P-3 Allens Made Physics Exercise Solution

JEE-Physics 3 6 .  = 5eV – 2eV = 3eV 1 1  hc  1 mv2 13.6  12 52   2 6eV  3eV 44. (a) E =    –V =  V = –3V Se S nh h 3 7 . Jn  2  nn = 2(r0n2) (b) 0 = mv – So J     = (2r )n  nn n 0 4 5 . Excitation upto n=3 is required so that visible length is emitted upon de-excitation. 38. E=  13.6 (Z 2 ) So required energy = 13.6 1  1  = 12.1eV n n2 9  13.6  13.6  Z2  r1  n 2 1  n1 1 T1  n 3 1  (2n)2 4  n2  r2 1  n2 2 T2 1  46. E2n  Z2  n 2 4 n 3 8 2 2  13.6  Z2    Z2  47. nth excited state = Cn 1  n(n  1) = n  4  n2  13.6  n2   2 2 E –E =   2n n 13.6 Z2 Z2 48. L   nh  ; r   n2  f   Z2   2   0.53 Z  ;  n3  E – E= n2  13.6 n2 2n n 4 (f r L)  Z Constant for all orbits. E– E= Z2 13.6  13.6   Z2  4 9 . K  K & K, K are type of atom. 2n n 4   (10.2)  n2  n2 5 0 . Energy released = (BE) – (BE) product reactant (E )(E ) = (13.6)2  Z4  5 1 . A1  Ne t1 , A 2  2Ne t2 2n n 4  n 4   Z2  2A1  e (t1 t2 )  n 2A1  t  t Tn A 2  A2 A2 2A1 (10.2)  n2   (E2n  En )  n2   (13.6)2  Z4  (E2n )(E n )   Z2  n2 (E2n  En )  (E2n )(En ) 4  n4  5 2 . Radioactivity law is valid for large samples 39. 10 = nC  n = 5; then 5 orbits are involved upon 5 3 . N1  N 0 e T1 ; N2  N 0 e T2 2 coming to second excited state so nth excited state is 6th [2nd, 3rd, 4th, 5th, 6th] R = N1 ; R = N2 1 2   (R1  R2 )  N1  N2  4 0 . On coming from 4  1 energy is greater than,  (N – N ) = less than or equal to energy corresponding to 12 24 T = loge 2 ;   loge 2 T V2  V02  V02 a1  n 4 24  16 (R1  R2 )T r n2r0 n2 r0n 4 a2 2  34 81 41.  n14 (N – N ) = (loge 2) 12 (N – N )  (R – R )T Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 12 12 4 2 . T  n3 Ti 1  n  3 n 1 5 4 . Final product state Tf 27  m  m 3     55. dN dt  1N1  2N 2  1N10e 1t  2N 20e 2t 4 3 . No of spectral lines n(n  1)  10  n2  n  20  0  n=5; n=–4 2 E = 13.6 1  1  =13.056; = hc 25  =95nm E 93

JEE-Physics EXERCISE –III Comprehension–1 Match the column 1 . For Balmer series, n =2 (lower) ; n =3,4 (higher) 12 v 11 ,J v  In transition (VI), Photon of Balmer series is 1.  ,KE   n,   n absorbed. n2 r 11 2 . In transition II : E = –3.4 eV, E = –0.85 eV but r n2 and v     n3 24 n hc hc E = 2.55 eV; E =    = E = 487 nm. 2 . For given atomic number, energy and hence 3 . Wavelength of radiation = 103 nm = 1030Å frequency of K–series is more than L–series. In one series also –line has more energy or frequency 12400 compared to that of –line.  E = 1030Å  12.0eV 3 . Consider two equations So difference of energy should be 12.0 eV (approx) eV = 1 mv2 = hv – 0 ...(i) Hence n =1 and n =3 s max 1 2 (–1.51)eV (–13.6)eV 2  Transition is V. no. of photoelectrons ejected/sec  Intensity ...(ii) 4 . For longest wavelength, energy difference should be minimum.So in visible portion of hydrogen atom, h minimum energy is in transition VI & IV. (A) As frequency is increased keeping intensity constant. 1 Comprehension–2 |V | will increases, m(v2 ) will increase. 1. i= qf =  2e v  ev 2s max  3   2r  3 r (B) As frequency is increased and intensity is decreased. 1 evr |V | will increase, m(v2 ) will increase and 2. M = 2s max u3 saturation current will decrease. evr evr evr 2evr (C) Its work function is increased photo emission 3 . Mnet  M u  M d  M d  3  6  6  3 4 . Net magnetic moment in that case will be zero. may stop. (D) If intensity is increased and frequency is decreased. Saturation current will increase. 4 . (A) In half life active sample reduce = R 0 Comprehension–3 2 1 . Q = CV  ne  0 A V     Decay number of nuclei is = R 0 d 2 2.85  10 12  10 (B) N = N e–t n  0.5  10 3  1.6  10 19  16 0 n = 8.85 × 109 n(2) where  = decay constant,  = 2 . Equivalent resistance Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 t1 / 2 N  N e  n2 t1/2 N  N0 V 16V t1/2 n 2 e R  I  2  106 A = 8 × 106 0 3. P  nhc where n = number of photons incident N0 N  N0 e  1   N  N0  1e   e  N0 e Ihc N0 N0  N0 per unit time. Also I = ne  P  e (2)t / T1/2 23 / 2 22 (C) N  N = (2  106 )(6.6  10 34 )(3  108 )   (4  10 6 )(1.6  10 19 ) 94

JEE-Physics = 9.9  107 m 9900 3.(i) For metal 1 threshold wavelength is 1  Å = 6187 Å 1.6 1.6 1 1 = 0.002 nm–1 or 1 = 500 nm = 5000Å Which came in the range of orange light. For metal 2 threshold wavelength is 2 4 .  Stopping potential V = 8V and KE = eV SS 1  KE = 8eV 2 = 0.005 nm–1 or 2 = 200 nm = 2000Å Comprehension–4 (ii) = hc 1 0 or   1 : 2 = 2 : 5 1 . Rate of production of B depends on the decaying 0 rate (A) : dNA  N A  N1 (iii) Metal 1 because 1 lies in visible wavelength range. dt hc h  4. We have eV0,    h    V0   B is decaying simultaneously with two rates  e e dNB  2N B  2N2 From graph,  dt  2V    2eV e dNB  3N B  3N2 2h dt Also slopes of the graph = 0.49  1015  e Number of nuclei of 'B' is = 1N1 – 2N2 – 3N2 2  1.6  1019  h 0.49  1015  6.536  1034 J / sec 2 . B will increase when N > ( +  )N 1 2 32 as initially N = N = N 1> 2 + 3 1 2 0 3 . N = 0, N =0 as both will decay completely : 5 . Maximum kinetic energy of photo electrons 21 1 hc N 2N02 Kmax = 2 mv2max=  – 0 3 = 2 therefore B is incorrect Now let 3000Å =  then 6000Å = 2  N3 hc 2   0 1 EXERCISE –IV(A)    v max 9  v2 max 2  hc 1 1 . Number of photons falling/s 2  0 1 7hc 7  6.62  1034  3  108  1.81 eV n  r2 [for point source]  0   16  3000  1010 16  1.6  1019 n  So for new distance n' = 6. KEmax = h   , Also Ephoto/time = N hc 9  Is  Is  18mA  2mA (i) When intensity of light is decreased numbr of 9 9 photons decreases but KEmax remains same Also saturated current  n (ii) When emitting surface is charged,  changes. So Vd is independent of n. KEmax changes . If emitter is changed then +ve no. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 18  0.22 of photoelectrons becomes zero. (i) VS = 0.6V (ii) Is= 0.6 2  2mA (iii) Increasing  , hc      12400 10 2 . E330 = 3300  3.7575eV 7 . Energy incident / area of sphere / sec = 4 0.12 12400 E220 = 2200  5.636eV Energy incident on atom Let work function be  = 10    0.05  109 2  6.25  10 19 J eV0 = (3.7575 – ) e, 2eV0 (5.6363–)e V0 = 1.87 V 4 0.12 Energy of one photon = hc  12.525eV  95

JEE-Physics No. of photon striking atom/sec 1 2 . 1 E  hc    3hc when E' = E, '= hc   = 6.25  1019  0.3125 3 E E3 12.525  1.6  1019  No. of photons/ area So new wavelength = 3 10 1 1020  10  10  4  0.12  12.525  1.6  10 19  80 1 3 . From the given conditions : 1020 E –E =(10.2+17)eV=27.2eV Since n = 1% n  n2 80 and E – E = (4.25 +5.95) eV = 10.2 eV n3 Equation (1) – (2) gives hh  8.  =de–broglie 2mK mv 2(  1  1   4 9  E – E = 17.0 eV or Z 1 3 . 6 ) = 17.0 3 2 v= u + at = 0   eE  t    h d h  Z2(13.6) (5/36) = 17.0  Z2 = 9  Z = 3  m  eEt  dt  eEt2 9 . From the figure it is clear that From equation (1) 1  1   27.2 (P+1). /2 = 2.5 Å Z2(13.6)  4 n2   /2 = (2.5 – 2.0)Å = 0.5 Å 1 1   2 7 .2  1 1 = 0.222 or  = 1 Å = 10–10 m.  (3)2 (13.6)  4  n2  4  n2 de–Broglie wavelength is given by h  h  1/n2 = 0.0278  n2 = 36  n = 6 p 2Km 1 1 1 1 R K = kinetic energy of electron B B 4 h2 (6.63  1034 )2  2.415  10 17 J 14.  R  2 2      2m2  2(9.1  1031 )(1010 )2   K   2.415 1017  eV = 150.8 eV 1  R 1  1   7R ; 7  1.6  10 19  p  9 16  144 B : p  36   10.(i) Kinetic energy of electron in the orbits of hydrogen 1 5 . Assuming Bohr's model to be applicable to the He and hydrogen like atoms = |Total energy| atom too;  Kinetic energy = 3.4 eV Energy of electron = –13.6 × 4 eV= – 54.4 eV (ii) The de–Broglie wavelength is given by initial energy of electron = 0 hh Energy of photon emitted = 544 eV   hc P 2Km    22.8mm E K = kinetic energy of electron Substituting the values, we have   (6.6 1034 J  s) 16. (i)Operating voltage = 40 kV, 0.5% energy for x ray Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 2(3.4  1.6 1019 J)(9.1 1031 kg)  99.5  n  e  V  720 100  = 6.63 × 10–10 m or  = 6.63 Å ehn 720 = 1.1 × 1017 1 1 . (i) Magnetic moment  IA  n  1.6 1014  0.995  40 103 4 m Produced is nth orbit eh 1  (ii) Velocity of incident e–  mv 2 = eV for hydrogen n =1  Magnetic moment 2e 4 m   eh ehB 2eV 2 1.6 10 19  40 103 (ii)   M  B  × B × sin30° = v=  4m 8m e m 9.1 1031 = 1.2 × 108 m/s 96

JEE-Physics 1 7 . Given K = 71 pm = 0·71 Å Now at t/2 time, active fraction = (e–t)1/2 = (0.64)½ = 0.8 hc 12400 eV – Å So decayed fraction in t/2 time is 0·2 or 20% K  17.46 keV EK – EL = = 0 71Å Thus EL = EK – 17·46 keV 24. U238 234 Th 24 He = 23·32 keV – 17·46 keV = 5·86 keV 90 92 m= (238.05079 – 4.00260 –234.0 4363) u 1 E = mc2 = 4.24764 MeV 1 8 . Total mass annihilated 2m e  c2 MeV If it emits proton spontaneously, the equation is not 1 balanced in terms of atoms & mass number. produced=mc2= c2 Total energy MeV  c2 =1MeV U238 237 Pa 14 H 91 92 Energy of 1 photon = 0.5 MeV m =(238.05079–237.065121–1.007834)u hc 1.2  10 12 = –0.022165 u    2.4pm  m is negative, so reaction is not spontaneous. E 0.5 19. 2 H 12 H 24 He Q 2 5 . Let at t = 0, capacitor starts discharging then at time 1 t, activity of radioactive sample = R = R0e–t Q = [ 4 × 7 – 2 × 2 × 1.1] MeV = 23.6 MeV charge on capacitor = Q = Q0 e–t/RC hc 12400 R 0 e –t e j= 2 0 .  min = eV = V Å Now R = Q0e–t RC R0 –t  – 1Rc Q e 12400 Q0 At 40 kV :  min = 40000 = 0.31 Å It is independent of time if  = 1 Wavelength of K is independent of applied RC potential. = t avg 20  10 –3 3 13.6  Z  12  E  hc R= 1 C = 100  10–6 = 200  C For K X–ray : 4 K 1216 EXERCISE –IV(B) Z 1 2 Å b gK = and given that 1 . =hf/2  =hf/3 1216 12 b gK =3 min  Z  1 2 = 3 × 0.31  Z  12  1216  1308  Z  1  36  Z  37 Potential at which electron stop coming out 0.93  hf  hf   2  from sphere–1, V =  hf 21. N = N e–t and N = N e–3t 1e 2e 1 01 2 02 N1  N01et = e2t  N =N , N1 e N 2 N 02e3t   01 02 for = ]  hf  N2 1  hf  3   e  2hf 1 from sphere–2, V=  3e 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65  1 = 2t  t= 2 After connection 22. Activity of x = Activity of y or n x = n y (i) V + V = V + V (V= final common potential) x y 12 L O L O0.693 hf 2hf 7hf NM PQ MMN PQP 0.6 9 3  2V =  V  2e 3e 12e n =n (T1/2 ) y x (T1/2 ) x y nx = (T1 2 )x 31 (ii) For sphere–2 : kQ  2 hf  7hf  hf  ny (T1 2 )y = 27 = 9 R  3 e 12e   12e Q hfR 2 3 . After t time active fraction No. of electrons flows n =  e 12ke2 N = 1 – 0·36 = 0·64 = N0 = e–t 97

JEE-Physics 2 . Power of source = 2J/sec Power of source = 5 × 10–3W Energy of 1 photon = hc 12400   5  103  4.15  1015   2.067eV 7.53  1019  1.6  6000 N2 No. of photons emitted /sec  1  1.5  1018 4.14  1.6  10-19 = 2  1019  6.0474  1018 N1 for 1W = 2.067 16 Current = 4.8 × 10–3A No. of photons striking on sphere of 0.6 m  4.8  103 6.0474  1018 / m2    1.6  1019  3  1016 = 4 0.62 No. of photons passing through aperture   3  1016  2  102  2% 1.5  1018 6.0474  1018  0.1  2 4.2  106  2  4 = 4 0.62    photons  2  4.15  1015  0.02  8.3  1013 No. of photons incident / area on screen (assuming  aperture to act as a secondary point source) Current  n e  13.3 A 4.2  1016 Also vmax 1650 = 2Vmax 5000  KE max 1650  4KE max 3000  4  4 5.42 = photon flux 4( 4.14 – ) = (7.53–) 16.56 – 7.53 = 3 = 1.1462 × 1014 photons / m2 sec 9.03 hc No. of photons incident on detector /sec   = 3  3.01eV  th  E  4126Å 0.50 hc 12400 = 1.1462 × 1014 ×  eV  3.1eV 4. 1 4000 10000 hc 12400   eV  2.48eV Photo current = N e  2.063  1010 A 2 5000 When the lens of f = –0.6 m is used, 1  1  1  1  1  v  0.3m hc 12400 v u f v 0.3  eV  2.06eV Thus image will be at 0.3 m from the lens in the 3 6000 direction opposite to the screen. Distance between screen & image = 5.7 m  Light having wavelength 6000 Å will not be able No. of photons striking lens to eject electrons. = No. of photons striking the aperture  = 1.05 × 1016 photons. Photons transmitted through the lens  Photo–current = (n1  n2 )e = 0.8 × 1.05 × 1016=0.84 × 1016 photon/s This new situation, A point source emitting where n 1 = no. of photons of light incident having 0.84 × 1016 photons/ sec of  = 6000Å wavelength 4000Å is kept at 5.7 m away from the screen. n 2 = no. of photons of light incident having Thus photons striking / area of screen wavelength 5000Å. Here n 1  hc   n 2  hc   n 3  hc   I0  3  10 3  103  1   2   3  3 3 0.84  1016 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 Area intercepted = 2cm2 = 2 × 10–4m2 = 45.72 = photon/sec =2.0574 × 1013 0.5  Photo–current = (n 1  n 2 )e Electrons emitted=0.9 × 2.0574 × 1013× (10 3  2  104 )(5000  4000)  1010  1.6  10 19 10000  6.63  10 34  3  108 Current = 1.4813 × 10–10 A = 0.144 A E = 2.067 eV ;  =1eV p K.E. = 1.067 eV; V = 1.067 V hh max S & 5. Momentum of particles = 1 2 hc Since particles are identical, mass of both particles 3 . Energy of photo with 1  3000Å  1  4.14eV are same. hc Energy of photo with 1  1650Å   7.53eV 2 98

JEE-Physics hh   v1 ˆi  v2 ˆj 1  1 v COM 2 2   mv = 1 & mv = 2 ; = k = k = 14.4eV= 23.04 × 10–19J 1 2   v1 ˆi  v2 ˆj    v 2  v 2   (iii) Emax = 4  1 = 13.5 eV v1COM 2 2 v1COM 1 2 v 2COM Emin = 4  3 = 0.7 eV 2 mv2 1 e2 h e2 r 4  0 r2  mvr= 2 – h 2h 2h 8.(i)   E = total 8  0    r mv m v 2  v 2 mv1 2  mv2 2 1 2 dE (ii) dt  P (loss of energy per sec)  2h  212 d  e2  P0  e2  dr P0 2 2 12  22 dt   8 0 r  r4  8  0 r2  dt r4  h    h         1  2 6. (i) E = 47.2 = 13.6 Z2  1  1  Z = 5 rt  4 9    e2 r2dr  8 0 P0 dt r0 0 1 1 6 P0 ( 4  0 )t  3cre2 t 1/ 3 25  9 16 e 1 r03 (ii) E = 13.6 ×    r3 = r3  2  r =r     0 0    13.6  25  7 eV = 16.5 eV (iii) For r=0, (to collapse and fall into nucleus) 9 16 1– 3cre2 t  0 r03 hc (iii) E1 = 13.6 Z2eV=  r03  10 30  10 10  100 sec 3 c re2 3  3  108  9  10 30 81 12400eV  A  t=   13.6  25eV   = 36.4Å nh mv2 4e2 m vrn  and  4  0 rn2 1 9.(i) 2 rn (iv) (KE) 1st orbit = mv2 2 0 h2  n2  n2 h2 0 9.1  10 31  (2.2  106  5 )2  rn  4 e2  m   4 0 0 m e 2  2  1.6  10 19 = 344 eV (PE) = –2KE = –688eV (iii) E nth   13.6   Z2m  1st orbit  n2  nh h L =  (n=1) 2 2 n2 0.53  12 E  13.6  42  1 0 0  1  1   408eV  22 42  (v) Radius = (R )  = 1.06×10–11m 0z 5 7 . Let n = no. of level of excited state 1 0 . Let the age of Earth be t and initially both were year present as N ; 0  nC = 6 (spectral lines) N N0 0 2 2  2t / t1/ 2 t / 4.5 103 Nu = Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65  n = 3 (3rd excited states) 238  No. of excited state level = n+1 = 4 2.7eV n=4 (Level B) Nu235 N0 n=3  2 t / 4.3 108 n=2 (Level A) n=1 Nu238 140  2 t1.18 109   (i) Principal quantum no. of initially excited level B =4 Nu235 1 t= 6.0418 × 109 years (ii) 2.7 = 1  1  k  16  2.7  14.4eV 1 1 . X  A   B k  4 16  3 No. of neucleus disintegrated = xt  x 1  et   Ionisation energy  99

JEE-Physics For 1 14. Let the amount of R 222 be N t x 0  N0 N0 22 4 Disintegration = x 1  1  e1   x N7.6 =   e Nuclei left after 7.6 day = N0 4 Energy released = E0x e Decay constant Energy utilized for melting = 0.5  E0x n2 0.693 e  = t1/2 3.8  24  3600 /sec E0x = 2.11 × 10–6 sec Mass of ice melted = 2eL F Activity  N0   No. of  particles 4 No. of neutrons produced 2 1 2 .2D  3 T + 1 P = N0  2.11  106  1.2  106 1 1 1 4  4000 Mass defect M =M –M N = 9.095 × 1015 nuclei= 3.354 × 10–6g Product Reactant 0  3.016049  1.00785  2 2.014102 = 4.023899–4.028204 1 5 . N = M e–t  m = 4.3 × 10–3 amu and 1 amu  931.5 10 E = mc2 = 4.01 MeV Thus momentum of this mass 3 T1 12 D 4 He 10 n = Mv, v= velcity of mass at time t = M e–tv 2 0 Let at time dt, defect mass is ejected with relative m(mass defect) = Mproduct – MReactant velocity v = [4.002603+1.008665] – [3.016049+2.014102] 0 v =eject/gram – v0 + v = [5.011268 – 5.030151] Linear momentum of M at time (t+dt) m = 0.018883 = (M–dm) (v+dv) + dm(v–v0) Since fext=0 E = m(931.5)  17.58 MeV (M–dm) (v+dv) + dm (v–v0)= Mv  (M–dm) dv = dmv0 E = E total  7.2MeV  dv  v 0 dm  dv  dm  v0   r v0 deutron 3 dt dt  M  dm  M  rdt M  dm Total Energy  n  m1  m 2 v  v 0n M0 Total M ass 312 D M0  rt  n   0.004305  0.018883   100 But M –rt = M e–t; v  v 0n M0  v 0t  3(2.014102)  00 M 0 e t = n = 0.384% 1 6 . Production of radioactive nuclei = /sec Disintegration= NA 1 3 . (Take a sample of 1020 Cm atoms) = dNA dt –decay Cm248  Pu244 + He4 At any time t    N A 96 96 2  NA dN A t   1  e t  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 Here m= (248. 072220 –244.064100 – 4.002603) u = 0.005517 u   dt  N A   E =mc2 = 0.005517 × 931 = 5.136 MeV 0   N A 0  t  N A    t  1 1  e t  Energy released in the decay of one atom   E = Efission + E = 0.08 × 200 + 0.92 × 5.136 Nuclei disintegrated =  = 20.725 MeV t    Total energy released from the decay of all 1020 atoms Total energy produced = E0  1  e t = 20.725 × 1020 MeV = 3.316 × 108J Energy used in water heating Power output = Total energy released  0 .2 E 0 t   1  e t   m sT  mean life 3.316  108 T  0.2E 0 t   /  1  et  = 1013 = 3.3 × 10–5 watts mS 100

JEE-Physics EXERCISE –V(A) 1 1     ATOMIC STRUCTURE & X-RAY 10. Rquired energy = 13.6 (Z)2  n 2 n 2  eV 1 2 1 . Energy required to ionize a atom from nth orbit is 1 1  12 32  13.6 = 13.6 (3)2  = 108.8 eV =+ eV n2 13.6 (4 )(3 ) E2  22 eV  3.4eV 1 1 . Number of lines = nC2 = 4C2 = (2) = 6 L2 (n)2 2 . Energy required to remove an electron from an orbit 1 2 . Energy = 2I  2(r2 ) 13.6 Z2 m1m2 where µ = reduced mass = m1  m2 is + eV . So, to remove the electron from n2 the first excited state of Li2+ is so energy = n22 (m1  m2 ) 13.6  32 2m1m2r2 E  22 =3.4 × 9 = + 30.6 eV 1 3 . Energy of radiation emitted 3 . Ionization potential will be lowest for the atom in E = h  E 0Z2  (n 1  1   E0Z2  2n  1  which the electrons are the farthest from the   1)2 n2   n2 (n  1)2  nucleus. So, the atom with the largest size will have the electron the farthest from the nuclei, hence to h  E0Z2 2n     1 remove the electron from this atom will be easiest.  n4  n3 So, the atom with least ionization potential is 133 C s . 55 PHOTOELECTRIC EFFECT 4 . The wavelengths involved in the spectrum of  deuterium 2 D are slightly different from that of W0  hf0  hc  WNa  Cu 1 0 WCu Na 14. Work function hydrogen spectrum; because masses of two nuclei are different. Na  WCu  4.5  2 6 . The energy of emitted photon is directly proportional Cu WNa 2.3 to the difference of the two energy levels.This differences is maximum between level (2) and level 1 5 . Covalent bonding electron cloud overlapping (1) hence photon for maximum energy will be region electron  pro bability density  liberated for this transition only. wave nature of electron  7. Whenever an atom gets de-excited from higher to 16. hf1  0  1 m v12 , hf2  0  1 m v 2  v12  v 2  2h f1  f2  the lower orbit, it emits radiation of a given frequency 2 2 2 2 m Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 E  hf  f  E 1 7 . Total change i n momentum from a reflecti ng The photons of highest frequency will be emitted, surface = 2P for the transition in which E is maximum, i.e., from 2 to 1. P 8 . k  mv2  mv2  k (independent or r) P rr n  h   m vr  r  n and T  1 mv2 is The momentum of incident radiation P 2E  2  2  c independent of n. 2E Hence, momentum transferred = c 101

JEE-Physics 1 8 . According to Einstein's equation of photoelectric 2 7 . 2d cos i = n  2d cos i  h V = 50 volt effect hf=W0+Kmax Kmax = (h)f–W0 meV Kmax 2 8 . 2d cos i = ndB q t hc 12400 W0 2 9 . E k    0  1.68  4000  0 By solving it 0 = 1.42 eV The slope of K vs f graph is h which is a fundamental 3 1 . No. of photons emitting per second from a source of power P is n = (5 × 1024) P constant and same for all metals at all intensifies. hc n P or nhc  W0  max 5  1024 P  1 9 . W0=4eV, max = ? ;   wavelength emitting    4eV  1240 nm  eV  hc= 1240 nm-eV  1020 5  1024  4  103 max   0.5  109 m  50Å max = 310 nm. And this wavlength comes in X ray region. 2 0 . Photocurrent  Intensity Radioactivity Nuclear Physics 1 N  1  n Intensity due to point source at r m is I  r2 N0  2  As distance is reduced to half, the photocurrent due 35.  where n are is number of half-lives. to point source will increase by a factor of 4. If T1/2 = 5 years then in 15 years, n=3 N  1  3 1 N0 N0  2  8 8 2 1 . The de-Broglie wavelengths and kinetic energy of    N  a free electron are related as  h  2  K1  1  2  1 3 6 . Though the compound can emit all the four particles 2mK 1 K2 2 2 namely electrons, protons, He2+ and neutrons. But the particle neutron can't be deflected in the 2 2 . According to Einstein's theory of photoelectric effect magnetic field, since it is a neutral particle. Hence, the deflectable particles are protons, electrons and Energy of incident photon = W0 + eV0 He2+. W0 = 6.2 eV; eV0 = 5 eV Energy of incident photon = 11.2 eV 3 8 . U238  U234 2 He4 Let recoil speed be V then by COLM The wavelength corresponding to this energy is 110 234V  4u  V  4u nm which falls in the ultraviolet region. 234 2 3 . The photoelectric phenomenon is an instantaneous phenomenon, hence the time taken by an electron to come out of metal is approximately 10–10 sec (found experimentally). 2 4 . With the increase in wavelength energy of the 3 9 . Rate of disintegration at any instant is directly Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 photon decreases. Therefore the KE of the electron proportional to the number of undecayed nuclei at coming out from the cathode also decreases. Due to which there will be a small decrease in plate that instant, i.e., dN current and once  becomes more than threshold   N wavelength electron will not come out from the dt cathode and hence current will become zero. Where  = decay constant We are given that at t=0 2 5 . The relation between energy (E) of a photon and dN disintegration momentum (P) associated with the photon is  dt  N 0  5000 minute E = pc at t =5 minute E hv dN disintegration  -  N  1250 dt minute The corresponding momentum p cc N  N 0e t 102

JEE-Physics On multiplying both sides by  we get 4 6 . Intensity of gamma radiation when it passes through (N) = (N0)e–t 1250 = 5000 e–t 1  e t x is I  I0ex 4  I0 I0 Taking logarithm on both sides we get 8  I0e36 ...(i) and 2  I0ex ...(ii) ln1–ln4=e–(5)  0–ln4=–5    1 ln 22 1  e 36 and 1  e x  e 3x  e 36  x  12 mm 5 82  = 0.4 ln2. N  1  n N0  2  4 0 . 8 have been emitted 47. We know that,  where, N are the 4– have been emitted 2+ have been emitted radioactive nuclei left after n-half lives  reduces atomic number by 2 N0 are the initial number of nuclei and n is number – increases atomic number by 1 + decreases atomic number by 1 N0 / 8 1 13  1  n So, Zeff = 92 – (8×2) + (4×1)–(2×1)= 96–18= 78 N0 8  2   2  of half-lives,     n 3 4 1 . A nucleus during decay emits (He2+),  (electrons),  or neutrino; it does not emit protons during decay. 15 Therefore T1/2 = =5 min. 3 4 2 . In order to fuse two nuclei against repulsion, the 4 8 . Radius of a nuclei  (Mass Number)1/3 repulsive potential energy has to be supplied by kinetic energy i.e., PE=KE R Te  125  1/3 5 5 R Al  27  3 3    R Te  R Al   6 fermi  7.7 × 10–14 = kT 3 2 4 9 . X n, 73 Li 3 1 n  A X 73 Li 24 He  7.7 × 10–14 = × 1.38 × 10–23T 0 Z 2 On conservating atomic number and mass numbers  T = 3.7 × 109 k on both sides, we get 4 3 . Applying conservation of momentum, we get A = (7+4) –1 = 10 Z = (3+2) – 0 = 5 m1v1  m2v2  m1  v2 1 Hence, A X 150 B m2 v1  Z 2 Also, R  m1/3  m  R 3 5 0 . Charge on -particle = 2e Charge on target nucleus = Ze m1 1  R1  3 R1 1 1/3 m2   R2  R2 2 When the -particle is projected towards the target       1 : 21/3 nucleus, then at r=r0, the -particle comes to 2   momentary rest. This position r0 from target nucleus is known as distance of closest approach. 4 4 . Given that 2 H1 2 H1 4 He2  energy Applying law of conservation of energy, we get Total B.E. of deuterium nucleus = 2.2 MeV Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 Total B.E. of He nucleus = 28 MeV 1 mv2  K ze2e  r0  1 , r0  Ze, r0  1 On conservating energy on both sides we get 2 v2 m (Energy)Deutron × 2 = (Energy)He+ Energy released r0 4.4 = 28 + E  E = 28 – 4.4 = 23.6 MeV 5 1 . 7 Li3 11 P 84 Be   The particle that will be emitted with Beryllium will 4 5 . At the closest point of approach be Gamma radiations. Initial KE = Final PE 5 2 . The energy spectrum of -particles emitted from a 5 × 106 × 1.6 × 10–19 kq1q2 radioactive source is = r0 N(E) 9  109  92  1.6  1019  2  1.6  1019 E r0  5  106  1.6  1019 E0 = 5.3 × 10–14 m = 5.3 × 10–12 cm. 103

JEE-Physics 5 3 . In a nuclear reaction the energy remains conserved 6 1 . ZXA Z–6( )A–12 2(+1 0) Z–8( )A–12  p7Li  2 4 H e 3(2 4) 3 2  No. of neutrons  A  12  Z  8  A  Z  4 energy of protons + 7(5.60) =2(4×7.06) No. of protons Z 8 Z 8 energy of proton = 17.28 MeV N 1 t/T t2 t1 N0  2 5 4 . Nuclear binding energy = 62.    1  1  T & 2  1 T [Expected mass of nucleus-Actual mass of nucleus]c2   2  3 2  Expected mass of nucleus = 8Mp + 9Mn Actual mass of nucleus = M0 3 Nuclear binding energy = [8Mp+9Mn–M0]c2 1 5 5 . Gamma ray is an electromagnetic radiation, due to 1 (t2 t1 ) T the emission of gamma ray, neither the mass number  1  2  1  (t2  t1) T = t2 – t1 not the atomic number changes. Though the 2 T daughter nucleus is same as parent nucleus but still there is a difference in the two, i.e., the daughter  t2 – t1 = 20 min. nucleus so obtained is present in one of the excited states and not in the ground state. 6 3 . 5m M+ mN m '  p p' 5 6 . Given that According to Conservation of linear momentum P' = P. Therefore ' =  0.693 1 1 6 5 . Released energy   Y 0.693 X = [1.6747 × 10–27 – 1.6725 × 10–27 – 9 × 10–31] Y X Y × (3 × 108)2 J = 0.73 MeV    T1/2 Tav  x  Y  1.44X  R Y  1.44 R X EXERCISE –V-B As decay rate of Y is more than the element X; hence Y will decay faster than X. 1 . The continuous X–ray spectrum is shown in figure. 5 9 . Total kinetic energy of products All wavelengths > min p2 p2 are found, where = Total energy released  E 2m 2m M  min  = (mass defect) c2 (where m = given) 2  p2   (M  m )   M  M    c2 12375  2  2m   2 2   min  Å V(in volt) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 p2 Here, V is the applied voltage. M  2    2   m  c2       2  2. 12375 12375  (in Å) = W (eV) = 4.0 Å  3093 Å   309.3 nm   = 310 nm 2  M  2  2 v  2 m 3. Number of nuclei decreases experientially   m  c2 v c MM  dN  6 0 . Because energy is releasing Binding energy per N = N e–t and Rate of decay   dt  =N nucleon of product > that of parent  E2 > E1. 0 Therefore, decay process lasts upto t=. Therefore a given nucleus may decay at any time after t=0. 104

JEE-Physics 4 . Since, the wavelength () is increasing we can say 9 . Radius of a nucleus is given by that the galaxy is receding. Doppler effect can be R = R A1/3 (where R = 1.25 × 10–15 m) given by– 00 = 1.25 A1/3 × 10–15m Here, A is the mass number and mass of the '   cv  706 = 656 cv ...(i) uranium nucleus will be m  Am p cv cv m = mass of proton = A(1.67 × 10–27 kg) p c  v  706 2 Density  = mass m A (1.67  1027 kg)  c  v   656   1.16    c + v = 1.16 c – 1.16 v volume 4 R 3 A (1.25  1015 m )3 3    2.0 × 1017 kg/m3  v  0.16c 0.16  3.0  108 m/s  2.16 2.16 1 0 . Energy is released in a process when total binding energy of the nucleus (= binding energy per nucleon v  2.2 × 107 m/s × number of nucleons) is increased or we can say, If we take the approximation then equation (i) can be when total binding energy of products is more than the reactants. By calculation we can see that only in v case of option (c), this happens. Given : W  2Y written as  =   c  ...(ii) Binding energy of reactants = 120 × 7.5 = 900 MeV      706  656 (3  108 ) m/s and binding energy of products From here v     .c  656  = 2(60 × 8.5) = 1020 MeV > 900 MeV.  v = 0.23 × 108 m/s Which is almost equal to the previous answer. 11. In hydrogen atom E =  Rhc . Also E  m n n So, we may use equation (ii) also. n2 5 . Atomic number of neon is 10. where m is the mass of the electron. By the emission of two – particles, atomic number Here, the electron has been replaced by a particle will be reduced by 4. Therefore, atomic number of whose mass is double of an electron. Therefore, for this hypothetical atom energy in nth orbit will be the unknown element will be Z = 10 – 4 = 6 given by E =  2Rhc Similarly mass number of the unknown element will n be A = 22 – 2 × 4 = 14 n2  Unknown nucleus is carbon (A = 14, Z=6). The longest wavelength max (or minimum energy) photon will correspond to the transition of particle 6 . From law of conservation of momentum, from n=3 to n=2. P = P (in opposite directions) hc 1 1 18 12  max  E 3  E 2  Rhc  22  32  max = 5R h  Now de–Broglie wavelength is given by p n  11 12.  KE  (with positive sign) [h = Plank's constant] n n2 Since, momentum (p) of both the particles is equal, Potential energy U is negative and therefore 1 = 2   /  2 = 1 1 1 Ze2  . Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 7 . Both the beta rays and the cathode rays are made up U  1    1 of electrons. So, only option (a) is correct. n rn  U n 4  0 rn  n2  (b) Gamma rays are electromagnetic waves. [because r  n2] (with negative sign) n (c) Alpha particles are doubly ionized helium atoms and Similarly total energy E  1 (with negative sign) (d) Protons and neutrons have approximately the n same mass. n2 Therefore, when an electron jumps from some 0.693 1 excited state to the ground state, value of n will  decrease. Therefore, kinetic energy will increase 8. (t ) = (t )  x y   = 0.693  (with positive sign), potential energy and total energy 1/2 x mean y xy x < y  Rate of decay = N will also increase but with negative sign. Thus, finally Initially number of atoms (N) of both are equal but kinetic energy will increase, while potential and total since y >  , therefore, y will decay at a faster rate energies will decrease. x than x. 105

JEE-Physics 1 3 . Minimum wavelength of continuous X–ray spectrum q ne it 12375 20. i    n  is given by min (in Å) = E(in eV ) tt e Substituting i = 3.2 × 10–3A Here, E = energy of incident electrons (in eV) e = 1.6 × 10–19 C and t =1s = energy corresponding to minimum wavelength min of X–rays  E = 80 keV = 80 × 103 eV we get, n =2 × 1016 12375  n 2 1 . R  R 0  2  ...(i)   (in Å) = 80  103  0.155 min Also the energy of the incident electrons (80 keV) is Here R = activity of radioactive substance after n more than the ionization energy of the K–shell electrons (i.e. 72.5 keV). Therefore, characteristic half–lives  R0 (given) X–ray spectrum will also be obtained because 6 energy of incident electron is high enough to knock out the electron from K or L–shells. Substituting in equation (i), we get n =4  t = (n)t = (4) (100 s) = 400 s 1/2 14. N x1 (t) 1 N 0e 10t 1 2 2 . During–decay atomic number (Z) and mass number N x2 (t)   N 0e t  (A) does not change. So, the correct option is (c) because in all other options either Z,A or both is/are e e changing. (Initially both have same number of nuclei say N ) r dU  eV0 0   e =–t/e–10t  e = e9t,  = 10 and  =  x1 x2 1 23. U = eV = eV n  r0   | F | dr r  9 t=1  t = 0 9 this force will provide the necessary centripetal force. 1 5 . Energy of infrared radiation is less than the energy of ultraviolet radiation. In options (a), (b) and (c), energy mv2  eV0 v eV0 ...(i) released will be more, while in option (d) only, energy Hence, m released will be less. rr 1 6 . Wavelength k is independent of the accelerating nh voltage (V), while the minimum wavelength c is Moreover, mvr  ....(ii) i nversely propor t ional to V. T herefore, as V is increased, k remains unchanged where as c 2 decreases or k – c will increase. Dividing equation (ii) by (i). 1 7 . During –decay, a neutron is transformed into a  nh m proton and an electron. This is why atomic number We have : mr =  2  eV0  rn  n (Z = number of protons) increases by one and mass number (A = number of protons + neutrons) remains  m2  m2 unchanged during beta decay. 24. (r ) = z  (0.53 Å) = (n × 0.53)Å  =n m  z 18 . The total number of atoms can neither remain constant (as in option a) nor can ever increase (as in m =5 for Fm257 (the outermost shell) and z=100 options b and c). They will continuously decrease 100 with time. Therefore, (d) is the appropriate option. n  (5)2 1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65  100 4 2 5 . Nuclear density is constant hence, mass  volume m  V. 1 9 . In second excited state n=3, 2 6 . Given that K + K = 5.5 MeV ...(i) 12  3h From conservation of linear momentum, p = p 12 So,  = =  2  2K1 (216m )  2K 2 (4m ) as P  2Km H Li while So, E  Z2 and Z = 1, Z = 3  K = 54 K ...(ii) H Li 2 1 |E | = 9 |E | or |E | < |E | Solving equation (i) and (ii). Li H H Li We get K = KE of –particle = 5.4 MeV 2 106

JEE-Physics 2 7 . Saturation current is proportional to intensity while h 3 5 . Momentum of striking electrons p  stopping potential increases with increase in  frequency. Hence, f = f while I < I  Kinetic energy of striking electrons ab ab h 28. 1  2mE 1  E1/2 p h2 2 hc or 2 K = 2m  2m2 This also, maximum energy of X–ray photons. E 2 9 . Activity reduces from 6000 dps to 3000 dps in Therefore, hc h2  0 2m2c 140 days. It implies that half–life of the 0  2m2  radioactive sample is 140 days. In 280 days (or h 1 3 6 . Rest mass of parent nucleus should be greater than two half–lives) activity will remain th of the the rest mass of daughter nuclei. 4 hc 3 7 . As for continuous X–rays min= eV so cut off initial activity. Hence, the initial activity of the sample is wavelength depends on the accelerating potential 4 × 6000 dps = 24000dps and is independent of nature of target. 3 0 . The first photon will excite the hydrogen atom (in ground state) in first excited state 38. Activity = N & T = n2 (as E –E = 10.2eV). Hence during de–excitation a 1/2  21 n2 n2 photon of 10.2eV will be released. The second T1 (2N 0 ) T2 (N 0 ) photon of energy 15eV can ionise the atom. Hence the balance energy i.e.,(15 – 13.6)eV = 1.4 eV is retained by the electron. Therefore, by the second photon an electron of energy 1.4 eV will be released. 12 12 So 5 = & 10 = T1 = 4T 1  1  2 1  (Z 1)2  1  Z2  1  Z2  31.  2   Z1     11  4 Solving this, we get Z = 6 3 9 . 550 nm light cannot emit electron for metal of work 2 function   3eV , so saturate current decreases for P to Q to R. Also|V |>|V |>|V | 3 2 . 4( He4) = O16 28 sp sq sr Mass defect 4 0 . Energy incident = Power(time) m = {4(4.0026) – 15.9994} = 0.011 amu = (30 mW) (100 ns)  Energy released per oxygen nuclei = 3 × 10–9 = (0.011) (931. 48) MeV = 10.24 MeV So momentum = E 3  109  1017 kg m / sec . 3 3 . After two half lives 1/4th fraction of nuclei will remain c  3 108 undecayed. 3/4th fraction will decay. Hence, the probability that a nucleus decays in two half lives is 3/4. 3 4 . The series in U–V region is Lymen series. Longest wavelength corresponds to minimum energy which occurs in transition from n=2 to n=1. 1 MCQ Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 122  R 1 . Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus   1 1  ...(i) is always less than the sum of masses of its  12  22  constituent particles. 20 N e is made up of 10 1.0 The smallest wavelength in the infrared region corresponds to maximum energy of Paschen series. protons plus 10 neutrons. Therefore, mass of 20 N e nucleus, 10 1  R M < 10 (m + m )   1 1  ...(ii) 1 pn  32   Also, heavier the nucleus, more is the mass defect. Solving equation (i) and (ii), we get  = 823.5 nm Thus, 20 (m + m ) – M > 10 (m + m ) –M np 2 pn1  10 (m + m ) > M – M 1  M < M + 10(m + m) p 1 p n n 2 2 Now since M1 < 10 (mp + mn)  M2 < 2M1 107

JEE-Physics 2 . Time period, nh  3h 2 2 2 rn rn 5 Angular momentum =  n = 3 T= vn (in nth state) i.e. T  n n n n2 32 1 Also r = 2 a0  9a0  Z a0  Z2 2 But r  n2 and n  n n For de-excitation Therefore, T  n3 Given T = 8T 1  Rz2 1  1   4R 1  1  n n1 n2        Hence, n = 2n  n 12 n 2  n 2 n 2 12 2 1 2 For n = 3 to n = 1: 3 . From the relation, 1  4R 1  1  9 hc  hc  1   1 9  32R eV =     V   e      e For n = 3 to n = 2: This is equation of straight line. 1  4R 1  1    9   4 9  5R hc Slope is tan  = e For n = 2 to n = 1: hc hc hc 1  4R 1  1    3 1:  : 3 = : : =1:2:4  1 4  R 01 02 03 2 1  0.001nm 1 Match the column 2 . (i) Energy of capacitor is less 1/2 cv2 therefore p. 01   = 10000 Å 01 (ii) work is done on the gas hence energy increases therefore q. 1  0.002nm 1 (iii) When mass decreases its energy increases. (iv) when current flows energy of magnetic field is 02   = 5000Å generated therefore t. 02 (B) work is done on the gas (C) mass is reduced and mass defect is converted 1  0.004nm 1  03 = 2500 Å into energy. 03 (D) mass decreases due to mass defect. Violet colour has wavelength 4000 Å. Matching List Type 1 . Completing reaction in list II So violet colour can eject photoelectrons from metal–1 and metal–2. 4 . (A) For 1 < A < 50, on fusion mass number for (1) 15 O 175 N  + decay) compound nucleus is less than 100 8  Binding energy per nucleon remains same (2) 238 U 29304 Th  (Alpha decay)  No energy is released 92 ( B ) For 51 < A < 100, on fusion mass number (3) 185 Bi 18824 Pb  (Proton emission) for compound nucleus is between 100 & 200 83  Binding energy per nucleon increases (4) 239 Pu 15470 La  (Fission) 94 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65  Energy is released. Comprehension Based Question (C) For 100 < A < 200, on fission, the mass Comprehension#1 number of product nuclei will be between 50 & 100 1 . 10.2eV –3.4eV –3.4eV 10.2eV –6.0eV  Binding energy per nucleon decreases –13.6eV H –13.6eV  No energy is released –54.4eV ( D ) For 200 < A < 260, on fission, the mass He+ number of product nuclei will be between 100 & 130 For H atom E = 10.2 eV, This goes to excite He+ ion from n = 2 to n = 4   Binding energy per nucleon increases   Energy is released. 108

JEE-Physics 2 . For visible region E < 10.2 eV and in this case Comprehension 4 E = –3.4 – (–6) = 2.6 eV I2 nh / 22 hc 12400eV  Å  4.8× 1. En  1 I2    n2h2 = 2 2I 2I 8 2 I =  4800Å 10–7 m E 2.6eV  3 . 2 2. h  E2  E1  h  h2 4  1 K1  Z1 n1 8 2 I Ratio of kinetic energy K2 Z2 n2 2  h  h2 4  1 8 2 I Since n = n = 2 & Z = 1 12 1 for H, Z = 2 for He+  K1 1 I 3h  3  2  1034  1.87  1046 kgm2 2 K2  82  82  4  1011 4 Comprehension#2  3 . Moment of inertia of CO molecule about centre of 1 . Due to the high temperature developed as a result mass : I  r2 where = m1m 2 of collision & fusion causes the core of fusion reactor m1  m2 to plasma. 2.  3 KT 2 = Kq1q2 3KT = Kq1q2 r  I I 1.87  10 46  2 r r  m1m 2  12  16  5 / 3  10 27  m1  m2 12 16 T = 1.44  109  1.3  1010 m 4  1015  3  8.6  105 Paragraph 5 T = 1.39 × 109 K 1  109  T  2  109 K 210 206 24 84 82 1. Po Pb He Q 3 . nt = 8 × 1014 × 9 × 10–1 = 7.2 × 1014 > 5 × 1014  Total energy released = MPo  MPb  MHe C2 0 nt = 4 × 1023 × 1 × 1011 = 4 × 1034 > 5 × 1014 = [(209.982876) – (205.974455 + 4.002603)] × 0 932 MeV nt = 1 × 1024 × 4 × 1012 = 4 × 1036 > 5 × 1014 = [0.005818] × 932 MeV = 5.422376 MeV 0 Kinetic energy of  particle = Comprehension#3 1. E = p2 .... (i) h  A 4  Q   206  5.422376 MeV = 5.319 p = ... (ii)  A   210  2m  By equation (i) and (ii) MeV = 5319 KeV h2 h2 (n2 ) 2 . Only in option (C); sum of masses of product is less E=  2m (4a2 ) then sum of mases of reactant 2m2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 for reaction 2 H 24 He 36 Li . As M < M + 1 Li Deutron [ n = a for stationary wave M 2 alpha on string fixed at both end] E  a–2 Subjective 1 . (i) Let at time t, number of radioactive nuclei are N. h2 (6.6  10 –34 )2 1 Net rate of formation of nuclei of A 2m4a2 2(1.0  1030 )  4  (6.6  109 )2 e 2. E   dN N dN t =dt N0   N dt   N E = 8 meV  dN  0 =–N  dt h h hn Solving this equation, we get 3. mv = v =  v n  m m (2a ) 1 N =  [–(–N0)e–t]...(i) 109

JEE-Physics n(2) E53 = E – E = 3.84 eV 5 3 (ii) (i) Substituting  = 2N0 and t =t = in E52 = E – E = 11.5 eV > 4 eV 1/2  5 2 E43 = E – E = 2.64 eV 4 3 3 equation (i) we get, N = N E42 = E – E = 10.2 eV > 4 eV 4 2 20 Hence, the energy of emitted photons in the range (ii) Substituting  = 2N0 of 2eV and 4eV are and t   in equation (i), we get 3.3 eV during combination and N=  = 2N  N = 2N 3.84 eV and 2.64 after combination. 0 0  2 . Given work function W = 1.9 eV 3 . Let ground state energy (in eV) be E Wavelength of incident light,  = 400 nm 1 Then, from the given condition E – E = 204 eV 2n 1 hc  E1  E1  204eV  Energy of incident light, E   3.1eV 4n2  (Substituting the values of h, c and ) 1   E1  4n2  '1 = 204 eV...(i) Therefore, maximum kinetic energy of photoelectron K = E – W = (3.1 – 1.9) = 1.2 eV E1 E1 4n2 n2 max Now the situation is as shown below : and E – E = 40.8 eV   = 40.8 eV 2n n e–  n=5  3  Kmax=1.2eV E5= –2.1eV  E1  4n2  = 40.8 eV...(ii) -particles He+ in fourth 1 excited state or From equation (i) and (ii), 1  4n2  5 He+ n=5 3 (Z=2) 4n2 Energy of electron in 4th excited state of He+ (n=5) 1 15 4 Z2  1  4n2  4n2  n2  1  n=2 From equation number (ii), will be E 5  13.6 n2 eV (2)2 E = –(13.6) (5)2 =2.2 eV 5 44 E1   n2(40.8) eV = – (2)2 (40.8) eV Therefore, energy released during the combination 33 = 1.2 – (–2.2) = 3.4 eV Similarly energies in other energy states of He+ will  E = –217.6 eV  E = – (13.6)Z2 1 1  Z2 = E1 2 1 7 . 6 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65   16  Z=4 (2)2 13.6 13.6 be E= –13.6 (4)2 = –3.4 eV 4 E =E –E min 2n 2n–1 (2)2  E1  E1 2  E1  E1  7 E1 4n2 (2n  1)2 16 9  E = –13.6 (3)2 =–6.04 eV 144 3 (2 )2 7 E =–13.6  13.6eV    144  (–217.6) eV 2 22 The possible transitions are  E = 10.58 eV min E54 = E – E = 1.3 eV < 2 eV 5 4 110

JEE-Physics 4 . Energy of incident photon, (iii) The population of X at this moment, E =10.6 eV=10.6× 1.6× 10–19J = 16.96 × 10–19 J N = N e–Xt = (1020)e–(0.1) (16.48)  N = 1.92 × 1019 i X 0 X Energy incident per unit area per unit (intensity) = 2J N= NXX [From equation (iv)] Y  No. of photons incident on unit area in unit time Y 2 (0.1)  16.96  1019  1.18 × 1018 = (1.92 × 1019) (1 / 30) = 5.76 × 1019 Therefore, number of photons incident per unit time N = N – N – N = 1020 – 1.92 × 1019 – 5.76 × on given area (1.0 × 10–4 m2) Z 0X Y = (1.18 × 1018) (1.0 × 10–4) = 1.18 × 1014 1019 But only 0.83% of incident photons emit photoelectrons  N = 2.32 × 1019  No. of photoelectrons emitted per second (n) Z 6.(i) Given mass of – particle, m=4.002 amu and mass of daughter nucleus M=223.610 amu, de–Broglie wavelength of –particle, n   0.53   1014 ) n = 6.25 × 1011 =5.76 × 10–15 m  100  (1.18 So, momentum of –particle would be K = 0 and K = E – work function h 6.63  10 34 min max i p    5.76  1015 kg–m/s = (10.6 – 5.6) eV = 5.0 eV  p= 1.151 × 10–19 kg–m/s  Kmax = 5.0 eV From law of conservation of linear momentum, this should also be equal to the linear momentum the 5.(i) Let at time t=t, number of nuclei of Y and Z are N daughter nucleus (in opposite direction). Y and N . Then Rate equations of the population of Z X,Y and Z are  dNX  =–X N ...(i) Let K and K be the kinetic energies of –particle   X 1 2 dt and daughter nucleus. Then total kinetic energy in  dNY  = X N – YN .. .( ii ) p2 p2   X the final state is: K = K1 + K2 = 2m  2M dt Y  dNZ  = YN .. .(i ii) p2  1 1  p2  M  m      2  m  M  = 2  Mm  dt Y 1 amu = 1.67 × 10–27 kg Substituting the values, we get N0X X  Y (1.151  1019 )2  M  m   (ii) K  2   M  m  Given N (t) = eYt  eXt Y For N to be maximum dN Y (t)  0 Y dt i.e.  N =  N ... (iv) (from equation (ii)) p2 (4.002  223.610)(1.67  1027 ) XX YY  N0X 2 (4.002  1.67  1027 )(223.61  1.67  1027 ) X  Y Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65  X (N0eXt )  Y [e Y t  eXt ] 10 12 1.6  1013 K  1012 J  MeV = 6.25 MeV X  Y eYt  1 X  e(X Y )t  K=6.25 MeV  Y  eXt  Y (ii) Mass defect, m= 6.25 amu = 0.0067 amu  (X – Y)t n (e) = n  X   t  1 n   X  931.470  Y   Y   Y  X Therefore, mass of parent nucleus Substituting the values of X and  , we have = mass of –particle + mass of daughter Y 1  0.1  nucleus+mass defect (m) 1  / 30  t  (0.1 / 30 ) n 1  15n(3)  t=16.48 s. = (4.002 + 223.610 + 0.0067)amu = 227.62 amu 111

JEE-Physics 7 . The reactor produces 1000 MW power or 109 W (i) Number of photoelectrons emitted upto t=10 s are power of 109 J/s of power. The reactor is to function (ii) for 10yr. Therefore, total energy which the reactor (No. of photons falling will supply in 10yr is E = (power) (time) in unit area in unit time) n   (area  time) = (109 J/s) (10 × 365 × 24 × 3600 s) = 3.1536 × 1017 J 106 But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it 1 or 3.1536 × 1018 J. One uranium atom = 106 [(10)16 × (5 × 10–4) × (10)] = 5.0 × 107 liberates 200MeV of energy or 200 × 1.6 × 10–13 or 3.2 × 10–11 J of energy. So, number of uranium At time t=10s charge on plate A, atoms needed are q =+ne=(5.0 × 107) (1.6 × 10–19) = 8.0 × 10–12 C 3.1536  1018 A n = 3.2  1011 = 0.9855 × 1029 or number of kg–moles of uranium needed are and charge on plate B, q = (33.7 × 10–12 – 8.0 × 10–12) = 25.7 × 10–12 C B (qB  qA )  Electric field between the plates E = 2 A 0 0.9855  1029 E=2 (25.7  8.0)  1012  2  103 N n  6.02  1026  163.7  (5  104 )(8.85  1012 ) C Hence, total mass of uranium required is (iii) Energy of photoelectrons at plate A m = (n)M = (163.7) (235) kg = E – W = (5–2) eV = 3eV  m  38470 kg  m = 3.847 × 104 kg Increase in energy of photoelectrons 8.(i) Total 6 lines are emitted. = (eEd) joule = (Ed) eV = (2 × 103) (10–2) eV = 20eV n(n  1) Therefore 2  6  n=4 Energy of photoelectrons at plate B = (20 + 3)eV = 23 eV So, transition is taking place between mth energy 1 1 state and (m + 3)th energy state. E =–0.85eV  10. E = h = Rhc (Z–b)2  n12 m  Z2  n 2  2 Z  –13.6  m 2  =–0.85  = 0.25...(i) m 1 1 Similarly E = –0.544 eV For K– series, b=1   = Rc(Z–1)2  n12  n 2  m+3 2 z2 z Substituting the values,  13.6 (m  3)2  0.544  (m  3)  0.2 ..(ii) 1 1 Solving equation (i) and (ii) for z and m. 4.2 × 1018 = (1.1 × 107) (3 × 108)(Z–1)2  1  4  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 We get m = 12 and z=3  (Z–1)2 = 1697  Z –1  41  Z = 42 (ii) Smallest wavelength corresponds to maximum 11 . Maximum kinetic energy of the photoelectrons would be K = E – W = (5 – 3) eV = 2eV difference of energies which is obviously Em+3 – Em. max   E = – 0.544 – (–0.85) = 0.306 eV max Therefore, the stopping potential is 2V. Saturation current depends on the intensity of light incident. hc 1240 When the intensity is doubled the saturation current  min =  E max  0.306 = 4052.3 nm. will also become two fold. The corresponding graphs are shown in figure. 9 . Area of plates A = 5 × 10–4 m2 distance between the plates d = 1cm = 10–2m 112

JEE-Physics 1 3 . Wavelengths corresponding to minimum wavelength (min) or maximum energy will emit photoelectrons having maximum kinetic energy. ( ) belonging to Balmer series and lying in the min given range (450 nm to 750 nm) corresponds to transition from (n=4 to n=2). Here, 13.6  0.85eV 13.6  3.4eV E = – (4 )2  E = (2)2 4 2 1 2 . Let n be the number of radioactive nuclei at time  E = E – E = 2.55 eV 0 4 2 t=0. Number of nuclei decayed in time t are given by n (1–e–t), which is also equal to the number of K = Energy of photon – work function 0 max beta particles emitted during the same interval of time. For the given condition, = 2.55 – 2.0 = 0.55 eV n = n (1–e–2) ...(i) (n + 0.75n) = n (1–e–4) ...(ii) 1 4 . Let N be the initial number of nuclei of 238U. 00 0 Dividing equation (ii) by (i), we get After time t, N = N 1n U 0   1  e 4 2 1.75= 1  e 2  1.75 – 1.75 e–2 = 1 – e–4 Here n = number of half–lives 3 1  1.75 e–2 – e–4 = ...(iii) t 1.5  109 1 13 4  4.5  109  Let us take e–2 = x = t1 / 2 3  N = N  2  Then the above equation is x2 – 1.75 x + 0.75 = 0 U 0   1 1 / 3  N 1  2   1.75  (1.75)2  (4)(0.75) 3 and N = N – N =    x  x=1 and Pb 0 U 0 4 2 3  1 1/ 3  From equation (iii) either e–2 = 1, e–2 =  2   NU  13  3.861 4 NPb but e–2 = 1 is not accepted because which means 1   2  3 1 5 . (i) From the relation r  A1/3 =0. Hence, e–2 = Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 r1   A1 1/3   A2 1/3  (14 )1 / 3 4 We have r2  A 2   4   – 2 n(e) = n (3) – n(4) = n(3) – 2 n(2)   = n (2) – 1 n(3)  A = 56 2 2 Substituting the given values, (ii) Z = A – no. of neutrons = 56 – 30 = 26 22 1  = 0.6931 – × (1.0986) = 0.14395 s–1  f = Rc 1  1   3Rc (Z–1)2 ka (Z–1)2  12 22  4 2 Mean–life t = 1 Substituting the given values of R, c and Z.  mean = 6.947 s We get f = 1.55 × 1018 Hz  ka 113

JEE-Physics 1 6 . For 0 < x < 1, PE = E  h  0  e  e  Kinetic energy K = Total energy – PE 19. eV = h –   VS    1 s 1  h = 2E – E = E  2m(2E0 ) ...(i) 0 0 0 vs For x > 1, PE = 0  Kinetic energy K = Total energy =2E 2 0 h   2  2m (2E0 ) ...(ii) From equation (i) and (ii), we have 1  2 2 17. h h  p  2mqV  slope = = constant  ratio = 1  e  P  mq  4  2  8  3 2 0 . Given T = 1386 sec.  mPqP ½ 18. dN  N 0 e t  n dN  n N 0  t dt dt dN  103 dt Fraction  n dN n N0 1 = N 0  N  N 0  N 0e t  1  e t dt 2 N0 N0   t    slope   year 1    1 yr 1 =  ln 2 80 2 1  e1386 4  1  1  4  = 4  4% 100  100 = 1  e100  t1/2  0.693  1.386 given time is 3 times of t 1/2   value of p is 8. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\05.Modern Physics.p65 114

JEE-Physics UNIT # 02 (PART – III) ELECTRONIC (SEMICONDUCTOR & LOGIC GATE) EXERCISE –01 EXERCISE –02 17. I I = Vin 5 = 250 mA 1 .  diode is in F.B. =   = 50 × 1  103 0 in R in 90 100 30V 10R 30V 10R 23. I = 90% I = I  I = × I = 11 mA A A C E 100 90 C 10R  E E 10R 5R I = I – I = 11mA – 10 mA = 1mA B B BEC n 2 (1019 )2 V =  5R  × 30 = 10V i = 1023 = 1015 AB  5R  10R  25. n= h ne V 0.1 I 500 IL 2 7 . E = d = 106 =105 V/m 2 . 10V I1 1k 5V 28. n = n 2  (1019 )2  1017 m 3 5V 10  5 e i 1021 I = = 10 mA nh IL = = 5mA, 500 1k I = I – I =10 mA – 5mA = 5mA 1L 3 1 20 2 9 . I = 100 = 20 mA  0.98 = 49 55 30 32.  =1 = I= = A 20 5V 1  0.98 3. 20  30 50 IC 5. R0 24 3 4 .  = IE  IC =  IE = 0.95 × 100 mA = 95mA Voltage gain= R in = 6 × 3 = 48 IC IE  IB 25 1 24 6. IC 3 5 .  = IE = IE ==  = IB  IC = IB = 49 × 5 µA = 245A 25 25 I = I + I = 5 + 245 = 250 A EBC 4 9 . Y = A  B = A.B = A.B = AND gate VBC 0.5 V 7. R= IC = 0.05mA = 10 k 0 7 2 . R = 2hR = 2  240  6.4  106 = 55 km 100 100 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\05 Semiconductor.p65 13. I = 90% I  I = 90 ×I = × 20 = 22 mA C 90 CE E 75. F =9 Nmax =9 1011 =2 MHz I = I – I = 22 – 20 = 2 mA C BEC 11 7 6 . N = fc2 = (10  106 )2  1.2 × 1012 m–3 14. = = enee 1.6  1019  1013  1200 max 81 81 = 520.9 /cm 5 15.  = ene e  ne = = 8 3 . R= 2Reh = 2 × 6400 × 103 × 500 m = 80 km ee 1.6  10 19  5000 = 6.25 × 1015 cm–3 70 E

JEE-Physics 1 6 . I = V  Vb  5 × 10–3 = (2  0.5 )  R = 300  Y  A.A  A.B  B.A  B.B RR Y  A.B  A.B 12400(eV  Å) 12400 when A = 1, B = 0 17. E =   = 0.75 = 16500 Å (Å ) then Y = 1.0  1.0 = 0.0 + 1.1 = 1 when A = 1, B = 1 V0 R L 2 0 . Vin =  R in then Y = 1.0  1.1 = 0.1 + 1.0 = 0+0=0 when A = 0, B = 0 then Y = 0.0  0.0 = 1.0 + 0.1 = 0+0=0 = RL 10k × 1mV = 1V  V  × V = 100 × 0 R in in 1k EXERCISE –04 VEB 7 9. A  hfe   50  C = –  1  hoe R L  = –  1  25  106 21. R= IB = 35  106 = 200 k  (103 )  b = – 48.78 VCC  VCE 10  5 12400(eV  Å) 12400 22. I = = 1k = 5 mA 1 2 . Energy band= = =0.5 eV C RC (Å) 24800 VCC = IBRB = IC RB IC 5.488  103  1 5 .  = IB = (5.6  5.488)  103 = 49  RB  .VCC  100 10  200 k 17. I = neAv  v  I IC 5mA d d n 2 4 .  = ene(e + h) vd1 I1 × n2 7 × 5 = 5 = 1.6 × 10–19 × 1.072 × 1010 (1350 + 480) vd2 = I2 n1 = 4 7 4 = 3.14 × 10–6 mho/cm 3 1 . Y = (A+B).A = A.A + AB = A+A.B = A(1+B) = A 18. I = 12 =2A 4 2 3 3 . Y = (W + X)·(W + Y) = W.W + W.X + W.Y + X.Y 1 = W + W.X + W.Y + X.Y 3 2 . For how pass filter, frequency greater than can't = W(1+X+Y)+X.Y=W+X.Y C pass through filter 3 4 . Y = (A + B) . ( A.B ) = (A + B) . ( A  B ) 11 = A. A + B. A + A. B + B. B  f <  f < = 4 × 104 Hz = 40 KHz = A. B + B. A = XOR gate C RC in given options maximum frequency below 40 KHz is 10.62 KHz 4 3 . Y = ( A + B).A = A. A + AB = AND gate 3 3 . LED is use in forward bias so I–V graph is to increase frequency of light emitted from LED; potential node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\05 Semiconductor.p65 barriar to diode is increased so A A A.P=Q Y=Q.R P Q 49. B A A.B=P R BP B P.B=R Here Y  Q.R  Q  R  A.P.  P.B 71 Y = P.(A + B) = A.B  ·(A+B) Y = A  B  ·(A + B) E

JEE-Physics UNIT # 12 (PART - II) PRACTICAL PHYSICS EXERCISE -I D2h V 2D h 1 8 . V = r2h    4 V Dh K m v 2 . K  mv2    V  100     0 .0 1    0.1   100 = 3% V 2  2.00     K mv  5.0  % error in kinetic energy = 2 + 2(3) = 8% mm    m  3 a 1 9 . Volume of 25 spheres = 25 × 1.76 = 44.00 Density  = v  a3 m a 3.  42 2 0 . T = 2 g  g = T 2 ab2 X  a  2 b c  4. X= c3  X   a b  3 c     g  2 T = 2x + y g T D  R 5. D = 2R  DR V  R  100  V  100  I  100 IR V I 21. R= 6 . (ABC)  A  B  C = a + b + c  5  100  0.2  100 =7% ABC A B C 100 10 8. T  2   g  4 2  V  b t g T2 V  b t 22.  V = bt      g   2 T V = (10.0) (1.00) (0.100) = 1.00 cm3 g  T P P F P  V = (1.00) 0.10  0.01  0.001   0.03cm 3 1 0 . F  A  2  F  P  2  10.0 1.00 0.100  4 2 3 . Thickness of the wall 1 1 . Area(A) = 4R2, Volume(V) = R3 = (4.23 ± 0.01)–(3.89 ± 0.01)=(0.34 ± 0.02) cm 3 2 4 . Area of disk = R2 = (3.14) (1.2)2= 4.5216 = 4.5cm2 A  2 R , V  3 R A RV R 25.  V = bt  V    b  t 1 2 . Average time period V  b t  2.63  2.56  2.42  2.71  2.80 V  100   0.01  0.01  0.01   100 5 V  15.12 10.15 5.28  13.12 = (0.066 + 0.098 + 0.189) %  = 2.624= 2.62 s = (0.07 + 0.10 + 0.19)% = 0.36% 5 Average absolute error Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\06.Practical physics.p65  0.01  0.06  0.20  0.09  0.18  0.54 26. T= rhg  T  r  h  g 55 2 Trhg = 0.108 = 0.11s T  0.01  102 0.01  102 0.01  1 5 . X = MaLbTc T× 100 =   × 100  1.25  10 2 1.45  10 2 9.80   X  a  M   b  L   c  T  = a+b+c = 0.80 + 0.69+ 0.10 = 1.59% = 1.6% X  M   L   T  2 7 . T  100  1 / 5  100  0.8% 1 6 . Circumference  = 2R = D    D T 25 17. = m    m  2 r     × 100 2 8 . Area = 1.2 × 2.345 = 2.884 = 2.9 cm2 r 2   m r   0.003 0.005  0.06  100 =4% 2 9 . 1 MSD – 1VSD = Vernier Constant = ×100+2× 0.3 0.5 6 115

JEE-Physics 3 0 . Least count = 1MSD – 1VSD 1   1   10  90   29  (1+ 1T)  = (1 + 2T)  but 30 VSD = 29 MSD  VSD =  30  MSD Therefore L.C. = 1 MSD – 29 M SD   30  1 + 1T – 10 = 1 + 2T + 90  MSD  0.5   1  1'  (1 – 2)T =    = 9 (1 – 2)T 30 30  60  9 3 1 . Reading = 2.30 mm 3 7 . Deflection is zero for 324 so value of unknown 3 2 . Least count = 1 MSD – 1 VSD 324 resistance = 100 = 3.24  1 1  (N–1) MSD = N(VSD)  VSD =  N  MSD Comprehension Least count = 1MSD – 1  1  MSD 0.5 N  1 . Least count = 50 = 0.01 mm 1 1 2 . ID = 321 + 7(0.5) + 17(0.01) = 324.67 mm =  N  (1mm)= 10N cm 3 . Zero error = –(50–41) (0.01) =–0.09 mm 3 3 .  L.C. = 1 MSD – 1 VSD Subjective Questions 1 . Least count = 1MSD – 1VSD m  m  0.08  0.02 = 0.1 – n n 9 = 0.5 – 10 (0.5) =0.05 mm 3 4 . L.C. = 1MSD – 1VSD but (N) MSD = (N+M) (VSD) 2 . Least count = 1MSD – IVSD  L.C. = 1MSD –  N N m  MSD     cos = –  = m   1  MSD  1  cos   N  m   N   cos    m 1  =  For minimum least count, m should be minimum so m=1  1mm  3 5 . Let 1 MSD = S & 1 VSD = V 3 . Zero error = –(50 – 44)  50  = –0.12 mm Given, 10V = 9S or S = V...(i) 1 Thickness of plate = 3+26× +0.12mm=3.64mm Also given (N× S( +2 (S–V) = NS' + 2 (S'–V') 50 or (N+2) (S'-S) = 2 (V'-V) 4. 100  1110   = 40m A 0.45 400 2 i AA i or S1T =   V 2 T  5 .  = 2i – 2r = 2i – A 22 N 2 1   A 60  30 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\06.Practical physics.p65 2  2  V  2  9  1.8 i= 2 = 2 =45° or   N 2   S N  2 10 N  2 A But sin i = sin 2 36. X  90  X = 10  sin    A  = sin A 10 90  2  2    2  1 0(1  1T )  90(1  2T ) sin   A  10   90       2 A sin  1  1 2 1 10  1 90   (1+ 1T)   = (1 + 2T)   116

JEE-Physics     cos ec A  cos    A     B |v|  2   2   2  = cos ec 30  cos 45   1 = 1   5 . 2ƒ 2  2  180 ƒ   100    1 100  5 %   2  18 ƒ 2ƒ |u| 2 180 6 . Index (Bench) error 7. Least count = 1mm = 0.01 mm = observed distance – Actual distance 100 Diameter of the wire = 0 + 52 × 0.01 mm = 0.52 = (xL –xo)– 10 cm = –0.2 cm mm = 0.052 cm u = (x – x ) – (–0.2) = 10.8 –(–0.2) = 11 cm 8. 1VSD  29  0.5   29o and Lo 30  60  v = (x –x ) – (0.2) = 22.5– 11.4 – 0.2 = 10.9 cm 1o  30  o Lo  2   60  111 1 1 1MSD   f  v  u   11  f = 5.5 cm and 10.9  1 o f u v  0.1 0.1  Least count = 1MSD – 1VSD =  60  f2  u2  v2  f =  (11)2 (1 0 .9 )2   [5.5]2=0.05  1 o  Reading = 58.5 ° + 9x  60  = 58.65°  f = (5.5 ± 0.05) cm 3 9. R  V  R  V  I  3%  3%  6% 7 . S = x cos  = (2) cos 53° = (2)  5  = 1.2 I R VI S  x  (cos )  x + (tan )   S = 1.2 EXERCISE -2(B) S x cos  x 0.2   4  2    1. V =a3 = (1.2 × 10–2)3 =1.728× 10–6m3 =1.7 × 10–6m3  3    2 1 80   = 0.12 + 0.06 = 0.18   m  m r   S =(1.2 ± 0.18) cm 2. = r 2     m  2 r     100  0.003  100  2  0.005  0.06  100  4%  0.3 0.5 6 EXERCISE -2(A) 4. v = f  = 2f ( – ) 2 1 3 . Diameter = M.S.R. + C.S.R × L.C. + Z.E. = 3 + = 512 × 2 (63.2– 30.7) × 10–2=332.8 ms–1 35 × (0.5/50) + 0.03 = 3.38 mm v     2  1  v     2  1    Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\06.Practical physics.p65 4 . Least count of vernier callipers  v  v   2  1   332.8   0.1  0.1   2.048 m/s L.C. = 1MSD – 1VSD   2  32.5     1 29 5. From 1 1 1 11 1  f = 5cm But here 29 MSD = 30 VSD  1 VSD = MSD f     30 v u 10 10 5 From graph u = 0.1 cm, v = 0.1 cm 29 1 1 f v u  L.C. = 1MSD – MSD  = MSD = × But f2  v2  u2 30 30 30  1 o so f=  0.1  .1  (25) = 0.05 0.5° =  60 =1 minute.  100 100   f = (5 ± 0.05) cm 117

JEE-Physics 42 g  2T cos  6 . g = T2  g    T Here T  2 sec as  increases, sin2  decreases so For (A) g  0.5  0.2(0.1) = 0.6 Absolute error |(d)| decreases g 1 2 Also fractional error For (B) g  0.5  0.2 =0.6 d   cos    g 2  2 sin2      cot   g 0.02 d For (C) g  0.5  2 =0.51 2 sin  For (D) g  0.1  0.1 = 0.15 as  increases, cot decreases, so fractional error g 2 decreases g MCQ's  g is minimum for (D). Also number of observations are maximum in (D). 11 1 1 . By using mirror formula   0.5 7 . least count  = 0.01 mm vu f Set 1  u = – 42 cm  v = – 56 cm 50 Set 2  u = – 48 cm  v = – 48 cm Diameter of sphere Set 3  u = – 60 cm  v = – 40 cm = 2 × 0.5 + (25 – 5) × 0.01 = 1.2 mm Set 4  u = – 66 cm  v = – 37.7±0.2 33 cm Set 5  u = – 78 cm  v = – 34.67±0.2 39cm F Mg (1)(9 .8 )(2 ) 2 . For a longer air column, absorption of energy is 8 . Y  A  A  (0.2  103 )2 (0.8  103 ) = 1.95 × 1011 = 2.0 × 1011 Nm–2 more. Due to end correction   e   . 4 Y  () 2 r  0.05  2  0.01 Y  r 0.8  0.4  Subjective 1 . Least count = 1MSD – 1VSD = 0.0625 + 0.05 = 0.1125 na Here n(MSD) = (n+1) (VSD)  1 VSD = n 1 Y = (2.0 × 1011) (0.1125) = 0.2 × 1011 na a  Least count = a–  n 1 n 1  Y = (2 ± 0.2) × 1011 Nm–2 F 4F 2. Y =  1 0 . Since object and image move in opposite direction,  D 2  D 2 the positioning should be as shown in the figure.  4   Object lies between focus and centre of curvature f < x < 2f Y   2 D  () 0.1  2  0.001   0.001 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ Y  D  = 110  0.050  0.125 Image Maximum percentage error Object right Y  100  1 +4+0.8 = 4.89% left Y 11 4 . Least count = 1mm = 0.01 nn 1 1 . 50 divisions = 2.45 cm 100 Diameter D=1mm+(47) (0.01)mm=1.47mm Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\06.Practical physics.p65 2.45 = 0.147 cm  1 division =  0.049cm 50 Curved surface area = 2r = D  least count = 1 MSD –1VSD = 0.05 – 0.049 = (3.14) (0.147) (5.6) = 2.6 cm2 = 0.001 cm. So vernier reading = 0.001 × 24 = 0.024 cm. 5 . Least count ofverimer callipers Therefore diameter of cylinder 1 9  10  = 5.10 + 0.024 = 5.124 cm. =  mm = 0.1 mm  Side of cube = 10mm + 1 × 0.1 mm 12. d  = 10.1 mm = 1.01 cm 2 sin   d   2   cos    constant Density 2.736  sin2    (1.01)3 = 2.66 g/cm3 118

JEE-Physics UNIT # 04 (PART–II) FLUID MECHANICS EXERCISE –I 1 5 . Increase in energy (  ) 1 F2L (5  10)2  0.2 2.5×10–5 r 1  10 2   2  10 4 = J 3 .     2  0.8 = 0.004 radian 2 AY  1011 4 . Stress () = F 1 7 . Fex = 2T = 2 × 7.5 × 1.5 = 22.5 N A (for breakingthecopperstressshould besamei.)e.18. Fex = 4rT T  Fex  4  1N/m 4 r 4 1  F1  F2  F  F2  F2  4F 1 9 . Initial surface energy ( )    A1 A2 AR2 4 R 2 = 2×T × 4r2 = 8r2T 6.   FL    L AY r2 Final surface energy (  )  8 . Volume = constant  A × L = constant = 2 × T × 4(2r)2= 32 r2T So energy neeeded (   ) = 32 r2T – 8r2T= 24r2T  A  1  = FL    L L2 2 0 . SE = 4R2T (n1/3–1) L AY A D2 = 4 T [(27)1/3–1] = 2D2T 2 4 2 9.  = FL  1  L1  r22  1  1   1 22. 4T 4T AY 2 L2 r12 2   Pexcess1  R 1 ; Pexcess2  R 2   FL (1  10)  1.1 Pexcess 1  R2  R2 1.01 1 = AY  1  10 6  1.1  1011   10.  = 0.1 mm  Pexcess 2 R1 R1 1.02 2 1 1 . Increment in length due to own weight v1  R 3  8 v2 1 1 ()  So ratio of volume R 3 2  = mgL gL2 1.5  9.8  (8  102 )2 2T   2  5 108 2 3 . Pin = Patm + r 2AY 2Y 2  70  103 = 1.013 × 105 + 10 3 =1.0144 × 105 Pa =9.6 × 10–11 m V r1 r2 V r2  r1 V 0.004 P    2 4 . 12.  V  100  B=  P = B    rcommon = r2  r1 here r1 = r2, so rcommon =    V   V  2 5 . rnew  r12  r22  32  42  5cm  0.004  g1 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 = 2100 × 106 ×  100  = 84 kpa 2 9 . at moon g' = 6 , height h  g P hg 200  103  9.8 30. h= 2T cos   h1  T1  d2 60 0.6 9 13. K   rdg h2 T2 d1    V   V   0.1   V   V   100  50 0.8 10   = 19.6×108 N/m2 31. geff = 0, h  2T cos  rdg 14. W= 1 F = 1 F 2L  W  L; W1  1  1 so water rises maximum height i.e. length of the 2 2 AY W2 2 2 capillary = 30 cm (  =30 cm) 62

JEE-Physics 3 3 . Density of water at 4°C is maximum so water rises 4 2 . Pressure at point A = Pressure at point B in capillary is minimum by h  2T cos  (A=B) rdg  hoilg = 25 cm waterg (4°Ch2Trdcogs)hhe=igh2t05d.8if1feren3c1e.25 cm h oil 34.   h  h  2h ( ) cos  cos 45 = 31.25 – 25 = 6.25 cm 3 5 . Mass of water M = volume × density = r2h 4 3 . Total force = P × A = hg × (h × L) M2 2r 2 M1  2 hr = constant  M  r  1  103  9.8 r = × (1× 2) = 9.8 × 103 N 2 2 AT 2  10 2  70  10 3  28N 4 4 . Work = PV = (3 × 105 – 1 × 105) × 50000 = 1010J 36. F =  t 0.05  103 4 5 . Barometer read atmospheric pressure. 37. 2T 2T 2  75 cm ()  hg= r h= rg  0.05  101  1  1000 =30 44 46. P1V1 = P2V2  (Patm + hwg) 3 r3=Patm × (2r)3 3 3 8 . Let mass of gold is m then mass of copper =210–m  hwg = 7 Patm (m210–m)  Patm = Hwg  hwg = 7Hwg  h = 7H upthrust = loss of weight (=  )  = 210g –198g  Vinwg = 12g  Vin = 12 cm3 4 8 . Weight = upthrust  mg = (3 × 2 × 10–2)× 103× g  m = 60 kg Total volume ( ) = m  210  m  12  m  210  m  12 4 9 . Upthrust = Vinwg = 100 g-wt gold cu 19.3 8.5 weight of water and jar= weight + Th = 700 + 100 = 800 g-wt  m = 193. 5 0 . In balanced condition ( )   So weight of gold ( )=193g V 3 9 . Force on bottom = Pressure × area Mg = Th  6g = 3 wg...(i) and (6+ m)g = V wg ...(ii) ( =× )  from equation (i) and (ii) 18 = 6+m  m = 12 kg  d2  51. Density of metal= wA  210 =7g/cm3 =hg ×  4  ...(i) wA  ww 210 180 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 force on vertical surface (  ) = Pressure × area (× )   hg    2 dh  = h2g  d ...(ii) density of liquid =  2   2  2  according to question () = wA  wL  210 120  90 =3g/cm3 wA  ww 210 180 30 d2 h2g  d h d hg ×  2 42 5 2 . Let mass of cube is m and side is a then 4 1 . Pressure on the wall (  )= hg (ma) 2 (m+200)g = a3wg ...(i) Net horizontal force (  ) m g = a2(a–2)wg ...(ii) = P × area = hg  (h)  h2g  a2(a–2) w + 200 = a3 w 22  a2 = 100  a = 10 cm 63

JEE-Physics 5 5 . Reading of spring (  )  EXERCISE –II = Mg – Th = Mg – Vinwg = 12 – 1000  10 6 × 103 × 10 = 7N F  W1  W   4  2 1. Stress ()= = AS 5 8 . Force due to pressure difference = P× A ()  In balanced condition ( )   = mg = P× A mg 3 104 10 Lw  P= = = 2.5 kPa A 120 5 9 . For horizontal motion (  ) L/4 w1 11 P1 + 2 V12=P2+ 2 V22 2. Tension in wire at lowest position 1 ()  3 × 105=105 + 2 × 103 V22  V22 = 4 × 102   V2 = 20 m/s T = mg+m2r So elongation ()  = FL (mg  m2L )L dV r 4  6 0 . dt  8  AY r 2 Y 6 1 . Velocity of efflux ( )  3. Tension in wire (  )=mg = 10N so elongation ()  = 2gh  2  10  5 = 10m/s F.L 10  3 rate of flow  = AY  106  10  1010 = 0.3 mm = Av = (1 × 10–4)× 10 = 10–3 m3/s 6 3 . V22 = V12+2gh=(2)2 + 2 × 1000 × 5.1 × 10–1=1024 Spring constant of wire ()=YA V2 = 32 cm/s 4. L 6 4 . Rate of flow ( )=Av = r2 × 2gh So effective spring constant ( )  = 3.14 × 1 × 2  1000  10 = 444 cm3/s YA k 6 7 . Viscous force ( )= 6rv k1k2 L  kYA = 6 × 3.14 × 18 × 10–5 × 0.3 × 10–1 × 102 = k1  k2 k  YA kL  YA = 101 .73 × 10–4 dyne L 6 8 . m1 = m2  V1d1 = V2d2 Time period ()  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 Rate of flow dV1  r 4  n1  V2 t1  d1 t1 m m(kL  YA) dt 8  n2 V1 t2 d2t2  2  2 k eff kYA 7 2 . Radius of big drop () FL  L AY F/A Y R = (n)1/3r = (2)1/3r v 'T R2 5.     = Slope of curve vT  r2 vT  r2   (2)2 / 3  41/ 3  v'T = 41/3 × 5 cm/s  L  (4  2)  103  1  10 9 Y (8000  4000)  103 2  L = 1  Y = 2 × 109 N/m2 64

JEE-Physics Acceleration ()a = F 1 1 . Potential energy ( ) m 6. H H g = mg =(r2h)g = (rh)2 2 22 dx TT F according to Zurin law rH = constant  u1 = u2 x 13 . For uniform radius tube in balanced condition mx F Fx ()  then tension in dx = =  m  r Extension in dx element = Tdx Fxdx 2T AY  AY h1 h1  rg  Fxdx F but weight of liquid in tapered tube is more than uniform tube of radius r then for balanced condition total extension  = 0 AY  2AY (r 7.  1 = 2  F1L 1  F2L 2 ) A 1 Y1 A 2 Y2 2T  F1  30  102  F2  20  102  F1 32 h < h1  h < rg 16  2 106 10 106  15 F2 1 4 . For spring balance A (A) in balanced condition F + 2F = 5000 g 12 = Mg –Th = 2g – Th 5000 kg for balance B (B) = Mg + Th = 5g + Th F2 F1 F2 1 5 . Due to extra water, extra upthrust act on the steel ball so ball move up. 2 15  F1 + 32 F1 = 5000 g  F1 = 2580 g (  )  So stress in steel rod ()  1 7 . Acceleration of ball in water (  )  F1 2580g = 161.2kg/cm2 = A1  16cm2  net force  Th  mg  V(d  D)g  (d  D)g m m VD D Velocity at the surface (  ) 8. S(urfacetensiondoesnotdependonsurfa)cearea.v = 2ah  2 (d  D) gh D node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 When ball come out from water then g act on the 9 . In balanced condition ( )   ball so height in air ( mg = 2rT  g)  2r mg 75  104  12.5  102 m h' = v2  2(d  D)gh  d  1 h =T  2g D  2g  D 6  102 2T 2T 2T  1 1  1 8 . Let V1 volume of the ball in the lower liquid then 1 0 . h = h1 – h2 = r1dg  r2 dg  dg  r1  r2  (V1)  V  g = V12g + (V – V1) 1g  2  72  2  2  = 0.293cm Vg(– 1)=V1g(2 – 1) V1    1  1   1  980  0.5 1  V 2  1 1  2 65

JEE-Physics 1 9 . When the ball is pushed down, the water gains 2 3 . Given COD=Q potential energy, whereas the ball loses potential energy. Hence, gain in potential energy of water (    ) ( V)rg   V   3 r  g B  2  8  (When half of the spherical ball is immersed in O 3r 45° water, rise of c.g. of displaced water = ) 45° 45° 8 (  A ) G1 G2 = Vrg 1  3   4 r3rg  13  13 r 4g 16  3 16 12 m1g m2g DC 4 G & G be the center of gravities of two liquids, Loss in PE of ball = V'rg = r4'g 12 3 then ( )  13 4 AOC = 90° = COB Work done = r4g – r4'g AOG1 = 45° 12 3 G1OD = 45°– COG2 = 45°  r 4g 13  4 ' G2OD = 45 + 12 3 Net torque about point O is zero    13 4  5  rm g sin(45°–) = rm2gsin(45+) 12 3  12 1  r 4g   0.5  r 4g svsin(45–) = vsin (45+) 2 0 . According to equation's of continuity s  sin 45    sin 45   () A1v1= A2v2 v  R2 s sin 45 cos   cos 45 sin  (R2) v = n(r2)v'  v' = n  r    sin 45cos  cos 45 sin  s    cos   sin   cos   sin  s   cos   sin   cos   sin  2 1 . P = (g sin  – µgcos) ...(i) h g sin–g cos P = g cos  h ....(ii)  s    tan  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65s Both should be same h   s    = tan  –  tan  = tan  –   s      tan 1 2 2 . Velocity of efflux of water ( )  2 4 . From right Limb = 2g  h   gh PA = Patm + hg h 2   2  A Force due to ejected water    PC = PA + a  2  + (2) a () 2 C B dp dm =P + 3 a = P + hg + 3 a ....(i) =  v=(av)v=av2 A 2 atm 2 dt dt Torque of these forces about central line From left limb PC = Patm + 2gh ....(ii) ()  P + hg + 3 a = P + 2gh h = 3a  atm 2 atm 2g = F × 2R + F × 2R = 4av2 × R = 4 aghR 66

JEE-Physics 2 5 . In pure rolling acceleration of the tube = 2a Transferred length of liquid () () = h – h1  h2  h1  h2 P = P + (2a)L (from horizontal) 12 2 A atm (from vertical) Transferred mass ( ) PA =Patm + gH A gH =  h1  h2  A.  a = 2L  2   Fb Loss in gravitational potential energy 2 6 . Torque about CM Fb  4 = I cm ( )   = Fb  (r2g)  r22g = mgh =  h1  h2 2 Ag  2 mg   4I 4I 4I Mass of the entire liquid () 27. v0  2gh, v  h  v0 = (h1 + h2 + h) A 2g If this liquid moves with a velocity v then its KE 2 42 (v)  2 8 . Rate of flow = Av Volume of water filled in tank in 15 s 1 = 2 (h1 + h2 + h)Av2 15 V  0 A  10 1  sin  t  dt   h1  h2 2 Ag  1 30   2  2 (h1 + h2 + h) Av2  10A  t  cos  / t 15  10A 1 5  30  g  / t 0    v  2(h1  h2  h) (h1 – h2) height of water level = V  15  30  m 34. v1  2gx and v = 2g(x  h) . 10A   2 Let cross section area of hole is a then rate of flow 29 . The free liquid surface between the plates is (a)=av cylindrical and curved along one axis only so radius of curvature (force = v(av) =av2  F1 = av12 and F2 = av22 x v1 h ) Net force   v2 d s 2s 2s = (F2 – F1) = a (v22 – v12) = a(2g (x+h)–2gx) r= and P0 – P =    P = P0– d = 2agh 2 r d 30. As the cork moves up, the force due to buoyancy 36. In floating condition () remains constant. As its speed increases, retarding weight = upthrust (=) force due to viscosity increase. The acceleration is variable, and hence the relation between velocity   A L  Dg   A L 2dg   A 3L  dg and time is not linear.  5   5 4   5 4  (   d  3d  5d D 2 4 4 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 40.Viscous force = weight (  = ) )  vv  A = mg sin   a2 = a3g sin  tt 31. PC – PA = a a+ndPaB = PC + h g agt sin  PB – PA = hg v   3 2 . When the levels equalize then the height of the liquid 4 1 . Net viscous force (  ) in each arm = h1  h2 dv Fv 2 = 2Fv = 2A dx Fv (F=1N1=2 × (0.5) × 0.5 1.25  102  = h1  h2 ) 2   = 2.5 × 10–2 kg–s/m2 67

JEE-Physics 4 3 . vT  (B – L) EXERCISE –III  v 'T  (10.5  1.5)  9  1 v T (19.5  1.5) 18 2 Comprehension # 1 v T  0.2 = 0.1 m/s 1. dy  ax  g / 2  1 22 dx ay  g g / 2  g  v'T = ....(effective g will be g – a = g/2)  = 45° mg 2 . As the slope of free surface is 45°. 4 6 . Tension in B = TB = 3 4mg (   45° ) Tension in A = TA = TB + mg = 3 y(0,4)  TA = 4TB (2,2) FT 2m Stress = A  r2 1m 45° Wire breaks when stress = Breaking stress for rA = rB (4,0) x 6m (=rA=rB ) Thus free surface passes through centre of box and  sA = 4sB having co-ordinates (2,2) at top of box.  A breaks before B ( TB    (2,2)) for rA = 2rB  sB = rB2 sA  TA  4 TB Length of exposed top part = 6–2=4m. rA2 (2rB )2 (   )   stresses are equal so either A or B may break (AB) 3. P = Pa + gh =105 + 1000 × 10 × 1 4 7 . The angle of contact at the free liquid surface inside =(105+104) N/m2 =0.11 MPa the capillary tube will change such that the vertical component of the surface tension forces just balance 4 . p = (105 + 103 × 10 × 4)N/m2 the weight of the liquid column. = [0.1 + 0.04] MPa = 0.14 MPa ( 5.Asmaximum slope of free surface is 1/3 for the ) condition of non-exposure of bottom of box, then ( 5 1 . If one surface is pushed down by x the other surface  1/3 ) moves up by x.( x x ) ax 1 as ax = g/2, 3ax = ay + g  Net unbalanced force on the liquid column ay  g 3 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 () =2xAg a =g/2, thus g/2 upward. y mass of the liquid column = A ay   –2x Ag = (A)a  a =   2g  x 2 ax    6  a = –2x    2g 2  2   T =  2g 68

JEE-Physics Comprehension # 2 2 . Rewriting the equation as 1. F = A (V – 0)2[1– cos 180°] P+ (20)g = P + 10(1.5)g + h(2)g. 0 0 0 = 2Av2=2 × 1000 × 2 × 10–4 × 10 × 10 = 40N From here we can see that h will decrease. 2. F = 2A (V –u)2 u = speed of cart Comprehension # 4 0 u du 2A t 1 . When the string is cut, tension becomes zero i.e.,  du dt net upward force on the block becomes W/2 or net upward acceleration of the block will become m dt = 2A(v0–u)2; 0 g/2 or 5 m/s2. 0 (V0  u)2  m 2A 2  103  2  104 4  (      m 10 100   W/2  u    g/25m/s2 )  1  2At     V0  u 0 m 2s 22  2 s Now, t   55 11 = 2At  4t a  m 100 ...(i) V0  u V0 1 4 11 2 . If weight is doubled then obviously upthrust will also  become two times, because weight can be increased at t = 10 sec  V0  u 10 10 2 only by increasing the volume by two times. When two out of three forces acting on the block have V – u = 2 u = 8 m/sec. doubled then tension will also become two times 0 to keep the block in equilibrium. 3. F = 2A ( V – u)2 ( 0   = 2 × 103 × 2 × 10–4(10–8)2            ) =2 × 103 × 2 × 10–4 × 4 F a   0.16 m/sec2 M 1 1 4t 1 1 4t     F 2F =3W 4. V0  u V0 100 8 10 100 2 4t W T= W 2W T'   , t = 1.6 sec. 2 80 100 W 3W Before F = W +  5 . F = 2A(V0–u)2 = 2 × 103 × 2 × 10–4 × 25 = 10N 22 P = F.u = 10 × 5 = 50 W. node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 Comprehension # 3 After 3W = 2W + T'  T' = W = 2T 1 . Equating the pressures at the same level of third When string is cut in second case, net upward liquid at the boundary of first and third liquids on acceleration will be (       left hand side.    ) 3(2WW/2gW)  g , 2 (       )  so time taken will not change. Pressure on left hand side = pressure on right hand ( ) side (  = )   P + (20)g = P + 10(1.5)g + h (2)g. 0 0 Solving this equation, we get h = 2.5 cm 69

JEE-Physics Comprehension#5 1 . When the muscles of the heart relax, as they do 4 . F = 6rv= 6× 10–3× 10–3× 3 = 5.65 × 10–5N during diastole, the heart is not exerting any force on the blood. 5. 6rvT = mg (     diastole mg 105  9.8 v T  6r  6   10 3  10 3 = 5.2 m/s       )   2 . Volume flow rate (  ) EXERCISE –IV (A)  Pressure difference ()  1. (i) Material A has greater value of Young's  (Radius of vessel)4 (  )4 modulus. Because slope of A is greater than B. If radius is increased by 10% volume flow rate would (A be increased by a factor (1.1)4  1.44. AB (10% (ii) A material is more ductile because there is (1.1)4  1.44      a large plastic deformation range between the elastic limit and the breaking point. 3 . Gravitational potential energy ( ) (A         energy   volume = (gh) (volume)   vo lum e      )   PE = 1050 × 9.8 × 0.3 × 8.0 × 10–6 (iii) B material is more brittle because the plastic = 2.46 × 10–2J region between the elastic limit and breaking 4 . W = mgh = (200 × 10–6 × 1050) (9.8) (0.5)  1.0J point is small. (B      5 . Power = blood pressure  volume of blood pumped  )    time  (which blood is pumped) (iv) Strength of a material is determined by the amount of stress required to cause fracture. Factor by which power increased = 7 × 1.2 = 8.4, Material A is stronger than material B. 20% increases means increase by a factor of 1.2. ( (      =7 ×1.2= A 8.4, 20%   1.2  ) B) Comprehension # 6 2 . (i) The area of the hysteresis loop is proportional to the energy dissipated by the 1. Q  1 wh en volume flow rate is multiplied by material as heat when the material  undergoes loading and unloding. A material for which the hysteresis loop has larger area density, it becomes mass flow rate. Both rates are would absrob more energy when subjected inversely proportional to . to vibrations. Therefore to absorb vibrations 2. From Q = R 4 (P2  P1 ) one would prefer rubber B. node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 8 L (    we have,   R 4 (P2  P1 )  8LQ   Substituting the value we get   4  103 Pa  s   3. From R = 2 vR v  R e B) e ; 2R (ii) Rubber A, to avoid excessive heating of the Flow remains laminar till R = 2000 car tire. e 4  10 3  2000 ( A )  v  2  1000  8  103 = 0.5 m/s 70

JEE-Physics 3. Maximum stress = F m(g  a) W 2W  from equation (i) T + 2T = W  T= , T= Area rm2 in 1 1 1 3 2 3 3 900(9.8  2.2) for rotational equilibrium  rm2 in   108  W 2W Tx = T (2 – x)  x 3 (2  x) 1 2 3  rmin  900 12 4 3 108 = 6mm x= m from steel wire 3 FL (4  6)  10  1.5 8 . In equilibrium mg = 2T  r2g = 2T 4 . ()steel = AY  (0.125  102 )2  2  1011 = 1.49 × 10–4m FL 6 10 1 r  2T 2  0.045 ()brass = AY  (0.125  102 )2  0.91  1011  3.14  8.96  103  9.8 = 5.7 mm = 1.31 × 10–4 m g  diameter = 2r = 1.14 mm 9. 4T T  hwater gr r =h water g  4 5. (a) F ' = m'a =  A L d  F  8  101  1  980  0.35  2  m   = 68.6 dyne/cm 4   A L d  dALg   ALdg  2  2  ALd  4 1 0 . P V + P V = PV 11 22 F' F' F    4T  4 R 3    4T 4 R 3 x  P R1  3 1  P R 2  3 2 F ' Ldg   P  4T 4 R3  P  4 R 3  4 R 3  4 R 3  Stress =   R  3  3 1 3 2 3  A4 Y= stress stress Ldg  4T 4 R 2  4 R 2  4 R 2   strain =  3  3 3 1 3 2  strain Y 4Y 6 . Compressive strength  Fmax  V = 44 4 Area 3 R3 – 3 R13 – 3 R23 and  Fmax = 7.7 × 108 × 3.6 × 10–4 = 2.772 × 105N S = 4R2 – 4R 2 – 4R 2 12  applied force < F  bone will not break. 4T max  P[–V] = 3 [S]  3PV + 4ST = 0 (ii)  = FL 3  104  20  102 1 1 . When the tube is taken out, a convex meniscus is AY  3.6  10 4  1.5  1010 = 11.11 × 10–4 = 1.11 mm formed at the bottom then. The total upward force node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 due to surface tension is 7 . For translatery equilibrium \\\\\\\\\\\\\\\\\\\\\\\\ ( T + T = W ....(i) T1 T2  ) 12 Steel brass F = 2rT + 2rT= 4rT for equal stress T1  T2 x (2–x) This balances the weight of water column of length H A1 A2 W (H)   T1  A1  0.1  10 4  1 T = 2T T2 A2 0.2  104 2 2 1 4T  4rT = (r2H) g H= rg 71

JEE-Physics 2T Final height = h1  h2 but h = rg therefore H = 2h 2 The length of the liquid column remaining = 2h Final potential energy (  ) ( )  h1  h2   h1  h2 2  4   2  = m g = A  g F 3000 10 Work done by = Initial PE – Final PE 1 3 . Pressure = A  425  104 = 7.06 × 105 Pa  Ag  h 2  h 2   Ag  h1  h2 2  1 2 2   2  600 10 F 1 4 . In equilibrium 800  104  25  104 + hg Ag (h1  h2 )2 = 4 F  60 104 – 8 × (0.75 × 103)× 10  25  104 8 18 .  hw + 8cm + ho = 22 cm + 22 cm F  hw + ho = 36 cm 25  104 = 1.5 × 104  F = 37.5 N PC = PB  hoog = hwwg  ho × 0.8 = hw × 1 1 5 . Pressure an the water surface  ho= hw = 1.25 hw  hw + 1.25 hw = 36 0.8 = Mg  [16 3 10  104 ] 36 A  104  1  hw= 2.25 = 16cm so BE = 22 – 16 = 6cm  30 104  2 104 Pa 2 0 . PP = Patm + hg  15  = 1.013 × 105 + 3 × 800 × 9.8 = 124.9 KN/m2 According to Pascal law = Pr = hg Pp = PQ + (1.5 + 3)g  h  Pr  2 104  2m  PQ=124.9 × 103–4.5 × 800 × 9.8 = 89.5 KN/m2 g   PR = PQ = 89.5 KN/m2 103 10 Ps = PR –(3 + 2.5)g = 89.5 × 103 – 5.5 × 800 × 9.8 = 46.4 KN/m2 Mass of water in the pipe 2 21.  A  1  w g   2 1  cos  = mg (2 cos ) = (r2h)  =  × 10–4 × × 10–3 = 0.2 kg  sin   sin   25  104  103 Th mass of water in cylinder=750–200=550 g = 0.55  2 sin2   2.5  2  0.55 = (R2H) 0.55 11 W 1m  H    16  104  103 m 1 1 32  sin2  =  sin  =   = 30° 42 1 6 . P1V1 = P2V2  (Patm + Hwg)V = Patm × 2V For minimum depth of water let water height is h node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65  Hg = Patm= 76cm× Hg× g =76cm × 13.6 wg  H = 1033.6 cm = 10.34m (       h)     A  s h   w g  h  mg  2  in 90   2  1 7 . Initial potential energy ( )   2.5  2 h1 h2  h2   1  h  2m 2 2  2  = m1g + m2g 25  10 4  103   Ag h 2  Ag h 2  Ag  h12  h 2  23. Specific gravity of block WA 15N 5 1 2  2 2  = WA  WW   2 2  15 12 72

JEE-Physics 2 4 . Let cross-section area of the plank is A then weight work done per unit volume by gravity froce of plank W= (1 × A) 0.5 × g length of plank inside ()  the water = 0.5 (     =g(h1–h2) = 103 × 10(2–5) = –30× 103 J/m3 cos  A     W=(1 × A) 0.5 × g,   2 7 . (i) Reaction force ( )    =0.5 ) = vdm  v2 A   moa = 2gh × A dt 100 100 cos  B  (A × h × )a 2ghA  a  2g =0.2 m/s2 0.5m = 100 100 F C (ii) m0 = Ah'  h' = m0 4 4A  D v= 2gh '  2g  m0  m0g W 4A 2A A So upthrust on the plank () 11  0.5  28. P1 + 2  v 2 = P2 +2 v22 =  cos  A × 1 × g 1 torque about point A (A   ) 1  2 (v22 – v12) = P1–P2 = hg W × AC sin  = Th × AD sin (1 × A) × 0.5 × g × 0.5 sin v22 = v 2 + 2gh 1 v2  (2)2  2  1000  0.51 = 32 cm/s  0.5   1  0.5   cos  A × 1 × g  2  = × cos   sin 29. (i) Velocity of flow (  )   11 = 2gh  2 10  3.6  6 2 m/s  1 = 2 cos2   cos  = 2   = 45° (ii) Rate of flow 2 5 . When beaker half full with water then it float with  4  10 2  2  Av       6 2 completely immersed. (      )  =9.6 × 2 × 10–3 m3/s (iii) Bernoulli's theorem between surface and A So weight = upthrust   390g  500 1  g = Vin × 1 × g= 640 cm3 (A     )   2 1 Patm = P + v2 + gh So volume of glass beaker 2 ()  1  P =Patm – v2–gh = 640–500 = 140 cm3 2 density of beaker = 390 = 2.78 g/cm3 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 140 1 103 (6 = 105– × 2 )2 – 103 × 10 × 1.8 2 2 6 . Work done per unit volume by pressure = change in energy = 4.6 × 104 N/m2 (=  3 0 . Let v' be the horizontal speed of water when it emerges from the nozzle then from equation of ) 1 continuity  2 (v22 – v12) + g(h2 – h1) (v'           1     )  = × 103[(0.5)2 – (1)2] + 103 × 10(5–2) Av 2 Av = av'  v' = a   3 103  30 103 = 29.625 × 103 J/m3 Let t be the time taken by the stream of water to strike the ground then vertical distance 8 73

JEE-Physics (t EXERCISE –IV (B)  ) 2h 1. V = A(a –x) g 1 Open h = gt2  t = 2  horizontal distance ( ) 2h Av 2h a–x PB R = v' g = a g B 3 1 . (i) v = 2gh (Acc. to Torricellis law of efflux) x P2 (ii) Reaction of out flowing liquid (F) = Mass coming A x –x out per second × velocity ((F)=   × Final pressure, P  P0 V0  P0a or pressure at B, F Idm dm V ax GH KJF = v dt Ma = v dt (A2h)a = vA1v b g  P0a  xg dm d A1x dx P2 = P + xg ax dt dt =A1 dt = = A1v] force exerted by pressure difference is  A2ha = v2A1  A2ha = 2ghA1 ()  [ v  2gh ]  a = 2gA1 f1 = (PB – PA) s = (P2– P0) s   P0 x  x g  s A2  a  x  h gh Mass of horizontal arm A(B) of liquid is 32. vA  2g   4 2 (A(B)) m=A(–x) x x gh 3h r=x+   2 22 (Range)A = vA× t = 2 4g ....(i) Bernoulli's theorem between surface and B {A  (–x)}    x   2   P0 x  x g  A  2  0  a  x  (B) x= 0.01 m  x=1 cm hh 1 ( 2 )v2+  2 g h  v  gh length of air column in sealed arm (a–x) = 6–1=5cm 2g 2 + g 2 = 2  4  3. (i) As for floating W = Th Vg= V1d1g + V2d2g (Range)B = gh  2h  RA  3 4g RB 2  A   3   A   1   A  or L  5     4 L   5  d   4 L   5  2 d dv 7  10 2 3 4 . F = A = 1 × 100 × 10–4 × 103 =0.7N dx 3 25 d+ d d 3 5 . Velocity at surface = terminal velocity i.e.,  = 4 44 (=)  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 (ii) Total pressure = p0 + (weight of liquid + weight  2gh  2 r2 (  )g of solid) A i.e., 9 P=P0+ H dg+ H 2dg+ 5 A  L  × g × 1 2 2 4 d×  5  A 2 (3  10 4 )2  (104  103 )  9.8 2gh  = 180 9 9.8  10 6 31 1 i.e. P=P0+ 2 H dg + 4 L dg = P0+ 4 (6H+L)dg  h  (180)2  180  180  1.65  103 m 2g 2  9.8 (b) (i) By Bernoulli theorem for a point just inside and outside the hole ( )  74

JEE-Physics 11 9 x9 P1 + 2 v12 + gh1= P2 + 2 v22 (iii) tan  = 10 =  4 10 x H H  h 1 20x2  40 3  2 dg+  2 P0+ 2 y i.e., P0 + 2dg = (2d)v2 60 – 9 g forces on front wall is 0 y or g(3H –4h) = 2v2 or v  (3H  4h)  3   5  x tan  9 (4dx) 2 0  3 (ii) As at the hole vertical velocity of liquid iszero so time taken by it to reach the ground, (36×103   5 x   9x2  36 × 103 + 15  81   3  20   3 2  )   t  (2h / g) So that     9  5  g h  12 ( 15 + 15) = 360  3 2  g 2h x= vt (3H  4h)   h(3H  4h) 2 g (iii) For x to be maximum x2 must be maximum, 7. W + Nv = gh R2 dd dx r i.e., (x2) = 0 or (3Hh – 4h2) = 0  dyx= cos dh dh or 3H – 8h = 0, i.e., h = (3/8)H 3H 3 3 x and Xmax = (3H  H)  H R 8 24 4 . Upthrust on the block W   gx(2r) dx sin  cos  2  g 3  g = 5 V  1500  g  2   5 V  1000   g  2  h dy  gx2h  (9  x tan ) = 1800 × 10–3 × 10 = 18N cos  0 Weight of the block = 10–3 × 800 ×  g  g  =12N h dx h x2  2  cos  0 g2  ax tan dx 0    so Tension in the string = Th–mg = 18–12 = 6N a  ah2 h3  5 . (i) tan  = g 1m g2   tan   1m   2 cos  3  a = 4 m/s2   g2h2 9R  h sin   h tan   2 cos  3 (ii) New a = 4.8 5 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 4.8 P W g2h2 R h tan  tan  =   2.4  2 cos   6 10 5m = volume of water left 1.2 g h 2 R h   cos 3     tan  1 1.4   2 × (2.4) (5) 30.6 W × 4 + 0.6 × 20   R h   cos  3   24 + 12 = 36 m3 g h 2  tan  Vf = 36 m3; Vi = 5 × 4 × 2  40 m3  Vf  Vi   100  10%  Vl  75

JEE-Physics 8 . Pressure at A, PA = P0 + h2g + (h–y)1g (ii) Particle starts oscillating in the fluid Pressure at B, PB = P0 () According to Bernoulli's theorem,  Work done by person pressure energy at A = pressure energy at B + kinetic energy at B 1 = Total energy of oscillation work = M2A2 1  PA = PB + 2 1v2 2 2 1 dL g 1 dm  2 2 (AH)dm dm L   v = 4ms–1 1 B  H   F = (Av ) (v – 0) = Av2 A or F = 7.2N 1  dm 2 Total mass of the liquid in the cylinder is work = 2 AH2 1  dL  g  2kg f-m () m = Ah1 + Ah2 = 450 kg 1 0 . Intially : mg = fB  mg = VdLg = AhdLg Limiting friction = mg = 45N when pulled slightly up by x then  F < Limiting friction, therefore, minimum force fnet = mg – fB = mg – A(h–x)dLg required is zero. = mg – AhdLg + AxdLg (F<)fnet= AxdLg Consider free body diagram for maximum value of force directry propotional to x therefore if will force.Considering vertical forces, N =mg perform S.H.M. ( x  (FB D )  N =mg ) (ii) ma = (mg – VdL(g)) Now considering horizontal forces, Fmax = F + N orFmax = 52.2 N fB 9.(i) AH dmg = AhdLg fB  /2  /2 mg dm 1m mg h= H dL ; fnet = mg – fB  fnet = AHdmg – AxgdL  AdL xg  If will perform SHM about its position  A (H )dm  a = g  x  dM H, with  = dLg dL dM 2gx  Hdm  a = g– 2  dL  x dx fnet = (Ag) d L   d2x 0.8 H  x2 0.8H dt2  gx  g can be compered with  d = fnet dx = Agd L  Hdm x 2  d2x dL 0 dt2 0  2 x  g  0  w  g  x2 0.8 2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65  Ag H(0.8)x  T =  and time required is = T/2, t = 1 sec    2  0  (0.8)2 H2 0.8 AgH2dm2 11. mg b sin   fB  b  x  sin   Ag H(0.8)(0.8)H   2 2  2   2  0 A = 4000 × 10–4; g =10, H=50 × 10–2; b  bx  5 b2 x2 dmg 2  2  9 2 2 dm = 8 × 10+2 (Ab) = (A x ) d Lg ;  bx  4000  10 4  10  2.500  10 4  .64 b = By solving It x = 3 2  = .32 × 104  32kgC 76

JEE-Physics 1 2 . fnet = f2 – f1  PA 1 4 . fg  fS,T   ghA mg v1 dp dm h mg + S(4a) = gha2 f2 = dt  v2 dt = sv22 f1 dp dm f2 v2 mg  4aS dt dt h   ga2 f1 =  v1  sv 2 1 fnet = s(v22 – v12) = s2g(h2–h1) dd fnet  0.51 Newton 1 5 . From diagram r cos  = 2  r= 2 cos  1 3 . We consider a ring element of radius r and thickness P0A = PT(A) (–h) –h PT dr whose centre is at the centre of disc. The velocity r of fluid at distance r from axis is v = r P0   P0  2T  0–2rT h  h  r  h PT = ; PA = (rdr PB = P0   2T  gh  P0 r) h r dv dr   P0 h  g h   2T (2 cos )  dx   dx  h d Where dx is the thickness of layer of liquid. T   P0 h  g h  4 d  h cos  (dx)  The area of the considered element is dA = (2 rdr) ()  the viscous froce on the considered element is () dv dF = (2 rdr) dx Here, velocity gradient is ( )  dv  v  r dx h h  r  2 r2dr dF = (2 r dr) hh node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 The power developed on the considered element by viscous force is ()  dP = v dF = (r ) 2 r2dr  22 r 3dr hh  Total power developed due to viscous force is ()  rR P  2  dP (on both sides) r 0  2 R 22 r3dr 2R 4  0h h  3.14  0.08  101  (60)2  (1 0 1 )4  9W 1  103 77

JEE-Physics EXERCISE-V(A) 7. Excess pressure inside a soap bubble is 4T P 1 r 1 . Elastic energy = × F× x Air will flow from the bubble at high pressure to the 2 F = 200 N, x = 1mm = 10–3 m bubble at lower pressure as P  1 , hence bubble r 1 of smaller radius will be at higher pressure, hence E = × 200 × 1 × 10–3 = 0.1 J air will flow from smaller to the bigger sphere. 2 (P4T 2 . Work done 1 kx2  1 k2 where  is the total r 22  extensions.  1 (k)  1 F   P 1r ,     22   3 . Energy density Energy 1 8 . Water will rise to the full length of capillary tube = Volume = 2 × Stress × Strain ()  Energy 1   9 .vs s    vs  0.1 m/s Volume = × Stress × Strain vg g   2 1 Stress 1 S2  1 0 . = Stress × = 1 Vg  2 Vg  k v 2  vT  Vg 1  2 2 Y 2Y T k 1 S2 1 1 . As liquid 1 floats above liquid 2, 1<2 Energy density =  2Y The ball is unable to sink into liquid 2, 3<2  5 . Velocity of efflux through a small hole = 2gh The ball is unable to rise over liquid 1, 1<3 where h is the position of the small hole from the   top of the vessel. Thus 1<3 <2 ()= 2T cos  2gh )h12.Capillaryrise gr . As soap solution has lower T, h will be low (T h  ) F/A YA 2   Y     F  A 20m 14. node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 YA 2  F  here v = volume of wire v vefflux  2  10  20  20 m/s F2 2  3A  2 F1  A  6 . The viscous force experienced by the spherical ball F  A2    A2   9 is expressed as  A1  (     )   F2 = 9F F  6rv  f  r  F  v 78

JEE-Physics 1 5 . In equilibrium ball will remain at the interface of EXERCISE –V(B) water and oil. () 1.Fromequation of continuity ( ) 1 6 . According to equation of continuity v1A1= v2A2 ()  and r12v1 v 2  u2  2gs ; v 2 1  2 10  0.15  v2 = 2 m/s v2 2 2 A1V1 = A2V2 or r2  v1A1 1  104 v2  Velocity of stream at 0.2 m below tap. Hence A2 = = 5 × 10–5m2 V22 = V12 + 2as = 0.16 + 2 × 10 × 0.2 = 4.16 m/s 2 3 . If we apply Newton's law to find the force exerted r2  r12 v1  16  106  0.4 3.2  103 m by the molecules on the walls of the container, we v2  will have to apply a pseudo force (the frame of 2 molecules is an accelerated frame). This pseudo so diameter = 2 × 3.2  103 m force acting on gas molecules will act in opposite to = 2 × 1.8 × 10–3 the direction of motion of closed compartment. The = 3.6 × 10–3 m result will be more pressure on the rear side and 1 7 . W = 8T [(r22) – (r12)] less pressure on the front side. = 8 ×  × 0.03 [25 – 9] × 10–4 =  × 0.24 × 16 × 10–4 ( = 3.8 × 10–4   = 0.384  mJ  0.4  mJ  )  1 8 . By volume conservation  4 R3 = 2  4 r 3   3  3  1 4 . Equating the rate of flow R = 2 3r () Surface energy E = T (A) = T (4 R2) (2gy )  L2  (2g  4 y )R2 = T (4 22/3 r2) = 28/3  r2 T L  L2 = 2R2  R = db  d  2 19. Terminal velocity V 5 . According to archemedes principle P1 h V1 7.8  1 13.2 () P V V2  8.5 10 –4 × 7.8  1.2 Upthrust = Wt. of fluid displaced. 10 (= V2 = 1.6 × 104  Fbottom = Ftop + V  g node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 10 V2 = 1.6 104 = 6.25 × 10–4 = P1 × A + V  g = (h  g) × (R2) + V  g =  g [R2 h + V] 2 0 . weight = mg = 1.5 × 10–2 N (given) 6 .  decreases as the block moves up. h will also length =  = 30 cm (given) decreases because when the coin is in the water it = 0.3 m will displace euqal volume of water, whereas when 2T = mg it is on the block an equal weight of water is mg 1.5  102 displaced. (  h T  2  2  0.3 = 0.025 N/m.    79

JEE-Physics 7. Y F   20 1  2  1011N / m2 . (      P2  A  6  104  10 = P2 × [(4r)2 – (2r)2] + P0 × (2r)2 8. K  P  1.165  1.01  105  1.55  105 Pa = [P0 + (h1 + h)g] ×  × 12r2 + 4r2P0 At the verge of rising   v  10 3  v  [P0 + (h1 + h)g]  × 12r2 + 4r2P0 9 . The square of the velocity of flux  =  × 10r2h × 3 g + [P0 + h1g] ×  × 16r2 ()  v2  2gh 2 10  2.475 16h 5h  = 50 m2/s2  12h1 + 12h = 3 + 16h1 3 = h1 2 2.(b) Again considering eqilibrium of wooden block.  A0  1  0.12 1   A  () A Total Downward force = Total force upwards 2.475m=h (= ) 3m Wt. of block + force due to atmospheric pressure = Force due to pressure of liquid + Force due to A0 v atmospheric pressure 52.5 cm (+ =  11. P2 – Patm = 4T 4T +  , P1 – Patm = R  R2 1 (16r2) +g+ P0 × 16r2 3 Here R2 > R1. So P2 < P1 Air will flow from end 1 to end 2. = [h2g + P0]  [16 – 4r2] + P0 × 4r2 1 2  (16r2)h g g = h2g ×  × 12r2 Comprehension h4 1. (a) Consider the equilibrium of wooden block.  16 3 =12h2 9 h=h2 the height ()3.(a)When h2 of water l evel is further Forces acting in the downward direction are upwar d force acting on the decreased, then the ( wooden block decreases. The total force downward Weight of wooden cylinder remains the same. This difference will be compensated by the normal reaction by the tank () wall on the wooden block. Thus the block does not P1× (4r)2 moves up and remains at its original position. (h2     P2 P0 W P2   ) node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65  h Subjective = (4r)2 × h × × g =  × 16r2 g 1 . For the wooden stick-mass system to be in stable 33 (b) Force due to pressure (P1) created by liquid equilibrium the center of gravity of stick-mass system of height h1 above the wooden block is should be lower than the center of buoyancy. Also (h1   in equilibrium the centre of gravity (G) and the center P 1 of buoyancy (B) lie in the same vertical axis. = P1 × (4r2) = [P0 + h1g] × (4r)2 ( =[P0 + h1g] × 16r2  Force acting on the upward direction due to  (G)(B)  pressure P2 exerted from below the wooden  block and atmospheric pressure is 80

JEE-Physics The above condition 1 will be satisfied if the mass is 2 . (i) As the pressure exerted by liquid A on the towards the lower side of the stick as shown in the cylinder is radial and symmetric. The force due figure. The two forces will create a torque which to this pressure cancels out and the net value will bring the stick-mass system in the vertical is zero. position of the stable equilibrium. Let  be the length (A  of the stick immersed in the liquid.  (1)  (ii) For equilibrium ( )  Buoyantforce= weight of the body   )  hAAAg + hBBAg = (hA + h + hB)A Cg FB=R2hg (where c = density of cylinder) (c = )  C h   h A A  h B B   (hA  hB) = 0.25 cm h/2   c   L/2 (iii) a  FBuoyant  M g ( R2L )g M mg   h AA  B (h  hB )  (h  hA  h B )C  g  C (h  h A  hC )    OB = . Let OG = y g 2  upwards For vertical equilibrium () 6 FG = FB (M + m) g = FB  R2Lg + mg = R2g   R 2L  m ...(i) 3 . When the force due to excess pressure in the bubble R 2 equals the force of air striking at the bubble, the bubble will detach from the ring (Nowyusingth)econceptofcentreofmasstofindy.( )  Then y  My1  my2 M m  Av2 = 4T × A   r  4T r v 2 Since mass m is at O the origin (m O)  y2=0 4 . When the tube is not there, using Bernoulli's theorem  y  M(L / 2)  m O  ML ( Mm 2(M  m)  )  ( R 2 L )L ...(ii) P  P0  1 v12  gH  1 v 2  P0 2(R2L  m ) 2 2 0    P 1 2 v 2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 Therefore for stable equilibrium > y  gH  2 v 0  1 2 But according to equation of continuity () ()  R2L  m (R2L)L A1v1  A2v2 or v1  A2v0  2R2  2(R2L  m ) A1  m >  R2L (   )  2  v   Minimum value of m is   R2L (   ) P  gH  1   2   A2 v 0   2 0  A1  (m)   2  P  gH  1 v 2 1   A2   2 0   A1  81

JEE-Physics Here P  gH  P The centripetal force required for the mass to rotate According to Poisseuille's equation ()   P  a 4  P  a4 Q= 8     8Q = (dm) x2  The total centripetal force required for the mass  P  gH a4  1   A  2  8Q 2 1  A   of length L to rotate (L     8Q 2  2  a 4 ) 0 1     v    L (dm )x2 0 where A2  b2 d2 A1 D2 Here, dm =  × × dx 5 . The free body diagram of wire is given below. 4 (FBD)  Total centripetal force ( )  FF  L d2   x2 0    4  dx    d2  2 L xdx 40  mg    d2  2  L2 ...(ii) 42 If  is the length of wire, then for equilibrium ()  This centripetal force is provided by the weight of 2Fsin=W. liquid of height H. (H F = S ×   2S ×  × sin  =  ×  × g ) or S  g also sin  = y/a From (i) and (ii) 2 sin  d2 d2 2  L2 2 L2  S  g  ag 4 H g    ; H  2g 2y / a 2y 4 L ag Integer Type 2y  Surface tension ( ) S = 4S 4  .04 1. (P ) = rA +P = 0.02 + 8 = 16 N/m2 in A 0 6 . From law of continuous A1v1 = A2v2 (P ) = 4S 4  .04 + 8 = 12 N/m2 in B +P =  v2    (4  103 )2  0.25 rB 0 .04   (1  103 )2 = 4m/s    n = nB Pin 3  A nA = Pin 2h Pin A VA ; B  rB  =6 and x = v × t = v × g =2m RT  rA  × A 7 . Weight of liquid of height H (H)  d2  H    g ....(i) 2. 500 H node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\04-Fluid Mechanics.p65 4 (500 – H) P = 300 (P – rg × 0.2) Let us consider a mass dm situated at a distance x 00 from A as shown in the figure. (0.5 – H) × 105 = 0.3 [105–104 × 0.2) (axdm ) 0.5 – H = 0.294 H = 206 mm wd A B A H x dx P dm B L 82