider a particle of mass m executing S.H.M.

with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by y = asin wt dy \ Velocity, v = = a w cos wt. dt 1 1 Kinetic energy, E K = mv 2 = ma 2w 2 cos 2 wt 2 2 Potential energy, 1 1 E P = ky 2 = mw 2a 2 sin2 wt (... k = mw2) 2 2 \ Average kinetic energy over one cycle EK

T

T

1 1 1 = ∫ E K dt = ∫ ma 2w 2 cos 2 wtdt T T 2

av

0

=

There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, DV = Ay change in volume Volumetric strain = original volume

0

\

T

(1 + cos 2wt ) 1 ma 2w 2 ∫ dt 2T 2 0

T 1 sin 2wt  ma 2w 2 t + =  4T 2w  0

=

=

1 1 ma 2w 2(T ) = ma 2w 2 4T 4

... (i)

Average potential energy over one cycle is EP = av

=

T

T

1 1 1 E dt = ∫ mw 2a 2 sin2 wtdt T∫ P T 2 0

0

T

(1 − cos 2wt ) 1 mw 2a 2 ∫ dt 2T 2

T 1 sin 2wt  mw 2a 2 t −  4T 2w  0

=

1 1 mw 2a 2[T ] = ma 2w 2 4T 4

From (i) and (ii), E K

av

−p − pV = Ay / V Ay

Here, negative sign shows that the increase in pressure will decrease the volume of air in the chamber. −EAy \ p= V Due to this excess pressure, the restoring force acting on the ball is F = p×A=

0

=

DV Ay = V V Bulk Modulus of elasticity E, will be stress(or increase in pressure) E= volumetric strain =

... (ii)

= EP

av

OR

Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m fitted smoothly in the neck at position C as shown in the figure. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D, where CD = y.

−EAy −EA2 ⋅A = y V V

... (i)

Clearly, F ∝ y. Negative sign shows that the force is directed towards equilibrium position. If the applied increased pressure is removed from the ball, the ball will start executing linear S.H.M. in the neck of chamber with C as mean position. In S.H.M., the restoring force, F = – ky ... (ii) Comparing (i) and (ii), we have k = EA2/V, which is the spring factor. Now, inertia factor = mass of ball = m As, T = 2π \

inertia factor m 2π mV = 2π = 2 spring factor E EA / V A

Frequency, υ =

1 A E = T 2π mV


nn 63

class-Xi

ParagraPh based questions Paragraph-1 A projectile is fired with speed v0 at t = 0 on a planet named ‘Increasing Gravity’. This planet is strange one, in the sense that the acceleration due to gravity increase linearly with time t as g(t) = bt, where b is a positive constant. 1. If angle of projection with horizontal is q, then the time of flight is (a)

6v0 sinq b

(b)

2v0 sinq b

2v0 3v0 sinq (d) b b 2. If angle of projection with horizontal is q, then the maximum height attained is (c)

3/2

(a)

1 (v0 sin q) 3 b

3/2

(b) 3/2

1 (2v0 sin q) (c) 3 b

4 (v0 sin q) 3 b

(d) None of these

3. At what angle with horizontal should the projectile be fired so that it travels the maximum horizontal distance? 1 1 (a) q = tan −1 (b) q = tan −1 2 2 –1 − 1 (c) q = tan (d) q = tan 2 2 Paragraph-2 One end of a light string of length L is connected to a ball and the other end is connected to a fixed point O. The ball is released from rest at t = 0 with string horizontal and just taut. The ball then moves in vertical circular path as shown. The time taken by ball to go from position A to B is t1 and from B to lowest 64


 position C is t2. Let the velocity of ball at B is v B and  at C is vC . O

A

 B

C

  4. If vC = 2 v B then the value of q as shown in figure, is 1 1 (a) cos −1 (b) sin −1 4 4 1 1 (c) cos −1 (d) sin −1 2 2   5. If vC = 2 v B then (a) t1 > t2 (b) t1 < t2 (c) t1 = t2 (d) Data insufficient    6. If vC − v B = v B , then the value of q as shown in figure, is 1/3

1 (b) sin −1   4

1/3

1 (d) sin −1   2

1 (a) cos −1   4 1 (c) cos −1   2

1/3

1/3

Paragraph-3 Block A in the figure hangs from a spring balance D and is submerged in a liquid C contained in a beaker B. The weight of beaker is 1 kg, and the weight of the liquid 1.5 kg. The balance D reads 2.5 kg and balance E reads 7.5 kg. The volume of the block is 0.003 m3.

figure. Initially, the spring is compressed by 5 cm and then the electric heater starts supplying energy to the gas at constant rate of 100 J s–1 and due to conduction through walls of cylinder and radiation, 20 J s–1 has been lost to surroundings.

D C

A

B k E

7. What is the density of the liquid? (a) 1666.7 kg m–3 (b) 1333.3 kg m–3 –3 (c) 2500 kg m (d) 1750 kg m–3 8. When A is taken out, what will be the reading of D? (a) 7.5 kg (b) 5.5 kg (c) 3.5 kg (d) 2.5 kg 9. When A is taken out, what will be the reading of E? (a) 2.5 kg (b) 2 kg (c) 1.5 kg (d) 3 kg Paragraph-4 A source S of acoustic wave of the frequency u0 = 1700 Hz and a receiver R are located at the same point. At the instant t = 0, the source starts from rest to move away from the receiver with a constant acceleration a. The velocity of sound in air is v = 340 m s–1. 10. If a = 10 m s–2, the apparent frequency that will be recorded by the stationary receiver at t = 10 s, will be (a) 1700 Hz (b) 1.35 kHz (c) 850 Hz (d) 1.27 kHz 11. If a = 10 m s–2, for 0 < t ≤ 10 s, and then a = 0 for t > 10 s, the apparent frequency recorded by the receiver at t = 15 s will be (a) 1700 Hz (b) 1310 Hz (c) 850 Hz (d) 1230 Hz

m

[Take k = 1000 N m–1, g = 10 m s–2, Atmospheric pressure, P 0 = 10 5 N m –2 , Cross-section area of piston A 0 = 50 cm 2 , Mass of piston m = 1 kg, R = 8.3 kJ mol–1 K–1] 13. The initial pressure of the gas is (a) 1 × 10–5 N m–2 (c) 1.10 × 10–5 N m–2

14. Increase in temperature of gas in 5 s is (a) 6.9 × 10–3 K (b) 6.9 × 10–2 K –4 (c) 9.6 × 10 K (d) 9.6 × 10–2 K 15. Work done by the gas in 5 s is (a) 300 J (b) 400 J (c) 114.5 J (d) 153.6 J solutions dv y −bt 2 1. (a) : = −bt ⇒ v y = + (v0 ) y dt 2 bt 2 or v y = − + v0 sinq 2 \ ⇒

dy bt 2 =− + v0 sinq 2 dt bt 3 y=− + v0 sinqt 6

Putting y = 0 for total time of flight T, we get

–2

12. If a = 10 m s , the apparent frequency recorded by the stationary receiver just at the instant when the source is exactly 1 km away from the receiver will be (a) 1700 Hz (b) 1310 Hz (c) 850 Hz (d) 1200 Hz Paragraph-5 2 kmol of an ideal diatomic gas is enclosed in a vertical cylinder fitted with a piston and spring as shown in the

(b) 1.02 × 10–5 N m–2 (d) 1.12 × 10–5 N m–2

T=

6v0 sinq . b

2. (c) : For maximum height dy bt 2 =0=− + v0 sinq dt 2 \

y is maximum at t =

2v0 sinq b


65

90 – 

 bt 2  ymax =  − + v0 sin q  t  6  vC

 2v0 sin q  b 2v sin q = − × 0 + v0 sin q    6 b b

\

3/2

=

(2v0 sin q) 2 v0 sin q 2v0 sin q = 3 3 b b

6v0 sinq b For maximum horizontal distance, dR =0 dq

3. (b) : R = vxt = v0 cosq ×

7.

If

vC = 2vB , then

8.

2 gL = 2 2 gL sinq or

2gL = 4 (2gL sin q)

1 1 or q = sin −1 4 4 5. (b) : Tangential acceleration is at = g cosq, which decreases with time. Hence, the plot of at versus time may be as shown in graph. or sinq =

at

A

t1

B

t2

C

t

Area under graph in time interval t1 = vB – 0 = vB Area under graph in time interval t2 = vC – vB = vB Hence area under graph in time t1 and t2 is same. \ t1 < t2.   6. (b) : v B − vC = v 2B + vC2 − 2v B vC sinq = v B ⇒

vB2 + vC2

or

vC = 2vB sinq

– 2vBvC sinq =

10.


1/3

r at 2 t=t + =t + 2v v

This is a quadratic in t, which gives at =

vB2

⇒ 2 gL = 2 2 gLsin q sin q 66

9.

1/3

1 1 ⇒ sin q =    4 4

1 q = sin −1   4 (a) : Buoyant force on block A, Fb = Vrg = 0.003 × rg, where r is the density of liquid. For liquid + beaker system, WC + WB + Fb = SE, where SE is upward reaction of the spring of the balance on the system and this is the reading of scale E. \ 1 g + 1.5 g + 0.003 rg = 7.5 g \ 0.003 r = 5 or r = 1666.7 kg m–3 (a) : Let WA be the weight of A. Considering the equilibrium of A in liquid, WA – Fb = SD , where SD is the reading of D 5   g  + 2. 5 g = 7. 5 g \WA = Fb + SD =  0.003 ×  0.003 Reading of balance D = weight of A = 7.5 kg (a) : When A is taken out, D will read the weight of A and E will read the weight of beaker + liquid. Reading of E = 1 kg + 1.5 kg = 2.5 kg (b) : Source frequency, u0 = 1700 Hz. Source (coinciding with receiver at t = 0) moves away with uniform acceleration a. Consider the wave which is received by the observer at instant t = t. It will have left the source at an earlier instant of time, say t(< t) when the distance of source was r say. If u be velocity of source at instant t, then 1 r = at 2 and u = at 2 We then have the relation, between t and t or

⇒ 6v02 sin2q – 3 v02 cos2q = 0 1 or q = tan −1 2

4. (b) : v B = 2 gL sinq and vC = 2 gL

sin3 q =

vB

⇒

−2v + 4v 2 + 8vat 2

 2at  u = at = v  1 + − 1 v  

⇒ P = P0 + kx0 + mg A A −2 1 × 10 5 1000 × 5 × 10 + or P = 10 + −4 50 × 10 50 × 10−4

 2 × 10 × 10  = 340  1 + − 1 340    27  = 340  − 1  17   v  u Then apparent frequency, ua =  v +u 0 Putting values,

= 1.12 × 105 N m–2 14. (a) : Net heat supplied in time t is

Q = ∫ (100 − 20) dt = 80t. As heat has been supplied to the gas, it expands under constant pressure, thus increasing the internal energy. \ dQ = nCPdT ⇒ Q = nCPDT

17  340  1700 = 1700 × = 1.35 kHz ua =   340 + u  27 11. (b) : Since a = 0 for t > 10 s, the source will move at constant speed of u = 10 × 10 = 100 m s–1 after t = 10 s. At t = 10 s, its distance r from the observer 1 1 is r = at 2 = × 10 × 100 = 500 m. Since the wave 2 2 500 will take a time t = 10 + which is less than 340 15 s, the wave that is received by the observer at t = 15 s would have left the source after 10 s (i.e.,) when its speed was constant at 100 m s–1. Hence, apparent frequency

⇒

15. (c) : From first law of thermodynamics, dW = dQ – dU = n(CP – CV)dT = nR dT W = nR × DT = 2 × 103 × 8.3 × 6.9 × 10–3 = 114.5 J 

 340  ua =  1700  340 + 100   17  =  × 1700  ≈ 1310 Hz  22  1 2 12. (d) : Now when r = at = 1 km = 1000 m 2 2 2 3 a t = (2 × 10 × a) Hence, u = at =

n

n

n

n

3

2 × 10 × 10

= ( 2 ) (100) = 141.4 m s–1 Hence, apparent frequency 340   ua =  1700 Hz = 1200 Hz  340 + 141.4 

13. (d) : At t = 0, let P be the pressure of gas. Free body diagram of piston would be as shown in figure. P0A kx0

n

n

n

n

n

n

PA

\

Q Q 80 × 5 = = 7 7 nCP n R 2 × 103 × × 8.3 2 2 = 6.9 × 10–3 K DT =

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mg

P0A + kx0 + mg = PA Physics for you | January ‘16

67

PRACTICE PAPER 2016 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 attempt only one of the choices in such questions.

section-A 1. Figure shows three point charges +2q, –q, +3q. Two charges +2q and –q are enclosed within a surface S. What is the electric flux due to this configuration through the surface S? +2q +3q

–q

S

2. Which is the best method of reducing current in an ac circuit and why ? 3. Why is an AM signal likely to be more noisy than a FM signal upon transmission through a channel? 4. Is Ohm’s law true for all conductors? Name two types of commercially available resistors? 5. How does focal length of a lens change when red light is replaced by blue light? section-B 6. Two conductors are made of the same material and have the same length. Conductor A is solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB. 68


7. Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index? 8. Two particles A and B of de Broglie wavelengths l1 and l2 combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional). 9. (i) Identify the logic gates marked P and Q in the given logic circuit. A B

P

Q

X

(ii) Write down the output at X for the inputs A = 0, B = 0 and A = 1, B = 1. 10. With the help of an example, explain, how the neutron to proton ratio changes during alpha decay of a nucleus. OR Calculate the energy released in MeV in the following nuclear reaction: 238 234 4 92U → 90Th + 2He + Q

[Mass of 238 92U = 238.05079 u, 234 Mass of 90Th = 234.043630 u, Mass of 42He = 4.002600 u, 1 u = 931.5 MeV c–2] section-c

11. A concave lens made of a material of refractive index m1 is kept in a medium of refractive index m2. A parallel beam of light is incident on the lens. Complete the path of the rays of light emerging from the concave lens if (i) m1 > m2 (ii) m1 = m2 and (iii) m1 < m2. 12. An electron and a proton are accelerated through the same potential. Which one of the two has (i) greater value of de Broglie wavelength associated with it and (ii) less momentum? Justify your answer. [mass of proton is 1837 times the mass of electrons] 13. In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavelength 450 nm. The screen is 1.0 m away from the slits. (a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum. (b) How will the fringe pattern change if the screen is moved away from the slits? 14. What is a carrier wave? Why high frequency carrier waves are employed for transmission? 15. Where on the earth’s surface is the value of vertical component of the earth’s magnetic field zero? The horizontal component of the earth’s magnetic field at a given place is 0.4 × 10–4 Wb m–2 and angle of dip is 30°. Calculate the value of (i) vertical component, (ii) the total intensity of the earth’s magnetic field. 16. In the figure, a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of emfs e1 and e2 connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) e1/e2 and (ii) position of null point for the cell e1. How is the sensitivity of a potentiometer increased?

17.

18.

19.

20.

21.

22.

OR Using Kirchhoff ’s rules determine the value of unknown resistance R in the circuit so that no current flows through 4 W resistance. Also find the potential difference between A and D. Name the three different modes of propagation of electromagnetic waves. Explain the mode of propagation used in the frequency range from a few MHz to 40 MHz using a proper diagram. In a diode AM detector, the output circuit consists of R = 1 kW and C = 10 pF. A carrier signal of 100 kHz is to be detected. Is it good? If yes, then explain why ? If not, what value of C would you suggest? Show that the first few frequencies of light that is emitted when electrons fall to the nth level from levels higher than n, are approximate harmonics (i.e. in the ratio 1 : 2 : 3…) when n >> 1. Define the term electric potential due to a point charge. Find the electric potential at the centre of a square of side 2 m , having charges 100 mC, –50 mC, 20 mC and – 60 mC at the four corners of the square. A long solenoid S has n turns per meter, with diameter a. At the centre of this coil we place a smaller coil of N turns and diameter b (where b < a). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as a function of mt2 + C. In an intrinsic semiconductor, the energy gap Eg is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration ni is  Eg  given by ni = n0 exp  −  , where n0 is constant.  2kBT 

kB = 8.62 × 10–5 eVK–1.

section-D

23. Mr. Sharma purchase heater marked with 80V – 800 W. He wanted to operate it on 100 V – 50 Hz ac supply. He calculated inductance Physics for you | January ‘16

69

of the choke required for operating that heater. (a) Specify the nature of Mr. Sharma. (b) How Mr. Sharma could have calculated the value of inductance? Explain. section-e 24. Describe an astronomical telescope. Derive expression for its magnifying power when final image is (i) at infinity (ii) at least distance of distinct vision. OR State Huygen’s principle. Use Huygen’s construction to explain refraction of plane wavefront at a plane surface. Draw diagrams to show the behaviour of a (i) convex lens, (ii) concave mirror when a plane wavefront falls on it. 25. Derive an expression for a potential at a point due to an electric dipole. OR Obtain the expression for the capacitance of a parallel plate capacitor. Three capacitors of capacitances C1, C2 and C3 are connected in parallel. Derive an expression for the equivalent capacitance.

26. State Biot-Savart’s law. Using this law derive an expression for the magnetic field at a point situated at a distance of x metre from the centre of a circular coil of N turns and radius r carrying a current of I A. OR Explain the difference between diamagnetic, paramagnetic and ferromagnetic substances. solutions

1. Total charge within a surface S = +2q + (–q) = +q According to Gauss’s law, q Electric flux φ = e0 2. The current in an ac circuit can be reduced by using an inductor coil or a capacitor in the circuit, as the power dissipation in these elements is negligible. 3. In AM, the carrier wave instantaneous voltage is varied by modulating wave voltage. On transmission, noise signals can also be added and receiver assumes noise as a part of the modulating signal. However in FM, the carrier wave frequency is changed as per modulating wave instantaneous voltage. This can only be done at the mixing/ modulating stage and not while signal is transmitting 70


in channel. Hence, noise doesn’t affect FM signal. 4. No, it is true only for metallic conductors. Commercially available resistors are carbon resistors and wire bound resistors. 5. According to lens maker’s formula,  1 1 1 = (m − 1)  −  f R R  1 2

As mb > mr ,  fb < fr i.e., focal length of lens decreases. 6. Resistance of conductor A, ρl RA = π(0.5 × 10−3 )2 Resistance of conductor B, ρl RB = −3 2 π[(1 × 10 ) − (0.5 × 10−3 )2 ] RA (1 × 10−3 )2 − (0.5 × 10−3 )2 0.75 3 = = = RB 0.25 1 (0.5 × 10−3 )2

\

7. Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle. m2 i.e., tanqB = where m2 < m1. m1 When light travels in such a medium, the critical m angle is sinqC = 2 where m2 < m1. m1 As |tanqB| > |sinqC| for large angles, qB < qC. Thus, polarisation by reflection shall definitely occur. 8. Here, If pA, pB > 0 or pA, pB < 0, pC = |pA| + |pB| ⇒

l AlB h h h + = \ lC = l + l l A l B lC A B

If pA > 0, pB < 0 or pA < 0, pB > 0,

|l − l A | h \ l = l AlB p =h B = C l A − lB C l AlB lC

9. (i) The logic gate P is NAND gate and the logic gate Q is OR gate. (ii) A B

P

—

A·B B

Inputs A 0 1

Q

X

Output B 0 1

A·B 0 1

A⋅B

X = A⋅B + B

1 0

1 1

10. Consider the a-decay of nucleus.

So, pe < pp, i.e., lesser momentum is associated with electron as compared to proton.

A → ZA−−24Y + 24 He Z X  238 4 → 234 92 U  90 Th + 2 He

13. Here, d = 0.15 mm = 0.15 × 10–3 m = 15 × 10–5 m, l = 450 nm = 450 × 10–9 m = 4.5 × 10–7 m, D = 1.0 m (a) (i) Distance of the second bright fringe, nlD  2 lD  x2 =  xn =  d d  −7 2 × 4.5 × 10 × 1.0 2 × 4.5 = = × 10−2 −5 15 15 × 10 = 0.6 × 10–2 m = 6 mm (ii) Distance of the second dark fringe, l D  3lD x2 =  xn = (2n − 1)  2 d 2d 3 × 4.5 × 10−7 × 1.0 = = 4.5 mm 2 × 15 × 10−5 lD (b) Fringe width β = . d When screen is moved away, D increases, therefore width of the fringes increases but the angular separation (l/d) remains the same.

Before decay, neutron to proton ratio, 238 − 92 n/p = = 1.58 92 After decay, neutron to proton ratio, 234 − 90 n/p = = 1.60 90 Thus ratio increases. OR 238 92U

4 → 234 90Th + 2He + Q Energy released Q = Dmc2 = (mU – mTh – mHe)c2 = (238.05079 – 234.043630 – 4.002600)c2 = 0.00456 × 931.5 MeV Q = 4.25 MeV

11. Let m1 be the refractive index of lens m2 be the refractive index of medium 1  2 1

1

2

2

(i) (ii) (iii) (i) m1 > m2, it will act as a diverging lens. (ii) m1 = m2, it will simply act as glass plate. (iii) m1 < m2, it will act as converging lens. 12. When a charged particle of charge q, mass m is accelerated under a potential difference V, let v be the velocity acquired by particle. Then 1 qV = mv 2 or mv = 2qVm 2 h h (i) de Broglie wavelength, l = = mv 2qVm 1 or l ∝ qm \

q pmp e × 1837me le = = >1 lp qe me e × me

So, le > lp, i.e., greater value of de Broglie wavelength is associated with electron as compared to proton. (ii) Momentum of particle, p = mv = 2qVm or p ∝ qm \

72

pe qe me me e = = × <1 pp q pmp e 1837me


14. A carrier wave is an electromagnetic wave of high frequency and of constant amplitude, which is employed to carry the audio signals from one location to other on the surface of earth. For the transmission of audio signals, the high frequency carrier waves are used, because these high frequency carrier waves travel through space or medium with the speed of light and they are not obstructed by earth’s atmosphere. 15. At the equator, the value of vertical component of earth’s magnetic field is zero. Given : BH = 0.4 × 10–4 Wb m–2, d = 30° (i) The vertical component of earth’s magnetic field BV = BHtand = 0.4 × 10–4 × tan30° = 0.23 × 10–4 Wb m–2 (ii) Total intensity of earth’s magnetic field, B= 16. (i)

−4

0.4 × 10 BH = cos d cos 30°

= 0.46 × 10–4 Wb m–2

Let f be potential gradient of the wire. Applying Kirchhoff ’s loop rule to the closed loop ACA, we get f(120) = e1 – e2 ...(i) Again, applying Kirchhoff ’s loop rule to the closed loop ADA, we get f(300) = e1 + e2 ...(ii) Divide (i) by (ii), we get e1 − e2 120 2 = = e1 + e2 300 5 5e1 – 5e2 = 2e1 + 2e2 or 3e1 = 7e2 e1 7 = ...(iii) e2 3 (ii) Let the position of null point for the cell e1 is l3. \ e1 = fl3 ...(iv) Divide (i) by (iv), we get e2 120 e1 − e2 120 or 1 − = = e1 l3 l3 e1 3 120 1− = (Using (iii)) 7 l3 4 120 = or l3 = 210 cm 7 l3 Sensitivity of a potentiometer is increased by increasing the length of the potentiometer wire. OR As no current flows through 4 W, the current in various branches as shown in the figure. Applying Kirchhoff ’s loop rule to the closed loop AFEBA, we get – I – I – 4 × 0 – 6 + 9 = 0 or 9 – 6 – 2I = 0 or 2I = 3 ...(i)

3 or I = 2

Again, applying Kirchhoff ’s loop rule to the closed loop BEDCB, we get 6 + 4 × 0 – IR – 3 = 0 or IR = 3 R=

3 2 =3× =2W I 3

(Using (i))

Potential difference between A and D = Potential difference between A and E \

VAD = 2 I = 2 ×

3 =3V 2

17. Three modes of propagation of electromagnetic waves are : (i) Ground waves (ii) Sky waves (iii) Space waves

In the frequency range from a few MHz to 40 MHz, the sky wave propagation is used. In the ionosphere of the earth’s atmosphere, there are a large number of ions or charged particles. It extends from a height of about 65 km to about 400 km above the earth’s surface. Ionisation occurs due to the absorption of the ultraviolet and other high-energy radiation coming from the sun by air molecules. The ionosphere is further subdivided into several layers, D, E, F1 and F2. The degree of ionisation varies with the height. The density of atmosphere decreases with height. At great heights, the solar radiation is intense but there are few molecules to be ionised. Close to the earth, even though the molecular concentration is very high, the radiation intensity is low so that the ionisation is again low. However, at some intermediate heights, there occurs a peak of ionisation density. The ionospheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz). Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere and escape. These phenomena are shown in the figure.

18. Here,

1 1 = = 10 −5 s uc 100 × 10 3

RC = 103 × 10 × 10–12 = 10–8 s

Here, 1 is not much smaller than RC. uc

1

Whereas for demodulation, << RC. So it is not uc good. For the circuit to be good, RC >> or or

1 uc

103 × C >> 10–5 or C >> 10–8 F C >> 10–8 F or C = 0.01 mF

19. Here, umn = cRZ 2 

1

2  ( n + p)

−

1 , n2 

where m = n + p, (p = 1, 2, 3, …) and R is Rydberg constant. 1  n2

For p << n, umn = cRZ 2 

 p  1 +  n

−2

−

1  n2 


73

2 p  2cRZ 2   1 2p 1  umn = cRZ 2  − −  , umn = cRZ 2 3   p  n3  n  n2 n3 n2 

Thus, umn are approximately in the order 1, 2, 3 …… 20. Electric potential at a point is the amount of work done to bring a unit positive charge from infinity to that point against the electrostatic forces. AC = BD =

1 4 pe0

 1.2eV   − 2k (600)  B

... (i )

s1 = s 0 e Conductivity at 300 K, 2m

( 2) + ( 2) 2

2

s2 = s0e

 qA q q q  + B + C + D   AO BO OC OD  

= 9 × 109 × 10 × 10–6 = 9 × 104 V

21. Magnetic field due to a solenoid S, B = m0nI Magnetic flux in smaller coil, f = NBA, where A = pb2  dB  −df −d (NBA) = − N pb2   So, e = =  dt  dt dt d dI = −N pb2 (m0 In) = −N pb2m0n dt dt 2 d ( 2 = − Nnpm0b mt + C ) = −m0 Nnpb2 2mt dt Negative sign signifies opposite nature of induced emf. The magnitude of emf varies with time as shown in the figure.

|| t

22. Conductivity is given by s = e(neme + nh mh) For intrinsic semiconductor, ne = nh = ni Also, mobility of holes (mh) << mobility of electrons (me) So, conductivity s = eneme Temperature dependence of intrinsic carrier concentration, ni = n0e

 Eg   − 2k T   B 


 1.2eV   − 2k (300)  B

 0.6eV 

=2m

 100 × 10 −6 50 × 10 −6 20 × 10 −6 60 × 10 −6   V = 9 × 109  − + − 1 1 1 1  

74

 Eg   − 2k T   B 

Hence, s = s 0e Conductivity at 600 K,

\ AO = OC = BO = OD = 1 m Potential at the centre of the square O is V=

So, conductivity, s = eni me = emen0e where emen0 = constant s0

 Eg  − k T   2 B 

... (ii ) 0 .6

 1 

+ + s1  −5  600kB  =e  = e 8.62×10  600  s2 s1 11.6 or =e = 1 × 105 s2 So, s(600K) = 105 s(300K) Conductivity increases rapidly with the rise of temperature.

23. (a) He seems to be a techno friendly person, having confidence and knowledge of electrical gadgets. P 800 (b) Current required, I = = = 10A V 80 E V 2 80 × 80 100 R= = = 8W , Z = V = = 10W P IV 10 800 Z = R2 + X L2 ⇒ X L = Z 2 + R2 X L = 102 − 82 = 6 W X 6 L= L = = 0.0191 H 2 pu 2 × 3.14 × 50 24. Refer point 6.9(2) page no. 382 (MTG Excel in Physics). OR Refer point 6.10(6), 6.11 and 6.12 page no. 443, 444 and 445 (MTG Excel in Physics) 25. Refer point 1.5(10) page no.9 (MTG Excel in Physics) OR Refer point 1.11 (4, 8) page no. 15, 16 (MTG Excel in Physics). 26. Refer point 3.1(1) page no. 169 and point 3.1(3 (vi)) page no. 170 (MTG Excel in Physics). OR Refer point 3.8(8) page no. 180, 181 (MTG Excel in Physics). nn

chapterwise McQs for practice

Useful for All National and State Level Medical/Engg. Entrance Exams oscillations

1. The displacement of two particles executing SHM are represented by equations y1 = 2sin(10t + q), y2 = 3cos 10t. The phase difference between the velocity of these particles is (a) q (b) –q p p (c) q + (d) q 2 2 2. The bob of a simple pendulum of length L is released at time t = 0 from a position of small angular displacement. Its linear displacement at time t is given by L g (a) X = a sin2p (b) X = a cos2p ×t ×t g L g (c) X = a sin g × t (d) X = a cos ×t L L 3. A horizontal platform with an object placed on it, is executing SHM in the vertical direction. The amplitude of oscillation is 3.92 × 10–3 m. What must be the least period of these oscillations, so that the object is not detached from the platform? (a) 0.1256 s (b) 0.1356 s (c) 0.1456 s (d) 0.1556 s 4. A particle of mass m is executing oscillations about the origin on the x-axis with amplitude A. Its potential energy is given as U(x) = ax4, where a is positive constant. The x-coordinate of motion where potential energy is one-third of the kinetic energy of particle is A A (a) ± (b) ± 2 3 A A (c) ± (d) ± 2 3

5. On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm support. If each spring has force constant k, the period of oscillation of the body (assuming the springs as massless) is

M 

(a) 2p[M/2k]1/2 (c) 2p[Mg sin q/2k]1/2

(b) 2p[2M/k]1/2 (d) 2p[2Mg/k]1/2

6. In damped oscillations, the amplitude of oscillations is reduced to one-third of its initial value a0 at the end of 100 oscillations. When the oscillator completes 200 oscillations, its amplitude must be (a) a0/2 (b) a0/6 (c) a0/9 (d) a0/4 7. The x-t graph of a particle undergoing SHM is as shown in figure. The acceleration of the particle at t = 4/3 s is

(a)

3 2 p cm s -2 32

(b) -

p2 cm s -2 32

(c)

p2 cm s -2 32

(d) -

3 2 p cm s -2 32


75

2

2

2

2

(a) y = x (1 – x /A ) (c) y2 = 4x2 (1– x2/A2)

2

2

2

2

13. A simple pendulum has a length l. The inertial and gravitational masses of the bob are mi and mg respectively. Then the time period T is given by mg l (a) T = 2p mi g

(c) T = 2p 76

 ba  (b) T = 2p    dg 

g  (c) T = 2p    dc 

 dc  (d) T = 2p   g 

15. Which of the following figures represents damped simple harmonic motion?

(a)

(b)

(c)

(b) y = 2x (1 – x /A ) (d) y2 = 8x2 (1 – x2/A2)

12. A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force, F = F0 sin w0t. If the amplitude of the particle is maximum for w = w1 and the energy of the particle is maximum for w = w2 then (a) w1 = w0 and w2 ≠ w0 (b) w1 = w0 and w2 = w0 (c) w1 ≠ w0 and w2 = w0 (d) w1 ≠ w0 and w2 ≠ w0

(b) T = 2p

 abc  (a) T = 2p   g 

mi l mg g mi × mg × l g


Displacement

10. Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy? 1 1 s s (a) (b) 6 12 1 1 s (c) (d) s 3 4 11. Two linear SHMs of equal amplitude A and angular frequencies w and 2w are impressed on a particle along the axes x and y respectively. If the initial phase difference between them is p/2, the resultant path followed by the particle is

l mi × mg × g

14. A piece of wood has dimensions a, b and c. Its relative density is d. It is floating in water such that the side c is vertical. It is now pushed down gently and released. The time period is

Displacement

9. A particle is performing SHM along x-axis with amplitude 6.0 cm and time period 12 s . What is the minimum time taken by the particle to move from x = + 3 cm to x = + 6 cm and back again? (a) 1 s (b) 2 s (c) 4 s (d) 6 s

(d) T = 2p

Displacement

y1 = 5 (sin 2 pt + 3 cos 2 pt ) , y2 = 5 sin (2 pt + p / 4 ) The ratio of the amplitudes of two SHMs is (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1 : 3

(d)

Time

Time

Time

Displacement

8. Two simple harmonic motions are represented by

Time

Waves

16. Two vibrating tuning forks producing progressive waves given by y1 = 4sin (500pt) and y2 = 2sin (506pt) are held near the ear of a person. The person will hear (a) 3 beats/s with intensity ratio between maxima and minima equal to 2 (b) 3 beats/s with intensity ratio between maxima and minima equal to 9

(c) 6 beats/s with intensity ratio between maxima and minima equal to 2 (d) 6 beats/s with intensity ratio between maxima and minima equal to 9. 17. The transverse displacement y(x, t) of a wave on a string is given by:

(

y ( x , t ) = e - ax

2

+bt 2 +2 abxt )

.

This represents a (a) standing wave of frequency (1/ b ) (b) wave moving in +X direction with speed

(

a /b)

(c) wave moving in –X direction with speed

(

b/a)

(d) standing wave of frequency

( b).

18. A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg m–3 and 2.2 × 1011 N m–2 respectively? (a) 178.2 Hz (b) 200.5 Hz (c) 770.7 Hz (d) 188.5 Hz 19. The phase difference between two points separated by 0.8 m in a wave of frequency 120 Hz is 0.5 p. The wave velocity is (a) 144 m s–1 (b) 384 m s–1 –1 (c) 256 m s (d) 720 m s–1 20. A wire under tension vibrates with a frequency of 450 per second. What would be the fundamental frequency if the wire were half as long, twice as thick and under one fourth tension? (a) 225 Hz (b) 190 Hz (c) 247 Hz (d) 174 Hz 21. Equation of a stationary wave is px y = 10 sin cos 20 pt . 4 Distance between two consecutive nodes is (a) 4 (b) 2 (c) 1 (d) 8 22. The speed of sound in hydrogen at NTP is 1270 m s–1. Then the speed in a mixture of hydrogen and oxygen in the ratio 4 : 1 by volume will be (a) 317 m s–1 (b) 830 m s–1 –1 (c) 635 m s (d) 950 m s–1

23. An object of specific gravity r is hung from a thin copper wire. The fundamental frequency for transverse standing waves in the wire is 400 Hz. The object is immersed in water so that one third of its volume is submerged. What is the new fundamental frequency? (a) 200

r -1 Hz 3r

(b) 400

3r - 1 Hz 3r

(c) 400

r -1 Hz 3r

(d) 200

3r - 1 Hz 3r

24. A closed organ pipe has fundamental frequency 100 Hz. What frequencies (in Hz) will be produced, if its other end is also opened? (a) 200, 400, 600, 800.. (b) 200, 300, 400, 500... (c) 100, 300, 500, 700.. (d) 100, 200, 300, 400... 25. A bus is moving with a velocity of 5 m s–1 towards a huge wall. The driver sounds a horn of frequency 165 Hz. If the speed of sound in air is 335 m s–1, the number of beats heard per second by the passengers in the bus will be (a) 3 (b) 4 (c) 5 (d) 6 26. A string is under tension so that its length is increased by 1 times its original length. The ratio n of fundamental frequency of longitudinal vibrations and transverse vibrations will be (a) 1 : n (b) n2 : 1 (c) n :1 (d) n : 1 27. A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. The speed (v) of wave pulse varies with height (h) from the lower end as shown in v

v

(a)

(b) h

h v

v

(c)

(d) h

h


77

28. A source of sound gives 5 beats s–1 when sounded with another source of frequency 100 Hz. The second harmonic of the source together with a source of frequency 205 Hz gives 5 beats s–1. What is the frequency of the source? (a) 105 Hz (b) 205 Hz (c) 95 Hz (d) 100 Hz 29. A transverse wave is described by the equation x  y = y0 sin 2 p  ut -   l The maximum particle velocity is equal to four times the wave velocity if (a) l = py0/4 (b) l = 2py0 (c) l = p/y0 (d) l = py0/2 30. The equation of a spherical progressive wave is (a) y = a sin wt (b) y = a sin (wt – kr) a a sin (wt - kr ) (d) y = sin (wt - kr ) (c) y = r r solutions dy1 = 2 × 10 cos (10t + q) ; 1. (d) : v1 = dt dy p  v2 = 2 = -3 × 10 sin10t = 30 cos 10t +   dt 2 p  \ Phase difference = (10t + q) – 10t +  2 p = q2 L 2. (d) : Time period, T = 2p g 2p g and w = = T L Displacement, x = acos wt = a cos

g t L

3. (a) : Here, a = 3.92 × 10–3m The object is not detached from the platform if a 4 p2 mg = mw2a = m 2 × a or T = 2 p g T = 2 × 3.14

3.92 × 10-3 = 0.1256 s 9. 8

4. (b) : Energy of oscillation, E = aA4 Kinetic energy of particle of mass m at x axis is K = E – U = a (A4 – x4) 1 As U = K ⇒ K = 3 U 3 78


\

a(A4 – x4) = 3ax4 or A4 = 4x4 ⇒ x = ±

A 2

5. (a) : It is a system of two springs in parallel. The restoring force on the body is due to springs and not due to gravity pull. Therefore slope is irrelevant. Here the effective springs constant = k + k = 2k. 1/2

Thus time period, T = 2 p [ M / 2k ] 6. (c) : at = a0e–bt, where b = damping coefficient. If T is the time period of oscillation, then according to the given condition, At the end of 100 oscillations, 1 -100bT a0 = a0e -b×100T or = e 3 3 At the end of 200 oscillations, a = a0e–b × 200T = a0[e–100bT]2 = a0 (1/3)2 = a0 / 9 7. (d) : The given curve is a sine curve with A = 1 cm and T = 8 s, i.e., 2 p p s-–11 w= = s T 4 If it is represented by x = A sinwt, then acceleration, a = –w2 A sin wt p 4 4 = –(p/4)2(1) sin  ×  [... t = s ] 4 3 3 2 3 2 p = - ( 3 / 2) = p cm s -2 16 32 8. (c) : As y1 = 5 (sin 2 pt + 3 cos 2 pt ) = 10 (sin 2 pt ) (1 / 2) + (cos 2 pt ) ( 3 / 2) = 10(sin 2pt cos p/3 + cos 2pt sin p/3) = 10 sin (2pt + p/3) and y2 = 5 sin (2pt + p/4) \ A1 = 10 = 2 : 1 A2 5 2p 9. (c) : As x = r sin wt = r sin t T T -1  x  So t = sin   [where r = 6 cm, T = 12 s] 2p r At x = 3 cm T  3  T p T 12 t = sin -1   = × = = =1 s  6  2 p 6 12 12 2p At x = 6 cm t′ =

T  6  T p T 12 sin -1   = × = = =3 s  6  2p 2 4 4 2p

So time taken in going from x = + 3 cm to x = + 6 cm will be 3 – 1 = 2 s Total time taken = 2 + 2 = 4 s 10. (b) : As per question, 75 75 1 2 2 K.E. = E= × ma w 100 100 2 \

1 2 2 75 1 2 2 ma w cos2 wt = × ma w 2 100 2

3 3 p or cos wt = = cos 4 2 6 p p p 1 or wt = or t = = = s 6 6 w 6 × (2 p / 2 ) 6 [... T = 2 s] or cos2 wt =

11. (c) : x = A sin (wt + p/2) = A cos wt 2 2 \ cos wt = x/A and sin wt = 1 - ( x / A ) y = A sin 2wt = 2 A sin wt cos wt or y2 = 4A2 sin2 wt cos2 wt

= 4 A2 ×

 A2 - x 2   x2  2 × = 4 x   1 - 2  A2  A 2  A x2

12. (c) : Amplitude resonance takes place at a frequency of external force which is less than the frequency of undamped natural vibrations. Velocity resonance takes place (i.e., maximum energy) when frequency of external periodic force is equal to natural frequency of undamped vibrations. 13. (b) : Torque acting on the bob = I a = –(mg)l sin q or (mil2)a = –(mgg)l q or

 mg g  2 a = -  q = -w q  mi l 

mg g \ T = 2 p = 2 p mi l mg g w mi l 14. (d) : Let the vertical side of dimension c of block be pushed in liquid when block is floating, then buoyancy force = ab x dw g = ab x g [... dw = 1] The mass of piece of wood = abcd g  So, acceleration = –ab x g/abcd = -   x  cd  where w2 =

Hence, time period, T = 2p

dc g

15. (a) : Figure (a) represents damped SHM.

w 16. (b) : He re, w 1 = 5 0 0 p , u 1 = 1 = 2 5 0 Hz ; 2p w w2 = 506p, u2 = 2 = 253 Hz 2p Number of beats/s = u2 – u1 = 3 Further, A1 = 4 and A2 = 2 A Amplitude ratio, r = 1 = 2 A2 2 ( ) I max r +1 \ = =9 I min (r - 1)2 17. (c) : y ( x , t ) = e -(ax

2

+bt +2 abxt ) 2

=e

 b  -a  x + t   a 

2

Comparing it with travelling wave equation b y(x, t) = f(x ± vt), v = a Further, +ve sign between x and t shows that the wave travels in the – X direction. 18. (a) : As u = [as \

1 T 1 T 1 T/A = = 2L µ 2L Ar 2L r

 1  T = stress = Y × strain = (2.2 × 1011 N m–2)  ] 100  A u=

1 ( 2 1. 5 m )

(2.2 × 1011 Nm -2 ) / 100 7.7 × 103 kg m -3

= 178.2 Hz 19. (b) : Here, u = 120 Hz x = 0.8 m, f = 0.5 p 2p 2 px 2 p × 0.8 x; l = = = 3.2 m From f = l f 0.5p \ v = ul = 120 × 3.2 = 384 m s–1 20. (a) : Frequency of fundamental note 1 T 1 T pD 2 , ; where µ = u= r \ u= lD pr 2l µ 4 T l when l ′ = , D ′ = 2D and T ′ = . 4 2 1 Frequency u becomes , i.e., 450 = 225 Hz 2 2 21. (a) : Compare the given equation with the standard equation of a stationary wave 2 px 2 pt y = 2a sin cos l T 2 px px = ; l = 8 units. l 4 Distance between two consecutive nodes l 8 = = = 4 units 2 2 Physics for you | January ‘16

79

22. (c) : Suppose V is volume of oxygen. Therefore, volume of hydrogen = 4 V. Density of mixture, 4V × 1 + V × 16 rm = =4 5V vm rH 1 1 = = = vH rm 4 2 v H 1270 = = 635 m s -1 2 2 23. (b) : The fundamental frequency for transverse stationary waves when hung object is in air is 1 T 1 V rg u= = ... (i ) µ 2l µ 2l vm =

When the object is immersed one third in water, T′ = Mg – upthrust V  = V rg -  rw × g  3  As rw = 1 g cc–1, therefore V  1 T ′ = V rg - g = Vg r -  3  3 Vg T′ = (3r - 1) 3 New frequency of vibration, Vg (3r - 1) 1 T′ 1 3 u′ = = 2l µ 2l µ Divide (ii) by (i) u′ = u

... (ii )

u′ Vg = (3r - 1) u 3Vg r

3r - 1 3r - 1 = 400 Hz 3r 3r

24. (a) : When other end of pipe is opened, its fundamental frequency becomes 200 Hz. The overtones have frequencies 400 Hz, 600 Hz, 800 Hz.... 25. (c) : Frequency of sound, u = 165 Hz. v  v + v × As u′ =  s u vs - v   v 335 + 5 335 × × 165 = 170 Hz 335 335 - 5 \ Number of beats per second u′ - u = 170 – 165 = 5 26. (c) 27. (a) : Let m be the total mass of the rope of length l. Tension in the rope at a height h from lower end = weight of rope of length h is \

80

u′ =


mg (h ) l T mg (h ) = gh As v = \ v= (m / l ) l (m / l ) v2 = gh which represents a parabola. Therefore, h versus v graph is a parabola. Option (a) is correct. 28. (a) : Two possible frequencies of source are = 100 ± 5 = 105 or 95 Frequencies of 2nd harmonic = 210 or 190 5 beats with source of frequency 205 are possible only when 2nd harmonic has frequency = 210 \ Frequency of source = 105 Hz x  29. (d) : Given; y = y0 sin2p  ut -  ,  l After differentiation, T=

dy x  = y0 cos 2 p  ut -  × 2 pu  l dt

 dy  \ Maximum particle velocity =    dt  max = 2puy0 × 1 Wave velocity = ul 2 py0 py0 = As 2puy0 = 4ul, \ l = 4 2 30. (d) : The amplitude of a plane progressive wave = a, that of a spherical progressive wave is a/r. nn

NOTIFICATION! JAMIA MILLIA ISLAMIA (A Central University by an Act of Parliament) (Accredited grade ‘A’ by NAAC) Maulana Mohd. Ali Jauhar Marg, New Delhi- 110025 Admission for B.Tech. & B.Arch.: 2016-17 Admission in B.Tech. & B.Arch. (2016-17) of Jamia Millia Islamia will be based on the scores of JEE Main-2016 conducted by CBSE. all aspiring candidates are required to fill online application Form at www.jeemain.nic.in. In addition, the candidates are also required to fill online application Form of Jamia Millia Islamia at http://jmi.ac.in/admissions. LAST DATES JEE Main-2016 online application Form: 31.12.2015 Jamia Millia Islamia online application Form: 31.01.2016 Dated: 16.12.2015

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solution set-29

1. (d) : the centre of mass of a uniform thick triangular sheet is at the centroid which divides a median in a ratio 2 : 1. the required centre of mass must be on the line of symmetry passing through O and the vertex to the right. Consider the shape of arrowhead to be obtained by cutting the triangular part to the left (with base of 6 cm and height 3 cm) from the uncut triangular sheet with the base 6 cm and height 9 cm. the ratio of masses of these two is 27 : 9, the thickness being uniform. 0 × 27 m − (−2 cm) × (9 m) = 1 cm \ xcm = 27 m − 9 m Hence, centre of mass of given shape is at a point 1 cm to the right of O. 2. (d) : the lens maker’s formula is 1  m L − 1  1 − 1  where m and m represent = L m f  mm   R1 R2  the refractive indices of the liquid filled and the surrounding medium respectively. Note that for a 1 1 double concave lens  −  is negative. Now,  R1 R2  for the lens to not act as a diverging lens, that is, for f to be positive, the first bracket on the right hand side of the relation must be negative. this requires mm to be greater than mL. this is true for option (d) as m2 > mw > m1 > ma. 3. (b) : Two particles will meet at P. After they will stick together, momentum m′v will remain same. But charge is doubled, so radius is halved. B

y

P (–r, r/2, 0) r

x

Finally, both move in dotted circle. 4. (d) : Volume of the balloon at any instant, when radius is r, 4 V = pr 3 3 Time rate of change of volume,

dr dV = 4 pr 2 dt dt Time rate of change of radius of balloon, 1 dV dr = dt 4 pr 2 dt Flux through rubber band at the given instant, f = B (pr2) −df dr d = − (Bpr 2 ) = −2 prB Induced emf, e = dt dt dt B dV  1 dV  =− = −2 prB   4 pr 2 dt  2r dt As volume of the balloon is decreasing, dV is dt negative. \

e= −

(0.04) × (–100 × 10–6) = 20 mV 2 × 10 × 10−2

5. (d) : As current at any instant in the circuit will be I = Idc + Iac = a + b sin wt 1/2

 T I 2 dt  1/2 ∫  1 T 2 dt  So, Ieff =  0 = + ( sin w ) a b t   T  T ∫0   ∫0 dt  1/2 1 T  =  ∫ (a2 + 2ab sin wt + b2 sin2 wt ) dt  T 0  T T 1 1 1 As ∫ sin wt dt = 0 and ∫ sin2 wt dt = 2 T 0 T 0 1/2 1   So, Ieff = a2 + b2  2   1  1 6. (c) : As 1 = (m − 1)  −   R1 R2  f \ For equiconvex lens 1 3  1 1  =  − 1  +  10  2   R R  ⇒ R = 10 cm \ Focal length of left part 1 3  1 1  =  − 1  +  f1  2   ∞ R  f1 f2 ⇒ f1 = 20 cm f  Similarly, f2 = 20 cm 1 1  1 = (3 − 1)  −   f′ −10 10  −10 cm ⇒ f ′ = 4 \ 1 = 1 + 1 + 1 = 1 − 4 + 1 = 2 − 4 f eq f1 f ′ f 2 20 10 20 20 10 or feq = –10/3 cm Physics for you | January ‘16

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7. (c) : (i) P.E. = Mg

I = 3.36(1 + 2t) × 10–2 A dI = 2 × 3.36 × 10–2 A s–1 dt Magnetic induction at every point on the loop, mI B= 0 2 pd Magnetic flux linked with loop at any instant, mI f = BA = 0 ⋅ pr 2 2 pd df m0 r 2  dI  Induced emf, e = =   2d  dt  dt Induced current, m r 2  dI  e I= = 0   R R × 2d  dt 

L = 0.5MgL 2

(ii) P.E. = Mg L = 0.5MgL 2 2 Mg L 2 2 (iii) P.E. = Mg R = = MgL ≈ 0.2 MgL p p p p2 4 (iv) P.E. = 2 MgL ≈ 0.4 MgL p 2R  (v) P.E. = Mg  R −   p  p−2 p−2 =  MgR = 2 MgL  p  p 1.14 = 2 MgL = 0.11 MgL p \ (i) = (ii) > (iv) > (iii) > (v) 8. (a) : the weight of the removed cylinder is 15 N. If a symmetrical hole was drilled on the other side, the uniform cylinder would have zero torque about P. this implies that the torque due to excess weight (15 N) on the other side has caused instability. Hence for equilibrium the torque of T must balance the torque due to excess weight. \ T(2a) = (15)  2 a  3  or T = 5 N We have considered torque about P so that torques of unknown forces N and f are zero.

−7 −3 2 −2 = 4 p × 10 × (10 ) × 2 × 3.36 × 10 8.4 × 10−4 × 2 × 1 = 5.024 × 10–11 A

nn solution of December 2015 crossworD 1

3

G O

6

9

D

T

F

14

A C D R A S T 20 E S A

a 2a/3 N P

f

9. (d) : As l / 3 = v0t1 ⇒ t1 = l 3v0 t t 2 l     2 2 Also = v1   + v2   2 2 3 4l ⇒ t2 = 3(v1 + v2 ) 3v (v + v ) l \ Mean velocity = = 0 1 2 t1 + t 2 v1 + v2 + 4v0 10. (a) : As the loop is very small, the distance of every point in its plane can be taken to be equal to d = 1 m. 82

I

C


23

B

  

  

C I 2R C L A Y C E L 4 A L 5 C R E R V I T 7 I S F U S I V I 11 12 L E A H 15 C E S S L I O L O N I A 17 V U D I E M 18 C V T O K E S 21I 22 B R E D O O T S T R A N 24 E L E C T 26 E

U I T L L U V I A L Y O M E T E R O G A M 10 Z T Y 13 D H Y N E D R L A T O M G R N F R A S P H A P

8

P E R O R I G 16 J E E E T E R

19 F O U N D L C R R O N 25V I O L E T C H O G R A M

Winners (December 2015) Akshat Puri (new Delhi) Karan Verma (UP) Harsh Gupta (Haryana) solution senders (november 2015) Mohammad naderi (iran, Khalkhal) Anuj sen (WB) Poonam shah (WB)

A

Discovered: A new Venus-like planet

stronomers have discovered a new Venus-like rocky exoplanet 39 light years away , which may be cool enough to potentially host an atmosphere. If it does, it is close enough that we could study that atmosphere in detail with the Hubble Space Telescope and future observatories like the Giant Magellan Telescope, researchers said. “Our ultimate goal is to find a twin Earth, but along the way we’ve found a twin Venus,” said astronomer David Charbonneau of the Harvard-Smithsonian Center for Astrophysics in US.”We suspect it will have a Venus-like atmosphere too, and if it does we can’t wait to get a whiff,” he said. The planet-GJ 1132b -orbits a red dwarf star only one-fifth the size of our Sun. The star is also cooler than the Sun, emitting just 1/200th as much light. GJ 1132b circles its star every 1.6 days at a distance of 1.4 million miles. As a result, it is baked to a temperature of about 232 degrees Celsius. Such temperatures would boil off any water the planet may have once held.

S

L

Lasers to make materials hotter than Sun’s core

Witnessed: Black hole swallowing a star 300m light years away

cientists have for the first time witnessed a black hole swallowing a star in a galaxy 300 million light years away and ejecting a flare of matter moving at nearly the speed of light. The finding tracks the star -about the size of our Sun -as it shifts from its customary path, slips into the gravitational pull of a super massive black hole and is sucked in, said Sjoert van Velzen, a Hubble fellow at the Johns Hopkins University in US. “It’s the first time we see everything from the stellar destruction followed by the launch of a conical outflow, also called a jet, and we watched it unfold over several months,” van Velzen said. The first observation of the star being destroyed was made in December last year. Researchers used radio telescopes to follow up as fast as possible. They were just in time to catch the action. By the time it was done, the team had data from satellites and ground-based telescopes that gathered X-ray, radio and optical signals, providing a stunning “multi-wavelength” portrait of this event. It helped that the galaxy in question is closer to Earth than those studied previously in hopes of tracking a jet emerging after the destruction of a star.

asers could heat materials to temperatures hotter than the centre of the Sun in only 20 quadrillionths of a second, according to new research that could revolutionise energy production. Physicists from Imperial College London have devised an extremely rapid heating mechanism that they believe could heat certain materials to ten million degrees in much less than a million millionth of a second. The method, proposed for the first time, could be relevant to new avenues of research in thermonuclear fusion energy, where scientists are seeking to replicate the Sun’s ability to produce clean energy. The heating would be about 100 times faster than rates currently seen in fusion experiments using the world’s most energetic laser system at the Lawrence Livermore National Laboratory in California.

Courtesy : The Times of India


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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. Is light visible or invisible?

–Sajal Sengupta

Ans. Light is invisible until it hits an object. It can only be visible to us whenever it get scattered or bounced off the dust particles in the intervening medium. We can see the source of light like stars at night as they are too dim to light up much but that light does hit our eyes and therefore we see the light. The above discussion on visibility of light is only confined to the electromagnetic radiation of wavelength from 400 nm to 700 nm which is called visible light, as these electromagnetic radiations gives the sensation of sight. Q2. Why various colours seen in a film of soap?

–Bidhan Banerjee

Ans. The colours in a soap film are caused by interference (internally and externally) of light waves. The reason for various colours which are seen in a film of soap is due to interference of light reflecting off of two nearby surfaces, the outer surface and the inner surface or interference between the transmitted portions of the incident light. In case of white light of range 400 nm to 700 nm, the interference enhances some wavelengths and suppresses other, which leads to the splitting of white light into visible colours. The wavelength that got enhanced or suppressed depends only on thickness of the film and the refractive index of soap film. Q3. It is said that while passing through turnings one should bend at maximum angle to have maximum speed, but in reality if one bends more than a specific value of angle, one falls down. Why? –Suryakant Khilar (Odisha)

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Ans. A cyclist while going round a curve on a horizontal track has to bend himself a little from his vertical position in order to avoid overturning. When he bends himself inward a component of the reaction of road provides him the necessary centripetal force for circular motion as tanq =

v2 rg

 2 −1 v or q = tan   , therefore, to take a safe turn  rg  cyclist should go slow and bend through a small angle q. This balances the torque about the wheel contact patches generated by centrifugal force due to the turn with that of the gravitational force. And the over bending of a cyclist on a turning road results a shift in the point of centre of gravity of the cyclist which increases the torque due to gravitational pull downward and it breaks the balanced condition. Hence over bending results in a fall down of a cyclist. Q4. Why does the earth rotate?

–Apratarkya Banerjee

Ans. The reason for the rotation of earth and planets arise from the concept of formation of our solar system. In the beginning solar system was a big cloud of dust and gases. This cloud began to collapse, flattening into large disk, rotating faster and faster continuously, as they collapsed, their gravitational orbit set those dust and gases to spin. As a result, the clumps of massive particle formed within that disk were going to naturally have some sort of rotation. Due to conservation of angular momentum spinning went faster and faster for e.g., skaters speed up their rate of spin as they brought their arms closer to their body. Also due to gravitational pull from all directions, the clump of massive particles became a round planet, our earth is also one of those clumps. Inertia of motion keeps the planet spinning continuously on its axis unless any external agent resists this rotation. Actually earth’s rotation is also affected by tidal pull of moon, it slows down at the rate of about 1 millisecond per year. Earth’s spinning was faster in the past. At that time, a day was about 22 hour long only. nn

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

1. A numerical description of how far apart objects are. (8) 6. The point on a wave with the maximum value or upward displacement within a cycle. (5) 9. A term used to describe a large, circular plain. (4) 10. Radiation, such as ultraviolet, able to produce a chemical change in exposed materials. (7) 11. The angular distance of an object around or parallel to the horizon from a predefined zero point. (7) 18. A stream of atomic nuclei that are observed to strike the Earth atmosphere with extremely high amounts of energy. (6, 3) 20. A line on a map joining points at which the acceleration of free fall is constant. (6) 24. The theoretical time reversal of a black hole, which arises as a valid solution in general relativity. (5, 4) 25. A series or chain of craters. (6) 26. A term used to describe a point directly underneath an celestial sphere. (5) 27. The outer edge or border of a planet or other celestial body. (4) 28. The chemical decomposition of materials into ions, excited atoms and molecules etc. by ionizing radiation. (10)

down

1. The surface of the Sun or other celestial body projected against the sky. (4) 2. The point of greatest separation of two stars, such as in a binary star system. (8) 3. An imaginary line in the sky traced by the Sun as it moves in its yearly path through the sky. (8) 4. A structural element that is capable of withstanding load primarily by resisting bending. (4) 5. A unit used to measure solar radiation and equivalent to 4 .18 × 104 J m–2. (7) 7. A mechanism that converts rotational motion to linear motion, and a torque to a linear force. (5) 8. An electrical device possessing reactance and selected for use because of that property. (7) 12. An extension of string theory in which eleven dimensions are identified. (1, 6)

13. The inner coat of the eye, having enormous number of light sensitive cells in the form of rods and cones. (6) 14. The amount of potential energy stored in an elastic substance by means of elastic deformation. (10) 15. A derived unit of energy, work or amount of heat in the international system of units. (5) 16. A mechanically commulated electric motor powered from direct current. (1, 1, 5) 17. The rate of mass flow per unit area. (4, 4) 19. A region of space where the density of matter, or the curvature of space-time, becomes infinite and the concepts of space and time cease to have any meaning. (11) 21. A term used to describe an exceptionally bright meteor. (6) 22. Meteor, generally brighter than magnitude-4, which is about the same magnitude of planet Venus in the morning and evening sky. (8). 23. Unit equivalent to coulomb per second. (6)

 Physics for you | january ‘16

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Physics for You - January 2016

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