P-3 Allens Made Physics Exercise Solution

JEE-Physics  x  ()  15  (iii) v=–4 (96) sin sin (96t) f = f'  v  vs   f0  v  vs   v   v  vs  at x = 7.5 cm, t = 0.25 sec 1 1 v = –4 × 96 sin  7.5  sin  9 6   =0  Beat frequency ( )   15   4   x  f = f0  v  vs  –f  256  330  5   256 (iv) Y = 4 sin  15  cos (96 t)  v  vs  0  330  5  = 7.87 Hz  2 sin  x  96 t + 2sin  x  9 6 t  v  vw  v0   3 40  10 10   15  15  v  vw   340  10  22. f = f  7 00 observer 0 = 680 Hz B 4000 106 14. v   2000 m / s  1000  v  330   v S   1 8 0  330  60   220Hz 23. f = f  v 0  24. f1 = f  v  440  330   484Hz 1 5 . 2 =1   = 2m 0   20   v  v S  330  1.5 v = f = 2.53 × 103 × 2 = 5.06 × 103 m/s 16. Let the pipe resonates in nth & (n+1)th harmonic f= f  v   440  330   403.3Hz 0   vS    20   (n(n+1)) 2 v 330 1.5  (n+1) 1944 = n(2592) v v 324  f0  v  3 25. f = f  v  v S   v S   n = 4  L = 1296 = 0.25 m 0 18. F > F pipe  330 330  string 3  340   330  v 330  vS  v= 1.5m/s   S F =2 T T S string 2L 4 1 0 2 = 40 T 2 6 . Frequency received by subnarine  () v 320 F=   200 Hz pipe  4  0.4 v f= f1  f0  v  v s   40 T  200  8  T = 27.04 N 1 1 9 . f = 305 – 300 = 5Hz Frequency reflected by subrnarine, (i)  Total beats produced in 5s = 5 × 5 = 25 ()  (5) f =f  v  vs   f0  v  vs  = 140  1450  100  2 1  v   v  vs   1450  100  (ii) Time interval in which max intensity becomes Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 = 45.93 kHz minimum ( )=12 1  1  1  0.1 sec 1gm 103 kg f 2 5 2 7 .  =  = 10–4 kg/m, T = 64 N 10cm 0.1m 2 0 . The frequency of B < frequency of A T v 1 64 (B<A ) v=  f = 2L  104 = 5 × 8× 100 0 2  0.1  f – f = 5  427 – f = 5  f = 422 Hz = 4000 m/s  v – v = 1 AB BB string fork 21. Frequency reaching the wall f1 '  f0  v  300   300   ()   v  v s   4000 – 4000 v  1  v = 0.073 m/s Frequency received by the observer 75

JEE-Physics  v  vw  EXERCISE –IV (B)   vw   2 8 . (i) f = f v v Hill 0 s  v  v  1. f = f 0  v  v 0  max s  1200  40  = 580  1200  40  40  = 599 Hz  5   vsound = 1240km/h  340  2   6      = 34 0  340 10  3  (ii)   t1 = 438.7 Hz 1240 t = 1  t = 1 hr  v  v0   340  60   v  vs   340  30  1 1 1240 f = f  3 40 = 257.3 Hz min 0 where t = time the sound to reach the hill 1 (t =)  1 2 . I = 4I0cos2 where  = (1 – 2)t = 103t Let t = time for the echo to reach the train (2t = ) (i) For successive maxima 2 () v = speed of echo (  )  echo = 1200 – 40 = 1160 km/hr 2 t = 10 =6.28 × 10–3 sec  v + v = d – v t train echo train 1 40 1200 (ii) For detection of sound ()  (40 + 1160) t = 1 –  1240 1240 2A2 = 4A2cos2 2 1  cos  = ± 1   ,  t2  1240 hr 2 44  Distance from the hill where echo reaches      4  2 the train  103t = 2  () = d–v(t + t )=1– 40  2 =0.935 km  t =   103  1.57  103 s 2 12 1240 Frequency reaching the hill 3. Frequency reaching the wall f' = f v () ()  1 0  v  v b   v  vw  Frequency reaching the motorist after reflection  vw  v t  f' = f  v from wall ( 1 0 Frequency of echo ()  ) f = f'  v  vw  vt  f =f '  v v m  = f  v  vm  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65  v  v w  1 1  v   v  vb  1 1 0  v  vw   v  vw  vt  Frequency directly reaching the motorist   vw    v  vw   f0 v v () t 1240 1200 f = f  v  vm  = 580 ×  = 620 Hz 0  v  vb  1200 1160 2  v  vs   Beat frequency ( )   v  vs  29. f= f = f  2v  o  2  guard o f = f –f = f (v + v) v b 2 0 m b 12 v 76

JEE-Physics 4 . (i) For the particle P (P) 6 . For air () y  v  y   20 3  v( 3) 1 = L1 1 = 2L t  x  2 1  v =–20 cm/s v = 330  v = f  330 = 500 (2L ) (along negative x-axis) 1 1 1 (x-)  L = 33 cm (ii) Equation of wave ( ) 1 y = A sin (t + kx + ) For CO (CO ) 2  L2 22 4 2 = 4L , v = 264 2 2  v = f  264 = 500 (4L ) 22 2 at t =0, x = 0, y = 2 2 , A= 4  L = 13.2 cm. 2  7. (i)   = v  330 = 1.65 m, d  4 = 2.4242  4 = 2 2 sin   = 4 ,  = 5.5 – 1.5 = 4 cm f 200  1.65 f = v  20cm / s  5Hz At infinity, path difference = 0  4cm ()  10t  x   As the man approaches, the path difference changes as 2 4   y = 4 sin ()  (iii) Energy carried in one wavelength 0,  , , 3 ,2 22 ()    Hence only minima will appear to the man. 1 E = A22 () 2 1 50  (4  102 )2  (10)2  4 (ii)For =   d2  x 12  x1 and 3  d2  x 2  x2  2 2 2 2 1000 100  x = 9.28 m , x = 1.99 m 12 = 162 × 10–5 J 5. Amplitude of reflected wave () 9 . 2  x   = 20cm  10 L 100 (i) Total no of wavelength =  5  k  k1   25  50   20 A= A  k 2  k1   2  25  50   10 3 () r i 2 = 0.667 × 10–3 = 6.67× 10–4  Number of loops formed = 2 × 5 =10 Amplitude of transmitted wave () () 5 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 (ii) Maximum displacement at x = 3 (x)  At   2k1  Ai  2  50  2  103 A = 6 sin    5 = 3cm  k1  k2  50  25  10 3  = 2.67 × 10–3  Equation of reflected wave  x  (iii) y = 6 sin  10  cos (100 t) ()  = 6 sin (10 x) cos (100 t) y = 6.67 × 10–4 cos  (2x + 50 t) v = 6 sin(10 x) sin (100 t) m/s r Equation of transmited wave () 11 14 6 sin(10x) 2 dx 0 2 v2dx  20    = KE max y = 2.67 × 10–3 cos (x – 50t) t 77

JEE-Physics T 100  10 EXERCISE –V-A where  ==0.4  KE = 36J  10 max 1. The fundamental frequency for an open pipe is  x  ()  (iv) y = 6 sin  10  cos (100t) = y + y fv 1 2   y =3sin  x  100t , y = 3 sin  x  100t    ;  2 l 1  10 2  10 2 l 1 0 . (i) Combination of waves producing standing v (ii) wave : Z + Z fopen  2 12 ()   Combination of waves prodcuing a wave 4 travelling along x = y line :Z + Z open pipe close pipe 13 (x=y  ; fclosed = vv 4   4 )   (iii) Position of nodes in case (i)x = (2n + 1) fopen  fA v 4 2k fclosed fB  2  v  2 :1 ((i))  4 . As the string is tied between two rigid supports hence case (ii) x–y = (2n+1) k there will be nodes at both ends. The longest wavelength for nodes at both ends will be the one Y 2  1011 12. v   5000 m / s for which) (  8000  ) 5 2L  2 1 = 2= L   = 5 5 0.4 m  k  2  2  5       2 40cm  0.4 2  = 2f = 2v  2  5000  2,5000  longest = 2 × 40 = 80 cm  0.4 (i) Equation of the wave (  ) 5. y  10 4 sin  6 0 0 t  2x    3  = 2 × 10–6cos(5x) sin(25,00 t) (ii) y = 10–6 sin (25000t – 5x) On comparing the given equation with the general 1 y = 10–6 sin (25000 t + 5x) equation of wave, we get ( 2 ) 1 3 . Amplitude after reflection y  y0 sin t  kx   Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 ()  = 600 ; k =2 A = A  k 2  k1   2.5  5  Wave speed (  )=   600 =300 m/s r i  k 2  k1  = 0.3  2.5  5  = –0.1 k2 Amplitude after transmission () 6 . The frequency of the vibrating string with respect to tuning fork is either (256+5) Hz or (256–5) Hz A= A  2k1   0.2 cm t i  k1  k2  (           (256+5) Hz  (256–5) Hz)  1T fwire = 2l  78

JEE-Physics On increasing tension; the beat frequency decreases 1 0 . Intensity change in decibel to 2Hz, so probable (f r eq uencyofthewi re wi th(  )   respect to fork none is 2Hz= 10 log I2 =20  log I2 =2 I2 = 102 =100 )  I1 I1 I1 256 11. n  1 RT , xn  1 RT , x  T 256 4x M 4M 258 But on increasing the tension of wire, the frequency 1 2 . y = 0.005 cos (x - t) of the wire must have to increase. So, if the original comparing the equation with the standard form, frequency of the wire is assumed to be 261 then () it reduces to 258 whereas if it is assumed to be 251 it has increased to 254. As we were expecting y  A cos  x  t  increase, so the correct frequency of the piano wire   T  2   is (256–5) Hz. (2/=  and 2/T =    = 2/0.08 = 25.00 and  =      261    258  251Hz 254 C 13.A B +1 –1     (256–5)Hz )  Between A & B 1 b/s 1 7 . If the frequency of fork 1 is 200 Hz then probable B & C 1 b/s 1 frequencies of fork 2 is either 196 Hz or 204 Hz. As on attaching some tape on fork 2, be at C & A 2 b/s  2 frequency increases, this is possible only if the  2 b/s 2 2 frequency of fork 2 is 196 Hz. ( 1 20 0 Hz 2 196Hz  204 Hz 214. n' =vv0 n  94 n 2196 Hz)  v 100 v sound from (iii) eqn. of motion v2 = u2 + 2as 5 8. Given that ( )vobserver =  v 2 = 0 + 2as  v0 = 2as 0 Applying Doppler's effect, we get 94 n   v  2as  n 100  v  ()   s = 98 m. f  f v  v0  ; f   f 6v / 5   6 ; f  6  v   v  5 f5 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 f  f  6 1  1  20% 15. y = 0.2 sin    0 t 4 – x  2  .0 0.50   f5 5 v 1 9 .  n' =  v  v s  n T    T  0.04 v= mk 0.04 1  300  0.50  10000 =  300  v s  (9500)  300–v = 300  9500  285  v = 15 ms–1 T=  0.50 2  0.04 = 6.25 N s s  0.04  10000 79

JEE-Physics 1 6 . y(x1t) = e[ ax bt]2 EXERCISE –V-B v = /K = b in –ve x direction. 1 . Mass per unit length of the string, a ()  1 7 . y1(x, t) = 2a sin (wt – kx) 1 0 2 y2(x, t) = a sin (2wt – 2kx) m  = 2.5 × 10–2kg/m But Intensity ( ) 0.4  Velocity of wave in the string. I1  2a n  2 1 () I2  a 2n   I = 22n2a2v    1 Intensity depends on frequency and amplitude v T 1.6  2.5  102  v= 8m/s So statement-1 is true statement-2 is false m (1 For constructive interference between successive )  pulses 1 8 . y1 = A sin (wt – kx) & y2 = A sin (wt + kx) ()  By superposition principle ( )  2 (2)(0.4) tmin =  = 0.10s y = y1 + y2 v 8 = A sin (wt – kx) + A sin (wt + kx) (After two reflections, the wave pulse is in same = 2A sin wt cos kx phase as it was produced, since in one reflection its phase changes by , and if at this moment next Amplitude ()= 2A cos kx identical pulse is produced, then constructive in- terference will be obatined.) At nodes displacement is minimum ( () 2A cos kx = 0  cos kx = 0 kx  (2n  1)   2 x  (2n  1)         2 2 2   x  (2n  1) where n = 0, 1, 2.... )  4 A  v  340   340  f  306  2. f = f  v  v s  f1 = f  340  34  1 19. and f =  340   340 f1 323 19 f  340  f  323   f2  2  17  306 18 v  1T 3. Fundamental frequency is given by 2  v vv ()  n0 = 2 nc = 4( / 2)  2 (with both the ends fixed) () 2 0 . Fundamental frequency ( ) 1 f  V  2 1 T1 y  strain  S Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 2  1.5  S  Fundamental frequency ( )v   eq 3 [for same tension in both strings] (S  cross – section Area) (  )  [] =1 2.2  1011  1 where  = mass per unit length of wire= .A 100 = 178.2 Hz (==.A) 3 7.7  103 ( = density ()) = (r2)  1   r  v  r  v1   r2   2    r   2L   1 v2  r1   1   2r   L  80

JEE-Physics 4 . Energy E  (amplitude)2 (frequency)2 harmonic while in the second case it corresponds ( E  ()2 ()2 to third harmonic. Amplitude (A) is same in both the cases, but ( frequency 2, in the second case is two times the ) frequency () in the first case. ((A)  2 8. The motorcyclist observes no beats. So, the ap- parent frequency observed by him from the two () sources must be equal. Therefore () E = 4E ( 21 5 . After two seconds both the pulses will move 4 cm ) towards each other. So, by their superposition, the f = f  330  v  =  330  v resultant displacement at every point will be zero. 2  176  330 22  165   Therefore, total energy will be purely in the form 1  330 of kinetic. Half of the particles will be moving up- Solving this equation, we get v = 22m/s wards and half of downwards. 9 . Let  be the end correction. ACB A CB ()  Given that, fundamental tone for a length 0.1m= first overtone for the length 0.35m. (    4m  ( 0.1     =0.35m )      v 3v 4(0.1  )  4(0.35  ) ) Solving this equation (  ) (  ) f'f  v  v0  we get = 0.025m = 2.5 × 10–2 m  v  6. Using the formula we get, 5.5  v  vA  ...(i) 1 0 . The frequency is a characteristic of source. It is 5  v  independent of the medium. (    v  vB  ) 5  v  and 6.0  ...(ii) 1 1 . f = f (both first overtone) Here () c0 v = speed of sound (  )  3  vc   2  v0   4L   20  v = speed of train A (A ) A  0  4  v 0  L  4 1 L 1 v = speed of train B (B  ) 3  v c  3 2 as v  B Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 Solving equations (i) and (ii) vB  2 1 2 . The frequency is a characteristic of source. It is vA independent of the medium.(    7 . Let f0 = frequency of tuning fork. ) (f0=) 13. v f1   (2nd harmonic of open pipe) 5 9g () f= 2   v 0 f2  n  4 (nth harmonic of closed pipe) 3 Mg (n)  ( = mass per unit length of wire)  2  ( =)  Solving this, we get M = 25 kg Here, n is odd and f > f (nf>f ) In the first case frequency corresponds to fifth 21 21 It is possible when n=5 because with n = 5 81

JEE-Physics n=5 n=5  MCQ 5v 5 1 . Since, the edges are clamped, displacement of  f2  4     4 f1 the edges u(x,y) = 0 for ( 1 4 . The question is incomplete, as speed of sound is u(x,y)=0) not given. Let us assume speed of sound as 330 m/s. Then, method will be as under. y ((0,L) C (L,L) 330 m/s) B  O A x = (63.2–30.7)cm or  = 0.65 m (0,0) (L,0) 0<x<L 2  speed of sound observed ( ) Line OA i.e. y =0 v= f = 512 × 0.65 = 332.8 m/s AB i.e. x =L 0<y<L 0 BC i.e. y =L 0<x<L  Error is calculting velocity of sound ()  OC i.e. x =0 0<y<L = 2.8m/s = 280 cm/s The above conditions are satisfied only i nalternatives (b) and (c). (   (b) (c))  15. f  v  T Note that u(x,y)=0, for all four values e.g. in alternative (d), u(x,y)=0 for y=0, y=L but it is not f = 2f zero for x=0 or x=L, Similarly in option (a) u(x,y)=0 AB CD at x=L, y=L but it is not zero for x=0 or y=0 while in options (b) and (c), u(x,y)=0 for x=0, y=0 x = L  T = 4T ....(i) and y=L. AB CD Further p=0 (–x)  T (x) = T  AB CD  – (as T = 4T ) 4x = x AB CD  x = /5 (u(x,y)=0 (d)y=0, y=Lu(x,y)=0 x=0 1 6 . Take y=Asin(t–kx) y x=L P (a) x=L, y=L  u(x,y)=0  x=0  x y=0 (b)(c) x=0, y=0 x = L y=L  u(x,y)=0  2. Maximum speed of any point on the string =a y () so vP = t = –Acos(t – kx) = a(2f)  2v  v 10 (Given : v=10m/s)  vP =  A 2  y2 =    A 2  y2    1  2(10  102 ) 10 10 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65  2af = 1 = (10)2  (5 )2  10 2 = 3 ms–1 0.5 50 f  1  a = 10–3m (Given) 2 a 1 T 3v 1 103 17 . 2L m  4  f  2  103  Hz 2 where v = 340 ms–1,  = 75 cm = 0.75 m Now according to given condition Speed of wave (  )v= f 1T 1T 3v   103 n– = 4 so n = m + 4 =  4  4  2  2L m 2L  (10 m/s) = s1   3 340  4 = 344 Hz.   = 2 × 10–2m = 4 0.75 82

JEE-Physics 3 . For a plane wave intensity (energy crossing per Speed of pulse (  )  unit area per unit time) is constant at all points. At t=1s and x=–1.25 m ( y y 0.16m  0.16m  12 –x x 12 x=–1.25 m x=0 (b) t=0 But for a spherical wave, intensity at a distance r t=1s from a point source of power P (energy transmit- ted per unit time) is given by : value of y is again 0.16 m, i.e., pulse has travelled a distance of 1.25 m in 1s in negative x- direction (Pr or we can say that the speed of pulse is 1.25 m/s  ) and it is travelling in negative x-direciton. Therefore, it will travel a distance of 2.5m in 2s. S The above statement can be better understood from figure (b). S (t = 1s x = –1.25 m y0.16m  x1s1.25m 1.25m/ s x-2s 2.5 m(b) ) I  P or I  1 5 . In case of sound wave, y can represent pressure 4 r2 r2 and displacement, while in case of an electromagnetic wave it represents electric and magnetic fields. 4 . The shape of pulse at x=0 and t=0 would be as (y  shown, in figure (a).  (x=0t=0(a) )  )  y 6 . Standing waves can be produced only when two 0.16m similar type of waves (same frequency and speed, 0.8 –x 0 x but amplitude may be different) travel in opposite y(0,0) = = 0.16m t=0 (a) directions. 5 (  From the figure it is clear that y = 0.16m ) max Pulse will be symmetric (Symmetry is checked Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 about y ) if at t=0 max (   y = 0.16m   Comp rehensi on#1 max 1 . In one second number of maximas is called the beat frequency.  y t=0 max y(x) = y(–x) (1s) From the given equation and () 100 92 0.8  Hence, f =f –f   = 4Hz b 1 2 2 2 y(x)   y(x  16x2  5  at t=0  y(x) = y(–x) 2. Speed of wave (  )  0.8  )   100 92 16x2  5 v= = v  0.5  = 200 m/s R 0.46  Therefore, pulse is symmetric.(  )  83

JEE-Physics 3. At x=0, y=y + y = 2A cos 96t cos 4t SUBJECTIVE 1 2 Frequency of cos (96t) function is 48Hz and that 1 . (i) Frequency of second overtone of the closed pipe of cos (4f) function is 2 Hz. ()  (cos (96t)  48Hz  cos v (4f) 2Hz)  5  4L  = 440 In one second cos function becomes zero at 2f times, where f is the frequency. Therefore, first function will become zero at 96 times and the sec- x=0; P=0 ond at 4 times. But second will not overlap with first. Hence, net y will become zero 100 times in 1s. L=5 x=x; P=±P0sin Kx 4 (1scos2ff  964  y1 100 ) L  5v m 4  440 Comprehension#2 B Substituting v= speed of sound in air =330m/s 1 . v = 340 + 20 = 360 m/s (v==330 m) SA 5  330 15 v = 340 – 30 = 310 m/s L  m SB 4  440 16 A 20m/s 340m/s 340m/s 30m/s  15  4  16  2 . For the passengers in train A, there is no relative   4L   3 m 5 54 motion between source and observer, as both are (ii) Open end is displacement antinode. Therefore, moving with velocity 20m/s. Therefore, there is it would be a pressure node or at x=0; P=0 no change in observed frequencies and correspondingly there is no change in their ( intensities. x=0 P=0) (APressure amplitude at x=x, (x=x   ) 20m/s  can be written as  P = ± P0 sin kx ) 2 2  8 m 1 k  where  3/4 3 3 . For the passengers in train B, observer is reced- Therefore, pressure amplitude at ing with velocity 30m/s and source is approach- ing with velocity 20m/s. ()  (B  30m/s     L 15 /16 20m/s)  x  m or (15/32) m will be 22 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 P = ± P sin  8 15  =±  P s i n  5 0    32    f1'   340  30  775Hz 3 0 4 800  340  20   and f2'  340  30  1085Hz P0  1120  340  20   P = ± 2 (iii) Open end is a pressure node i.e. P=0  Spread of frequency (  )  (P=0) =f ' – f ' = 310 Hz Hence, P = P = Mean pressure (P ) 21 max min 0 (P = P =  (P )) max min 0 84

JEE-Physics (iv) Closed end is a displacement node or pressure i.e., the amplitude of reflected wave will be 1.5cm. antinode.(Negative sign of A indicates that there will be a r ) phase change of  in reflected wave. Similarly. Therefore, P = P + P0 and P = P – P0 (1.5cm A max 0 min 0 r   2 . Amplitude of incident wave ()  )  P QR  2  32  A t   32  80  (3.5)  2.0cm A = 3.5 cm L1=4.8m L2=2.56m i Mass = 0.06 kg Mass = 0.2 kg i.e., the amplitude of transmitted wave will be 2.0cm Tension () T = 80N Mass per unit length of wire PQ is (2.0cm) (PQ)  3 . Speed of sound (  )v = 340 m/s 0.06 1 Let  be the length of air column corresponding m=  kg/m 0 1 4.8 80 to the fundamental frequency. (   and mass per unit length of wire QR is 0 ) Then (QR)  v  212.5 0  v  340 40 4(212.5) 4(212.5 ) = 0.4 m  m2 0.2 1   kg / m 2.56 12.8 (i) Speed of wave in wire PQ is Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 (PQ) 3.2m v1  0.4mT 80  2.4m 1.2mm11 / 80 =80m/s 1.6m 2.0m and speed of wave in wire QR is 2.8m (QR ) 0.8m T 80 In closed pipe only odd harmonics are obtained.  Now let  , ,  , etc., be the lengths v= m2 1 / 128 = 32 m/s 2 1 2 34  Time taken by the wave pulse to reach from P corresponding to the 3rd harmonic, 4th harmonic, to R is 7th harmonic etc. Then, (PR) (          , ,          1 2 34  4.8 2.56  4.8 2.56  v t    80  32  s   t = 0.14s V1 V2 3  41  = 212.5   = 1.2m 1 (ii) The expressions for reflected and transmitted amplitudes (A and A ) in terms of v ,v and A are v rt 12 i  5  212.5  2 =2.0 m as follows :(v v    42  12 (A A )Ai rt v2  v1 2v2 and  v  = 212.5   = 2.8m 7  43  3 A= A and A = v1 A r v2  v1 i t  v2 i Substituting the values, we get ( )  v 9  34   = 212.5  4 = 3.6m A  32  80 –1.5 r =  32  80  (3.5) = cm or heights of water level are ()  85

JEE-Physics (3.6–0.4)m, (3.6 –1.2)m, (3.6 – 2.0) m,(3.6 – 2.8)m. 4 . Velocity of sound in water is ()  Heights of water level are ()  2.088  109 v=  103  1445m / s 3.2m, 2.4m, 1.6m and 0.8m. w Let A and a be the area of cross-sections of the pipe and hole respectively. Then Frequency of sound in water will be (Aa)()  A f0  vw 1445  f = 105 hz w  14.45  103 Hz 0 (i) Frequency of sound detected by receiver (ob- server) at rest would be (    )  a Source f0 vs=10m/s Observer A = (2 × 10–2)2 = 1.26 × 10–3 m2 (At rest) and a = (10–3)2 = 3.14 × 10–6 m2 vr=2m/s Velocity of efflux ( )v 2gH f1  f0  vw  vr   (1 0 5 )  1445  2  H z Continuity equaiton at 1 and 2 gives :  vw  vr    1445  2   vs 1 0 (12)  dH  f = 1.0069 × 105 Hz a 2gH  A  dt  1  Rate of fall of water level in the pipe. (ii) Velocity of sound in air is () () va  RT (1.4)(8.31)(20  273)  = 344m/s M 28.8  103   dH   a 2gH  Frequency does not depend on the medium.  dt  A Therefore, frequency in air is also f =105 Hz Substituting the values we get ( )  0 dH 3.14  106 ( dt  1.26  103 2  10  H f =105 Hz) 0  Frequency of sound detected by receiver (ob- server) in air would be () dH f2  f0  va va  w vs   105  344  5     (1.11  10 2 ) H   w   344  5  10  Hz dt Between first two resonances, the water level falls from 3.2m to 2.4m. ( f = 1.0304 × 105 Hz 2  3.2 m2.4 m)  5 . (i) Frequency of second harmonic in pipe A = Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 dH frequency of third harmonic in pipe B  =–(1.11 × 10–2)dt ((i)A=B H ) 2.4 dH t    vA   vB  H =–(1.11 × 10–2) dt  2  2A   3  4B  3.2 0  2[ 2.4  3.2 ] = –(1.11 × 10–2).t vA  3  B)  A RTA 3  t  43s vB 4 A MA   (as =  B RTB 4 MB 86

JEE-Physics A MB 3  dE = 22 f2 a2 sin2 kx dx B MA   (as T = T )  T 4 AB MA A  16   5 / 3 16 v2      MB B  9    7 / 5   9  Here, f  2  (42 ) and k =   5 7 Substituting these values in equation (i) and inte-   A  3 and  B  5  grating it from x=0 to x=, we get total energy of string ((i)x=0 x=  MA  25 16 400 )E 2 a 2 T MB   21   9   4 189 (ii) Ratio of fundamental frequency in pipe A and  v f  v  vs  in pipe B is : (A  B   8. From the relation f'  )  fA vA / 2A va 2.2  300  fB vB / 2B vB  f 300   (as  = ) we have  vT  ...(i) AB   A RTA 1.8  f  300  and  30 0  v  ...(ii)   MA T  B RTB   A . MB (as T = T) Here, vT = vs = velocity of source/train B MA A B Solving equation (i) and (ii), we get v = 30m/s MB T MB 189 9 . Maximum particle velocity (   )  MA  400 from part (i), we get Substituting A = 3m/s ...(i) Maximum particle acceleration fA  25 189 3 ()   2A = 90m/s2 ...(ii) fB 21 400 4 Velocity of wave ( )  =20m/s 6 . Fundamental frequency ( ) k ...(iii) 0.6r From equation (i), (ii) and (iii), we get  =30 rad/s A = 0.1m and k=15m–1 v  Equation of waveform should be (  ) f = 4(  0.6r) y = A sin(t + kx + ) y=(0.1m) sin [(30 rad/s)t + (1.5m–1)x + ]  Speed of sound (  )v = 4f (+0.6r) 1 0 . L = 20 cm; m = 1 gm  v = (4) (480) [(0.16) + (0.6) (0.025)] = 336m/ m 1 gm / cm 1 10 3  20 20 s = = × g/m L 10 2  2  7.     = 2, k =     1   200  2  = kg / m ; T = 0.5 N Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 The amplitude at a distance x from x=0 is given by A = a sin kx (x=0xA= asinkx) 0.5 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\a v= 1 = 10 m/s; f = 100 Hz \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\  200  x=0 x=   v  10  1 m f 100 10 Total mechanical energy at x of length dx is   1 m = 5 cm (dxx ) 2 20 11 dE = (dm) A22  (dx) (a sin kx)2 (2f)2 22 87



JEE-Physics UNIT # 11 (PART - II) WAVE OPTICS (Nature of Light & Interference) EXERCISE  –I I1 17. Given  I2 =4    I  =  4I (let  the  I   =  I) 2 8 . I   a 2,  I     a 2 1 1  1 2 2 I  =  I   +  I  +  2 I1 I2 cos        Imax  ( I1  I2 )2  (2 I  I )2 = 9I resultant 1 2    2 2  2       I  =  I1  I2 =  2 I  I = I I   =  I1  I2 ,  When  cos    =  1,    =  0,  2,.... min max Imax  Imin  9I  I 8I  4   Imax  Imin 9I  I  5 9 . y   =  a  sint  =  a  cos  t  –/2), 1 101 y   =  a  cos  t 2 D 1 9 . For  minima  y   =  (2n–1) where  n  is  the  no.  of    3  n 2d 10. y   =  a  sin t    and  y   =  a  sin  t nD 1 2 d minimum.  For  maxima  yn  A a12  a 2  2a1a2 cos    where       9D D 2 y 5th dark 2 d ;   y1st max ima  d 3  a2  a2  2aa cos   3a By  Equation  y 5th min  y1st max =  7  ×  10–3 3 9D D 7  10 3  15  105 2 2d  d 7  103    7  50  102 1 1 . In  interference  pattern  we  can  see  that  resultant amplitude  of  super  imposed  wave  depends  on  the     =  600  nm phase  difference  of  waves  so  it  varies  from maximum  to  minimum  amplitude  by  redistributing D of  energy  but  total  energy  remains  conserved. 2 0 . Fringe  width    =  d ; ' =  'D '    2D  ,   '  4 D  4 d' d/2 d 12.  3/2 2 0  2 2 1 . On  a  given  screen  width  n=  car  tan,  Here  n  is  d number  of  fringes  and  = D   is  the  fringe  width When  path  difference  is    then  phase  difference  is  2 2 n    =  n      n  =n  11 22 21 22 When  path  difference  is  1  then  phase  difference  is     n2  n11 =  92  5898  99.360 2 5461 When  path  difference  is  x  then  phase  difference  is  2 x  Number  of  fringes  are  integers  so  n   =  99 2 13. Resultant  amplitude  A  =  a 2  a 2  2a1a2 cos  2 3 . Mica  sheet  of  thickness  't' Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 1 2 P   –  Angle  between  two  waves  or  phase  difference. Refractive  index  '' S1 So  A  depends  on  both  amplitude  &  phase S2 difference. XP =  S P  –[(S P–t)   +  t ] x 2 1 air sheet =  (S P–S P)  –  (t  –  t) 21 =  d  sin  –(–1)t 1 5 . From  the  definition  of  coherent  source. Additional  path  difference Position  of  nth  maxima 16. Sustained  interference  means  a  interference n  dyn  (  1)t   y  =  nD  D (–1)t pattern  (arrangement  of  fringes)  can  not  be  change D n dd with  time.  It  is  only  possible  when  the  phase difference  between  waves  at  a  point  does  not Shift  of  interference  pattern   D   1 t change  with  time. d 76

JEE-Physics 24. Central  fringe  is  that  fringe  where  the  path difference  x  =0,  for    all  wavelengths.  So  the central  fringe  is  white  followed  by  the  some coloured  fringes  and  after  that  there  is  very  much overlapping  in  fringes  so  equal  illumination  exists. 2 5 . Central  maxima  i.e. I  = ( I1  I2 )2    I  (2 I )2 =4I 0 0  I =  I0 •  When    increases,  the  fringe  patterns  shift 4 towards  the  slit  when  sheet  introduced  shift [Intensity  due  to  single  slit] D  1 26. For  maxima,  I   =  ( I1  I2 )2   from  the  above  two  fact  variation  in  I max t with  is When  I   =  I   then  I   =  4I 12 max  For  minima  I   =  2 min I1  I2   I   =  0 min  2    I2  I1 ,  then  I' I1  I1   4 I1 2 max 2   2 3 1 . There  is  a  phase  change  of    when  the  ray  enters I1  from  rarer  medium  into  denser  medium,  the I  '  =   I1   Imin boundary  is  called  rigid  boundary  and  also  no 2 change  in  phase  when  it  enters  into  rarer  medium. min 3 2 . As  distance  between  S   &  S   is  very  much  less  and 2 7 .   =  200nm,  d=  700  nm 12 equal  to  'd'  so  the  angle  made  at  'P'  is  very  much No.  of  maxima  =  2d  2  700nm  7 small  200nm S1 28. Fringe  width  remains  unchanged    d d P D S2 But  central  bright  fringe  (zero  path  difference) D Shift  downwards  therefore  whole  fringe  pattern shifts  downwards. 2 9 . When =1 i.e. no change in path difference due to Arc d optical path difference so the mid point of the   Angle  =  D    D screen is again the central bright fringe. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65   Fringe  width  =  D       d    B u t  w h e n   >  1  t h e  c e n t r a l  b r i g h t  f r i n g e  w i l l  s h i f t d  D  D 3 4 . Path difference for points A & C is O so constructive according to  d (–1)t interference  take  place. As    increases,  CBF  will  shift  upwards  from midpoint  i.e.  at  mid  point.  Less  bright  fringe 5 appears  and  when  the  shift  is  equal  to  value  of Path  difference  for  B  &  D  =  5m  =  2 fringe  width,  the  dark  fringe  (zero  intensity)  will So  distructive  interference  take  place. a p p e a r. OR •  Intensity  variation       O    central  bright  fringe        y    distance from  'O'  screen 77

JEE-Physics 35.  x+x P EXERCISE  –II Ax x  =  n 1 .   Path  differences    3 y S1 3 S2 x1  2  x2    x3  4  x 4  4 O D Therefore  I   >  I   =  I   >  I 2341 in  this  case  path  difference  is  d  cos  .  2 . Path  difference  =  1 So n =  = 3 cos   cos  =  3 6 D1   Phase  difference  =  2   6 3        3D  =  D2  y2 D2  y2 3   I  =  I  +  I  +  2 I1 I2 cos  1 2   y  8D =  2 2D If I   =  I   =  I  then  intensity  at  central 12 I   =  I  +  I  +  2 I I cos  0°  =  4I 3 6 . Let  t  be  the  thickness  so  corresponding 0 and  intensity  at  a  point  'P' x  =  t   3 t  .  Also  I   =  4I so  I'  =  2I I' =  I  +  I  +  2 I  I  cos  max 3 2 We  know  I  =  I +  I  +  2 I1I2 cos   33 I0 =  3I    4I   =  0.75  I R 1 2 4 4 0  2I  = I + I  +  2 I2 cos  3 t      cos  3 t  0 Phase  difference  =  2  Path  difference       3.    cos  3t   cos  or  cos 3  or  5    2 2 2 4 . Let  the  distance  'y'from  'O'  where  nearest  white   3 t     t    ;  3 t  3  t =   spot  occurs.  Then  the  path  difference  =  0  2 6  2 2 and  3 t  5  t  5  2 6 3 7 . At  the  central  Maxima  x=0 d 2d/3 y But  upward  shift   (2  1)tD   and D d Downward  shift   (  1)2 tD  2d  2  d  2 d  3   3  tD tD  D2   y  D2   y 0 d d So  net  shift  y  = [2–1  –  2 +  2]    y  3 d Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 3 8 . I'  =  4 (4I)=  3I.  y = 6 So 4I  = I +I  +2I  cos       cos  =  1        5. Path  difference  occurs  due  to  the  medium  of 2 3 variable  refractive  index  =  (–1)t But  value  should  lie  between  3  &  6.   1 a2 2    1  ax  dx  So  it  cannot be  3 For  second  minima    =  3 0 For  third  maxima  =  6 For  minima  at  '0'  the  path  difference  should  be (for  minimum  value  of  ) 2 a2   ,  a      2 2 78

JEE-Physics 6. For  reflected  light  waves  the  two  virtual  source 11. The  intensity  of  light  is  I()  =  I cos2   (images  of  source  in  mirrors)  at  a  distance  '3d'   0 2 Path  difference  at  O  =  S O–S O where    2 ( x )   2 (d  sin  ) 21     S1 d (i) For    =  30° S O c 3 108 2d D   =  300  m  and  d  =  150  m  106 S2    2  (1 5 0 )  1              300   2  2 2 4 D2  (2d)2  D2  d2    D2  +  4d2    =  D2  +  d2  +  2  + 2 D2  d2    I()  =  I cos2    I0  4  2 3d2 0 3d2 =  2D    (ii)  For    =  90° 2D    2  (1 5 0 )(1 )           and  I()  =  0 7 . By  Geometry,  path  difference  at  '0'  for  minima  300  2 2  d= 5cm S1 0  should  be  (2n–1) S2  /2 (iii)  For    =  0°  :     = 0    =0        I()  =  I 2 20  1 2 . From  the  snells  law   S O –  S O  =  (2n  –  1)   D=12 1  ×    sin  (90– )  =  sin 2  1 2 12  D2  d2  D  (2n  1)  bd b 2 1× c    c       d  13 . d =  2    2  2  (13–12)cm  =  (2n–1) 2 2 12  6 3 For  n=1,  2,3     =  2cm,  3 cm, d23  2   2  2 A 60°  2 A 8. As  path  difference  d sin   n  d  n sin   2  3A  2   2  As  sin   1   so  d  >  n,  where  n  is  an  integer A' 3A    3A 2   Therefore  d    and  d  2   I    A2  A 2   =3A2    I  =  3I R R 1 0 . Seperation  between  slits  =b,  screen  distance  d  (>>b) 14. fringes  to  be  more  closed  when  fringe  width  becomes  reduced  fringe  width    D d Path  difference  at  'O'  must  be  odd  multiple  of  As    decrease    also  decreases  (   <   ) 2 Blue green n for  missing  wavelengths      S O–S O= Ba 1 5 .  =  212 5 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 S1 O y QP (–a,0) O (2a,0) bd 3a S2 For  point  P,  x  =  3a  =  15  and  at  point  Q  it  is also  15  somewhere  at  point  B,  it  is  zero    thus  in  d2  b2  d  n half  part  of  the  circle  available  maxima 2 15  ×  2  =  30 n22 n Thus  total  maxima  =  2  ×  30  =  60  d2  +  b2  =  d2+ 2 d 42 b2    (n=1,3,5) nd 79

JEE-Physics 1 6 . The  maxima  at  P  becomes  minima  &  then  maxima Reflected (Source of alternately.  But  central  bright  fringe  is  always 2 . wave fronts  reflected  remain  at  O. will be wave  fronts) 1 7 . Here  the  shift  produced  by  mica  sheet  of  thickness  t  is  (t–t)  =  t(–1).  It  should  be  equal  to  extra  path  sin  45°  =  m sin   =30° traveled  by  the  ray  SS O i.e.    =  SS   –  d   2d  d 12 2  2 t S   2 2  1)     t=2d ( infront  of  1  d( 2  1) 3. (a  b) 1  b  2a(  1) a  21.  =     2(  1)  For  parallel  beam,  a   .  So    =  2(  1) EXERCISE  –  III 4. Tr ue/False 2 . For  interference  pattern,  sources  must  be  cohrent and  two  independent  light  sources  are  never  be coherent. 3 . Central  bright  fringe  (Path  difference  at  this  point is  zero  so  this  fringe  is  white  then  following  fringes are  coloured.) Match  the  Column 5. v1t 1 . Additional  path  difference  by  introducing  the  thin 1=1 sheet  is  given  by  (–1)t v2t 2 Resultant  path  difference  at  P  =  Geometrical  path difference  +  Optical  path  difference Comprehension–1 v1t  2  2  2  2 2 1 . Optical  path  length v2t 1 1 1 1  x3 1 4 6 . Angle  made  by  the  direction  of  light  with  the dx  0   =  1  x2 dx   x  3  0  m 22 3 y–axis is cos    12  22  32 14 2 . Optical  path  length  must  be  optimum  i.e.  mini- mum. Comprehension–4 P \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 3. A B \\\\\\\\\\\\\\\\ Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\ 1 . Third  order  bright  fringe  is  the  3rd    bright  fringe from  the  central  bright  fringe  having  3  path difference. For  any  point  AP  +  PB  =  constant 2 . At central bright fringe the waves from slits are in =  2  (semi–major  axis  of  ellipse) phase  and  following  bright  fringes  having  a d i f f e r e n c e  o f  2 .  4   =  2  ×  2    2 nd o r d e r  b r i g h t Comprehension–3 fringe 1 . Wave  front  of  point  source  is  spherical.  The  point 3 . CBF    Path  difference  =  0 source  is  at  origin  and  distance  travelled  by  wave X   At  1, in  't'  time  with  a  speed  of  light  'c'  is  'ct'.  Hence radius  of  wave  front  is  'ct'. A   Equation  of  sphere  is    x2  +  y2  +  z2  =  (ct)2  dark  fringe  having  path  difference  = 2 80

JEE-Physics Xc    At  3, EXERCISE  –IV(A) bright  fringe  having  path  difference=           600 nm  300nm 1.   2  2 2 (i)  I   =  I  +  4I  +  2  ×  I 4I cos  4 (|X |–|X |)=  result cA =  5  I  +  2 2 I  =  7.8  I Comprehension–6 (ii)  I  =  I  +  4I  +  2 I 4 I  cos  =  5I–  4I  =  I 1 . For  strongly  reflect  light  path  difference result  (2t  + 2 )  –  2 =     (for  minimum  thickness) (iii)  I =  I  +  4I  +  2 I ×  4I  cos  4  = 5I  +  4I  =  9I result =1 2. The  position  of  maxima  where  the  path  difference between  two  ways  is  integral  multiple  of  .  =1.5 And  positions  of  minima  where  the  path  difference between  two  ways  is  odd  multipleof  /2.  =1.8  2t  =      t  600nm  200nm 2 2 2 1.5 S1 S2 2 . Again  as  previous  question  path  difference  for  n,t 2    reflect  light  must  be  the  odd  multiple  of  3. I 2 As  given  reflection  cofficient  =  20%    AB  =  5 n  2t =  2 I B B' A I/5 4I × 4   t  n  640 nm =  n  ×  120    (n=1,3,5) 25 5 2  2 1.33 A' Hence  =  3  ×  120  =  360  nm 4I 4I 5 5×  5  16I 3 . Path  difference    2t  – 2 =  0 A'B' =  125 If  AB  and  A'B'  interference  than 2t =    t  2     2 2 4 I  =   I 2 =  I  81  5 25 4 . t  =  350nm,  n  =  1.35 max  16I  2nt  –   m     (m  =1,3,5,...) 5 125  22 2  I 16I  I 1 Imin     (m 1)  5 125  5 25 t  2  350nm   I /I =  81  :  1 2 1.35 max min 4. d  =  0.2  cm,    =  5896  Å,  D  =1m Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65   350 2.7 2  945 2 5. D 5896  10 10  1 Fringe width  =   =  0.3  mm (m 1) m 1 81 d 0.2  10 2 For  m  =  1,3,5,    =  945,  473... If  system  is  immersed  in  water  (=1.33),  then  the fringe  width  becomes 5 . t  =  1  m,  n  =  1.35,    =  600  nm 2 1.35 106  '    0.3 mm =  0.225  mm 2  1.3 Path  difference  =  2.7  ×  10–6  –  300  ×  10–9  =  2.4  m Shifting  of  fringe  pattern  due  to  plate  is  given  by D d (–1)t  [Towards  the  side  of  plate] Due  to  two  plates  introducing  infront  of  slits  then shifting  is  resultant  of  both

JEE-Physics D 5D EXERCISE  –4(B) [(1.7  –  1)2t  –  (1.4–1)t]   1 . I  =  10–15  W/m2,      =  4000 3 Å,  t  =  3mm. dd The  path  difference  due  to  glass  plate  3mm (Position  of  5th  bright  fringe) (t  –  thickness  of  one  plate) 3mm t  =  5  =  5  ×  4800  Å  =  2.4  m   Thickness  of  second  =  2t  =  4.8  m Path  difference=   ds   (n 1)dx 6 . Fringe  width    D ;   '  (D  5  102 ) 0 dd 3mm (D  5  102 ) D 3 x  1)dx  2 x3/2  2 x Given  |'–|   3 3mm     d d (1  0   5  10 2 Phase  difference   3  ×  10–5  =  =  4000 2 3  103 =    ×  107–3 d 2 3  105  103     5  102 =  6000  Å 3  10 10 7 . The  length  of  the  screen  for  the  fringe  pattern   2n  =  104  ×   =  2y At  point  I  there  is  point  of  maxima    n  =  5  ×  103 y   Intensity  =  I  +  I  + 2 I  I cos  104         =  4I  =  4  ×  10–15  W/m2 S1 max D 2 . Fringe  pattern  forms  on  a  screen  the  distance  of the  nth  maxima  in  x–direction  i.e.  x  cordinates  is S2 dsin nD'  and  y–position  is  decided  by  the  SHM  of spring.   d  sin  max  =  y =1;  y  =  D D Mg D'  =  D  +  (1  cos t)   No.  of  maxima k  2  D  2D 3 . Path  difference  =  d  cos  60° fringe width D / d  2d  2  5cm =  3.3  (say  3) A  3cm 8 . Due to the introduction of sheet in front of one slit 30° 0.1mm O D B (thickness t and refrective index ) the shift =  d (–1)t i.e.  the  path  difference  becomes  (–1)t  instead  of zero  at  centre  of  screen  (x    0)   Phase  difference  =  2  (  1)t As  given  at  0  the  Intensity  =  3I  3I  =  I  +  4I+ 2 I 4I   cos Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65   Resultant  intensity II I I cos 2 (  1 ) t    2 I0  4  4  2 4 4    3 I 2 d  (1.5  1)  20.4  10 6  1 t  106   2 Intensity  at  centre  I  =  4I'    I'  =  4   2 2  3 [I'  –  Intensity  due  to  one  slit] I0  2I   cos  2 (   1)t   0.1  ×  10–3  +  (20.4  –  t)10–6 4 1                               =  2/3  × 6000  ×  10–10  =  4  ×  10–7  I0  I cos2 2 (  1t)   I cos2  (  1)t   t  =  20.4  –  0.4  +  100  =  120  m   2     82

JEE-Physics 4 .   Fringe  width OA  =  3  =  3(1/3)  mm  or  OA  =  1mm D 5000  1010  80  102  m  150 m (ii)  If  the  gap  between  L   and  L   is  reduced,  d  will   12 d 4  2  103 3 decrease.  Hence,  the  fringe  width    will  increase or  the  distance  OA  will  increase.   Net  upward  shift    D D  w g  1 t1  d w y  1 t2  25m 1 S1 0.5mm A d S 0.25mm O Phase  difference  at  point  C 0.25mm 2 S2 0.5mm   2   25m    D=1.0m  150m  3 0.15m 0.3m 1.3 m IC  Imax cos2   Imax 3  IC 3 7 . Path  difference  between  rays  1  and  2  : 2  4  Imax  4 5 . Distance  between  two  sources  S   and  S Pr R Only 5000Å is coming out 12 ii r1 d  =  2  ×  2a cos    sin  =  2  a  sin  2 S1 acos QS 2  acos x  =  (QS)  –  PR  ...(i) da Further QS  sin i ;  PR =  sin  r 2acos PS PS S2   PR / PS  sin r        (QS)  =  PR QS / PS sin i Screen  distance  D  =  b  +  2a  cos2   Substituting  in  equation  (i),  we  get  x=0   D  (b  2a cos2 )  (b  2a)   Phase  difference  between  rays  1  and  2  will  be  0 d 2a sin 2 4a or  these  two  rays  will  interfere  constructively. (if    is  very  much  small) So  maximum  intensity  will  be  obtained  from  their interference    or 6 . (i)  For  the  lens,  u  =  –0.15  m;  f  =  +0.10  m    2 2 Imax  I1  I2  4I  I  9I 11 1 8 . Each  plate  reflects  25%  and  transmits  75°.  Incident Therefore,  using      we  have vu f 1 11 1 1 beam  has  an  intensity  I.  This  beam  undergoes v  u  f  (0.15)  (0.10)   or    v  =  0.3  m multiple  reflections  and  refractions.  The corresponding  intensity  after  each  reflection  and Linear  magnification  ,  m  =  v  0.3  2 refraction  (transmission)  are  shown  in  figure. u 0.15 Hence,  two  images  S   and  S   of  S  will  be  formed 12 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 at  0.3  m  from  the  lens  as  shown  in  figure.  Image S   due  to  part  1  will  be  formed  at  0.5  mm  above 1 its  optic  axis  (m=–2).  Similarly,  S   due  to  part  2  is 2 formed  0.5  mm  below  the  optic  axis  of  this  part  as shown. Hence,  d  =  distance  between  S   and  S   =  1.5  mm Interference  pattern  is  to  take  place  between  rays 12 1  and  2. I   =  I/4  and  I   =  9I/64   =  1.30  –  0.30  =  1.0  m  =  103  mm 12   =  500  nm  =  5  ×  10–4  mm Therefore,  fringe  width, D (5  104 )(103 ) 1  I2  2 d (1.5 ) 3     mm =  mm  Imin I1   1 Imax  I1    I2 49 Now,  as  the  point  A  is  at  the  third  maxima 83

JEE-Physics EXERCISE  –  V-A cc 7 . µ  =  v    v  =  µ 1 . To  demonstrate  the  phenomenon  of  interference  we Since  I  is decreasing  so µ  also  decreases  and  hence require  coherent  sources,  i.e.,  sources  with  same v  increases frequency  and  a  fixed  phase  relationship. So  v  is  minimum  on  the  axis  of  the  beam. 2. dsin  =  n   n= d sin  = 2 sin  =2sin 8.   x Axis converge  when  it  enter  in  the  medium. When  light  is  moving  and  as  it  enters  the  medium than  along  the  axis  velocity  is  decreasing  so  as  we  nmax  =  2 move  away  from  the  centre  (that  is  x  in  figure)  the  Maximum  number  of  possible wave  covers  less  distance  and  hence  shape  is interference  maxima  =  (2n   +  1)  =  5 convex. max 3. The  shape  of  interference  fringes  on  the  screen  is 9 . At P  :  x  =  0;    2  0  0  hyperbola. IP   I  I 2  4I 4 . The  intensity  at  a  general  point  with  respect  to  4 I  I0 cos2   At Q    :  x   2  maximum  intensity  is     2 IQ  I  I  2 II cos   2I ;  IP  4I  2    2 IQ 2I 1 Phase  difference,    (Path  difference)    2    10.  2 6 3 I coherent I  I  4 I Inoncoherent  =  I  +  I  =  2I I   60   2  2 3 Icoherent 4I 2 I0 cos    4  Hence,      3  Inoncoherent 2 I 1 2  2   1 1 . 2 Im 9 5 . Third  bright  of  known  light Imax  I 4I  9I  Im  I  X3  =  31D   .....  (1) IP  =  I  +  4I  +  2  (I)(4I) cos d =  5I  +  4I  cos 4th  bright  of  unknown  light =  I  +  4I  (1  +  cos) X4  =  42D   .....  (2) =  I  +  8I  cos2  d 2 Given X3  =  X4 Im 1  31  =  42 9 2  =   8 cos2 33 1 2 . For  bright  fringe  S1P  –  S2P  =  n Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 2  =  4 1  =  4   ×  590  =  442.5  nm 6 . Parallel  cylindrical  beam  gives  planar  wavefront P S1 S2 d D So  fringes  are  concentric  circles  (centre  of  origin) 84

JEE-Physics EXERCISE  –V-B points. 1. I   =  I  +  4I  +  2  Distance  S  =  D1  (2 n 2  1)  (2n1  1)  4I2 cos 2 =5I d  2  A I  =  I  +  4I  +  2 4I2 cos  =  I Now  putting  n   =  10  and  n   =  3 21 B S  =  4  ×  7  ×  10–3  m    S  =  28  mm So  difference  I   –  I   =  4I AB 2 . As  path  difference  due  to  slab  =  (–1)t 5. Given  I   =  4I =  I.  So  I  =  I  +  I  +  2I  cos     (–1)  t=  n R 4 for  minimum  thickness  t  of  plate,  n  should  be cos    =   1       =  2 minimum  i.e.  n=1    (–1)t  =   2 3    t =    1  t =  1.5  1   t = 2 Corresponding  path  difference  x  =  3 PR d   3 . In OPR ;  = cos   OP =  So d sin  =  3  = sin–1  3d  OP cos  OC 6. Inet  =  I0  +  I0  +    2I0  cos  =  Imax  2I0  cos    = in COP  cos  2 =   OP 2 OC= OP  cos 2   d cos2 0     =  n    2 x      x  =  2n  1  cos  2 4  So  Path  difference  =  CO  +  OP  +  2 MCQ  d cos2  d   1 . Given  d,  ,  I   =  4I ,  I   =  4I  &  I   =  I cos  cos  2 1 21 2 if  d=,  then  maximum  path  difference  (d  sin  )  will be less  than .  So  there will  be  only  central maxima d(2 cos2  1) d    2d cos    on  the  screen,  because  in  the  equation  d  sin    = =   cos  cos  2 2 n,  n  can  take  only  one  value. Now  for  constructive  interference  at  P  between If  <  d  <  2,  then  the  maximum  path  difference BP  and  OP,  path  difference  =  n will  be  less  than  2.  So  there  will  be  two  more maximum  on  screen  in  addition  to  the  central   1  maximum.  Intensity  of  dark  fringes  becomes  zero 2  2  2dcos  +  =  n    2d  cos    =  n   if  intensities  at  the  two  slits  made  are  equal. [So  C  &  D    are  not  correct]   cos    =   2n  1   ;  For  n  =  1,  cos    =   Comprehension  based  questions  4d  4d Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 4. At  the  area  of  total  darkness,  in  double  slit 1. apparatus,  minima  will  occur  for  both  the wavelength  which  are  incident  simultaneously  and normally.  2n 1 1  (2m  1) 2   2n 1  2 The  wave  fronts  in  both  the  media,  are  parallel,  2  2 2m 1 1 the  light  will  be  a  parallel  beam.   2n  1  560  7   or  10n  =  14m  +2 2. Point  c  and  d  are  on  the  same  wave  fronts, 2m 1 400 5 By  inspection,  the  two  solutions  are C  =   . d Also     =   .  So  clearly     –     =     –   (i)  If  m   =  2,  n   =  3  (ii)  If  m   =  7,  n   =  10 cf dfce 11 22   Distance  between  are  as  correspond  to  these 85

JEE-Physics 3 . As  the  ray  bends  towards  the  normal  so  medium For  minima,  above  centre  O (2)  is  denser. 1 y  =  tan  1=  15 =0.26  m 1 Match  the  column 1 . ( A ) P   central  maxima  so  has  highest  intensity. 3 0  y  =  tan  2  =  7 =  1.13  m ( B ) (P0) =  4 2 I(P )  =  2I   (P1)  =  0 0 For  minima,  below  centre  O, I(P1)    =  4I y '  =  –0.26  m y '  =  –1.13  m 12  3   (P )  =   I(P )  =  2I+  2I  ×  There  will  be  four  minima  due  to  interference  at 2 43 22 positions  ±  0.26  m,  ±  1.13  m 3  4 (ii) When  incident  beam  makes  as  angle  of  30°  with =  12 =  2I  +  3  I 2. x–axis  S1 y S1 =  –    =  I  (3.732)  =  3.732  I NM  12   M  S2  I  (P )  =  0 y ( C ) (P)  =  2 0 M S2  P1     I(P )  =  2I  x1= d sin – d sin x2= d sin + d sin 4 1 For  central  maxima,  path  difference  should  be  zero  P2      I(P )  =  3I x =0  o r   x   = 0 2 3 2 1 2 d  sin    =  d  sin    ==30° 3  2    y  =  D  tan     y  =  1  ×  tan  30°  =  0.58  m =    =   66 For first minima d sin – d sin =  2 (D)  P0   3 I(P )  =  2I  4 0  d sin  =  2 + d sin   sin  =  2d  + sin   P1   3   I(P )  =  2I  –  2I  =  0 0.5 3 4 4 1 sin   =  2  1 +  sin  30°    sin   =  0.75  = 4  P2   3   3 3 4 3 I(P )  =  2I  +  2I    y  =  D  tan    =  1  ×  7   =  1.13  m 22 2 =  9  4  5   =  3.732  I For  first  minima,  on  either  side, 12 12 d sin  =        sin    =    2 0.5  1 Subjective 2 2d  0.1 4 1 . (i)  When  the  incident  beam  falls  normally  :   tan   1 =  0.26  m Let  path  difference  =  x 15   x  =  S P  –  S P  =  d  sin   21 For  minimum  intensity  (2n  1)   y  =  1  ×  tan    =  0.26  m Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 d sin  = (2n–1)   sin  =  1 2 2d Therefore  y  coordinates  of  the  first  minima  on  sin  =  (2n  1)(0.5)  sin  =  2n 1 either  side  of  the  central  maximum  are  y   =  0.26 1 m  and  y   =  1.13  m 2 2 1 4 (i)  Location  of  central  maximum  on  y–axis.  Let  the central  maximum  be  obtained  at  a  distance  y  below Since  sin    <  1    n  can  be  either  1  or  2 point  O.  It  will  be  below  O  because  glass  sheet covers  the  lower  slit. 11 When  n=1,  sin    4 ,  tan  1  =  15 33 When  n  =  2,  sin  2  =  4 ,  tan  2  =  7  y  = D  tan   or  y  = 1  × tan   86

JEE-Physics S1 the  upper  surface  of  layer  as  R .  R   is  reflected 11 O y from  a  denser  medium.  It  undergoes  a  phase S2 P change  of  .  Part  of  AB  is  reflected  from  surface of  layer  as  R .  R   is  reflected  from  a  rarer  medium 22 as  a   =1.8  and  a   =1.5. mg There  occurs  no  phase  change  in  R . 2 x   =  S P –  S P  =  yd A R1 R2 air 1 1  2 D B Due  to  glass  sheet  x 2   g  t m=1.8  m  1 t C  g=1.5 For  zero  path  difference   x   =  x 12   yd   g    g  tD R   and  R   therefore  possess  an  initial  phase B  m  1 t   y  m  1 d 12 difference  of    before  they  undergo  interference.   y   3 / 2  1  10.4  106  (1 .5 ) Now,  for  construction  interference  net  phase  4 / 3  0.45  10 3  difference  should  be  2n  where  n  is  an  integer. y  =  4.33  ×  10–3m  =  4.33  mm    =  2n –   =  (2n–1)    x  =  (2n–1)  2 (ii) Light  intensity  at  O,  x =0,  x =  g  t Since  x  =  2(m)t  =  1.8  ×  2t  =  3.6t 12  m  1   Net  path  difference  =  x2  or  3.6  t  =  (2n–1)   Net  phase  difference  =  2 x2  2 For  least  value  of  t  is  n=1    3.6  ×  t  =       2  3 /2  1 (10.4  ×  10–6)= 13 min 2  107  4 /3 3 6 648 or  t   =  nm  or  t   =  90nm.    I() 3 min 3.6  2 min   I()  =  I  cos2  2    or    Imax 4 =  0.75  max (iii) Wave  length  of  light  that  form  maxima  at  O,  if 4 . (i)  O  is  the  middle  point  of  slits  S   &  S 12 600nm  light  is  replaced  by  400  to  700  nm  light  :  g Also  SS  =  d=0.8  mm  in  figure  tan    =  y1  m 12  D1 At  O,  x  =   1 t For  maximum  intensity  at  O,  x  =  n, AC where  n  =  1,2,3....... R S1     =  x , x , x ,..... P O 1 2 3 Q Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65 x   3 /2  1 (10.4  ×  10–6  m)  or  x  =  1300 S2  4 /3 S 10cm D nm, For  maximum  intensity  at  O, 1300 1300 1300   =  1300  nm,  2 nm,  3 nm ,  4 nm tan   40  1  sin   1  1  1  tan    =  1300  nm,  650  nm,  433.33  nm,............ 200 5 26 5.1 5 or    =  6.5  ×  10–7  m,  4.33  ×  10–7  m Path   d ifferenc e  x1  =  SS   –  SS 1 2 3 . AB  denotes  incident  ray.  It  is  partly  reflected  from 87

JEE-Physics 1 6. Let  the  n th  maxima  of  1  coincide  with  n th  maxima x1  =  dsin  =  0.8  ×  5 or  x1  =  0.16  mm ...(i) 1 2 of   . 2 Let  R  represents  the  position  of  CBF  i.e. n1 1D  n2 2D n1  2 d d    n2 1 Net  path  difference  should  be  0. Now  x2  =  S R– SR n1 700 n1 7 1   n2 500    n2 5 2 or  x2  =  dsin   x1  =x2  ...(ii)   For  central  bright  fringe x2  –  x1  =  0   dsin  –  x1  =  0 Minimum  integral  value  permitted  for  n   is  7.   d  sin  =  x     =  0.16  mm 1 1   Minimum  distance  =  n1 1D   where  D =  103 d d 0.16 1   (0.8)  sin  =  0.16    sin  =    =  0.8 5  7 500  109  103   tan   1  1  1  sin    =    =  3.5  mm 24 4.9 5 1 y2 1 DIFFR ACTION  &  POLARISATION D2 5 So tan        y  =  2  cm 2 Exercise  II  :  Previous  years  questions Thus  CBF  will  be  2cm  above  point  Q. 2 . Intensity  of  the  polarized  light  coming  out  of (ii)  When  liquid  of  refractive  index    is  poured. 2       Then  for  CBF  at  Q,  net  path  difference  =  0 polarizing  sheet  will  be  I   I0 cos2 d (–1)t  =  x     –1)100  =  0.16    –1  =  0.0016 0 1 A C On  solving,  we  get  I  I0 S1 Q 2 S S2 3. I  I0  sin 2   and      ay       D  For  principal  maximum  y=0,  =0 hence,  intensity  will  remain  same,  i.e.,  I=I0 BD 4 . In  polarisation  intensity  changes  as  the  crystal  is t 5 . (i) S is a point source, fringes formed will be circular. 1 (ii)  Ratio  of  minimum  and  maximum  intensities  : rotated.  In  scattering  for  Rayleigh's  law  I  4 Intensity  of  light  direct  from  source=  I   =  I   (say) is  minimum  for  Blue;  so  I  is  larget 10 Both  statements  are  correct  and  reason  is  also correct Intensity  after  reflection  I   =  0.36  I 20 I m in  I1  I2 2  0.4 2 1 5. Since  the  incident  regnt is  unporaizes,after  light Imax  I1  I2   1.6  16      =   passis  through  polaroid  A=  I0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\04.Wave Optics.p65  2 (iii)  Shift  of  AB  for  same  intensity After  this  light  passes  through  B, If  intensity  at  P  corresponds  to  maximum  it  means that  constructive  interference  occurs  at  P. Ienergent  =  I  Iin  cos2    =   I0  cos245°    = I0  2  4   Path  difference  between  direct  waves  from  S and  reflected  waves,  from  reflector  AB,  is  n Let  AB  is  shifted  by  x  (towards  P  or  away  from  P)   additional  path  difference  introduced  =  2x For  minimum  value  of  x   ;  n=1. 1   Path  difference  =  1  ×      =  600  nm   2x  =  600  nm    x  =  300  nm 88



JEE-Physics UNIT # 02 (PART – II) WORK, POWER, ENERGY AND CONSERVATION LAWS EXERCISE –I  3 m v 2  1 kx2 ; k  3mv2 8 2 4x2 1 . By applying work energy theoram change in kinetic 9. Total mass ; f  6m, f = 6m (20) = P C energy = W + W g ext.P To Drive 12m : f  14m  f = 14 m C 0 = mg( cos 37° –  cos 53°) + W ext P (14 m ) v = 6(m ) 20  8.57m/s CC To drive 6 boggie : force  8m = 50 × 10 × 3  4 W force = 8m  P = 8 mv 1 5 5  + ext P C C (8m )v = 120m  15 m/s CC W = 100 joule 1 0 . By applying work energy theoram ext   1 mv2–0 = W + W 2 . Work  F.dr , Work =–  (0.5)(5)Rd F=mN 2 g fr 0 for the second half work energy theorem change  [work] = (2.5) (R) (2) = –5 J in kinetic energy = W + W  T g fr 3 . W  f.d 0 = 100mg + W = –100 mg fr As work done for the first half by the gravity is Mg Mg g/2 100mg therefore work done by air resistance is mg  T  ; T  less than 100 mg. 22 Mg W    Mg x dx  2  1 1 . x = 3t – 4t2 + t3; v  = 3– 8t + 3t2 dt 4 . For conservation force work done is independent of the path a  dv = 0 – 8 + 6t dt W + W = W , 3+4 = W = 7 J AB BC AC AC  4 5 . By applying work energy theorem W   F.dx   3(6 t  8)(3  8 t  3t2 )dt 0 K E     v  1  v  t2   K.E.  mv2 t2 W= 528 mJ f .d m  t1  2  t1  2 t12 OR From work energy theorem 6 . Slope of v–t graph  Acceleration  –10m/s2 W= 1 m v 2  1 m v 2  1 3  103  2 2 2 1 Area under v–t graph  displacement  20 m 2 work =  = 2 (10) (20)  –400 J 3  8 4  3 42   32   528 mJ f.s 7 . By applying work energy theoram 1 2 . Power = constant, Fv = C node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 KE = work done by all the forces mvdv = Cdt  v2  2C t  v  2C t mm New kinetic energy = 1 m v 2  mv2 2 f 8  vf  v0  v = u – gt0   v0 as v  dx   dx  2C  tdt 2 2gt0 dt m 8 . By applying work energy theoram 2C t3/2 x   x  t3/2 1 m v2 1 1 kx2 – mv2=  m 2/3 2 42 2 1 3 . a = k2 rt2  v 2  k2rt2 cr 52 E

JEE-Physics dv Put the value of u2 in equation (i)  v2 = k2r2t2  v = krt  a =  kr m 5g T dt T – mg =   T = 6 mg P =   m (kr ).(krt )  mk2r2t m a T .v 2 0 . When the string is horizontal 1 4 . P.E.  Maximum  Unstable equilibrium mv2 T  P.E.  Minimum  Stable equilibrium ...(i) v T P.E.  Constant  Natural equilibrium v2 = u2 – 2g v2 = 5g –2g = 3g mg  None of these u So T  m  3g  3mg  1 5 . P.E.  Maximum  Unstable equilibrium P.E.  Minimum  Stable equilibrium So net force P.E.  Constant  Natural equilibrium = T2  mg2  3mg2  mg2  10 mg dU 2 1 . In case of rod the minimum velocity of particle is Force =   –(slope) zero at highest. dx 2 2 . As velocity is vector quantity [ slope is –ve from E to F ] Force = +ve repulsion v = v12  v 2  2v1v2 cos  [as  = 90°] 2 Force = –ve attraction v = v 2  v 2 1 2 1 6 . By applying work energy theoram By applying work energy theorem velocity at z KE = Work done by all the forces 11 mv 2 – mu2 = – mgL 0=W +W +W g spring ext agent 2 22 –W = (W + W ) v2 = u2 – 2gL  u= 2(u2  gL) g spring ext agent 2 U = (W + W ext agent ) [U = W ] spring g 23. By applying work energy theorem  KE = W g 1 7 . U = mgh 0 height w.r.t. ground = ( – h), U = mg (–h) (r–b) 1 8 . By applying work energy theorem K.E = W + W 1 S ext mv2 = mg(r–b)  v 2g(r  b) agent 2 1 2F mv2 0 = – Kx2 + Fx  x  2 4 . Net force towards centre equal = 2K r node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 2F2 mg cos  – N = m x v2  Work done = r K v  rg cos  1 9 . At lowest point By applying work energy theoram mu2 u 12 T  mg   ....(i) 2 mrg cos  – 0 = mgr(1– cos ) = cos = 3 at highest point T = 0 mv2 g and v2 = u2 + 2as 2 5 . P = P12  P22  2P1P2 cos  mg   , v  for cos  = maximum  P minimum  = 360°  g 2  u2  2 g  2 for cos  = minimum  P maximum  = 180° g = u2 –4g 53 u2 = 5 g E

JEE-Physics 2 6 . Tension at any point T = 3mgcos N+ m v 2 = mg cos ...(ii) Given 3mg cos = 2mg C 2 2 r  cos 3   = cos–1  3  3 at C, N = 0  cos = 4 EXERCISE –II 8. W=   U= U –U =  m  g    mg – 3mg man  2   4  –2 = 8 f i 1 . For body B : mg – T = m(2a) 2E 1 g 9. At x =  ; E = kx2 = U  KE = 0 For body A : 2T – mg = ma   a= 5 k 2total a = 2a and a = a 1 0 . Equation of motion : B A A  Velocity B distance  m gsin 37° – T = m a and 2T– m g = m a of after travelling A AA B BB   2as  4g gg 5 a = 2a = 2 × = AB 12 6  Velocity of A : v = v B  g vA  2a A .sA  gg 2  1  A2 5 6 3 2. COME  K + U = K + U vA g 1 1 2 2 2 3 11  vB   2 0+ k x2 + k x2 21 22 1 1  x2 1  x2 1 1 . COME : K + U = K + U = mv2 + k   + k   B BA A 2 1 2 2 2 2 11 2 0 + k (13–7)2 = mv 2 + 0 2 2A 1 11 2 k  62 2 (k + k )x2 = mv2 + (k + k )x2 m v A  12 28 12 N = =1440 N AR 5 v = 3 k1  k2  x2 12. W =KE   0  1 mv2  v = 2gA f 2 4m   .mg dr r 3 . Work done against friction = mgh = loss in P.E. 1 3 . Conservation of mechanical energy explains the  Work done by ext. agent K.E. at A & B are equal. =W + PE Acceleration for A = gsin f 1 = mgh + mgh = 2mgh Acceleration for B = gsin 2 4 . COME  K + U = K + U 112 2 sin > sin a >a 12 12 0+ mg (1–cos60°) = 1 = g F and displacements are in opposite directions. mv2 + 0  v ext 2 1 4 . COME : K + U = K + U 5 . COME : K + U = K + U A AB B 1 12 2 0 + mg × 25 = 1 m v 2  mg 15  mv 2 = 20mg node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 2 A A 1 0 + mg (4R) = mv2 + mg (2R)  mv2 = 4mgR 2 2 Forces at position 2 : Forces at B : N = mg – m v A =0 R = 20 m mv2 R N= –mg = 4mg – mg = 3 mg R 1 5 . Area of graph 6 . F = m g – m g     P = f .v = (m –m )gv  P.dx  mv.a.dx  m v.  vdv  dx ext 2 1 inst ext 21 =   dx  7 . COME : K + U = K + U v mv2dv  m v3  u3   10.v3  1 B BC C u 3 73 1 m v 2 + mgr = 1 m v 2 + mg rcos...(i) = 2 0 2 C Force equation at C 1 4 m/s = (4+2) × 10  v = 2 54 E

JEE-Physics 1 6 . Power = QgH = Av.gH=A 2gH .gH 2 5 . Conservative forces depends on the end points not on the path. Hence work done by it in a closed = 103  d2  2  10  40  10  40 ( d= 5 cm) loop is zero. 4 2 = 21.5 kW 2 6 . For equilibrium, F=0  x(3x–2)=0x=0  x= 3 1 27. v2 = v2 +2 (–g)L 1 7 . For upward motion : mgh + fh = m × 162 0 2 For v =0, v0  2gL 1 2 8 . For velocity to maximum acceleration must be zero. downward motion : mgh – fh = m× 82 h = 8m  mg – kx = ma = 0 2 W  3ˆi  4ˆj  8ˆi  6ˆj  8W mg 1 10 18. P   F.S  t t 6 x= k = = 5cm 0.2  Height from table = 15 cm 1 9 . For equilibrium : Ncos =mg & Nsin = kx  2 9 . 1 1 1 2  kx = mgtan (N = normal between m & M) W =KE= 2 mv2= 2 m(at)2 = 2 × 1× 10 N 3 =150 J  U  1 kx2  m 2g2 tan2  3 0 . Sum of KE and PE remains constant. 2 2k  3 1 .  mgx 1 0  1 kx2 0 1 m v 2  v=8 m/s 2 2 20. W + W = KE  – mgh – f.d = 0– mv2 g F 2 1 3 2 . K.E. = work done by all the forces – mg 1.1 –  mg d = – mv2 (= 0.6)  d=1.17 m  2 K.E. = m a.s When acceleration is constant 21. For motion P  0  K + U = K + U 1 O O P P K.E.  t2 [as s = at2] 2  For motion Q  0  K' + U' = K + U 33. F  3i  4j is a conservative force ie therefore O OQQ  K =U; K' = U = 2U = 2K W =W OP O2 P O 12  t = 2 2h / sin  3 4 . To break off reaction becomes O, QO mv2 v2 g sin   t1 i.e. mg cos = R  cos  = Rg ....(1)  t = 2 h / sin   t2  2 t1 PO g sin  A mv2 R a2t2 N P 22. v a s  ds Rcos mgcos s= mg dt 4 R O node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 W= 1 m v 2  0  1 m  a2s = 1 ma2 a2t2  = ma4 t2 2 2 2 48 But from energy considerations 2Mg 1 2 3 . Maximum elongation in spring = mgR [1– cos ] = mv2 K 2 Condition block 'm' to move is  v2 =2gR (1– cos ) using it in (1) 3 cos = 2(1– cos ) Kx  mg sin37° + mg cos 37°  M = 5 2  cos  = 2 – 2 cos  cos  = 3 2 4 . COME : K + U = K + U So sin = 4 5 11 2 2 1  1  mg 1  cos 60 93 2 mv 2 0  0  v = 7 m/s Now tangential acceleration g sin = g 5 0 0 3 E 55

JEE-Physics 3 5 . Given v=gr EXERCISE –III mv2/R Match the column R=0 E 1 . W = force × (displacement in the direction of force) mg g mg r C mg 1 R=W R=W W = [10 × × 2 × 16] = –160 joule mg A mg g2  w= N.s = m(g+a) cos 1 2  16  cos  N  2  NS R=mg+mv2/r=2mg=2W = (12)  3 (16) 3  90-a fr 22 mgsin\\\\\\\\\\\\\\\\\\\\\\ 2 r = 12 × 12 = 144 J  3 6 . In this case T = u [for 1 resolution] W= fr .s fr mgcos 1 gt2 2h = m(g+a)sin (16) cos (90 – ) mg 2 g Also h =  t = 1 = (12) × 16 × 4  48 joule 2h 2r u 2h W =W +W + W  32 joule net g N fr But t = nT  g = n u  n= 2r g du 2. f =– dx  30 Ni 1 3 7 . Given mv2 = as2....(i) conservative 2 change in kinetic energy =2 [Area under (a–x) graph] v2 2as2 as mass is 1 kg   [80 + 40] = 120, So a = = ....(ii) r R mR 1 KE = Mv2= 8 J 2initial dv dv ds dv (A) KE = 128 J Also a =  .  v f (B) W=  30 × 8  240 J t dt ds dt ds can f d 2a (C) W = KE  120 J But from equation (1) v = s Net m (D) W + W = 120; W = –120 J cons ext ext 2a  2a  2as 3 . By applying conservation of momentum wedge will put it above a = s m  m   m ....(iii) acquire some velocity =  mv x where v is velocity t M m x 2  2as2 of block w.r.t wedge in negative x-direction.   m  So that a = a 2  a 2   2as2  ( A ) Work done by normal on block is r t  mR 1  mv  2 –  M  2 = M x 2 2as  s  m i.e. a   R  1  ( B ) Work done by normal on wedge is m node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 1  mv  2 2  M   s  2 = M x is positive.  R  So force F = ma = 2as 1  m ( C ) Net work done by normal is = 0 1 (D) less than mgh as K.E. is < m2gh, 38. Tension w i ll be mg cos  at extreme s but it 2 mv2 KE > KE is positive. becomes mg cos  +  . f 4 . For v  5g , the bob will complete a vertical In the given situation by making diagram, we can circular path. shown that T – Mg cos = Mv2 and tangential For 2g <v< 5g , the bob will execute projectile motion. L acceleration = g sin For v < 2g , the bob oscillates. 56 E

JEE-Physics Comprehension#1 Comprehension#3 1. W=  W = – mg  1 a t 2  1 . By applying work energy theoram f .d s  2  0 2 . For the motion of the block in vertical 1 Mv2–0 = W mg – N = ma , N = m(g–a ) 00 2g W=  Na0t2   m(g  a0 )a0t2 1 Mv2 = mg  v 2g N 2 2 2 3 . For observer A pseudo force on the particle is zero 2 . 2g  5g(  x) W=0 3  1 ma2t2 2g = 5g(–x) 5x=3  x  5 4 . W = fnet.ds  W = ma 2 at2  2 5 . For observer A the block appears to be stationary 3 . Net force towards the centre will provide the  Displacement is zero hence w =0 required contripetal force Comprehension#2 mv2 kx kx – mg = Mv2 1 . N – Kx cos30° – mg cos 60° = R m2g R kx – mg=  As velocity of Ring = 0 N = kx cos 30° + mg cos 60° 3mg mg  kx = 3mg  x = k kn cos30° 60 Comprehension#4 30° N 1 . Particle will have some translatory kinetic energy mgcos60° 30° as well as rotatory energy the whole of the K.E. is 30° converted into potential energy h < 6 60° 2 . By applying conservation of mehanical energy kn mg 1  (2  3 )mg (2   3  mg 3  mu2 = mg(h)  u2 = 80 3R 3 )R  2   2 2  mg  mg  mg 1 22  mu2sin2 30 = mgh  h = 1m 2 Total height = 2 + 1 = 3m 2 . f = (kcos 60°) x + mg cos 30° Comprehension#5 net 1 . From the F.B.D. of the blocks :  (2  3 )mg 1 mg 3 upper block is –ve and lower block is +ve as (2  3 )R  3R 2 2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65  mg  1  3   2mg 1kg 6m/s 2  3  3 fr fr   2 a = 2a cos 60 = a = 2g horizontal rev 3 v = decreases, v = Increases upper lower 2 . By applying conservation of momentum 3 . By applying work - energy theoram 1 × 6 + 2 × 3 = 3(v) v = 4m/s By applying work energy theoram 1 mv2 – 1 1 1 (2  3 )mg (2  3 )2 R2 2 0= kx2; mv2= 2 3g 11 – (1) (36) + (1)(16) = w 2 2 2 2 fr 1 1 mg (2  3 )R  v  gR(2  3)  –18 + 8 = W  W = –10 J mv2 = 23 3 fr fr 2 and Work done on the lower block +10j Wnet = 0 E 57

JEE-Physics Comprehension # 6 3. Maximum power = F × V 4. max A B du 2A B 5. 1 . u = r2  r  dr   r3  r2 Maximum force applied by camel is during the accelerated motion. du 2A B r 2A We have V2 – U2 = 2as, 25 =02 + 2·a·50 f   r2 , F = 0  dr r3 B a = 0.25 m/s2 a 2 . As potential is minimum at r=r the equilibrium is for accelerated motion Fc f 0 stable.  F– f = ma c 3 . Given that  F = mg + ma c AB 2A AB2 BB B2 = 0.1 × 1000 × 10 + 1000 × 2.5 r2  r U   U= as r = B ; i 4A2 2A 4A = 1000 + 250 = 1250 N  Uf=0 W =U – U  B2 This is the critical point just before the point where f i it attains maximum velocity of almost 5m/s . 4A A B 3B2 Hence maximum power at this point is 4. K.E. + P.E. =T.E, 0 + r2  r  = 1250 × 5 = 6250 J/s. 16A By solving the above equation r = 2r0 We have W = PT,P = 18 × 103V + 104 J/s 3 103 104  P = 18 × × 5 + J/s and 5 Comprehension#7 1. (A) W + W= KE  WCL=KE – W 2000m CL f f T5 = 5 m / s =400s (a) During acceleration motion negative work is done P =18 × 103 × 104J/s 10 against friction and there is also change is kinetic energy. Hence net work needed is positive. 2000m and T10 = 10 m / s = 200s (b) During uniform motion work is done against friction only and that is positive. (c) During retarded motion, the load has to be W5 104 (9  1)  400  W10  104 (18  1)  200 stopped in exactly 50 metres. If only friction is considered then the load stops in 12.5 metres which is less than where it has to stop. Hence the camel has to apply some force so The time of travel in accelerated motion = time of travel in retarded motion. that the load stops in 50 m (>12.5 m). Therefore the work done in this case is also positive. DC BA 2. W| =  K E – W 50 m 2000 m 50 m CL accelerated friction m otio n where W is work done by camel on load. V5 CL T = T =  = 20 sec AB CD 0.25 1  a  2 mv2  =  0 – [–kmg.50] Now time for uniform motion =T = 2000 =400 s 1 ac 5 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 = × 1000 × 52+0.1 × 10 × 1000 × 50 440 2  Total energy consumed =  Pdt 125  =1000  2  0 similarly, W |retardation = KE – W 20 420 CL friction   [18.103 V  104 ] dt   [18.103.5  104 ] dt 0  1 mv2  – [– kmg.50] = 1000 75  2   2  0 20 WCL | accelerated motion 125 5  5 : 3 440  WCL | retarded motion = 75 3   [18.103 V  104 ] dt 420 58 E

JEE-Physics 20 20 From energy considerations [1 8 . 1 0 3 104 105 420   V dt  dt  t 20 1 mg R cos  = mv2  v2 = 2g R cos  00 2 440 440 putting this value in equation (i)   18.103 Vdt   104 dt we get T = 3mg cos  420 420 Also acceleration a = a 2  a 2 Total r t Putting Vdt = dx and changing limits appropriately it becomes  v2 2  R  60   g sin 2  2g cos 2  g sin 2 18.103 dx  1 0 4 t 20  105 [420  20] 0 0 = g 4 cos2   sin2  2100 1 0 4 440 420  aTotal = g 1  3 cos2  18.103 Now virtual component of sphere's velocity  dx  2050 = 18.103.50 + 104[20]  + 105·400 +18.103[50]+104[20] Joules vy=vsin = 2gR cos  sin v = 90 × 104 + 20 × 104 + 400 × 105 vsin + 90 × 104 + 20 × 104 J = 4.22 × 107 J Comprehension#8 Applying maxima-minima 1 . By applying work energy theoram change in kinetic dvy   sin sin   d cos  cos  1 2gR   2 cos   energy = w  0 – mv2 = W S 2s 2. As the kinetic energy of block is decreasing, 2gR   sin2   cos    cos   = 2 cos   1  sin2   cos2   tan2 = therefore work done by the normal is = – mv2 2 2 1 2 3 . W = – mv2   = tan–1 2  tan   2 net 2 So sin   2 and cos   1 5 . W = 0 as for the B change in velocity is zero. 33 net Thus tension T = 3 mg cos 6 . As there is no change in kinetic energy stored is 1 due to = 3mg × 3 = 3 mg node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 Comprehension#9 1 . Conservation of mechanical energy can only be applicable in absence of non conservative forces Comprehension # 10 Comprehension 11 Using work energy theorom Balancing the forces T = mv2  mg cos  ....(i) m  2g  R sin   mgR 1  cos   1 mv2 ..(i) R 92 O R m 2mg mv2  T Also mg cos   sin   Rcos 9R N mv2 +mgcos 2g  R v2 = gRcos – R sin  ...(ii) mgsin mg 9 E 59

JEE-Physics From equation (i) & (ii) EXERCISE –IV(A) 2mg R sin   mgR 1  cos   m  gR cos   2g R sin  A Mgcos 9 2  9  Mgsin  4sin + 18 (1–cos) = 9cos–2sin 1.  6sin + 18–18cos = 9 cos  6 sin – 27cos + 18 = 0  2sin – 9 cos + 6 = 0 m BC m2g 90 W =Mgsin × AC = Mg × AB 9 Mg v  W = Mgcos × AC × cos 180° f mg = –Mg × (BC) 4R 3 D Now let sin = x so cos = 1  x2 G Than 2x – 9 1  x2 +6 = 0 E BF 34 W = Mg(sin × DG + sin  × GF) = Mg × DE Solving x = 5 = sin so cos = 5 ;  = 37° Mg Now putting =37° W = – Mg (DG cos  + GF cos) = – Mg(EF) 4R 4 f in  = h+ Rcos = R  =– Mg × BC (BC = EF) 35 From WET, KE will be same in both cases. = 20R  12R  32R 2. v =v 15 15 3. CF 2g From equation (ii) v2 = gRcos Rsin Heat generated = work done against friction  (mg) (vt)= (0.2 × 2 × 10) × 2 × 5= 40 J 9 40 = cal = 9.52 cal 4.2 4 2g 3 Blocks are moving with constant speed. v2 = gR × – R  59 5 T=Kx = gR 4  2  = 10gR  2gR BT A node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65  5 15  15 3 f mAg 1 32R 2gR t + 1 gt2  m g = T = kx = f = m g Now using S = ut + gt2; = AB 2 15 3 2 mA 2 2  9.8  = =10 kg and x = 1960  m= B 0.2 2R 1 t can be obtained t = g  Energy stored in spring = kx2 2 1  19.6 2  = 2 × 1960 ×  1960  = 0.098 J 60 E

JEE-Physics 4. Work done by force =  Fdx 9 . Let extension in spring be x due to m 5. 01 6. 7. then m1gx0 = 1 k x 2  kx0 = 2m1g 0 8. 1/2  cos x 1 / 2 2 E W =  sin  xdx =  m 0 0 but kx  mg so 2m g  mg  m  0 1 1 2  = – cos + cos 0 = 1J m therefore minimum value of m1 = 2 2 Work done by external agent = – 1 J COME : K +U = K + U 1 0 .  = 3 (t + sint);  = 3 +3 cost;  = –3 sin t 11 2 2 3mgr 1 1 = mv2 + kr2 ...(i) 22 2   F= m2R 2  mR 2  t  2  = 9 10N mv2 Force equation kr = mg + r 2mg 1 1 . COME : mv2 = mgh Solving we get, k = r = 500 N/m 2 If resultant acceleration, a, makes angle  with v2 thread, then asin = gsin a = bt2 =  v= bR t  a = bR v2 2gh nR t acos =  =   P = FV = mbRt t sin   sin  0 Pdt mbR t2 / 2 mbRt <P>= t = =  tan = 2h   = tan–1  2h  0 dt t 2 As C falls down, A & B move up. 12. COME : K +U = K + U COME : K + U = K + U 11 2 2 1 12 2 AC x 60° N B 60° mv2 mg R C 1 mgR AB 0 + MgR = mv2 +  v = gR 22   4a Forces at B  N = mgcos 60° + mv2 15 3 = 0 + mgx = 0 + 2mg a2  x2  a  x = 3 R2 GFH JKIPotential energy U =1 × x2 x2 mv2 mv2 mv2/R x = –x 1 3 . T = mg+ , T = – mg 2 2 max R min R node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 For minimum U, mv2 T mg  mg dU 2x d2U Tmax  mv2 R  5 (R=2m) Tm in 3 T dx = 2 – 1 = 0 and dx2 = 1 = positive  mg R mg mv2/R 1  v = 4 5 m/s so at x = 1, U is minimum. Hence Umin = – J 2 1 4 . Here the bob has velocity greater than 2g and Total mechanical energy = Max KE + Min PE  Max KE = 1   1 = 5 smaller than 5g . Hence the thread will slack 2 mv2max = 2 –  2  2 after completing semicircle. 25  vmax =  = 5 ms–1 1 2 61

JEE-Physics Force equation : N + mv2 = mgcos...(ii) v mv2/R r 90- T B  h = rcos = 19 r  mg 27 O (c)   g sin 2  g cos   g anet  ar  a t  A 3g 1. EXERCISE –IV(B) COME : K + U = K + U a : Natural length 1 12 2 a : Initial elongation 1 m 3g + 0 = 1 mv2  mg    sin  ...(i) 2a : additional elongation 22 1 9a Force equation at B : COME : k(3a)2 = mgx  x = 22 (above point of suspension) T + mg sin = mv2 ...(ii) R 2. WET : W + W + W + W = KE N Mg f sp 1 gsin  11 Solving for T=0, we get sin =  v = 0 + 0 – k.mg (2.14 +x) + 0 – kx2 = 0 – mv2 3 B 2 2  The particle will execute projectile motion after  x = 0.1 m tension become zero. At x = 1m, F = kx = 2 × 0.1 = 0.2 N spring g 1 1  v = vsin =  min 3 3 F =  .mg = 0.22 × × 10 = 1.1 N S.F. S 2 Hence the block stops after compressing the spring. 1 5 . COME : K + U = K + U  Total distance travelled by block when it stops A AB B 11 = 2 + 2.14 + 0.1 = 4.24 m 0 + mg(2R) + kR2 = mv2 + 0 + 0( k = mg/R) 22 mv2 dU d 2r3  = – 6r2  = 5 mg  Force equation at B 3. Conservative force, F = – = – dr dr R mv2 This force supplies the necessary centripetal  T = mg + = 6mg acceleration. BR mv2 1 = 6r2  mv2 = 3r3 1 6 . For speed u , contact at top is lost. 0 r2 E = K + U = 5r3 = 5 × 5 × 5 × 5 = 625 J N+ m u 2 =mg  (N=0) u = gr 4. For part AB : (R=4a) node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 0 r0  v0    a   4a  2  2 v 0  r t1   t = 4 t= g 1 (a) For vertical motion; For part BC : (R=3a)  t =3  a   Horizontal distance 2  2 v 0  s = 2u .t =2 gr × 2r = 22r  a  0 g For part CD : (R=2a) : t =2  2 v0  3 (b) COME :  a  1 m u0 2  mgr  1 mv2  mgr cos  ...(i) For part DA : (R=a) =: t =  2 v 0  4 23 2 62 E

JEE-Physics 5 a = 0 + mg  5  1 + 1 Mu2  1 m u cos 2 2  t = t + t + t + t = v0 2 1 2 3 4 5 . At position B;  u = 3029 m/s mg = Tcos= k..cos 2mg a  a  a  cos  8. Initial elongation in each spring = sin   a 1 = Mg  Mg  20cm = 2mgcot  cot=  kx  kx0 2  2 0  2 a Total initial length of each spring (a) OB = acot = 2 = 50 + 20 = 70 cm (b) COME : K + U = K + U CCO O \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 1  2mg  2 a 2 = 1 1 ka2 0 + mga + ×  a  2 mv2 + 2 2 (i)  v = 2 ga (ii) K + U = K + U CCP P [ P is the point of greatest depth] 1  2mg  2 a 2 Equilibrium position = 2 kx = mg  mga + 2  a  100 1  2mg x   10 cm = – mgx + 2  a  (a2 + x2)  x = 2 a 2  500 and due to inertia it goes 10 cm also up = 20 m 6 . COME : K + U = K + U 9. B iif f ucos 10 a m A u 6 x  v 8 x For constant length of string =v = u cos M h= sin a COME : 11 10 mg × 5 = mv2 + mu2  u = 22 1.64   0 + mgx = 0 + Mg a2  x2  a 40  v = ucos = m/s 2mM 41  x  M2  m2 a node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 7. COME : K + U = K + U 11 i if f 1 0 . COME : mu2 = mv2 + mgL ( 1+ sin)....(i) 1m M 1m 22  mv2 mv2 For equationT + mgsin = ...(ii) vR L M=2kg L mg m=½kg Since the particle crosses the T=0 cos = 2/ 5 8  m u line at its half of its range ucos  v2 sin .cos   L cos   L ...(iii) u  0 + Mg × R g 8 1 E 63

JEE-Physics  cos   1    60 EXERCISE –V(A) 2 N From equation (i)  u = gL   3 3  1. Spring constant (k)= 800  2 2  2. 3. m Work done in extending a spring from 11. WET : W + W + W + W = KE X1 to X2 = Uf–Ui = 1 k X 2  1 kX 2 SP mg N f 2 2 2 1  1  h  2  m g h  1 1 0 2  sin    sin  2 2    k   sin     0 W  k X 2  X 2   × 800 [0.15)2–(0.05)2] 2 1 mgh cot   1 mv2  400  15  2   5 2 400 2  100   100   = [225–25]  10000 v= 2   1 k  h 2  m g h cot  400  200 m mgh 2  sin    = =8J   10000 k = 5 × 103 N/m 1 2 . WET  W + W + W + W = KE W  1 k  x 2  x12  mg N T f 2 2 NT T f    W 1  2  2 5  103  10  10 2  5  10 2  mgcos  W  1  5  103  104 100  25 mg  2 /2 – mgR+0+W +  mg sin .Rd cos180°= 0 75  5  10 1 75 T    18.75 N-m 0 24  W = mgR 1   Power = FV = constant i.e., mav = k T  a v=k1   dv  v  k1  vdv  k1dt  dt  1 =/2 1 3 . t2 = ( = )  t = 2 sec On integrating both sides, we get 224  Average velocity = 2R = 1 m/s  v2  k1t  v2  2k1t v  2k1 t1/2 2 t  ds  k2 t1/2dt  s   k2  t3/2  s  t3/2  3 / 2  1 4 . The string can break at the lowest point node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 m v 2 V0 4. z zHere F  x, by using work energy theorem H 5. T = mg + KE = F dx  KE  x dx  KE  x2 max R 0.5  v2 VH v1  45 = 5 + Given that acceleration a = t1 ...(i) 0.5 COME: v 2  v 2  2gR H 0 1 Power = Fv P=(ma)v 2 v 2  40 2 10  =30 P=(ma2t) [v=at] 0 H = v 2 = 30 3 P   m v12  t  replacing a= v1  max 0 = = 1.5 m  t12  on t1    2g 2 10 2 64 E

JEE-Physics 6. Work done in pulling the hanging part of the chain 10. V x  x4  x2  mg   4 2  7. upon the table= For minimum value of V, 8. 2 9. dV 4x3 2x E  0    0  x = 0, x = ±1 dx 4 4 0.6 m=l 1 1 1 So, Vmin (x=±1) =  = J 42 4 Now, Kmax + Vmin = Total mechanical energy where m = mass of the hanging part  K max   1  2 or Kmax = 9 l = hanging part of chain  4  4 W   4  0 . 6   10  0.6  3.6 J mv2 9 v 3 ms–1  3  2 or 24 or 2 According to work-energy theorem, 1 1 . Applying work-energy theorem, W=K Case I : F  3  1 m  v0 2  1 m v 2 2  2 2 0   where F is resistive force and v0 is initial speed. 45° l B Case II : Let, the further distance travelled by the AF bullet before coming to rest is s.  F 3  s  Kf  Ki  1 m v 2 2 0 1 2 3 s 1 2 Mg 8 0 2 0   m v    m v or 1 or 3  s =1 ors = 1 cm Work done by F from A to B (3+s)=1 = Work done by Mg from A to B 4 44  F( sin45°)=Mg [1–cos45°]  F=Mg(2–1) Momentum would be maximum when KE would be maximum and this is the case when total elastic PE 12. a  Fk 15  7.5 m/s2 . is converted into KE.  According to conseration of energy m2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 1 kL2  1 Mv2 Now, ma  1 kx2  2  7.5  1  10000  x2 22 22  kL2  Mv 2 or MkL2 = p2 ( p = Mv) or x2 = 3 × 10–3 or x = 0.055 m or x = 5.5 cm M  p  L Mk 13. Question is somewhat based on approximations. Let mass of athlete is 65 kg. Applying work-energy theorem at the lowest and Approx velocity from the given data is 10 m/s highest point, we get So, KE = 65 100  3250 J WC + WNC + Wext = K 2 WC + 0 + 0 = Kf–Ki So, option (d) is the most probable answer. 1 WC(Gravity) = 0 – 2 × 0.1 × 25 WGravity = –1.25 J 65

JEE-Physics ab EXERCISE –V(B) 14. U=  x12 x6 dm d Force = v  = v × (volume × density) dU a 6b  x=  2a /6 1. F = – dx  12 x13 – x7  0  b  2. dt dt U(x = ) = 0 3. = v d Ax   = v  A dx  Av2 4. a b b2 dt dt Uequilibrium =  2a2    Power = Force × velocity  b   2a 4a  b  = (Av2) (v) = Av Power  v 3  b2  b2 dU  U(x = ) – Uequilibrium  0    4a   4a F = – dx  dU = – Fdx 1 x U x  kx2 ax4 1 5 . m2  t  dU   2   kx  ax3 dx or 0 24 d 1 Let potential energy U(x) = 0    t  dt  t 2 1 1 x2  ax2  F = ma  t 2   t  0 = 2  k  2  k1 x2 x has two roots viz x = 0 and x = 2k .  k2 x1 16. Given same force F = k1x1 = k2x2  a 11 ax2 W1 = k 1x 2 & W2 = k 2x 2 If k < , P.E. will be – ve or 2 1 2 2 2 1 when x > 2k , P.E. will be negative. 2 W1 k1 x12 a W2 > 1 so As 1 1  F = – kx + ax3  At x =0, F=0, 2 k 2 x 2 Slope of U–x graph is zero at x=0. 2 F x1 k2 2k  Fx2 > 1  k1 > 1 Thus P.E. is zero at x=0 and at x=  k2 > k1 statement 2 is true a Slope of U–x graph, at x=0, is zero. OR Mechanical energy is conserved in the process. if x1 = x2 = x Let x=Maximum extension of the spring.  Increase in elastic potential energy = 1 kx2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 2 1 2 K x 2 W1 1  K1 Loss of gravitational potential energy = Mgx W2  1 K2 2 K 2 x2  Mgx = 1 kx2 or x = 2Mg 2 k  W1  K 1 <1 The gravitational field is a conservativefield. In a W2 K 2 conservative field, the workdone W does not depend on the path (from A to B). It depends on  W1 < W2 initial and final points. statement 1 is false  W1= W2 =W3 66 E

JEE-Physics 5 . For conservative forces, 9 . It is a case of uniform circular motion. xx kx2 Velocity and acceleration keep on changing their directions. Their magnitudes remain constants. Fdx   kx dx or U(x)–U(0) =– Kinetic energy remains constant.  U =– 2 1 1 . (i) For circular motion of the ball, the centripetal 00 force is provided by (mg cos–N) But U(0) = 0, as given in the question,  U(x) = kx2 or x2 = 2U x   mg cos–N = mv2 ...(i)  d  2k  R  2  It represents a parabola, below x–axis, symmetrical about U–axis, passing through origin.  d  2  By geometry, h = R  (1–cos) 6 . Energy conservation gives L 5gL By conservation of energy, v2= u2–2g(L–L cos) 5gL Kinetic energy= potential energy 5gL 1 mv2  mg  R  d  1  cos  or or 4 =5gL–2gL (1–cos) 2  2  or 5=20–8 + 8 cos or cos = – 7  3  v2  2  R  d 1  cos  g ...(ii) 8 4  2  From (i) & (ii), we get total normal reaction force N. 7 . Tsin=m2 (Lsin)  T = m2L N = mg(3cos–2) ...(iii) (ii) To find N and N NA AB \\\\\\\\\\\\\\\\\\\\\\\\\\\\ For graphs : h  From (iii), at A,  v Tcos T mg N = mg (3cos–2) ...(iv) BA R A O (i) If N =0, A i.e. At A, N =0, Tsin 0 = mg (3cos–2) mg 2 or 3 cos  =2 or cos  = 3 max  Tmax  324  36 rad/s When N becomes zero, the ball will lose contact mL 0.5  0.5 A with inner sphere A. After this, it makes contact with outer sphere B. When  – 0, N= mg A The N versus cos  graph is a straight line as shown A 8 . According to problem particle is to land on disc. in the figure. NA mg node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 2/3 cos t n (ii) To find N : Rsint B 2 R t Consider : cos > 3 The ball makes contact with B. If we consider a time 't' then x component of displacement is Rt N– (–mgcos) = mv2 or N + mg cos B B Rsint < Rt d Thus particle P lands in unshaded region. R 2 For Q, x-component is very small and y-component mv2 equal to P it will also land in unshaded region. = R  d / 2 ...(v) E 67

JEE-Physics Sphere B N v 1 4 . A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 d  W to the particle. If the initial speed (in ms–1) of the particle is zero, the speed (in ms–1) after 5 s  is. [IIT-JEE 2013] mg Ans. (5) R Sphere A P = Fv   mv dv   0.5  dt   v 5 1 0.2   v2  1 5 0 2  2  2 0 mvdv  dt   By energy conservation,  v2 = 25  v = 5 m/s 1 mv2  mg  R  d   R  d  1 5 . The work done on a particle of mass m by a force 2  2   2  cos   mv2  x 3/2 ˆi  y 3/2 ˆj  or d = 2mg (1–cos) ...(vi)    Kx2 y2 x2  y2  R  2 (K being a constant of appropriate dimensions), From (iv) and (v) when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius N + mgcos = 2mg–2mgcos a about the origin in the x-y plane is :- B N = mg(2–3cos) ...(vii) B 2 [IIT-JEE 2013] When cos  = ,N=0 2K K 3B (A) a (B) a When cos  = –1, N = 5 mg. B Thus the N – cos  graph is as shown in the figure. K B (C) 2a NB (D) 0 5 mg Ans. (D) mg Particle is moving in x-y plane so –1 2/3 cos   xˆi  yˆj    k x ˆi  y ˆj   k xˆi  yˆj    r F  r3 r3 r3 kr r3 1 2 . m g – T = m a ...(i) Force is central (i.e. conservative) so work done 11 by this force in closed loop = 0 T–m g = m a ...(ii) Paragraph for Questions 16 and 17 22 (m = 0.72kg; m = 0.36 kg) A small block of mass 1 kg is released from rest 12 at the top of a rough track. The track is a circular 10 From (i) and (ii) a = m/s2 arc of radius 40 m. The block slides along the track 3 1 10 5 TT without toppling and a frictional force acts on it a m2 m1 a d = × × 12 = m in the direction opposite to the instantaneous 23 3 velocity. The work done in overcoming the friction 10 10 up to the point Q, as shown in the figure below, node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 v = 0 + × 1 = m/s is 150 J. (Take the acceleration due to gravity, 33 51 100 g = 10 m s–2) [IIT-JEE 2013] W = 0.36 × 10 × + × 0.36 × T 32 9 y W =8J R P T 30° 13 . By using work energy theorem (W = KE) mgx  1 kx2  0  1 mV 2 QR 22  V 2  1.44  V  1.2  0.4  4  N  4 Ox 9 3 10 68 E

JEE-Physics 1 6 . The magnitude of the normal reaction that acts 1 7 . The speed of the block when it reaches the point on the block at the point Q is Q is (A) 7.5 N (B) 8.6 N (A) 5 ms–1 (B) 10 ms–1 (C) 11.5 N (D) 22.5 N (C) 10 3ms1 (D) 20 ms–1 Ans. (A) Ans. (B) Work energy principle from equation (i) mgRsin – W = 1 mv2 ..... (i) gR Wf  f 2 m  v= 2 sin   = 10 m/s  = 30° m = 1 kg R = 40 m Wf = 150 s  = 30° mv2 N N – mg sin =  R  mv2 N = mg sin  + =7.5N  mg R node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\04 Work Power Energy.p65 E 69

JEE-Physics UNIT # 12 (PART - I) MODERN PHYSICS (Atomic and Nuclear physics) EXERCISE –I e 6 V0  hc  e 2V0   e2 1 14.      hc  0  4 1 . P(D) = 1 – e–t = 1 – e =–× 2/ e2 hc 4 2 11  1 1 1 1 2 4  1 hr 15.  I  r2 intensity becomes th 2. T= hr, T = hr; T = 1 6 4 A 1 2B 4 A+B 24 v1  2h0  h0 1 v2 5h0  h0  2  v2 1 16.  8  106 ms1 so, first 2 hr = 1 half lives (by A) next 1 hr = 4 half lives (by B) 17. M = M e–t  M0  M e t  n  1  = –t 0 20  20  0 1  n(20)  T1 / 2  t  n(2 )  n (1 0 )  next 2 hr = 3 half lives (by A+B)   n2   n2  N0  T1 / 2  t 28 thus N = ( Total eight half lives)  t = 16.42 days 3. dp  F  (2P sin ) 18. hc  0  1 mv2 ; 4hc  0  1 m v 2 F  2 3 2 1 dt t  F = 2npsin [n = number of photon]  0  4 0  0  4 1 m v 2   1 m v 2 h 3 3  2  2 1  1N= 2(n) sin 30°  n = 1027  1 4 1  0 4 4 . In photo electric effect the maximum velocity of e–  2 m v 2  3  2 mv2   3  v1 v 3 will corresponding to KE & other are less than 1 max it.   0.1  n N 5. KE = h  19. (a) P = N hc (b) (c)  max  100 i ne  20. K2  hc    hc   K < K2 6 . For threshold frequency, h0=  0  h K1  2  2   1 2 2  hc  KE = h–= h(– ) hc  2  2 max 0 1 Isaturation  n where I= nh 7 . K = h –   V = Kmax  4eV  4 volts 2 1 . Greater work function greater intercept max se e 8. K1  h1  1  1  0.5  1 2 2 . De broglie waves are independent of shape & size K 2 h2  2 2.5  0.5 4 of the object. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 9 . 1  2  2.53      2948 Å 2 3 . De broglie waves are probability waves and are 2 1 5.06 5896 applicable for all the objects. Wave nature is observed for the small particles like electrons. 1 0 . eV = h  2 4 . K.E.  qV S 11. E = h  hh hh 12. k K.E. = h  1  p1  2m1K1 ; 2  p2  2m 2K 2 max frequency depends on of light properties m2 m1 of cathode  q  q1  q2 & V is same  1  2 1  2 13. 2 1 89

JEE-Physics h 1  R(Z  1)2 1  1  2 5 .  = p (so same)   38.  n 2 n 2   1 2  h K  3 For K line, n =1, n = 2 2mK kT  12 26. ; 2 1 1 1   3   12    4   R(43  1)2  R  4 2 2 ...(i) 22 h h 25.15    Å and 2m 3 kT 3mkT T 2 1  R (2 9  1)2 1  1   R 282 3 ...(ii)  12 22   4  2 7 . Maximum photon energy = 13.6 eV (emitted) Dividing eq. (i) by (ii), we get   9    9  So K = 13.6 – 4 = 9.6 eV 4 4 max Hence stopping potential is – 9.6 V So – 10V can stop 2 8 . No. of electron that can accommodate in nth shell = 2n2 1 R(Z–1)2 1  1  Total number of elements =  = 2(1)2 + 2(2)2 + 2(3)2 + 2(4)2= 60 39.  n 2 n 2  1 2 13.6eV 2 9 . E = 22 = 3.4 eV For wavelength of K , n = 1 to n = 2 1 2 1875 R = R (Z –1)2 1  1 ...(i) 4 A 4  and 675 R = R[Z –1]2 1  1 ...(ii) 1 B 4  1 1 7 30. E – E = Z2  9  16   144 Z2 By solving eq. (i) & (ii) we get 43  Z = 26 and Z = 31 1 1  3 Z2 AB 4  4 [Four elements lie between these two] E – E = Z2   1 1 21 4 0 . min  V so 6.22  10  min = 0.622 Å E –E = Z2 1  1   3 Z2 42  4 16  16 4 1 . Characteristic X–rays corresponds to the transition 31. E = h     1  1  of electrons from one shell to another.  n12   4 3 . N x (200 × 106 × 1.6 × 10–19) = 1000 n 2 2  max for n =2 to n =1 4 4 . Q = –7(5.6) + [4(7.06)]2  Q = 17.3 MeV 2 1 1 1 1  45. 4 He  2n  2p number  12 n2  2 32. Wave  1.097  107  BE = [2(1.0073 + 1.0087) – 4.0015] 931.5 MeV Now in series limit corresponds to n =  to n = 1 46. 2 Deuteron  4 H e 2 1  Wave number for series limit = =1.097× 107 Q = BE of product – BE of reactant  = [28 MeV – 2(2.2 MeV)] Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 13.6(2)2 = [28 – 4.4] = 23.6 MeV 3 3 . E = 24.6 + 12 = 79.0 eV A 4 47. K= A  4 Q  48 = A 50 A = 100  A 3 4 . N =nC = 5C = 10 22 3 5 . 2r = n() 4 8 . 0 n1 1 p1 1 e0   3 6 . E = hc  1242eV  nm = 59 keV 49. N = N – N  N 0 [e t1  e t2 ]  0.021nm 12 3 7 . High atomic no. and high melting point. 1 5 0 . 90 days  3 half lives, left 8 i.e. 12.5% Disintegrated  100 – 12.5 = 87.5% 90

JEE-Physics  N0 6. K = 10.4eV ;  = 1.7 eV 10 max N0e–t n(1) n(10) 51. N = N e–t = – = –t E = 10.4 + 1.7 = 12.1 eV  n=3 to n=1 0  t   n10  T  t =33 days 1242.eVÅ  n2  1/2 12.1 =  5 2 . 63% or nearly 2/3 1 7 . Stopping potential  frequency  wave length m   NA  Saturation current  rate of photoelectron 53. R = N =   M  w emission. Also, K.E. = h– , P= max 2mKE EXERCISE –II h h 8 . (1) A  2mTA (2) B  2mTB nh 13.6 Z2 (3) T = T – 1.5 eV (4)  = 2 1 . (A) L = 2 (B) E = n2 BA BA cZ 9. K = 1242 eV  4.5eV (C) v  137 n (D) K = U max 200 2 K = 1.7eV at cathode 2. 1  R 1  1  , n =4 max 0 12 n2  K = (1.7 + 2) eV at anode max If polarity is revered, no e– reach at collector. hc hc hc 1 1 1 10. N = N e–t   N0  N e T1/ 2 0 2 0 (a) 0    loge 2 1 1 2 0 1 2    T1 / 2  T = mean hc hc hc hc 1 1 1 1    Assuming  to be 0 or 0 1 2 3 0 1 2 3 11. +1 3. 13.6 1  1   13.6Z2 1  1  69 E172   69 D172  71 C176 1 9   9 n2  1 0  72 B176   74 A 180 1 1  1 2 . At t = 0 : N = N 13.6 12 22  10  4. |K|= = 10.2 eV At time t : N = N e–t 20 Decayed in time t (N – N ) = N (1–e–t) 12 0 1   Probability that a radioactive nuclei does not decay 12 22  |U|= 27.2  = 20.4eV N  N 0 e t  e t N0 N0 in t=0 to t : 2h h h L =  2 2 2 13. Add equation 3 2 H 24 He + p + n 1 ke2 U ke2 m v2 (A) Q=[– [m ( 4 He )+m(p)+m(n)]+3m( 2 H )]× 931 MeV 5 . U   3r3 ; F  r  r4  r 2 1 n2h2 Q mv2 r3 = ke2; m2v2r2= 42 (B) 1016W = t Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 1 4 . A is balanced both in mass number & atomic no. m (constant), Also V2  n6 (h  h0 ) r  n2 m4 e 15. h = h0 + eV ;V = ; eVs=h– s s U ke2 Here =frequency of incident light and  m3  , K.E. = 1 mv2  n6 = properties of emitter 3  n6  2 m3 hc  1  1   n6 n6 16. E =   n 2 n 2 U  m 3 , T.E.  U  K.E.  m 3 1 1 2 For second excited state to first excited state n6 E= hc 1  1   hc  5 then total energy  m 3 1   4 9    36  91

JEE-Physics For first excited state to ground excited state 1 2 6 . min  V if V  then (min)  E2  hc 1  1  hc 3   4   4  No. of collision per electron increase then intensity increases (A) E1 5 (B) E1   hc   2   2 27 27. E2 27 E2  1   hc  1  N 5 1 P1 5 BE/A   P2  (C) P  27 ZA 1 7 . P.E. = –2(K.E.) 28.  = 0.173 , N = N e–t T.E. = (P.E.) + (K.E.) 0 T.E. = –2(K.E.) + (K.E.) T.E. =–(K.E.)–T.E. = K.E. K.E. = 3.4 eV N = (N – N e–t) = Decayed amount  = 6.6 × 10–10 m 00 N = N 1  1  0 e  18. 122.4 = 13.6Z2 91.8 = 122.4 1  1  N  N 0 (1  0.37)  N 0 (0.63) 12  Z=3; 4  So an electron of KE 91.8eV can transfer its N   N 0 (0 .6 3 )   1 0 0  63% N0  N  energy to this atom. 0 1 9 . Room temperature  n=1 T = loge 2  4year Upon absorption excitations take place to many 1/2  higher states which upon de-excitation emit all U.V., infrared and visible light. 29. P= nrhc ,P = nbhc r b r b 2 0 . Room temperature  n=1 so lyman series if P = P Since  >   n > n rb rb rb 27.2eV 13.6eV   21. U   n2  C k  n2 30. (a) P = N hc (b) i =  e n (c) %    100%  n N 13.6eV E = 1  1 3 1 . We have work is done by only electric field. Thus if E   n2  C 13.6eV  n2 m 2   decreases, & thus momentum of electron E v, v here in these questions C = + 27.2 eV decreases & vice–versa, while in magnetic field  v remains constant. 2 2 . We have r  1 ,  = reduced mass 3 2 . For electron  = 150 =1Å  db 100  50 Kq2  m v 2  m v 2 nh r e N 2 ; mv r = re rN ee 3 3 . In photo electric effect only one to one Interaction. 2 3 . K > 20.4 eV for inelastic collision 34. N  N 0e0t , N '  N e 100t  1  e 9 0t 0 e Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 24.  decay : He4, so both Z & A decreases. 1 2 t = 90 decay : e0 , +1 so A will not change but Z will change (decreases)  decay : e0 , hc hc h0    eV2  2   3 –1 1 so A will not change but Z will change (increases) 35. eV = , , eV = 1 3 decay : no change in A & Z. 25. V = 12400  10 10 volt 2 11  m in   if 2V = V + V  2 1 2 2 1 3 12400  1010 12.4  104 harmonic progression 66.3  10 12 66.3 V = volt   18.75 kV 92