P-3 Allens Made Physics Exercise Solution

JEE-Physics 2 2 . N = F + mgcos, f = mgsin but f  µN 2 7 . Here µ = tan Retardation of block = gsin so mgsin  µ (F + mgcos) N from v2 = u2 + 2as f  F  mg  sin   cos    F v 2= 2(2gsin)s  s = v 2 mgsin mgcos 0 0 4g sin  1 / 2 3  2 8 . Let the mass of 'C' be M for 'A' remains stationary  Fmin = 2 × 10  0.5  2  = 20(1 – 0.866) = 2.68N Mg Aceleration of system a = M  2m  m A is stationary w.r.t. to 'B' f=N FBD of A 5N N = mg N=m× a 0.1× 9.8 2 3 .  Block is stationary so f = 0.98 N = 0.98 mg  gM  3m  = ma Mg mg 100N 3m  M–M = 3m  M =   1 10kg 24. Friction less 40kg 2 9 . Acceleration of car along slope Limiting friction = g sin – µgcos F = mg = 0.6 × 10 × 9.8 = 58.8 N 1  3 S = 10 × 2 – (0.5)(10)  2  = 5 – 4.33 = 0.67 ms–2 100 N > 58.8 N Now from v2 = u2 + 2as i.e. slab will accelerate with different acceleration. v = 62  2(0.67)(15 ) 100N 10 kg f= KN = 36  20.1 = 56.1 40 kg = 7.49ms–1 f= KN f =40a 0.4 × 10 × 9.8 = 40a  a = 0.98 m/s2 2 5 . Acceleration of box w.r.t truck 3 0 . Here 'M' is in equilibrium. f ma  µmg So net force on 'M' must be zero. f = Mg (upwards) Mg = =2 – (0.15)(10) = 0.5 ms–2 m The box will fall off at time t then from 31. tan = h   h = µR h R  11 R s = ut + at2 ; 4 = (0.5)t2  t = 4s 3 2 . Let forces acting on mass m in equilibrium are node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 22     Distance travelled by truck = 1/2 (2)(4)2 = 16m F, F1, F2 , F3 , F4 2 6 . Acceleration of system = 20  2 = 3ms–2     F  F1  F2  F3  F4  0 [equilibrium condition] 4 2   For upper block w.r.t lower block  F1  F2  F3  F4  F ...(i)  After cutting the string with force F , the net force F1 on mass m   2kg ma a Fnet     F [(from (i)] f = F + ma = 2 + 2(3) = 8N Fnet  F1  F2  F3  F4 1 mm 36 E

3 3 . For (A) : JEE-Physics  EXERCISE –II Fnet  Fˆi  Fˆj  Fˆi  Fˆj  2Fˆj For (B) : 1. Maximum tension in string T sin 30° = 40 max 2.    3.  T = 40 = 80 N max 1 Fnet  Fˆi  Fˆj  3  1 Fˆi  3Fˆi  Fˆj    2 For (C) : 80  Fˆi  Fˆj Fˆi  Fˆj  For monkey T – mg = ma  a = 5 – 10 Fnet 0 max    = 6 ms–2 For (D) :    2Fˆi  Fˆi  Fˆj Acceleration along the groove = (g sin 30°) (sin 30°) Fnet  Fˆi  Fˆj g 10 2s 25 = 2s =  = 2.5 ms–2; t =  F1 44 a 2.5 F2 3 4 . F2   FBD of block :  F4 where F4  F3  F1 N F3 mgsin30° 30° m   mg ma Fnet  F2  F4  30° a  mgsin 30° = macos 30° Fnet a g a = g tan 30° = 3 40sin30°=20N N = mg cos30° + ma cos 60° N 5kg FBD 5kg 40cos30°=203N 3 g1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ = mg × +m× × 2 32 35. N 30° mg 3mg  mg 2mg F= = 50N 1 23 3 Net vertical force acting on the body is equal to zero. 3mg N =F = mg cos 30° = 22  F1 2mg  24  3mg = 3 3 6 . 2T cos   W F2 3 W node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 T T 4. T W 2 cos  T  cos     2m1m2 T1 T1 m1  m m1 m3 37. T  g KK 10  2 1  m2 10  m2  1kg m2 m4 1  m2 T2 E 37

JEE-Physics T = (m + m ) g  m1  m2  1 1 12  m1  m2  2 T – T = (m + m) g  T = (m +m –m –m )g 10. a = g, d = at2 1 2 3 4 2 1234 Net force acting immediately after cutting x = T 2 T2   m 1  m2  m3  m 4  g t= 2d 2m1  m2 d m4  m4  = Acceleration = a m1  m2 g 5. Maximum value of f1 = 0.3 × 30 × 10 = 90 N 1 1 . In (A) T = kx = 2g 1 Maximum value of f = 0.2 × 40 × 10 = 80 N In (B) T = kx = 3g – 3 × g 12 2 2 5= 5g Maximum value of f = 0.1 × 60 × 10 = 60 N 3  Least horizontal force F to start motion = 60N In (C) T = kx = 2g – 2 × g4 3 = 3g f1 F 3 f2 f3 x1 = 5x2 = 3x3 2 12 4 6. Acceleration of system, a= FF 1 2 . Acceleration of B, aAB  2m  m  2m 5m 2F  a 1 mg g mAaB Contact force between B and C = (2m)a = 5 = m A g  2 µmAg  B mB 2m 4 To prevent downward slipping P & D of block A w.r.t. B µ  2 F  = mg  F = 5mg a= m A g  m A a B = µg +a = g  g  3g  5  2 AB mA B 24 4 7 . Velocity of Block 'A' at any time a= g 1 3 . Acceleration of B : a = mg  N2  sin   v = v – gt 1m 1 0 N1 N2 and velocity of 'B' is v = m g t 2M here v –t graph is a straight line of negative slope B N2 1 a1 mg and v –t graph is also a straight line of +ve slope. Acceleration of A, a2= mg  N2 a2 2 m 8 . Block A and C both move due to friction. Hence mg less friction is available to A as compared to C. a1 a2 a sin  = a node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 1 12 Maximum acceleraton of A = µg = 2 g but mDg  But acceleration of system a = 3m  mD g  mDg   mg  N2  sin2   mg  N2 2 3m  mD mm   m = 3m  N2  mg cos2  D a  sin2  9 . Acceleration of system  4 1 3 g cos2  2g sin2  =  4  1  g = 5 × 10 = 6 ms–2 1  sin2   Relative acceleration of blocks = 12 ms–2  a= g – Now 2 + 4 = 1/2 (12) t2  t = 1 sec 2 1  sin2  = 1 a 2 t2  g sin2  2 1  sin2   Displacement 38 E

JEE-Physics dx dy 1 9 . Let acceleration of blocks be 'a' then 1 4 . x2 + d2 = y2  x  y aT dt dt  xv = y(20)  v = 25 ms–1 a A A mgsin30° mgcos30° T y mg d mg 37° x T + mg sin 30° = ma , mg – T = ma 3 mg  a = g, T= 4 4 1 5 . ˆf  cos   270ˆi  sin   270 v ˆf  sin ˆi  cos ˆj f 2 0 . Acceleration of A = gsin down the plane  Acceleration of B = gsin down the plane And also the contact force between two is zero. 16. For man For plank 21. Let tension in string be T 17. 22. 18. a a T TT T T T f T E T f A aA aB aC T + f = 100 a T – f = 50a mAg T mBg T 2 100 4 mCg  2T = 150 a  a =  ms–2 150 3 Here (T)a + (T)a + (2T)a = 0  a +a + 2a =0 A B C AB C 4 m g – T = m a , m g – T = m a , m g – 2T=m a 3 A AA B BB C CC  T – f = 50 a  100 – f = 50 × gg g  T = 6.5 N, a = , a = , a = – 100  f = 3 N towards left A3B3C 3 5g – T = 5a Acceleration of block w.r.t ground A m g 2T – 8g = 8 a = =g=2ms–2 C m g g5 2T Acceleration of block w.r.t. plank a = ;a = = ms–2 2T ma  mg A 7 C 14 7 8kg = =a–g = 4–(0.2)(10) = 2ms–2 and a = 2a T T m AC 5kg 11 Here a = 0 as T < 10 g 100kg Now s = ut + at2 gives s = (2)(1)2 = 1m(w.r.t. B 22 5g ground & w.r.t. plank) a= 4 = a – (–a ) mR m R node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 23. Let acceleration of masses aR  a =  5g  a   w.r.t. pulley be a Ma0 Ma0 m  4  60kg a R Mg–T – Ma = Ma a 0 M M aR T + ma0–mg = ma Mg  (M–m)g –(M–m)a = (m+M)a Mg 0  5g   T – 600 = 60 ×  4  aR  100 a= M m (g–a )  M m  0  & T–1000 = 100 × a R But a0>g so a<0 and T<0  Tension in string will be zero By solving we get tension = 1218 N 39

JEE-Physics 2 4 . Let a  g (i.e. friction is static) 2.   2ˆi ;    4j then both the blocks are at rest w.r.t. plank. aA aB  0 ; aC Therefore spring will be in its natural length. Now let a > g (i.e. friction is kinetic) Pseudo force on A as observed by B = 0 then both the blocks are moving with same acceleration w.r.t. plank. Pseudo force on B as observed by C In this case spring force is equal to zero. = (4j) m (+ve y–axis) B Pseudo force on A as observed by C EXERCISE –III = (4j) m (+ve y–axis) A TRUE/FALSE Pseudo force on C as observed by A 1 . In case (a), tension is less than 2mg. = (2i ) m (–ve x–axis) 2 . Buoyant force pushes the balloon to the left. C 3 . Tension is greater than mgcos20° to provide ASSERTION & REASON necessary centripetal acceleration. 1 . For a non–inertial observer, pseudo force acts even 4 . Friction is responsible for forward movement. on a stationary object. 5 . Friction is zero if there is no tendency of relative 2 . Impulse applied by cement floor and sand floor are motion. same. 6 . Friction force always opposes the tendency of 3 . A sharp impulse breaks a brick. relative motion. 4 . At low altitudes, density of air is high. 7 . The tangential velocity about the centre of earth is different in both cases and hence normal reactions 5 . Static friction is generally greater than kinetic are different. friction. 6 . Rotation of the wheels stop but translation is present. 7 . In pulling case, normal reaction is smaller than the normal reaction in the pushing case. FILL IN THE BLANKS 1. f =  mg = 0.6 × 1 × 10 = 6N 8 . On the block only two forces act. One force is gravity max and the other is exerted by the incline. f= ma = 1 × 5 = 5N  f = 5 N 9 . Same tension propagates to either team but the Pseudo external force coming from ground help to decide the winner. L/2 L/2 2. N  N Comprehension#1 L FT8 for the left part, N = ma = 1.  sin(90  53) sin 90 sin(90  37) 2 MATCH THE COLUMN FT8 T   53° 3/5 1 4/5  T = 10N and F = 6N F 1.(A) For the entire system 37° F – F – (3 + 2 + 1) g sin 30° = (3+2+1) a node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 12  a 60 18  30 2. N sin 30° = N sin60° 8N 12 N1 = 2m/s2 N2 N1 = 3 N2 ...(i) 30° 60° N cos30° + N cos 60° = Mg 6 12 (B) Net force on 3 kg block =ma = 3 × 2 = 6N (C) Normal reaction between 2kg and 1kg Mg 60° 30° = F + 1 g sin 30° + 1a = 18 + 5 + 2 = 25N  N= 50 3 N and 2 1 (D) Normal reaction between 3kg and 2kg N = 50 N 2 = 2gsin30° + 2a + 25 = 39N 40 E

JEE-Physics Comprehension # 2 Comprehension #4 1 . Acceleration of the systme 1 . f = 30 N, 20N; f = 60 N, 40N T 1max 2max 2kg f = 90 N, 60N 200  (5  4  7)g 40 7kg 3max = = m/s2 5 4 7 16 for the lower half of the system 18N D T – 9g = 9a  T = 112.5 N 20N C B 20 N = f1 2 . For maximum acceleration, T = 4mg 40N 100 N for maximum retardation, T = 0 15N 40 N= f2 Equation of motion  T – mg = ma  a = 3g (max. acc.) & g (max. retardation) f3  a = 0 Comprehension#5 B 1 - 4 The hanging mass 'm' has the tendency to go up or 2. aC 100  40  20 = 4 m/s2 to go down or to remain stationary.  f 10 Acceleration of the system a = M0  M  m 3. a = 20 18 = 0.2 m/s2 D 10 T FBD of m : f Comprehension#3 a 1 . Static friction = 400 N (say) N Kinetic friction = F mg N = ma ...(1) mg = T ± f ...(2) Distance travelled = 1  (F  f)  12  F  f mg(M0  M  m) 2M 2M F= (M  m ) From table Hence F has a range of values for which M and m remain stationary with respect to block M . 500  f 600  f  2 700  f  2.5 2M 2M 2M 0 If friction is absent , then there exists only one value of F for the above said setup. :  1.5 ; & Comprehension #6  f = 200 N; M = 100 kg NN  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 400  S = = 0.4 mg 1000 200 mg 2. k = = 0.2 1. For equilibrium : 2 Nsin  = mg  N = 2 sin  3. 1000 If '' decreases sin decreases and N increases. E 700  f 700  200 If F = 700N, a =  = 5 m/s2 M 100 2. When  = R, 2 R cos  =   v = u + at = 0 + 5 × 1 = 5 m/s 1 mg mg  cos  = 2   = 60°  N = 2  33 2 41

JEE-Physics Comprehension#7 4. (i) T – 40g = 240  T = 632 N (ii) 392 – T = 160  T = 232 N 1 . N= 60–20 = 40 20N (iii) T = 392 N f = 0.1 × 40 = 4N N The rope will break in case (a) as T > 600 N. 20N  F= N2  f2 f TT T contact = 1600 16 = 1616 N 60N 0.2 × 1= 0.1kg a 2 5. (i) a T – 2g = 2a 1.9kg 2 . When F = 0,  = 0 2g 5g When F increases, friction increases gradually too  T = 2 (g +a) = 2 (9.8 + 0.2) =20 N limiting value and then decreases to its kinetic value. (ii) For midpoint of upper wire Hence  increases to a maximum value and finally settle to a value smaller than this value. T = 5(g +a) = 5 (9.8 + 0.2) = 50 N EXERCISE –IV (A) 20 10 6 . acceleration of the system a = 6  4 =1 m/s2 1 . Total force exerted by the sphere   m    = m    F dv N dv mg dt dt If tension in the spring is T then for 6 kg block = 2 5ˆi  2ˆj   2(10ˆj) = 10ˆi  24ˆj 20–T=6 × 1  T=14 N so reading will be 14 N Total force exerted by the sphere N Fsin30° F=20N  7. 30° = F  (10ˆi  24ˆj)N M Fcos30° FLim 2 . Acceleration of the blocks (2kg + 4kg + 6kg) Mg F = N = (Mg – Fsin30°)= 0.5(5 × 9.8 – 20 1 lim ) 2 a  24 12 1 m / s2 246 = 0.5 (49.0 – 10) = 0.5 (39) = 19.5 N For 6 kg block f2 – 12 = 6 × 1  f  18N F = F cos 30° = 20 3 = 17.3 N For 4kg block f1 – f2 = 4 × 1  f1 = f2 + 4 applied 2 = f1 = 18 + 4 [ f2 = 18N]  f1 = 22N Since F < F lim applied p  Force of friction = F = 17.3 N 3 . Average force F = applied t  8. For block B node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 where p  p2  p1 = 2mv T = f = m2 g & For block A T = m1g and time taken by the body in moving from A to B is t =  d2 m1 By solving above equations  = m2 v 9. Mg = µ Mg cos + µ Mg + Mg s in  1 2 3 2 p 2mv 4 mv 2  M1 = µ (M2cos + M3) + M2sin So F = t =  d 2 =  d = 0.25 (4 × 4 + 4) + 4 × 3 21 v 5 = = 4.2 kg 55 42 E

JEE-Physics 1 0 . N = mg – F sin; Fcos µN 1 3 . Let  be friction coefficient between A and B. As 12 N force on A is required for slipping so max force mg (F ) applied on B so that A & B move together. Fcos  µ (mg – F sin) F  cos  sin B  cos + µsin should be maximum for 4g  12    1 59 6 F  –sin + µcos = 0 min F1 m g    F  10  9  15N  tan = µ & Fmin = 1  µ2 96 1 1 . According to FBD – for vertical equilibrium 1 4 . For the motion of pully f = F sin30° – mg = 50 – 30 = 20 N F net F – 2T = 0  T = 2 = 50 N in upward direction. Since the tension is less than gravitational pull on 8kg. Hence block of 8 kg will not be lifted. Therefore a2 = 0. For 4 kg block T– m1g = m1a1  a1 = 2·5 ms–2 1 5 . For mass B N sin37° = m a 37° N B As block has tendency to slip up the wall, hence friction on it will act downwards. N = Fcos30° = 50 3 N  N = mBa = 1  3 = 5N But the limiting friction is, sin37 (3/5) 1 25 3 1 6 . Downward acceleration of bead (50 N = 21.65 N µN = 3 )N = 42 mg  N mg  µ(ma) =m = 1 2 . Since the string is under tension so there is limiting m friction acting between the block and the plane. = g – µa = 10 –1/2 × 4 = 8 m/s2 Now from s = ut + 1/2 at2, 1 = 1 × 8 t2 2  t = 1/2s EXERCISE –IV (B) 1. Once the block comes to rest, kinetic friction disappears and static friction comes into the node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65  Fx = 0  50 + N cos 45° = N cos 45° existence. N fL1 = 24 N or (1 – ) 2 = 50 ...(i) (Limiting friction force between 10 kg block and  Fy = 0  N cos 45°+N cos 45° = 150 incline) N fL2 = 3N or (1 + ) 2 = 150 ...(ii) (Limiting friction force between 5 kg block and 1  150 incline) Equation (ii)  eqn (i)  1 –  = 50 If we ignore the friction for the time being, then or 1 +  = 3 – 3  4 = 2 or =1/2 system has the tendency to move down the incline E 43

JEE-Physics (ii) When the string between A and B is cut, A and B face a downwardforce of 3mg as shown in the figure. and C faces no unbalancing force. Tendinogf DMioreticotnion  a = 2mg  2g() A m 2mg a =  2g() ; a = 0 Bm C So, we can say the friction force is acting opposite (iii) When the spring between B and C is cut, to direction of this tending motion. As system is not moving, f and f are static in nature. C faces a force of mg in downward direction & A and B a force of mg in upward direction. 12 mg g mg For equilibrium of both the blocks,  a = a =  () ; a =  g() A B 2m 2C m 6g = T + f and 4g + f = T 12 5. a TT T a 6. T T N1 other conditions are a1 2Ma1 Ma1 5M 2Mg f < 24 N and f < 3N N1 12 2g From hit and trial (better substitute f first) we can After solving equations we get a1 = 23 2 A 20 µ = 0.1 draw some conclusions. F B 30 µ = 0.2 If f = 0, T=40N, f = 20N C 40 µ = 0.1 21 If f2 = 3N, T=43 N, f1 = 17N So, f should lie between 0 to 3N 2 f should lie between 20 to 17N 1 T should lie between 40 to 43 N 2 . Here Acceleration of A = acceleration of C  m   tan   m  3 (i) Maximum friction on ground = 2  2 g sin    2   2 g cos  =  4  mg sin  (20 + 30 + 40)g(0.1) = 90N = f mm m 0  22 Maximum friction on between 30 & 40 kg blocks Acceleration of B = mg sin  = g sin  = (50)(0.2)(10) = 100 N = f B m Maximum friction on between 20 & 30 kg blocks 3 . a = T  0.4(4)(10) = T  4 44 = (20)(0.1)(10) = 20N = f A Maximum value of f in which there no slipping any where = 90N T (ii) F aA+B node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 A 100N B 90N aC 100N C For upper block µ g = a For this condition a = a s A+B C  100  90  F  100  F  112.5 N  0.6 × 10 = T T  10 T = 40N 40 50 4  4 4 4 . (i) When the spring between ceiling and A is cut, f A and B face a downward force of 3mg and C faces no unbalancing force. (iii) A F B 20 20 100 3mg 3  a = a =  g() , a = 0 A B 2m 2C F 120 20  F – 120 = 30 F = 150 N E 30 20 44

JEE-Physics 7.(i) (a) F = 160 N 9 . Arrangement will collapse when normal reaction between A & B becomes zero. f = µ m g = 0.5 × 20 × 10 = 100 N s max s1 C am2 = F 160 = 3.2 ms–2 AB m1  m2 = 20  30  for m 64N m1 160N (f < 100 N) 1 s Let N = normal reaction on A & B due to surface fs 3mg  a m1 = 3.2 ms–2 then 2 N cos  = 3 mg   N = (b) F = 175 N 2 cos  km1g (0 .3 )(2 0 )(1 0 ) N m2 30 ° 150° 90°– am2  = = 2ms–2 90° 30° N1 mg a m1 = F  km1g = 175  60 = 5.75 ms–2 For cylinder A m1 20 N mg sin 30  sin(150  )  tan  = 1 . 160  60 3 3 (ii) For m : a m1  20 = 5ms–2 1 1 ms–1 10. C A 60 160 =  10 m s2 B 30 3 For m : a m2  45° 30° 2 160N m1 1 A B 60N 1 1 45° 1 30° 3 3 dv velocity of A w.r.t. B = (1 + 3 ) ms–1 8 . At t = 1 sec; dt = 4t  smg = m(4 × 1) 11. Kx 3 Kx  N  mg  B 22 N  3 1mg   3 1 R A 60° 60° O R2 mg 44  N  mg 3A N=  3  s = = = 0.4 2 1  2  mg g 10 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 Velocity of car at t = 3 sec 1 2 . When the right string is cut, the body is constrained to move in the circular path. But when the right v = 2(2)2 = 8 ms–1 spring is cut, the body moves along and normal to the spring. Velocity of block at t = 1 sec is v = 2(1)2 = 2ms–1 0 37° 37° Velocity of block at t = 3 sec is v = v + kgt 1 0  v = 2 + k (10 × 2) 1 But v = 8 so 8 = 2 + k(20) T Kx 1 Mgsin37° 6 = k(20)  k = 0.3. Mgcos37° Mgcos37° E Mg 45

JEE-Physics For string, Mg cos 37° = Ma ...(i) 4 . Due to action reaction pair in body spring stretching 2 force is same.  Both will read M kg each. For spring, kx – Mg sin 37° = Ma' ...(ii) 1 5 . When lift is stationary Fspring and Mg cos 37° = Ma'' .... (iii) 1 m m But initially 2kxcos 53° = Mg 5 ....(iv) Kx = Mg 6 where : a = a12  a \"12 1 a1 a12  a12 25  a2 = a2 = 24 mg EXERCISE –V(A) 1 . The particle remains stationary under the acting of 490   Fspring = mg  49 = m × 9.8  m = =5 kg three forces F1 , F2 and F3 , it means resultant force 98 is zero. When lift is accelerating downwards     mg – F'spring =ma  F'spring = 49 – 5 × 5 = 24 N F1   F2  F3 Since, in second case F1 is removed (in terms of 6 . Initial thrust magnitude we are taking now), the forces acting are F2 and F3 the resultant of which has the magnitude = m(a+g) as F1, so acceleration of particle is F1 in the = 3.5 × 104 [10 + 10] m = 7 × 105N  direction opposite to that of F1 . 2 . The system of masses is shown in the figure. 7 . In this question T2 T1  so  = constant CBA Fsystem  0 v F rope 8. M TT M P T2 = ma = 2 × 0.6 = 1.2 N 3 . The free body diagram of the person can be drawn MP P = (M+m)a ; T=Ma = Mm as 9 . Weight of the block is balanced by frictional force T  Weight = N = 0.2 × 10 = 2N a  1 0 . fkinetic  N   mg f m v person 2 6m/s fkinetic  Fnet  m a node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 60g mg = ma  a=g Let the person move up with an acceleration a then,   a=g v  u  at T – 60 g = 60 a  a max  Tmax  60g 60   0  6ˆi  10t ˆi  a max  840  60g  4 m/s2    6  6  0.06s 60 10t 100 46 E

JEE-Physics p 15. d  1 a1 t12 ; d  1 a 2 t 2 1 1 . Use F = 2 2 2 t  144 = [40 × 10–3 × 1200]N  N = 3 dd 1 2 . g= 9.8 m/s2 45° Rough  m1  m2  45°  m1  m2  smooth a= g a1=gsin45°; a2 = g(sin45° – cos45°) m2=4.8kg  5  4.8  m1=5kg d  1 a1 t12  1 a2 t 2 ; t2 = nt1 (Given) a =  9.8  × 9.8 = 0.2 m/s2 2 2 2 1 On solving k  1  n2 f 1 6 . According to work-energy theorem,W=K = 0 Work done by friction 1 3 . 3m0g°sin30° + work done by gravity = 0  –( mgcos)  +mg sin  10 f = mg sin30°; m = g sin 30 = 2kg 2 1 4 . On drawing the free body diagram of block from  the frame of wedge, we get or cos = sin 2 or  = 2tan  Sp eed 2 1 7 . Stopping distance = 2  Retardation Retardation = g Stopping distance = 100 100 =1000 m 2  0.5 10 a p 150  103  20  18. F    30 N t 0.1 For the block not to slip on wedge F macos N 1 9 . The acceleration of the system, a = ma mgsin Mm mg  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 m MF mgsin= macos i.e., a = g tan force acting on m f  ma   m F E  m  M  2 0 . Relative vertical acceleration of A with respect to B = g (sin260° – sin230°) = 9.8  3  1 = 4.9 m/s2  4 4  47

JEE-Physics 3 . Forces on the pulley are F 2 1 . Minimum force required to push up a body. T=Mg F1 = mg sin  +  mg cos  Min. force required to prevent from sliding F = F12  F22 F2 = mg sin  –  mg cos  1 F=  m  M 2  M2  g mg Given  = tan    T=Mg 2 4 . For equilibrium in vertical direction for body B we F1 mg sin    mg sin  have The ratio F2 = mg sin  –  mg sin  = 3 : 1 2 2 . F(t) = F0e–bt Tcos Tcos dv  m dt = F0e–bt F0 Tsin B Tsin T mb T C  mdv  F0ebtdt v(t) A 2mg mv =  F0 e bt  C b mg mg at t = 0, v = 0 t  v =  F0 e bt  F0 2 mg = 2T cos mb mb v = F0 (1  e bt )  2 mg= 2 mgcos  mb  T = mg (at equilibrium) 1  cos = 2  = 45° EXERCISE –V-B 5 . The forces acting on the block are shown. Since the 1 . N  m2f L  ma  m2f L block is not moving forward for the maximum force F applied.  L  2f L  f   f mg 60° Fcos60° Now from f =0+ t  [ 0  0 ] N F t = f =   =   Fsin60° 2 . The two forces acting at the insect are mg and N. Therefore F cos 60° = f = N...(i) Let us resolve mg into two components. (Horizontal Direction) mg cos balances N and F sin 60° + mg = N ...(ii) From (i) and(iii)   13 F cos 60° =  [F sin 60° + mg] node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 f N  m g mgcos mg mgsin  F = cos 60   sin 60 mg sin is balanced by the frictional force. 1  3 10 5  N = mg cos 23 1 1 1  3  20N f = mg sin But f=N= mg cos 2 23 2 4 1  g cos  = gsin  cot=   cot =3 48 E

JEE-Physics 6 . The forces acting on the masses are shown. 8 . The acceleration of mass m is due to the force Applying Newton's second law on mass Q, we get T cos  F – f=ma ...(i) F a Q m T T F a a f f T m Tsin P m  Tcos m M.P. E.P. xx T cos  (Mean (Extreme  T cos = ma  a = m position) position) ...(i) Where a is the acceleration at the extreme position. Now applying Newton's second law on mass P F ...(ii) f=ma ...(ii) Also F = 2T sin  T = 2 sin  From (i) and (ii) [Acceleration is same as no slipping occurs between Q and P]. From equation (i) and (ii)  F  cos   a2  x2  a =  2 sin  m  tan    F = 2ma  a  F  kA  F  kA  2m 2m x  Substituting this value of a in eq. (ii), F Fx == 2m tan  2m a2  x2 kA kA 9 . y = kx2 We get f= m×  dy 2m 2 tan = dx = 2 kx ... (i) 7 . By equilibrium of mass m, T'=mg ...(i) Ncos = mg and Nsin = ma By equilibrium of mass 2m, T=2mg+T' ...(ii) a tan = g From (i) and (ii), T=2mg+mg=3mg ...(iii) ... (ii) When the string is cut : Ncos ma N  • Nsin mg T T  2m 2m 2mg 2mg m T' T' m node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 mg mg a from (i) & (ii) 2kx = a/g; x = 2gk For mass m : Fnet = mam  mg = mam  am = g 1 0 . f varies from mg cos to  mg cos. For mass 2m : F = 2ma  2mg–T=2ma net 2m 2m 2mg–3mg=2ma2m a = g  2m – 2 E 49

JEE-Physics 1 3 . Given m = 20 Kg, m = 5 Kg, M=50 Kg, Now since f < (f ) , there is no relative motion 12 1 1 max  = 0.3 an g = 10 m/s2 between m1 and M, i.e., all the masses move with (A) Free body diagram of mass M is same acceleration, say 'a'. T N1 f = 15 N and f = 30N T ƒ1 21 T Mg T Free body diagrams and equations of motion are as follows : N F a a f1=30N a T m1 m2 T MF f1=30N f2=15N (B) The maximum value of f is For m : 30 – T = 20a...(i) 1 1 (f ) = (0.3) (20) (10) = 60N 1 max For m : T – 15 = 5a...(ii) The maximum value of f is 2 2 (f ) = (0.3) (5) (10) = 15N For M : F – 30 = 50a...(iii) 2 max Solving these three equations, we get, Forces on m1 and m2 in horizontal direction are as follows : 3 F = 60N, T=18N and a = m/s2 5 1 4 . a= mg sin   kmg cos  m T m1 m2 T f1 f2 fk N Now there are only two possibilities. (i) either both m and m will remain stationary 45° 12 mg (w.r.t. ground) or (ii) both m and m will move (w.r.t. ground).  aA  g sin   kA g cos  ...(i) and 12 First case is possible when  T  f1 max or T  60N and  T  f2 max or T  15N  aB  g sin   kBg cos  ...(ii) Putting values we get These conditions will be satisfied when T  15N node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 say T = 14 N then f = f = 14N. 0.89 0.79 a = and a = 12 A2 B2 Therefore the condition f1=2f2 will not be satisfied. a is relative acceleration of A' w.r.t. B= a –a Thus m and m both can't remain stationary. AB A B 12 L=2m  L  1 aA / Bt2 2 In the second case, when m and m both move 12 E f = (f ) = 15 N 2 2 max Therefore f = 2f = 30N 12 50

JEE-Physics [where L is the relative distance between A and B] 1 6 . Force to just prevent it from sliding 2L 2L = mgsin – mgcos t2 =  Force to just push up the plane or aA /B aA  aB = mgsin + mgcos Putting values we get, t2 =4 or t=2s According to question Distance moved by B during that time is given by mgsin + mgcos = 3 (mgsin – mgcos) 1 aB t2 1  0.79  4  2  0.7 10  7 2m 2 2 2 2 S = = 1 1  1  2 2  2 2  .    3  Similarly for A = 82 m. 1 5 . Applying pseudo force ma and resolving it. 1 Therefore  = 2 N = 10  =5 Applying F =ma for x–direction. net x m a c o s  – ( f + f 2 ) = m a x 1 macos –N1– N2=max macos –masin– mg=max  a = a cos–a sin–g x =  2 5  4  2  25  3   2  10  10m / s2 5   5 5   5 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 E 51

JEE-Physics UNIT # 11 (PART - I) RAY OPTICS AND OPTICAL INSTRUMENTS EXERCISE -I 6 . Image formation by concave mirror when object is real & placed at focus f, forms image at infinity but not in other cases. 1. 7 . For M1 : u = – 10 cm f = – 20 cm; h0 = – 0.1 cm 90° –  = 180 ° – 70° – 70°   = 50° 2. 200 200 v = 10  20 = 10  v = 20 cm v hi 20 m = – u  ho = 10  hi = 2 × – 0.1 = – 0.2 ; hi = –2mm 20 For m2 = 10 =2  hi = 0.2 cm = 2 mm Distance between two images = 2 mm 3. L1 L2 8 . As convex mirror is placed at origin & reflecting I1 I2 I1' surface towards –ve x–axis. For u < 0  Real objects. For all cases of real objects, the image formation towards right of the origin v is +ve and also between pole & focus. a 9. b i x-z plane Second Image I1' (image of I1) = 5 +1 = 6 m r [from person] d 4. c From Snell's law : 1.5 sin i = 2 sin r a c a4 1.5 × a2  b2 = 2 × c2  d2  c = 3  a2  b2  c2  d2  1 unit vector  10.  1.014 1.02... 1.50  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65  vI1I2 = vI2 – vI1 = 0 For composite transparent IIr slabes =sin= constant sin = 1.6  sin x sin x = 5/8 sin. 5.   4ˆi  5ˆj  8kˆ ,   3ˆi  4ˆj  5kˆ 1 1 . Initially observer is mirror for image formation of vM v0 object O. As O is kept at the bottom, then it sees somewhat above the bottom.  Plane of the mirror is xy  10  vI  3ˆi  4ˆj and for kˆv IZ  Distance = 5 +  = 2 vMZ – vOZ = 2 × 8 – 5 =11 10  3   vI  3ˆi  4ˆj  11kˆ =5+ 4 = 12.5 cm (in front) Hence image from the mirror after reflection is also at 12.5 cm. 49

JEE-Physics 1 2 . Image formation by mirror at a depth 'h' from 1 8 . An object is seen when reflected light from object mirror.The distance (apparent) between these two enters into eyes when refractive index exactly same then no reflection from the object. 2h will be =  1 9 . As Speed of light in vacuum / air  = Speed of light in medium P c  = v  velocity is different in different medium h From v = n But frequency is the fundamental property. It never h change by changing the medium hence  is also I changed as v is change. 1 3 . Image position coincides with the object when an 20. object is placed on centre of curvature i.e. R = 0.5m when transparent liquid is filled then phenomenon occurs when the pin is placed 0.4 m from mirror i.e. effective distance must be 0.5 again 0.2 +  0.2 = 0.5 ;  = 3/2 1 4 . From Snell's law 2 sin i = 1 sin 90° i=c 45°30°  30° 30° 30° From Snell's law i' = 45° Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65  Again 1. sin 45° = 2 sin r 1. sin 45° = 2 sin r ; r = 30° r = 30° & 2 sin 30° = 1 sin r' 1 = 45° – 30° = 15° r' = 45°  = 180 – 2 × 30 = 120° Hence  = 180° – (30+45°) = 105°  = 180 – 2 × 30 = 120° & from Snell's law (refraction) 1 5 . When a ray enters into different medium & again reenter in a same medium there is no deviation 2 sin 30° = 1. sin r; r = 45° due to refraction (overall) but bottom silvered glass 4 = r – i = 45 – 30 = 15 surface provides 180° deviation from the reflection. so net = 1 + 2 +    = 270° 1 6 . Maximum possible deviation =    –C 2 1 . n1 > n2 But at here i > r not possible Reflected ray) Denser  Rarer c c 22. (i) For (i) u2 – 1 = u2 1 v1 6 1 7 . As given : r + r' = 90° Denser Rarer 3 (u1) v1 = 2 × 6 × 2 = 18 cm (u2) For plane surface : I1  object r u2 sinr = u1sin r'  90° n2 n2 r r' dapperent= n1 × dactual = n1 × (18–R)  u2 sin r = u1 sin (90–r) r u2 sin C = u1 sin 90° 1  tan r sin C = 1 1 18  6 2 2 3/2 = × = 8 cm (C: Critical angle)  sin C = tan r; C = sin–1 (tan r) 50

JEE-Physics 23. 3/2 4/3 = 3/2 4 /3 u = – 30, f = – 10  10 3/2 uf 300 300 vu 3 1 4 4/3 v= = = = – 7.5 2v 60 – 3u  u  f 30  10 40   2.5 cm in front of mirror. So v = 2  ve 3 29. So for any value of u, v is (–ve). So image is virtual. 2 4 . As given u2 > u1 > u3 (i) v Ix  m 2 v 0 x (same direction)  u2 > u1 lens act as a concave lens for upper (ii)  m = –ve, m = v half portion of lens & u3 < u1  Lower half portion act as a convex lens. , 2 v u 2 5 . m = 1 = –2 = u |v| = |2u| (v = –ve, u = +ve ) then v Iy  ve (90–u) = 2 u  u = 30cm 11 1 Hence lens location = –40 + 30 = – 10 cm 3 0 . v – 20 = 10  v = 20 cm 2 6 . u = – 30 cm; h0 = + 0.5 cm; f = 20 cm  R for mirror = 2f = 2 × 60 = 120 cm  x =20 + R = 20 + 120 = 140 cm 0.5 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ O uf 30  20 v= = = 60 cm u  f 30  20 v 60 = –2 31. Concave lens behaves as divergent lens and m= = diverges the beam of light as passes from lens. u 30 When the screen is brought close towards lens it receive more number of rays hence light intensity hi =mho = –2 × 0.5 = – 1cm from P axis increases. 1+ 0.5 = 1.5 cm below xy. 2 7 . From figure 1st behave as diverging and 2nd 33. Am = A1; A  1   A2 behave as converging  m  In 1st incident ray appear to pass at 5 cm distance Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 point and after refraction it becomes parallel to  A2 = A1A2  A = A 1 A 2 principle axis i.e. the focal length f = – 5 cm and for second f = 5 cm 3 4 . X = u2 – u1 = 24 28. v1 m = u1 =3  v1 = 3u1 or u2 = 3u1 I1 I2 3u1 – u1 = 24  u1 = 12 cm Distance between object & screen is Final image (by lens) object I2 D = v1 + u1 = 3u1 + u1 = 4 × 12 = 48 cm D2  X2 482  242 Hence f = = 4D 4  48 51

JEE-Physics 48  2448  24 72  24 EXERCISE -II = 4  48 = 4  482 = 9 cm 1. 3 5 . red sin i = 1 sin r 45° 45° 1 1.39 × 1.414 = sin r sin r < 1  possible (refraction is possible) 1 OAB  + 90 +  = 180°;  = 30° green × sin i  1.44 × 1.41 >1  not possible Same for blue 1.47 × 1 >1 2 . (R.R) y 1.41  It also not possible. 45° IR 1 45° (IR) O P 3 6 .  sin A = 1 × sin 90°  = sin A &  sin A 1A (R.R) y' = 1 × sine = × = sine 2 sin A 2 90° Field of view of image is decided by the reflected rays. As we can see that the reflected rays from the extreme end points of mirror at a height 'd' from P and below P all points on PY'. Hence (C). 3 . M2 M1 sin A  C = sin–1 1 A  2  2 2  sine = 2 sin A cos A sec 22 I2 O I1 I2 3 7 . min= 2i – A; 60 = 2 × 60 – Å; A = 60° L/3 L/3 2L/3 2L/3  A  min.   60  6 0  L L+L/3  2   2  sin sin = = Distance between any to images must be 2nL/3 A sin 60 sin 22 37° 4. 37° 3 /2  = 1/2 = 3 v53° 3 8 . i = 50° , equilateral prism A = 60°; e = 40° x Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 Total angle of deviation  = i + e – A = 50 + 40 – 60;  = 30° 10m Hence min <  min < 30° fR As we know the reflected ray = 2 mirror 39. 9 R = 2× 2 ×  = 18 × 2 = 36  vS cos 37° = v fR=y vS = v sec 53° = RR sec 53°= 10 (sec 53°)2 × 18 f   ; R  g  B (VIBGYOR)  52 y>z >x. = 10 ×  3  × 18× 2 = 1000 m/s 52

JEE-Physics 5 . vI = 2vM – v0 f In ACD : sin = 2f    30  = 2 × 0 – 5ˆi  6  2tˆj R R vI = 5ˆi  10ˆj ; vI0 = vI – v0 y PB = R – sec   R  3 2  hence vIo = 5ˆi  10ˆj – 5ˆi  10ˆj x PC = R CB = PC–PB = R–R + R 3 [at t = 2 sec] = 10ˆi = R = 2f  CB  2 6. 3 3 f 3 11. \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ Af u vIG = 2vMG – vOG uf vIG1 = 2v, vIG2= 4V .......... vIGn = 2nv u = – u, f = –f vA = f  u v cos Image of last end of rod vB = – f 7. uf f2  uf  uf Length = vB – vA = –f– f  u = v sin fu  f2 . Hence [D] Length = uf v cos v cos Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 V v sin V uf b \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ B v sinB 12. u = – u, f = –f vB = f  u BC u  bf vC = f  u  b u b 8 . Hence we can see from the images velocity in uf u  b f u mirror 1 & IInd vI2 = 2 vsin vB – vC = f  u – f  u  b ub uf bf  f 2 =  + =b fu fub fub  u  f  1 3 . When observer in rarer L medium and object is in A L denser medium then the length of immersed L portion will be shorter (apparent) A 2L+L Hence length of image B 9 . If the angle between mirrors is  then the deviation of rod in water from two plane mirrors will be = 2 –2 = 2L + L –L = L + L 300 = 2 – 2   = 30° Hence length of image of rod seen by observer in 360 Hence 30 = 12 air is = L  L = L  L . Hence [B] Number of images = m–1 = 12 –1 = 11   AD 1 4 . For parallel composite slab 2  sin = constant 1 10.  1  a 2  c r=ai+bj f a2  b2 = c2  d2 c CB P  ˆr1  ˆr2  1 53

JEE-Physics   a2  b2  c2  d2  1  1a  2c 1 9 . 1 =1; 2 = 3/2 d AB 1 5 . For parallel composite slab usin = constant 2 – 1 = 2  1 v  R x CD 0 × sin i = (n) × sin 90° 3 1 3/2 1 1 E 0 –= = 0 sin 30° = 0 + n  18 2 v  R 2R 11 11 v = 3R 2  1  4n  18  18  4n  2  n  4 dx ABE & CDE are similar 3R = 2R  upper layer of n=3 1 6 . From Snell's law 1.4 × sin x = 1 sin r 2 x= d 3 20. 1 1 m1  1 m1  u1  f m2  u2  f ; m2 liquid 53° =1.4  u2 – f = –(u1– f)  u1 + u2 = 2f sinx = 0.8  f  u1  u2  x = 53° 2 1.4 × 0.8 = sin r > 1 (which is not possible) 2 1 . From Snell's law   sin 2° = 2 sin r  Ray will not refrect but it has total internal   = 2° (small angle) reflection. Hence (C).  3 32 2 1 7 . Critical angle 2 sin c = 1; sin c = 3 2° Geometrical path length travelled by the light in the slab will be =2 +1 + 2 +1 =6m n2 sin 2°  2° ;  × 2° = 2 r° 1 8 . n1sinC = n2  sinC = n1  2 0 n1(denser) r = 1° & 2 × 1° = 3 r'; r' =  3  2 2 . 2 = (2 + e) –  e =  r1 + r2 =  sin 2 =  sin r1 ,  sin ( – r1) = sin  m(sin .cos r1 – cos  sin r) = sin   sin2 2 cos  sin 2   sin  1     sin   2   Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 C sin  2  sin2 2  cos  sin 2  sin  TIR n2(rarer) 2  sin2 2  2 cos2   1 2 – sin2 2 = 1 + 4 cos4 + 4 cos2  1  n 2 2 = 1 + 4cos4 + 4 cos2  + 4 sin2 cos2 2 2 = 1 + 4cos2 (sin2 + cos2) 1× sin = n1 sin (90–C) = n1cosC = n1 n12 2  1 cos     = sin–1 n12  n 2 ; sin < n12  n 2 2 2 8 54

JEE-Physics 2 = 120 ( = 60°)  Hence [A]. 23. 360 90° 2 4 m= = 6 (even) r2 3 60 1 5 90°-r1 1 sin 90° = 1 sin r1; 2 sin r2 = 1 cos r1 2 60° 2cos r2 = 3 sin r3 60° 1 4 sin r4 = 3 cos r3; 4 cos r4 = 5 sin r5  =60° 1. sin 90° = 5 cos r5 Number of images = n = m–1 = 6–1 =5 Square and add 1 + 22 + 42 = 12 + 32 + 52 [Hence (B) & (C). Reflected ray from 'M1' strikes r on mirror 'M2'. 2 4 . min = 38°;  = 44°; i = 42°   e = 62° Hence retrace its path. Hence [D].  = i + e –A ; 44 = 42 + 62 –A  A = 60° min = 2i–A ; 38 = 2i– 60  i = 49° y 28. 45° 45° 2 5 . 2 < 1 1 = i + e–A 45° = 53° + 37°–A = 90–A 2 < 90–Å x 50 + e –A < 90–A e < 40° 1 2 dy 2 dy 2  dx  cos x  dx  1 y= sin x; = 37° 50° 53° tan 45° = 1 = 2cosx 2 6 . Required length of mirror = M1M2 So the mirror 1 n 1 2 should be the length of half of the size of man = cos x = 2  x = 3  x = 3 and 3 170 2 3  sin   1  3 = 3 = 85 cm & it is placed such that its lower y= × 2 3 2   2  edge at a half the height of eye level height. M1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 29. 170cm  160cm  E  M2  80cm Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 u1 = –60 & u2 =–90 2 7 . If the angle between mirrors be – then the angle 11 1 of deviation after two reflections is  = 2 – 2    60 v1 60  v1 = 30 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 11 1  v2  36    90 v2 60  =60° v2 – v1 = 36 – 30 = 6cm 240 = 2 × 180° – 2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ m  v2  v1  6 1 2 = 360 – 240  v2  v1 30 5 for mirror vI = –m2v0 convex mirror makes virtual image. 55

JEE-Physics 33. v < 0 30.  2  1  2  1 ; 2   2  1  1   0 v u R v  R u  |ri| 1 K1 K0  K (A) Virtual image is formed for any position of 1 0 if 2 < 1. 2 (B) Virtual image can be formed if x > R & 2 < 1. 34. v > 0 1 i  2 = 2  1 + 1 > 0 and if 2 < 1  = 0 at K0  K0 = 2 =1 v R u 1 1  2 – 1 > 0; 1  2 > u1 2 = ?; K = 2 ; 1 :denser; K : increases Rx Rx  1R r=0,  3 -0, 2 = 3  x > 1  2  (B) if 2 < 1, 1 = ? 1 1R K = 2  Then virtual image is formed of x < 1  2 2 :denser; 1 : rarer; K : decreases  3 5 . 30cm 2 sin i = 1sin 2  1 3 sin = = 2 = K1 3 2 r    15cm 15cm At k1,  ; so, 1    2 2 3 6 3 1 . 1 = 60° 11 1    v = – 30 cm 54 60° v 30 15 sin 30°= 3 sin2 P 11 1 3 1 60° – =  v = 60 cm v 60 30 5 60° sin2 = 8 (B) Final image is at 60 cm from lens towards left of it. (C) final image is real 2 = sin–1 5 30°2 8 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 Also (C), TIR is at P ceases if 5 sin 60° =  × sin 90°; 5 3 3 = 3 2 36. 32. P 2 1 C O xR d = |f1| – 2|f2| Real image can't be formed always virtual. 3 8 . m1m2 = 1 & d > 4f (uv method) 56

JEE-Physics 4 0 . Height of object EXERCISE -III = H of image 1  H of Image 2 = I1I2 = 9  4 Fill in the blanks Height of object = 6 cm 2 . Rays starting from O will suffer single refraction from spherical surface APB. Therefore, applying m1 = I1 = v1  9 v1 A +Ve C O u 6 = u1 2 1 E 3 3 u1 O 2 u1 – u1 = 2 P u2  v1  v1  X= =X 2 B D 15cm 3 2 – 1 = 2  1 ; 1.0 – 2.0 1.0  2.0 D = 90 = u2 + u1 = 2 u1 + u1 = vu R v 15 10 36 11 1 u1 = 36  X = 2 = 18 =– or v = – 30 cm  Distance between two position of lens = X = 18 v 10 7.5 D2  V2 902  182 Therefore, image of O will be formed at 30 cm to the right of P. Note that image will be virtual. There will be no effect of CED. Focal length f = = = 21.6 cm 4D 4  90 41. 3 . Rays falls normally on the face AB. Therefore it will pass undeviated through AB.  r2 = 90 – 60 = 30° sin i2   2 = sin r2  i2 = 45° Deviation = i2 – r2 = 45° – 30° = 15° (Deviation at face AC only) 4 . P = 2PL + PM 1 2 1 2 1 2 II case : f = f + fm =  RR 43. 1  2  0 (condition for achromatism) 1 2 R f1 f2 = R  F = 2 = 7 ...(i)  f Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 1 : 2 = 2 : 1 (Given) 12 1  f1: f2 = –2 : 1 ......(i) (for fullfill condition) I case : P = 2PL + PM  F = fL + fM 1 2 1 1 R 1  1  1  1 f1   1 ...(ii) f = R +   F = 2   1 = 20...(ii) f1 f2 10 f1 1  f2  10 By Dividing (i) and (ii) R 2 1 7 by (i) & (ii) f1 = 10cm & f2 = – 5cm   20  20 – 20 = 7 2 R = 20  13 20  = 13 57

JEE-Physics I1 (A) Convex mirror   Real object  virtual image between focus pole 5. O (3hx) (B) We can see from (v–u) graph, the image distance v between pole & focus for all position of real objects. =4 2 3 (3hx)+2x (C) v Ix  m2vox ; vIx =  f f x  u     I3  dx =1cm/sec  R  2 I2 dt  2  vIx = R x u ; x object distance Apparent distance of image by mirror from bird (D) Mirror is convex 3h  x  2x 4. =  + (3h–x) 23h  x 2x 2x =  +   2 (3h–x) +   Velocity of image = 2  0  dx  + 2 dx 2  dt   dt = 2× [0–(–1)] +  × (–1) 31 E1 at H  H   H    H = 2 × 1 + (–1) = cm/sec.  2   2  22 E2 at  H  H  H  H    3   2   2  Match the column 1 . For (A) : All reflected rays pass through the two F1 at H  H   H  1  1  2   2 mirror upto infinite reflections. For (B) : After IInd reflection ray will not pass through any mirror For (C) : Only rays from object will be reflected by the mirrors and no reflected rays will be further reflected. For (D) : Similarly as C f  3H   13  2. m =  2   2  fu (A) Convex fmirror1 u = –ve, f = +ve F2 at  H  H (B) m = f f  2 f f   vf v f Concave  m =  f (C) Concave mirror u = – R (2f) 5. m= = 1  y = mx + c) f m= f f  f  1 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 f m = slope = f (magnitude of slope of line)  2f Intercet on y axis c = –1 (unity) (D) Convex mirror u = – 2f Intercet on x-axis  m = 0 f f1 m= =  f  2f 3f 3 v 0= –1  v = f (focal length) f 3 . v (+ve) f \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 7. Convex mirror  u (+ve) (A) (parallel to principle axis after reflection passes from focus) 58

JEE-Physics Diverging the ray (C) A = 70° : A–r 70° – 30° = 40° > C(30°) 70° - 0° = 70° > C All rays reflected back (B)  Concave lens (D) A = 50° : A–r = 50° – 30° = 20° < C(30°) Refrection 50° - 0° = 50° > C (Reflection) (C) Converge at focus Comprehension # 2  Convex lens 1 . In this case x, H and  remains same so required Reflects & intersects H H  x sin  (D) actually length of mirror =  concave mirror H  2x 1 1  1 sin 450 2 = =m 1 2 1 3 8 . (A) When object is at focus, image is also real 2 . In this case only x will change so equired length of from equi biconvex lens. 3 1 1  1 1 1 1  2 sin 450 =2 = 3 f  R1 m  mirror = 5 52  1 1 2 2      R2  3 . In this case x = 1–1 = 0 so length of mirror = R 1 1  0 sin 450 1 For equi biconvex f= 2   1 ...(i) As  f  120 = 2 Hence final image position v decreases Comprehension # 3 v 1 . On squeezing the lens, the focal length of lense hence m = u also decrease decreases. (B) From (i) R   f  2 . As v decreases, the magnification is also decreases. Radius of curvature increases i.e. now the object is not placed at focus and it is case 3 . As turnip away from the lens, distance increases when object is between focus and optical i.e. u increases hence for same v, f has to decreases centre. Hence image formation is virtual. so we should decrease the squeeze of lense. (C) Glass slab provides shift of the object in the Comprehension # 4 incident ray direction. So for lens object 1 . u = –25, f = f, v = 2.5 position becomes between focus and optical centre. 25  f 25  2.5 = 25  f  f= 11 (D) When medium is changed then relative . 9 . r > y > g > b  r < y < g < b 2 . Maximum focal length = distance of retina from  Cr > Cy > Cg > Cb (C : Critical angle) eye – lens = 2.5 cm Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 1 0 . Critical angle A 3 . Normal person power of eye–lens 2 × sin C = 1. sin 90° i r A-r 100 C = sin–1 1 =  P = 2.5 = 40 D; u = – 100 cm, v = 2.5 2 6 1. sin 90° = 2 × sin r 100  f 100  2.5 = 100  f  f = 41 1 sin r =  r= 2 6 100  0 < A – r < C for all rays refrection  Power = 100 =41D 41 (A) A = 15°  A – R =  ve hence (a) (B) A = 45°  45– 30°= 15° < 30° for same rays  Required power of lens = 40 = 41 + P2 P2 = – 1D 45° – 0° = 45° > 30° for same rays 59

JEE-Physics 25 EXERCISE -IV (A) 4 . Minimum focal length of eye lense from Q.1= 11 100 100 11 1. ABC & AB'C A  Power = 25 / 11 = 25 = 44D x  x3 0.2m 100 0.20 1 And u = – 100 cm, v = 2.5, f= 41 , x = 0.2x + 0.6 BC Power = 41 D 3 x 0.8x = 0.6  x  cm Required power lens = 44D = 41 D + P2 4 3m P2 = 3D B' 1m C' 5 . For an average grown up person minimum distance x = 75 cm of object should be around 25 cm So (D). Comprehension # 5 2. H min 40cm  80cm 20cm 1 . The focal length varies with wave length.  H = 160 cm min & H min 80cm  2. The combination will be free from chromatic 80cm 20cm observation if dF = 0. Hence [A].  H = 320 cm max 3 . Condition of achromatism 1 2 1 f1 3. f1 + f2 = 0  2 = f2  0 f1  f2 = – 2f1. Hence [D] = = ' 20 f2 4 . 1 = 0.02, 2 = 0.04 [Dispersive power] 1 f1 0.02 f1  2 = – f2  0.04 = f2  f2 = –2f1 11 1 111 f= –20 cm; v  60  20   F f1 f2 1  1  1  2 1  1 11 1 40  v  30 40 20 40 40 40  Hence f1 = 20, f2 = – 40. Hence [A] v 60 20 60  20 m   v  30  1  1 u 60 2 5 . This aberration related with lenses not for mirrors. so image coordinates are (–30, 3/2) At m : 2 11 1 1 1 1 50  70  20    v v 70 20 70  20 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 140 v= = – 28 cm 5 v 28 1 2 So magnification =   u  7 0  5 3 Object length = cm 2 image length =  3  2  3 cm  2   5  5 60

JEE-Physics Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65belowaxisofM:y=37 . Shift due to glass slab 2  5  \\\\\\\\\\\\\\\\\\\\\\\\\\\\ x coordinate = (40–28) = 12 = 6 1  1  = 6 × 1 2cm / 2  = 3 6cm 3 so coordinates = 1 2,  3  5  t = thickness of glass slab;  = 3/2 4 . Speed of ballon = 20m/s 20m/s Reflected final image concides with the object when the effective distance of the object in equal to radius 1 of curvature of mirror u = 40 + 2 = 42 cm. d = 20 × 4 – × 10 × 16 15cm 2 d=0 8. u =–15, R=–20, f=–10 \\\\\\\\\\\\\\\\\\\\\\\\\\\\ ub 150 x = t   1 =3cm v =  =–30 1   u  f 5 v 30 For mirror : m    =–2, v =–20 Object– I1 : u = –30, f=20 u 15 0 v  30  20 = 12 cm = 33+9 = 42 cm v = – m2v = –4 × – 20 = 80 30  20 I0 from object  4  2 5. x  45  3   3  40cm So total distance = 60 cm 9. QC f = – 20 cm Q 11 1  vu f 11 1 R 3 3 2  3 P 43  v  60  20 sinc 2  Q = 60° C 1 1 1 40  v  30 3 3 3 sin 30°  sin   1    sin 1 1  sin =    42 33 v 60 20 60  20 So image after reflection from mirror will be at d = 30 cm. So 10 cm above 'A' 1 0 . Let refrective index =  After refraction at A : It will be at 10 3 / 2  90  45 cm C C 8 4 4 / 3 2 so distance from 'B' layer is  45  3 3.2  4  4 45   45  So final image after refraction at 'A' will be at : 1 From snell's law  sin C = 1.sin90° = sinC =  3 2 45 45    from 'A' surface tan C  1 3.2  43 2 2  1 2 6. x1 =6  x =6× 3 x1 1  1 2 x2 C  10  2   2–1 356 2–1 =  16  ;   16 =1.17 and x2 =4, x = 4 × 3  2 2  t = x + x = 6 + 9 = 15cm 11. sin c  R 1 R 1 1 2  R  d 1.5 R  d 1.5 61

JEE-Physics 1 1  1.5 > 1+ dd 1 1 5 . From lens maker's formula  1  d / R  R  R  1 1 1 f  (  1)  R1  R 2  1.5 2 we have 1  3  1  1  1 0.3  2  R R  (Here R = R and R =–R)  R = 0.3 12 Now applying 2  1  2  1 at air glass vu R 3/2 1 3/2 1  surface, we get v1 (0.9) 0.3 4  v = 2.7m 1 1 2 . v = 16 cm/sec;   4 cm/sec i.e., first image I will be formed at 2.7m from the B 3f 1    lens. This will act as the virtual object for glass v Bf  v B  v f  –16 = v B – 4 ; v B = –12cm/sec water surface. v app  v act 12  v act Therefore, applying 2  1  2  1 at glass n obse n object 41 vu R  ; v = 9 cm/s act water surface, 4/3 3/2 4/33/2  13. we have v2 2.7 0.3  v = 1.2 m 2 P Q 2r 60° i.e. second image I is formed at 1.2m from the r 2 r 2r lens or 0.4 m from the plane mirror. This will act C as a virtual object for mirror. Therefore, third real image I will be formed at a distance of 0.4 m in 3 front of the mirror after reflection from it. Now From snell's law this image will work as a real object for water-glass 1. sin 60° =  sin r (at Pt 'P') again  sin r = 1. sin 2 r (at Pt 'Q') interface.  sin 2r = sin 60° r = 30° Hence applying 2  1  2  1 vu R  1. sin 60° =  sin 30°    3 3/2 4/3 3/24/3 14.  we get v4 (0.8  0.4) 0.3 in rr  v = 0.54 m i.e. fourth image is formed to the 4 R right of the lens at a distance of 0.54 m from it. Now finally applying the same formula for glass- air surface. 1 3/2 13/2  v 5 0.54 0.3  v = –0.9 m 5 n  1  n  1 v  nR i.e., position of final image is 0.9 m relative to the From (1) surface : v  R ; n 1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\ lens (rightwards) or the image is formed 0.1 m Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 1 n 1n 2 n n behind the mirror. R v  2R R R R v  2R   , 16. 1mm 2  n(v  2R  R)  2v – 4R = nv – nR (0,0) R R(v  2R) f=10 f=10 On putting the value of v 15 45 2 nR  4R  n  nR  nR Image formed by the lens v  15  10 = 30 n 1 (n  1) 15  10 4 (Right from lens) 2nR – 4nR + 4R = n2R – n2R + nR, n  Reflection from mirror u =–15 3 62

JEE-Physics v  15  10  30 ; m   (30)  2 20. 15  10 15 4mm Co–ordinate of Image = (30,0.3cm) 40cm Now again taking refraction from lens) v  15  10 = 30 (left from lens) m  30  2 100cm x 15  10 15  Final co–ordinate of image = (–15cm, –6mm) f 0.4 x 1 7 . For real image m = From similar triangles  ; x = 1 cm f  16 40 100 f 2 2 . A=60° Virtual m = i + e – 60 = 23° & e = i + 23° 2i + 23°– 60° = 23° f6 i = 30°  f f  f f 6  f + 16 = –f–6 1  16  e = 53°  2   sin r1 2f = – 22  f = 11cm 18. A C   sin 60  r1   sin 53  4 A' 5 O 1.5 x 3 1  4 B'    2  cos r1  2  sin r1   5 B 15 20 10 v = uf  15  10  30 3 1  sin2 r1  1   4 u 15  10    2 2 sin r1  5 f ABC & A'B'C are similar x  1.5  x  1  3 1 1 1 4    2 1  42   2   10 30 2 2 5 1  3 42 1 1  4 4  42   212 1 1  Area = x2 =  × = cm2 5  5  3 44     1 9 . Given– Distance between two position of the lens for image is formed 1 441 1  129 43  u v=(D-u) 4 25  3 75 5 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65Candle 2 3 . For critical angle  sin (90–r) = 1. sin90 ScreenD i r 90–r r cos r  1 x = u – u = 0.8 m (Given)  60 21  =30° D=u +v 11 m= v1 3  v1  3u1 90– (120–r) = (r –30) 1 u1 From snell's law  sin (r–30°) = 1 sin30° v2  1  v2  u2 1 u2 3 3  [sin r cos 30° – cos r sin30°] = 2 x = v –u = 3u –u = 0.8 11 11  2  1 3 1 1  1 u = 0.4      1  v= 3 × 0.4 = 1.2   2  2  2 1 D = 1.2 + 0.4 – 1.6 D 2  x2 (1.6 )2  (0.8 )2 u2 1  3 1  2 1  4     71/2 f  = 0.3m = 30cm 2 3  3  4D 4  1.6 63

JEE-Physics EXERCISE -IV (B) 16x2[36 + (d–x)2] = 9( d – x ) 2  x 2  1024   9  AO OB 1.  tan 30,  tan 30 16 × 36x2 + 16x2(d–x)2 = 9(d–x)2x2 + 1024(d–x)2 1m 1m d2 d2 d2 d2 d2 d2 16 × 36 + 16   9   1024 4 44 44 4 d 4  448 = 16 7 5 . As given in question 2 60° r || r 60° r  AO + OB = AB = 2 tan30° = m |||||||||| 3 2 . From sine rule O r| 1 60° 120° 1 (60–r)+(60–r)= 3 (60–r)+(180–2r)+(60–r) 2  I1 4r 2r  120–2r=100– 3  3 =20  r=30°  120° 30°  1 sin 60 =  × sin30°  u  3 30° 20cm I2 20     20  sin 30  20  1  cos 30 3 sin120 sin 30  v2  6. Case- I 3 . (a) v Im    u2  v Jm vIm = –1cm/sec  1m / s    1  v Jm  100  vJm = 100cm/sec = 1m/sec, vm = 20 m/sec  vJm = vJ – vm  vJ = vJm + vm = 1 + 20 = 21m/sec (b) m  v  dm  d  f  u dt dt  u  f   dm  f du  dm   u f f 2 1    du  1 11 1 1 dt (u  f2 )   dt dt   f  dt      f ' u v 60 30 dm  1  2 1 30  60 dt  10    f '   90   (1)  10 3  f' = –20cm 10 4. rr 17' 11 2 1  (  1)  1  1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 6'   f , f  60   r 6' 11' f ' fm d–x x 1 12 i   ( 1) 10'8\"= 32 20 30 60 3 4\" 2( 1) 1 1 60 20 30    4x  dx 3  32  2 36  (d  x)2 3  3  – 1 = 2  1  = 1.5 x2  64

JEE-Physics Case-II O 30cm  =4 3 20cm f=40cm When there is liquid in the mirror u = 2 f' &u = f'  water surface behaves like a plane mirror when there is no liquid in the mirror secondly image formed by refreaction from water surface followed by reflection from mirror and 1 1  1 u(u  30) again refrection out of the water surface.   30  Distance of the object from mirror u u  30 30 y 2u  30 4 = u = 20+ × 30  u =–60  u2 – 10x + (30)2 = 0 3 90  (10)2  4(30)2 60  40 u  2 ; u  45  15 5  v = 60  40  v = 24 cm But there u is +ve so that Now refrecation from water surface 45 15 5  1  4/3  14/3  v =–33 cm  u  45 15 5 & u  2f  f  2  2 v (20  24)  1 1 2 1 1 2 1 =   15 3  5 + 9. 90– x – 360° –[135 + 90 + 90–r]  x = 45 – r 2 30 60 f  fM f   1  1  4  3  5 90° 3 5 3 5  135° r 90° x   1  5 2  1  5 1 7. For totally internal reflection 2 sin (45–r) = 1 sin 90°  sin (45–r) = sin30°  r=15° from snell'a law 1 sin = 2sin15° 3 1  3 1  sin v = 2 × 2 2   = sin–1  2  11 Sinc  1  C  sin1 1  45  2 SinC =  2/ 3 10. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65  C  Sin 1 3  60 2 cos60  R  x  R, x  5cm Rx 8 . Image formed by partial reflection from water Required area as shown in figure = 4R 2  2R 2 surface at a distance 30cm below the air water 2 surface. 65

JEE-Physics n 1 n 1 2nR 13 10 cm   v1   2n  1    v  from L3 11. v1 2R R v 10 3 10 Ray converges at d= cm from L3 3 (b) 1  1  1  v  fx 5  fx  1 1  1 v x f xf xf v1 v f 2nR 4n 1 1   1   1 1   1   1  Now u2 = R-V1= R  2n  1  2n  1 R v1  fx f  v1 f fx  5   5    x f   x f  1 h 1n  v2  4n  1 2R  4n 1  fu  5u  5f  fu  2n  1  R  v2    3n  1   2R 5  v1  u   5  u f   v1  uf Total shift = 3R – |v2| 1 uf 1 f u  f    v1 5u  5f  fu f  fx  5u  5f  fx  4n 1  2R  n 1   5  x f  = 3R 3n 1 =   3n 1  R 1  5x xf fx   5x fu fu f  5f   5f  I v f 2dI  f0 du  12. 0 m  u fu ; rel = dt  (f  u)2 dt   2 Putting the value after solving x  50 cm 3 1  40  2  4  2  8 cm/sec 1 4 . Initially the system will act as simple concave mirror given 10 10 5 1 3 . (a) v = –60, m=–2  u = –30 When a parallel beam comes from '' from left, 111 1 1 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65  v  u  f  60  30  f after refraction from L1 it forms image at 10 cm from L1. f = – 20 cm  R = 40 cm This image will behave as a object for 'L2' So object distance form 'L2' = + 5 cm In second case : lens + mirror system 11 1 Then Pmeq = PL + PL + PM So image formed at v :   Pmeq = equivalent power of the mirror PL = power of the lens v 5 10 PM = power of the mirror 11 1 1 1 1 1 1   1  2  v  5  10  10  v  10cm Pmeq= fL  fL  fM fL  3  R Therefore L2 will form a image at +10 cm from L2 1   1  2  2 2   2   3  on the right side. This image will act as object for L3 fL  3  R 3R  R   3  Object distance for L3 = + 5 cm final image at v = ? 10 1 3R 1 1 1 1 1 1 1 1 1 Pmeq =   fmeq  10 v  f v v 5  3R fm   5  1 0    u 10 feqm = – 12 cm 66

JEE-Physics Now if object is placed at distance = x then EXERCISE -V-A 11 1 1. The formula for the number of images formed by v  u  feq , ( v = – 60 ) two inclined mirrors; inclined at angle  is 1  1  1  1  1  1  4 N  360 1 60 u 12 u 60 12 50   u=–15 cm So object displaced = 15 cm 360 If  even 15. 0,I f2=20cm  whereas 360  If 360  odd N  d=10cm f1=20cm =3/2 360 \\\\\\\\\\\\\\\\\\\\\\\\\\ For  = 60°, N = 60 =6 Hence, N = 6 –1=5 111 d 21 22 3. The working of optical fibres is based on the    4. phenomenon of total internal reflection. f f1 f2 f1f2 fL fm 20 120 120 60 The resolving power of an astronomical telescope f2  14  7 D 111 1 1 1 120 is = where D is the diameter of the objective      u v u f 120 u 20 7 1.22 lens. In order to improve the resolution of an 120  10  50   7 1  1  u = 100 cm astronomical telescope; one has to increase the  7 7  50 u 20 19 diameter, i.e., linear aperture of the astronomical telescope. 1 6 . (i) Given that r2=C  r1  45  C & sin e = n sin (45°–C) 5. The image formed by the objective of a compound sin e = n (sin 45 ° cos C - cos 45° sin C) 6. microscope is real and enlarged as object lies between focus and centre of curvature.   e  sin1 1 7. 2 n2  n12  n1 (ii) to pass refracted ray undeviated The number of images formed when the angle r2 = 0  r1 = 45° and sin e = n sin 45° between the mirrors is 90°  e  sin 1  1.352   sin 1 0.956 360  2  N = –1=4–1=3 60 17. As ray AB is incident at an angle of 45° and for this ray to suffer total internal reflection. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 C 45° A B 45° Here 45° > C  sin 45° > sinC Now 2 f  d    d 1 1 n 2  2f 2n 67

JEE-Physics 8 . A silvered lens behaves as a mirror. For the image 1 1 to be of same size as of object must the object be 1 3 . P1= f1  15D , P2 = f2 =5D held at the centre of curvature of mirror when 1 11121 When thin lenses are kept in contact with each other,   the effective focal length of the combination will be fsilvered lens f1 fm f1 f1 fm 1 11    15  5  10 1  1 1 2 f R    feff f1 f2 f1  fm 2 R R  feff 1 0.1m 10cm     Therefore  1  2  1  2  2 10 f R RR 1 5 . By using Snell's law we get   1  sin   The radius of curvature of this mirror R 30  2 1  sin2  1 = 2f =  = 1.5 = 20 cm  sin2 = 33 9 . Radius of circle 1 1  t tan c  t 12 36 sin = 3   = sin–1  3    cm 2  1 16  1 7 1 6 . Incident angle from given plane 19    A·k 1 = cos–1  A   –10   –1   = cos–1  20  = cos–1  2  =  3  t C Now by snells law n sin 1 = n sin 2 1 2  n1   2 3 1 3   2   2 1 0 . Let dots can be resolved by eye at maximum distance sin =  n2  sin =  R then 1.22  d 2 1 DR  2 = 45°     ƒ 2 R  Dd  3  10–3 1  10–3 1.22 1.22  500  10–9  1 7 . Velocity of image = –m2V0 = –  ƒ  u  V0 = 4.92 m ~ 5m 1 1  20 2 (15) = – 1 m/s 1 1 . Power of lens = 2  1  R1  R 2   In air power is – 5D = –  20  (280) 1 1 15  –5 = (1.5 – 1)  R1  R 2  ...(i) 1 8 . Focal length of a lens is given by Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 In other medium power is P 1  (  1 )  1  1 f  R1 R2  1.5   1 1 11 P   1 .6  R1  R 2    f   ...(ii) f Dividing (ii) by (i) P  0.1  1  P  1D R  B  fR  fB 5 1 0.5 Focal length will increase 1 2 . The relation between angle of minimum deviation and refractive index is D=(–1)A 68

JEE-Physics h 1  1 EXERCISE -V-B   19. Apparent shift = 2 1 . The ray diagram is shown in figure. Therefore, the Kerosene image will be real and between C and O. 1 h2 Normal at P will Water h1 C pass through C C image 1 1  1 1  \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 1  2  Total apparent shift = h1   h2  2. Let us say PO = OQ = X 1 11 1 1 f  240  1=1.0 20.   P f v u 12 (240) 21  1=1.5 O Q +ve Applying 2  1  2  1 vu R Substituting the values with sign 1.5 1.0 1.5  1.0  h 1  1  1 1  1   1 X X R Shift = µ  1.5  3 cm (Distances are measured from O and are taken as To get image at film, lens should form image at positive in the direction of ray of light) 1 35 1 1 2.5 0.5 3 3 v u   X = 5R distance = 12 –  ;  XR  21  3  1  u = –560 cm 3. R = – R, R = + R,  = 1.5 and  = 1.75 240 35 u 1 2g m R1 R2  1   g  1  1 f  m  1  R1 R 2  2 2 . From trignometry R Substituting the values, Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 1  1.5   1 1  1 we have   1.75  1  R  R   3cm 3min f 3.5R R-3mm R2 = (3)2 + (R–0.3)2  R  15 cm  f = +3.5 R 3 108 Therefore, in the medium it will behave like a lens  2  108  1.5 convergent lens of focal length 3.5R. It can be  from Lens maker formula, understood as, m > g, the lens will change its behaviour 4. The lens Maker's formula is : 1   L  1  1  0.5  1   1 69 1   nL  1 1 f  s  1  R1 R 2   15  30 f  nm  1  R1  R2  f = 30cm where n = Refractive index of lens and L n = Refractive index of medium. m In case of double concave lens, R is negative and 1 R is positive. 2 1 1 Therefore,  R1  R2  will be –ve.

JEE-Physics For the lens to be diverging in nature, focal length (r ) = 90° – (r ) and (r ) = C 1 max 2 min 2 min  nL  [for total internal reflection at AD]  n m  1 f should be negative or should be positive n2  n2  n1  n1  or n > n but since n > n (given), therefore the where sin C  or C  s in 1 Lm 21 lens should be filled with L and immersed in L . 21  (r ) = 90° –  1 max C Now applying Snell's law at face AB : 5. O n1  sin max  sin max  sin max n2 sin(r1 )max sin(90  C ) cos C  ii n1 AD or sin max = n2 cos C rr  r sin 1  n1 cos C   sin 1  n1 cos sin 1  n2  iB    n1  C Ei  max =  n 2   n 2   F  Divergence angle will remain unchanged because 8. Applying Snell's law at B and C, in case of a glass slab every emergent ray is parallel to the incident ray. However, the rays are displaced i iC slightly towards outer side. B (In the figure OA|| BC and OD|| EF) 6 . The ray diagram will be as follows : G  sin i = constant or  sin i =  sin i H 1 B4 C D I But AB||CD i = i or 1 = 4 J  B C S 9 . Figure (a) is part of an equilateral prism of figure (b) as shown in figure which is a magnified image E of figure (c). d HI = AB = d; DS = CD = 2 d Since, AH = 2AD  GH = 2CD = 2 = d 2 Similarly, IJ = d  GJ = GH + HI+ IJ = d + d + d = 3d Therefore, the ray will suffer the same deviation in figure (a) and figure (c). Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 7 . Rays come out only from CD, means rays after refraction from AB get total internally reflected at 10. P AD. From the figure : A r1 r2 D S Q  max n1 C ir N n2 RT B KM r + r = 90°  r = 90° – r 12 12 PQ = QR = 2h  i =45 70

JEE-Physics  ST = RT = h = KM = MN Applying Snell's law ( sin i = constant) at 1 and 2, So, KS = h2  (2h )2 = h  5 we have  sin i =  sin i 1 12 2 Here, 1 = glass, i = i; 2 =air = 1 and i = 90° 1 2 h1 1   g sin i = (1) (sin 90°) or g  sin i  sin r = h5 5 sin i sin 45 5 15. Critical angle C  sin 1  1     sin r  1 /     5 2 1 1 1 Wavelength increases in the sequence of 1 1 . f  (  1)  R1  R 2  VIBGYOR. According to Cauchy's formula refractive index () For no dispersion decreases as the wavelength increases. Hence the refractive index will increase in the sequence of d 1  =0  1 1  R =R ROYGBIV. The critical angle  will thus increase   d     0 12 f  C R1 R2  in the same order VIBGYOR. For green light the 0.2 incidence angle is just equal to the critical angle. 1 2 . d=0.2 tan 30°  3 For yellow, orange and red the critical angle will be greater than the incidence angle. So,these colours will emerge from the glass air interface.  23 1 6 . During minimum deviation the ray inside the prism   30 is parallel to the base of the prism in case of an equilateral prism. d 0.2 / 3 =2 3 0.2m 30° 30° 1 7 . When the object is placed at the centre of the glass sphere, the rays from the object fall normally on the surface of the sphere and emerge undeviated. Therefore, maximum number of reflection are 30. 33.25 18. Distance of object from mirror  15  1.33  40cm 1 3 . Image formed by convex lens at I will act as a 1 Distance of image from mirror  15  25  33.8cm virtual object for concave lens. For concave lens 1.33 I1 For the mirror, 1  1  1 I2 vu f 4cm  1  1 1  f = –18.3 cm  33.8 40 f 26 cm  Most suitable answer is (c) 111 11 1 1 9 . Let focal length of convex lens is +f, then length of   v  4  20  v = 5 cm vu f 3 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 v5 concave lens would be – f. From the given Magnification for concave lens m    1.25 2 u4 1 12 1 As size of the image at I is 2cm. Therefore, size of condition.     f = 10 cm 1 30 f 3f 3f image at I will be 2 × 1.25 = 2.5 cm. Therefore, focal length of convex lens = +10cm 2 and that of concave lens =–15cm Air 90° 2 11 1 r  14. 20. Refraction from lens : v1 20 15 Water Glass r  v = 60cm + ve direction 1 i.e. first image is formed at 60 cm to the right of i lens system. Reflection from mirror : After reflection from 71

JEE-Physics the mirror, the second image will be formed at a 11 1 11 11 1 distance of 60 cm to the left of lens system. 27.     and   v 30 15 v 30 v 10 15 11 1   11 1 Refraction from lens : v3 60 15  +ve     v=6 direction or v = 12cm v 15 10 3 Therefore, the final image is formed at 12 cm to the left of the lens system. 2 1 . r = f tan  r  f  r2  f2 2 8 . Here normal is along ˆj Angle between incident ray and normal f   1 r cos   2 2 2 . Since  < , both reflection and refraction will take ˆi  3ˆj  ˆj 3 c 11     30 place. From the figure we can see that angle between reflected and refracted rays  is less than 2 180° –2. 2 9 . Real image  v = +8 m Air m = v   1  u = –24 m u3 Water   24  8   f= uv  6m u  v 24  8 2 3 . Since object and image move in opposite direction, for plano convex lens the positioning should be as shown in the figure. Object lies between focus and centre of curvature 1  u  1  1  1  1  3  1 1 R = 3 m f < x < 2f f  R   6  2  R  \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 30.(P) at prism surface ray moving towards normal so Image (µ > µ )at block surface ray moving away from 21 Object right left normal so (µ < µ ) 32 (Q) No deflection of ray on both surface; so µ = µ 12 =µ 3 (R) At prism surface ray moving away from normal so 2 4 . In the position of minimum deviation angle of µ < µ . At block surface ray movign away from A 21 refraction r = normal so µ < µ but since on total internal reflection 2 32 not takes place on prism surface n  1  1  µ1 sin 45° < µ2 sin 90°  µ1 < 2µ2 8  8  (S) Total internal reflection takes place 25. n0sin =  0  sin90°  sin =   = sin–1 so µ sin45° > µ sin 90°  µ >2µ2 1 2 1 8 2 6 . v2 – u2 = 2as; v2 = 2(g)(7.2); v = 12 m/s MCQ Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 1 . For total internal reflection to take place : 7.2m Angle of incidence, i > critical angle, c 12.8m 1 1 1  sin i > sin c  sin 45 ° > n  2n X = µx  n  2  n > 1.414 b/f Therefore, possible values of n can be 1.5 or 1.6 in the given options. dX (B / f) dx 4 (12m/s) 16 m/s 2 . At point O (1) sin60° = 3 sin r  r=30° =µ  At point L : TIR occurs : because 3 sin 45° > 1 dt dt 3 72

JEE-Physics At point M : 3 sin (30°) = 1 sin     60 60 60° D r1  r2  2  30  r(say ) C 70° sin i i 20°  n1  sin r 60° 40° A  sin i = n sin 30° B 1 sin i   10.8  104  1   1.5  3 1.20  (600)2    2 4   2 ( = 0 = 600 nm)  i = sin–1 (3/4) 2. Incident ray  Match the column A 6 3ˆi  8 3ˆj  10kˆ  (6 3ˆi  8 3ˆj)  (10)kˆ 2 . (A)–pr, (B) –qst, (C) –prt, (D) –qs P z y zy Q'  A For (p) :  2  1 and 2  3  3  1 O x –10k x Q 63 i + 83 j For (q) :  1  2  3  QO  PQ (As shown in figure)  Note that QO is lying on x-y plane. Note, QQ' and Z-axis are mutually perpendicular. Hence, we can show them in two-dimensional figure For (r) :  3  2  1 as below. z P k A  1= 2 For (s) :  3  2  1 Q Q' O r=1 q  2= 3 For (t) :  2  3  1 R  Vector A makes an angle i with z-axis, given by SUBJECTIVE i  cos 1  10  cos–1 1  10.8  104  (10)2  (6 3 )2  (8 =   3 )2   2  1 . n1  1.20  2 and  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 1.80  104 i = 60° n = 1.45 + 2 Here,  is in nm. Unit vector in the direction of QOQ' will be 2 qˆ  6 3ˆi  8 3ˆj  1 (3ˆi  4ˆj) (a) The incident ray will not deviate at BC if n = n 12    2 2 5 10.8  10 4 1.80  1 0 4 6 3 8 3 20 20  1.20   1.45  (  0 ) 9 104 3 102 Snell's law gives 3  sin i  sin 60 20 0.5 2 sin r sin r   0.25  0  or 0 = 600 nm (b) The given system is a part of an equilateral  sin r  3/2  1  r = 45° 3/ 2 2 prism of prism angle 60° as shown in figure. At minimum deviation Now, we have to find a unit vector in refrated ray's 73

JEE-Physics direction OR. Say it is ˆr whose magnitude is 1. 15cm from the mirror (towards right) and since magnification is +½ , length of final image would be 1 [qˆ  kˆ] Thus, ˆr  (1 sin r)qˆ  (1  cos r)kˆ  2 3.6 1  1.8cm  2  1 1 (3ˆi  4ˆj )  kˆ  A'B' = 1.8 cm 2  5 Point B is 0.6 cm above the optic axis of mirror, 1 ˆr  1 (3ˆi  4ˆj  5kˆ) 1 52 therefore, its image B' would be (0.6)  2  = 0.3cm 3 . Applying 2  1  2  1 above optic axis. Similarly, point A is 3 cm below 1 the optic axis, therefore, its image A' will be vu R 1 2 First on plane surface 3  =1.5 cm below the optic axis as shown below: 1.5  1  1.5 1  0 AI1 mR   AI = –(1.5 mR) B' 0.3cm 1 Then on curved surface 1.5cm 1  1.5  1  1.5 A'  (1.5mR  R ) R A'B'=1.8cm [v =  because final image is at infinity] 15cm Total magnification of the image. 1.5 0.5 m = m × m = (–3)   1   3  1 2  2  2 (1.5m  1)R R 34 3 = 1.5m + 1 2 m=2 m  3  3 A'B' = (m)(AB)    2  (1.2) =–1.8 cm  4 . (a) Rays coming from object AB first refract from the lens and then reflect from the mirror. Note that, there is no need of drawing the ray diagram if not asked in the questions. Refraction from the lens : u =–20 cm, f = +15cm 111 1 1 1 5. (i) When angle of prism is small and angle of Using lens formula   ;   incidence is also small, the deviation is given by =(–1)A v u f v 20 15  v = +60cm and linear magnification, v 60 Flint m =   3 i.e., first image formed by A1 1 u 20 A2 the lens will be 60cm from it (or 30cm from mirror) Crown towards left and 3 times magnified but inverted. Length of first image A , B would be 1.2 × 3 = Net deviation by the two prism is zero. 11 3.6 cm (inverted). A So, 1 + 2 = 0 or (1–1) A + (2–1)A2 = 0 ...(i) 1 B1 Optic axis of lens Here, 1 and 2 are the refractive indices for crown and flint glasses respectively. B 1.51  1.49 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 Optic axis of mirror Hence, 1  2 = 1.5 A1 Reflection from mirror: Image formed by lens and  = 1.77  1.73  1.75 (A ,B ) will behave like a virtual object for mirror 22 11 A = Angle of prism for crown glass = 6° 1 at a distance of 30cm from it as shown. Therefore u=+30cm, f=–30cm Substituting the values in Eq. (i) we get Using mirror formula, (1.5 – 1)(6°) + (1.75 – 1)A = 0 11 1 or 11 1  v =–15cm 2    This gives A = –4°. Hence, angle of flint glass vu f v 30 30 2 m2 v 15 1 prism is 4° (Negatice sign shows that flint glass prism & linear magnification,    is inverted with respect to the crown glass prism.) u 30 2 (ii) Net dispersion due to the two prisms is  (b1  r1 )A 1  (b2  r2 )A 2 i.e. final image A'B' will be located at a distance of 74

JEE-Physics = (1.51 – 1.49) (6°) + (1.77 – 1.73) (–4°) = –0.04° (0.4)(0.09)  (1.2)(0.01) 6 . For refraction at first surface  (0.4)2 = – 0.3/s 2  1  2  1 ...(i)  Magnitude of rate of change of lateral v1  R magnification = 0.3 /s. For refraction at second surface, 9 . Critical anglest at 1 and 2 3  2  3  2 ...(ii) 1 1=2 v2 v1 R i i Adding equation (i) and (ii), we get  2=2 2 3=3 3  3  1 or v2  3R v2 R 3  1 3 C1  sin 1  1   sin 1  1   45 1  2   2  C2  sin 1  3   sin 1  3   60  2   2  Therefore, minimum angle of incidence for total Therefore, focal length of the given lens system is internal reflection to take place on both slabs should 3R be 60°. 3  1 i = 60° min 7 . Applying Snell's law on face AB, 1 0 . (a) At minimum deviation, r = r = 30° 12 (1) sin 45° ( 2 ) sinr  sin r  1 or r = 30°  From Snell's law   sin i1 or 3  sin i1 2 sin r1 sin 30 i.e., ray becomes parallel to AD inside the block. 3 Now applying, 2  1  2  1 on face CD,  sin i1  2 or i = 60° 1 vu R (b) In the position shown net deviation suffered by 1.514  2  1.514  2 the ray of light should be minimum. OE  0.4 Therefore, the IInd prism should be rotated by 60° (anticlockwise) Solving this equation, we get OE = 6.06 m B,D E 11 1 8 . Differentiating the lens formula   vu f with respect to time, AC 1 dv 1 du we get  v2 . dt  u2 . dt  0 (as f = constant) f 20 11. m  m 25   4 and fu 20  25  v2    dv    u2  du ...(i) 20 2 m 25  43 6  dt  dt  m 50 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\03.Rayoptics.p65 m 50    20  50 3 Further, substituting proper values in lens formula, we have 12. 1 1  1  f  R  10m  11 1 vu f 2   (u=–0.4m, f=0.3m) or v = 1.2 m 3 1 1  u1  50m v 0.4 0.3   Putting the values in equation (i) 1st condition 25 u1 10 Magnitude of rate of change of position of image = 0.09 m/s 7 1 1 Lateral magnification 2nd condition     u2  25m 50 u2 10 dv du v u.  v. m= dm dt dt  dt  u  25  18  3km / h u u2 Velocity = t 30 5 75



JEE-Physics UNIT # 06 (PART - II) WAVE THEORY EXERCISE –I    3  Waves on surface of water are combination of  1 . 3t  10. y =10sin ;y = 5 sin 3t  3 cos 3t 1 2 longitudinal and transverse waves. (    )10  1 sin 3 t  3   10 sin  3 t   2 2 cos 3t 3  1  100 2.  = 100m, v = 25m/s, T = = = =4 s f v 25  A /A = 10/10 = 1:1 12 v 600 / 2 3 3.    m f 500 5 1 1 . y = 0.25 cos(2t –2x) f = const. 1 Number of waves () f = f/2   = 2 = 600  600  5 =1000 y = 2 × 0.25 cos  2 t  2x  = 0.5 cos(t+x) 3 2  2 2  2waves 22A 2 v  A 2 4 . f = 1 sec = 2Hz ;  = 5m  v = f = 10m/s T2  T  12. I=  I   5. y = a sin t 1 y = a cos t = a sin (t + /2) I1  A 2  1 2  2 2 I2  A  2 1  T2     1 T1     1 :1 y lags y behind by phase . 2 12 2 6. y propagartes in +x-axis and y along –ve x-axis. 1 12 I  A2 and I  RL (y yx-1 3 . 2 12 )  2 1 1 I1  R2  A  A1  R2  25 7 . y = cos kx sin t = [sin (t + kx) + sin(t – kx)] I2 R1 A2 R1 5:3 A 2 12 2 9 (Stationary Wave) ( ) y= cos2(k +t) = 1 14. a a 2  a 2   a 12  a 2 3 2 [1 + cos 2(kx + t)] 1 2  2a1a2 cos 2 2 (Progressive Wave) ( ) I1  A  2  9  A1  3 I2  A  1 A2 1 15.  1 2  t 8 . y = 4 cos2  2  sin 1000t  A1  2   Imax  A1  A2  2  A2  1  3  1  2  4  2  A1  A2   A1  3  1   2  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 = 2 t   1 0 0 0  1  t  sin 1 000  1  t  Imin        4 :1 cos sin 2  2      A 2  1  2  = sin 1001 t + sin 1000t + sin 1000t + sin999t 1 6 . v  Tension  v  kx = sin 1001t + 2 sin 1000t + sin 999t 9. y = y sin 2  ft  x ; v = y (2f) cos 2  ft  x v'  k(1.5 x)  v'  1.5 = 1.22  v' = 1.22 v 0    0    v  v = y02f ; v = f max wave Given ( ):v = 4v 17. v  coeff of   30 = 30 m/s  I max wave coeff of k 1   y0(2f) = 4(f)    y0  T =  × 900 = 1.3 × 10–4 × 900 = 0.12 N 2 69

JEE-Physics 18. L = TL T = LAY LAY   Ay 1T  L L= 28. f ; 2L  AY  = dA  v  T AY f' 2 T 1  1  f'  f 2  1  dA = 70 m/s 2L  2 2  19. f =f (m  no. of loops in steel wire n  no. of 2 1 loops in aluminium wire) (m      29 .7 = 0.14 ;  = 0.02 n  ) mT n T f  v  3  108 = 1.5 × 1010 Hz   0.02 2L1 1A 1 2L 2 2 A 2 m 80 n 80 3 0 . d = vt = 330 × 5.5 = 1815 m   2  60 7800  106 2  45 2600  3  106 m  4 (minimum) () R T n 3 M  M H2 O2  M O2  VH2 O2  VO2 31. v m T4 80 f= 2L1  7800  106 = 337.5Hz 1A1 2  0.6 32. f = f   air  v air 330  5.5 mm air water fair  60 103 T xg 20. v    gx  v2 = gx  3 3 . v = f  = f   512 × 4L = f × 2L 11 22 2 (symmetrical about x) (x   )   f = 1024 Hz 2 2 1 . The right end will shoot up on the wire. 3 4 . v = f  = f   f × 2L = f' × 2L  f' = f 11 22 ( )  vv 4L 2 35. f =f,   1 = 1 = 2 = L 2 3 . The equation represents a progressive wave mov- 1 2 32 ing along x-axis of single frequecy. 12 (x  L1  2 1 )   L2 3  4 6  v 20 24. L= 5  10= 5  = 4m  f =  = 3 6 . f11 = v 2 2  4 5Hz  4L  3v v  3   ; f (2L)= v  f = f  v  f1 4L 2 2 2L 1 25. y = y + y' = 0 (at x =0)  a sin (kx – t) + y' =0 res put x = 0 and get y = 0 Given that f – f = 100 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 res 12 2 6 . Distance between position having 3 nodes and 2  3v  v  100  v  200Hz antinodes = wavelength = 1.21 Å 4L 2L 2L (32=)  2 7 . n  = n  = n  = n 3 7 . v = f  333 = 333    =1 11 22 33 = Length of pipe in second harmonic.   +  +  =  n n n  ()  1 2 3  n1 n2 n3 11 1 1 3 8 . v = f (/4= L)  336 = 20 × 4L  L = 4.2 m    n n1 n2 n3 70

JEE-Physics v 330 4 7 . 1 = 2L ; 2 = 2(L–y) f   1m 39. = 330 v v v 1 1  L = /4, 3/4, 5/4.... = 25cm, 75cm,125cm f = f –f = 2  1 =2  L  y  L  21 Minimum length of water column =120–75 =  vy  vy 45cm 2(L  y)L 2L2 () I2 400 48. dB = 10log I1 = 10 log = 10 log20 20  4 0 . =L + 0.6 r = 10(1 + 0.3) = 13dB 41 (closed organ pipe) ( ) I2 I2 I1  20 = 10 log I1 =L + 1.2 r 49. dB = 10 log 1 (open organ pipe) ()  I = 100 I (taking antilog) 21  4 (L + 0.6 r ) = (L + 1.2r )  r – 2r = 2.5 L v 1 2 21 50. f = f  v  v obs  o 41. f = 500 = 250 ; f = 506 = 253 (observer is approaching) (   )   1 2 2 2  f = 3 s–1 = 3 × 60 min–1= 180 min–1  450  330  = 500 Hz  330  33  4 2 . If f > f ; f = 260 Hz ; If f < f ; f = 252 Hz B AB B AB  v  v  v s  51. f = f – f  fo – f app o o 4 3 . For sonometer wire ()  f  v 1  vs 2.5  n × 100 = (n+1) × 95 fo v  v s v  vs 100 n = no of harmonics   vs  1  vs  v 8m/s  vs 40 41  n = 19  L   L  v  2   2   f  19  4  20  4  L = 16  L  52. f = f  v  v B   450  330  10  =425 Hz  2  B A  v  v A   330  30   f = 20  4 = 156 Hz 4 4 . v = f × 50 = f × 51 53. f =f  v v  =385  34 0 340  0.5  = 374 Hz 12 min o   R    20  vv Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65  f – f =   0.1  v = 255 m/s 1 2 50 51 4 5 . 2f = f × 15 × 8  f = 120 Hz 4 6 . C1 A C2 B f 256 Hz f 262 Hz 12 2(f –256) = (f – 262)  f = 250Hz 1 1 1 2(f – 256) = (262 – f )  f = 258 Hz 2 22 71

JEE-Physics EXERCISE –II 6 . Let wave equation be (  ) c  v z  e x v t t0 2 1 . No. of wave striking the surface f = f  c  At t=0, x + vt = x + 2  vt =2 () 00 At t=1s, x–v(1–t ) = x–2  v = +4 m/s 0 freqeuncy of the reflected wave 7 . y = sin (t – kx + ) v =  cos (t – kx + ) ()   c  c  v  f' = f  c  v  =f  c  v  at t =0, x = 0, y =–0.5, v >0   = – 6 Wavelength of the reflected wave y  sin  t  kx    6    = c  c  c  v 8. f  fo  v  gt  df  fo  0  g f' f  c  v   v  dt  v  2  1000  1000 10  v = 300 m/s 30 v 2 .  =10   = 0.2 m Node occurs at x=  , 3 , 5 = 0.05 m, 9. y = 2mm sin  2 x  1 0 0 t   44 4  3  0.15 m,.....; ()  Antinode occurs at x =  , , 3  .....= 0.1m,0.2m,  0 = 2 sin  2 x  100t   22  3  0.3m,......; ()  wave speed (  ) v = 50 5 m/s   n  8  100t  (n = 0,1,2,3,....) 10 = 3 3 . y=0 at x=0.  t m in  25  n   / 100  1 s  nx  3 3 300 100 This can be statisfied by the term sin  L  =  10. x L ( sin  nx  )  (at a distance 'x' from free end)  L  (x) vv 5  T  x  3 g x 2 4. f– f = 5  2L 2  16 2  16.2 dx(g  2g) 1 2 0  v  0.2  5 v = T 3gx2 3xg Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 2  16  16.2 wave   2L  x  2  L  v 5  16.2  f1  2  16  0.2  405Hz and f = v  5  16  400Hz  v2  3xg  2v dv  3g  a = 3g/4 2 2  16.2 0.2 2 dx 2 5 . For closed tube () (constant everywhere) (  )    V L and '  Vv  L    f'  V v 1 3g 4 4f 4 4f ' 4(L  ) Now S = ut + at2  L = 0 + t2 28  t  8L / 3g 72

JEE-Physics 1 1 . Total energy (  ) 2 0 . Let a = initial amplitude due to S and S each. 12 1 (a= s s ) = 2 2A2 = 22fn2A2= 22n2 f12 A 2 12 I = k(4a2), where k is a constant (k  ) 0 (f = fundamental frequency ( ) After reduction of power of S , amplitude due to 1 1 f = Frequency of nth harmonic (  ))  S =0.6a. 1 n (S SS=0.6a) 1 11 TE Due to superposition ()  <KE> = <PE> = 2 a = a + 0.6a = 1.6a, and max 1 2 . In the stationary waves, the particles in the alter- a = a –0.6a = 0.4a min nate loops are out of phase. I /I =(a /a )2 =(1.6a/0.4a) = 16 max min max min () 13. f  RT k k EXERCISE –III , fA  L 2 2L M L M Comprehension#1 2k 3k 1 . v = Point a  f= L 32 , f= 2L 28 2 . v = Points c,d,e B C  3k 11 3 . v = points b,f f = ,  f /f = =0 D L 44 C D 28 4 . v = points 0,d,h max 14. 3  2    4 m T w  xg  w 2 3 v   xg 5.  Pr. Amp. = P c o s  3 x  = P at x = 0 0  2  0  v2  w  xg  2v dv  g  a  g (x=0P cos  3 x  = P )  dx 2 0  2  0 Comprehension#2 and –P at x = 2/3= P at x = 4/3 1. 2  = 2n (n = 0,1,2,3,.....)  2 (R )  2n 00   2  Pipe may be closed at x = 0 and open at x = 3 m. (x=0x=23 m   R = R, R , R , R .......... n 2 3 4 17. 300m / s 2. 2   (2n  1) (n=0,1,2,....)    12m.  25Hz Separation between A and B = 6 m = /2  2 (R )  (2n  1)  (AB)  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 1 8 . Comparing with the equation    2R  2R, 2R , 2R ..... 2n  1 35 ()  nx   I0  2 y = 2Asin  L  cos(t) 2  2A = 2mm or A = 1mm 3. I   I0   2I0 2 nx n max L = 6.28x = 2x or L= 2 m. For n =1, L = 0.5 m. 4 . max to produce maxima at D = R (D=R ) max 5 . max to produce minima at D = 2R (Dmax=2R) 73

JEE-Physics EXERCISE –IV(A) FL 7 . For wire, L = L   = AY  F = AY  1. h= 1 gt 2  t1  2h and h = vt , t = h 300 106  2  1011  1.21  105  20 1  0.1 2 g 2 2 v 340 1 F1  f  2L  2 1  t = t + t = 8.707 sec = 11 Hz 12 7 15  2 v T 2 10 2 . 13 22  4.5  103  100m / s 2 8.  9 2.25 A = 2 2 = 2.828 mm  2.83 mm t  L  2  0.02 sec v 100 2  22 2 2  A  I1  A   T2   2   1  1 10  10m / s , t  L 0.5 3. I   T   I2  A 1   T1    1    2  1 9. v 10 3 / 10 2 v   0.05 sec 2 10 S1 6m P 10. y  5 sin  x  co s( 4 0 t )  3  4. = 2.5 sin  40 t  x   2.5 sin  4 0 t  x   3   3  6.4m S2 2   (2n  1) (i) Equation of incident wave  () for destructive interference () 2 f(6.4  6) = y = 2.5 sin  4 0 t  x  v 1  3  (2n+1) Equation of reflected wave f  (2n  1)v = 400(2n+1) (v = 320 m/s) ()  0.8 x =400 Hz, 1200Hz, 2000 Hz, y =–2.5sin  4 0 t  3  2   2800 Hz, 3600 Hz,4400 Hz 5. v T 90  9.8 10 m/s (ii) 2     = 6cm  8  103  105  3   Distance between adjacent nodes = 3cm f = 256 Hz; A = 5cm = 0.05 m () Equation of the wave (  )  x  (iii) v =–5(40) sin  3  sin(40 t)  y = A sin (t – kx) = 0.05 sin (2ft – 2f x )   1.5   9 v sin  3   8  = –200  sin 40   = 0 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\03.Wave Motion.p65 = 0.05 sin(1609 t – 4.84x) 11. = 340  3.4  10 4 m , = 1486 =1.49× 10–3 m air water T 2  10  2 10 m/s 106 106  6. V=  1/2 bottom  x   5  1 2 . (i) (y ) = 4 sin  15   4 sin  15  2 3 max x=5cm T 8 10 V=  1 / 2  4 10 m/s top  (ii) 2      30cm  15 f  v top  v bottom top= 4 10  0.06  0.12m  Position of nodes ()  top  bottom 2 10 = 15cm, 30cm......... 74