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Page 1 5. Here assertion is wrong since bdecay process is n p e v ® + + – but reason is true hence correct option is (d). 6. Here assertion is true but reason is false. Correct option is (c). 7. Here both assertion and reason are true and reason may or may not be true. Correct option is (a, b) 8. Both assertion and reason are true but reason is not correct explanation of assertion. Hence correct option is (b). 9. Here reason is true but assertion is false Q 1 amu = 931.5 MeV Correct option is (d) 10. Both assertion and reason are true but reason does not correctly explain assertion. Hence correct option is (b). 11. Here both assertion and reason are true and reason may or may not be correct explanation of assertion . Hence correct options are (a, b). Objective Questions (Level 2) ¢ Single option correct 1. Let initially substance have N i nuclei then N N e i t =  l dN dt N e i t =   l l At t t = we get dN dt N e N t t i t æ è ç ö ø ÷ =  = =  l l 0 …(i) At t t = 4 dN dt N e N t t i t æ è ç ö ø ÷ =  = =  4 4 0 16 l l …(ii) Dividing Eq. (i) and Eq. (ii) we get e t 3 16 l = …(iii) Now at t t = æ è ç ö ø ÷ 11 2 dN dt N e t t i t æ è ç ö ø ÷ =  =  11 2 11 2 l l =  ´   l l l N e e i t t 8 2 3 2 =  = ´  l l l N e e N i t t 4 3 0 16 16 From Eqs. (ii) and (iii) = N 0 64 Hence, correct option is (b). 2. We have l l l = + = + 1 2 2 30 2 60 log log Þ l = log 2 20 Now N N e t =  0 l Þ N N e t 0 0 2 20 4 =  log Þ log log 4 20 2 = t Þ 2 2 20 2 log log = t Þ t = 40 yr Hence correct option is (c). 3. From graph it is clear that number of nucleons in X is N 3 and binding energy per nucleon is E 3 for Y nucleon is N 2 and BE per nucleon is E 2 . Hence X Y E N E N + = + 3 3 2 2 Similerly W E N = 1 1 88 Page 2 5. Here assertion is wrong since bdecay process is n p e v ® + + – but reason is true hence correct option is (d). 6. Here assertion is true but reason is false. Correct option is (c). 7. Here both assertion and reason are true and reason may or may not be true. Correct option is (a, b) 8. Both assertion and reason are true but reason is not correct explanation of assertion. Hence correct option is (b). 9. Here reason is true but assertion is false Q 1 amu = 931.5 MeV Correct option is (d) 10. Both assertion and reason are true but reason does not correctly explain assertion. Hence correct option is (b). 11. Here both assertion and reason are true and reason may or may not be correct explanation of assertion . Hence correct options are (a, b). Objective Questions (Level 2) ¢ Single option correct 1. Let initially substance have N i nuclei then N N e i t =  l dN dt N e i t =   l l At t t = we get dN dt N e N t t i t æ è ç ö ø ÷ =  = =  l l 0 …(i) At t t = 4 dN dt N e N t t i t æ è ç ö ø ÷ =  = =  4 4 0 16 l l …(ii) Dividing Eq. (i) and Eq. (ii) we get e t 3 16 l = …(iii) Now at t t = æ è ç ö ø ÷ 11 2 dN dt N e t t i t æ è ç ö ø ÷ =  =  11 2 11 2 l l =  ´   l l l N e e i t t 8 2 3 2 =  = ´  l l l N e e N i t t 4 3 0 16 16 From Eqs. (ii) and (iii) = N 0 64 Hence, correct option is (b). 2. We have l l l = + = + 1 2 2 30 2 60 log log Þ l = log 2 20 Now N N e t =  0 l Þ N N e t 0 0 2 20 4 =  log Þ log log 4 20 2 = t Þ 2 2 20 2 log log = t Þ t = 40 yr Hence correct option is (c). 3. From graph it is clear that number of nucleons in X is N 3 and binding energy per nucleon is E 3 for Y nucleon is N 2 and BE per nucleon is E 2 . Hence X Y E N E N + = + 3 3 2 2 Similerly W E N = 1 1 88 The reaction is W X Y ® + The energy released is ( ) E N E N E N 3 3 2 2 1 1 +  Hence Correct option is (b). 4. Energy = ´ + ´  ´ ( ) 110 200 8.2 90 8.2 7.4 = ´  200 (8.2 7.4) = ´ = 200 160 0.8 MeV Hence correct option is (d). 5. The reaction is 1 2 1 2 2 4 H H He energy + ® + Energy = ´  ´ (4 7 2 1.1) MeV =  = (28 4.4) 23.6 MeV Hence Correct option is (b). 6. Total energy released per second = ´ 16 10 6 W = ´ 16 10 6 J/s Energy per fission = 200 MeV = ´ ´ ´  200 10 10 6 19 1.6 = ´ ´  2 10 11 1.6 J Q Efficiency = 50% Hence power (energy converted per second) = ´ ´ ´ = ´   2 10 50 100 10 11 11 1.6 1.6 J Number of fission = ´ ´ =  16 10 10 10 6 11 18 1.6 /s Hence correct option is (d). 7. dN dt A N = l Q After time N become conservation Þ dN dt = 0 Þ N A A T AT = = = l log log 2 2 Hence correct option is (d). 8. By conservation of momentum M v M M H H H = + ¢ ( ) n Þ v v ¢ = 2 Let initial KE of Hatom = K Final KE of each Hatom = K 2 For excitation K E E 2 4 13 6 2 1 =  =  + æ è ç ö ø ÷ 13.6 eV . Þ K 2 = 10.2 eV Þ K = ´ ´ ´  2 10 19 10.2 1.6 Þ 1 2 2 10 2 19 M u H 10.2 1.2 1.6 = ´ ´ ´ ´  Þ u H 1.2 10.2 1.6 1.673 = ´ ´ ´ ´ ´ ´  2 2 10 10 19 27 = ´ 6.25 m/s 10 4 Hence correct option is (c). 9. Let us suppose just before the death no radioactive atoms were present hence original activity A 0 is given as A N 0 0 = l …(i) After death the radioactivity decreases exponentially ie, A dN dt N N e t = = =  l l l 0 …(ii) Dividing eq. (ii) by eq. (i) we get A A e t 0 =  l or lt A A = log 0 or t A A = 1 0 l log Now A 0 15 = decay/min/gram A = 375 200 decay/min/g but l = 0.693 yr 5730 89 Page 3 5. Here assertion is wrong since bdecay process is n p e v ® + + – but reason is true hence correct option is (d). 6. Here assertion is true but reason is false. Correct option is (c). 7. Here both assertion and reason are true and reason may or may not be true. Correct option is (a, b) 8. Both assertion and reason are true but reason is not correct explanation of assertion. Hence correct option is (b). 9. Here reason is true but assertion is false Q 1 amu = 931.5 MeV Correct option is (d) 10. Both assertion and reason are true but reason does not correctly explain assertion. Hence correct option is (b). 11. Here both assertion and reason are true and reason may or may not be correct explanation of assertion . Hence correct options are (a, b). Objective Questions (Level 2) ¢ Single option correct 1. Let initially substance have N i nuclei then N N e i t =  l dN dt N e i t =   l l At t t = we get dN dt N e N t t i t æ è ç ö ø ÷ =  = =  l l 0 …(i) At t t = 4 dN dt N e N t t i t æ è ç ö ø ÷ =  = =  4 4 0 16 l l …(ii) Dividing Eq. (i) and Eq. (ii) we get e t 3 16 l = …(iii) Now at t t = æ è ç ö ø ÷ 11 2 dN dt N e t t i t æ è ç ö ø ÷ =  =  11 2 11 2 l l =  ´   l l l N e e i t t 8 2 3 2 =  = ´  l l l N e e N i t t 4 3 0 16 16 From Eqs. (ii) and (iii) = N 0 64 Hence, correct option is (b). 2. We have l l l = + = + 1 2 2 30 2 60 log log Þ l = log 2 20 Now N N e t =  0 l Þ N N e t 0 0 2 20 4 =  log Þ log log 4 20 2 = t Þ 2 2 20 2 log log = t Þ t = 40 yr Hence correct option is (c). 3. From graph it is clear that number of nucleons in X is N 3 and binding energy per nucleon is E 3 for Y nucleon is N 2 and BE per nucleon is E 2 . Hence X Y E N E N + = + 3 3 2 2 Similerly W E N = 1 1 88 The reaction is W X Y ® + The energy released is ( ) E N E N E N 3 3 2 2 1 1 +  Hence Correct option is (b). 4. Energy = ´ + ´  ´ ( ) 110 200 8.2 90 8.2 7.4 = ´  200 (8.2 7.4) = ´ = 200 160 0.8 MeV Hence correct option is (d). 5. The reaction is 1 2 1 2 2 4 H H He energy + ® + Energy = ´  ´ (4 7 2 1.1) MeV =  = (28 4.4) 23.6 MeV Hence Correct option is (b). 6. Total energy released per second = ´ 16 10 6 W = ´ 16 10 6 J/s Energy per fission = 200 MeV = ´ ´ ´  200 10 10 6 19 1.6 = ´ ´  2 10 11 1.6 J Q Efficiency = 50% Hence power (energy converted per second) = ´ ´ ´ = ´   2 10 50 100 10 11 11 1.6 1.6 J Number of fission = ´ ´ =  16 10 10 10 6 11 18 1.6 /s Hence correct option is (d). 7. dN dt A N = l Q After time N become conservation Þ dN dt = 0 Þ N A A T AT = = = l log log 2 2 Hence correct option is (d). 8. By conservation of momentum M v M M H H H = + ¢ ( ) n Þ v v ¢ = 2 Let initial KE of Hatom = K Final KE of each Hatom = K 2 For excitation K E E 2 4 13 6 2 1 =  =  + æ è ç ö ø ÷ 13.6 eV . Þ K 2 = 10.2 eV Þ K = ´ ´ ´  2 10 19 10.2 1.6 Þ 1 2 2 10 2 19 M u H 10.2 1.2 1.6 = ´ ´ ´ ´  Þ u H 1.2 10.2 1.6 1.673 = ´ ´ ´ ´ ´ ´  2 2 10 10 19 27 = ´ 6.25 m/s 10 4 Hence correct option is (c). 9. Let us suppose just before the death no radioactive atoms were present hence original activity A 0 is given as A N 0 0 = l …(i) After death the radioactivity decreases exponentially ie, A dN dt N N e t = = =  l l l 0 …(ii) Dividing eq. (ii) by eq. (i) we get A A e t 0 =  l or lt A A = log 0 or t A A = 1 0 l log Now A 0 15 = decay/min/gram A = 375 200 decay/min/g but l = 0.693 yr 5730 89 t = ´ ´ = ´ 5730 15 200 375 5730 200 25 0.693 0.693 log log Þ t = ´ 5730 8 0.693 log = ´ = ´ 5730 3 2 5730 3 0.693 log Þ t = 17190 yr Hence correct option is (c). 10. N N e P t t =  + 0 1 l( ) …(i) and N N e Q t =  0 l Now A N P P = l and A N Q Q = l Þ A A N N e P Q P Q t = =  l 1 Þ lt A A Q P 1 = æ è ç ç ö ø ÷ ÷ log Þ t A A T A A Q P Q P 1 1 = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ l log log Hence correct option is (b). 11. The given reactions are 1 2 1 2 1 3 H H H + ® + p 1 2 1 2 4 H H He 3 + ® + n Þ 3 H He 1 2 2 4 ® + + n p Mass defect Dm = ´    ( ) 3 2.014 4.001 1.007 1.008 Dm = 0.026 amu Energy Released = ´ = ´  0.026 MeV 3.87 J 931 10 12 This energy produced by the three deutronatoms. Total energy released by 10 40 deutrons = ´ ´ = ´  10 3 10 10 40 12 28 3.87 J 1.29 J The average power P = = 10 10 16 16 W J/s Therefore total time to exhaust all deutrons of the star will be t = ´ » 1.29 s s. 10 10 10 28 16 12 Hence correct option is (c). 12. N N e N e t t t 1 0 0 2 1 1 = =   l log …(i) N N e N e t t t 2 0 0 2 2 2 = =   l log …(ii) R N 1 1 1 = l …(iii) and R N 2 2 2 2 = = l l …(iv) Let after time t, R R 1 2 = then Þ R R 1 1 1 = Þ l l 1 2 1 2 1 ´ = N N Þ l l 1 2 2 1 2 2 1 1 2 = =  é ë ê ù û ú ´ N N e t t t t t log Þ log log t t t t t t t 2 1 2 1 1 2 2 = ´ ´  æ è ç ç ö ø ÷ ÷ Þ t t t t t t t 1 1 2 2 1 2 1 =  ´ 0.693 ( ) log Hence Correct option is (a). 13. The given reaction is Z A X Y 232 90 ¾® + a Þ Z A X Y 232 90 2 ¾® + He 4 Þ Z = 92 and A = 228 Q Initially X is in rest hence momentum of aparticle after decay will be equal and opposite of Y. Þ M v M v Y Y = a a Þ v M M v Y Y = a a Total kinetic energy K M v M v T Y Y = + 1 2 2 2 ( ) a a Þ K M v M M v M T y Y = + é ë ê ù û ú 1 2 2 2 2 2 a a a a 90 Page 4 5. Here assertion is wrong since bdecay process is n p e v ® + + – but reason is true hence correct option is (d). 6. Here assertion is true but reason is false. Correct option is (c). 7. Here both assertion and reason are true and reason may or may not be true. Correct option is (a, b) 8. Both assertion and reason are true but reason is not correct explanation of assertion. Hence correct option is (b). 9. Here reason is true but assertion is false Q 1 amu = 931.5 MeV Correct option is (d) 10. Both assertion and reason are true but reason does not correctly explain assertion. Hence correct option is (b). 11. Here both assertion and reason are true and reason may or may not be correct explanation of assertion . Hence correct options are (a, b). Objective Questions (Level 2) ¢ Single option correct 1. Let initially substance have N i nuclei then N N e i t =  l dN dt N e i t =   l l At t t = we get dN dt N e N t t i t æ è ç ö ø ÷ =  = =  l l 0 …(i) At t t = 4 dN dt N e N t t i t æ è ç ö ø ÷ =  = =  4 4 0 16 l l …(ii) Dividing Eq. (i) and Eq. (ii) we get e t 3 16 l = …(iii) Now at t t = æ è ç ö ø ÷ 11 2 dN dt N e t t i t æ è ç ö ø ÷ =  =  11 2 11 2 l l =  ´   l l l N e e i t t 8 2 3 2 =  = ´  l l l N e e N i t t 4 3 0 16 16 From Eqs. (ii) and (iii) = N 0 64 Hence, correct option is (b). 2. We have l l l = + = + 1 2 2 30 2 60 log log Þ l = log 2 20 Now N N e t =  0 l Þ N N e t 0 0 2 20 4 =  log Þ log log 4 20 2 = t Þ 2 2 20 2 log log = t Þ t = 40 yr Hence correct option is (c). 3. From graph it is clear that number of nucleons in X is N 3 and binding energy per nucleon is E 3 for Y nucleon is N 2 and BE per nucleon is E 2 . Hence X Y E N E N + = + 3 3 2 2 Similerly W E N = 1 1 88 The reaction is W X Y ® + The energy released is ( ) E N E N E N 3 3 2 2 1 1 +  Hence Correct option is (b). 4. Energy = ´ + ´  ´ ( ) 110 200 8.2 90 8.2 7.4 = ´  200 (8.2 7.4) = ´ = 200 160 0.8 MeV Hence correct option is (d). 5. The reaction is 1 2 1 2 2 4 H H He energy + ® + Energy = ´  ´ (4 7 2 1.1) MeV =  = (28 4.4) 23.6 MeV Hence Correct option is (b). 6. Total energy released per second = ´ 16 10 6 W = ´ 16 10 6 J/s Energy per fission = 200 MeV = ´ ´ ´  200 10 10 6 19 1.6 = ´ ´  2 10 11 1.6 J Q Efficiency = 50% Hence power (energy converted per second) = ´ ´ ´ = ´   2 10 50 100 10 11 11 1.6 1.6 J Number of fission = ´ ´ =  16 10 10 10 6 11 18 1.6 /s Hence correct option is (d). 7. dN dt A N = l Q After time N become conservation Þ dN dt = 0 Þ N A A T AT = = = l log log 2 2 Hence correct option is (d). 8. By conservation of momentum M v M M H H H = + ¢ ( ) n Þ v v ¢ = 2 Let initial KE of Hatom = K Final KE of each Hatom = K 2 For excitation K E E 2 4 13 6 2 1 =  =  + æ è ç ö ø ÷ 13.6 eV . Þ K 2 = 10.2 eV Þ K = ´ ´ ´  2 10 19 10.2 1.6 Þ 1 2 2 10 2 19 M u H 10.2 1.2 1.6 = ´ ´ ´ ´  Þ u H 1.2 10.2 1.6 1.673 = ´ ´ ´ ´ ´ ´  2 2 10 10 19 27 = ´ 6.25 m/s 10 4 Hence correct option is (c). 9. Let us suppose just before the death no radioactive atoms were present hence original activity A 0 is given as A N 0 0 = l …(i) After death the radioactivity decreases exponentially ie, A dN dt N N e t = = =  l l l 0 …(ii) Dividing eq. (ii) by eq. (i) we get A A e t 0 =  l or lt A A = log 0 or t A A = 1 0 l log Now A 0 15 = decay/min/gram A = 375 200 decay/min/g but l = 0.693 yr 5730 89 t = ´ ´ = ´ 5730 15 200 375 5730 200 25 0.693 0.693 log log Þ t = ´ 5730 8 0.693 log = ´ = ´ 5730 3 2 5730 3 0.693 log Þ t = 17190 yr Hence correct option is (c). 10. N N e P t t =  + 0 1 l( ) …(i) and N N e Q t =  0 l Now A N P P = l and A N Q Q = l Þ A A N N e P Q P Q t = =  l 1 Þ lt A A Q P 1 = æ è ç ç ö ø ÷ ÷ log Þ t A A T A A Q P Q P 1 1 = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ l log log Hence correct option is (b). 11. The given reactions are 1 2 1 2 1 3 H H H + ® + p 1 2 1 2 4 H H He 3 + ® + n Þ 3 H He 1 2 2 4 ® + + n p Mass defect Dm = ´    ( ) 3 2.014 4.001 1.007 1.008 Dm = 0.026 amu Energy Released = ´ = ´  0.026 MeV 3.87 J 931 10 12 This energy produced by the three deutronatoms. Total energy released by 10 40 deutrons = ´ ´ = ´  10 3 10 10 40 12 28 3.87 J 1.29 J The average power P = = 10 10 16 16 W J/s Therefore total time to exhaust all deutrons of the star will be t = ´ » 1.29 s s. 10 10 10 28 16 12 Hence correct option is (c). 12. N N e N e t t t 1 0 0 2 1 1 = =   l log …(i) N N e N e t t t 2 0 0 2 2 2 = =   l log …(ii) R N 1 1 1 = l …(iii) and R N 2 2 2 2 = = l l …(iv) Let after time t, R R 1 2 = then Þ R R 1 1 1 = Þ l l 1 2 1 2 1 ´ = N N Þ l l 1 2 2 1 2 2 1 1 2 = =  é ë ê ù û ú ´ N N e t t t t t log Þ log log t t t t t t t 2 1 2 1 1 2 2 = ´ ´  æ è ç ç ö ø ÷ ÷ Þ t t t t t t t 1 1 2 2 1 2 1 =  ´ 0.693 ( ) log Hence Correct option is (a). 13. The given reaction is Z A X Y 232 90 ¾® + a Þ Z A X Y 232 90 2 ¾® + He 4 Þ Z = 92 and A = 228 Q Initially X is in rest hence momentum of aparticle after decay will be equal and opposite of Y. Þ M v M v Y Y = a a Þ v M M v Y Y = a a Total kinetic energy K M v M v T Y Y = + 1 2 2 2 ( ) a a Þ K M v M M v M T y Y = + é ë ê ù û ú 1 2 2 2 2 2 a a a a 90 Þ K M v M M T Y = + é ë ê ù û ú 1 2 1 2 a a a Þ K K T = + é ë ê ù û ú a 1 4 228 Þ K K T a = 232 228 Hence Correct option is (b). 14. Energy of emitted photon = 7 MeV = ´ ´ ´  7 10 10 6 19 1.6 J = ´  11.2 J 10 13 Momentum of photon = ´ ´  11.2 J 10 m/s 10 3 13 8 = ´  11.2 kgm/s 3 10 21 Q Initial nucleus is stationary Applying conservation of momentum principle O = + Þ =  ® ® ® ® P P P P nuc photon nuc photon Þ     P P ® ® =  nuc photon Þ P nuc 11.2 kg m/s = ´  3 10 21 Mass of nucleus = 24 amu = ´ ´  24 10 27 1.66 kg But P mK nuc nuc 2 2 = Þ K P m nuc nuc 11.2 11.2 10 1.66 Joule = = ´ ´ ´ ´ ´ ´   2 42 27 2 9 2 24 10 Þ K c nuc 11.2 1.66 1.6 eV = ´ ´ ´ ´ ´ ´ ´    112 10 18 24 10 10 42 27 19 . » 1.1 keV Hence correct option is (b). 15. Let time interval between two instants is t 1 then N N e t t 1 0 1 =  + l( ) and N N e t 2 0 2 = l A N N e t t 1 1 0 1 = =  + l l l( ) A N N e t 2 2 0 2 = =  l l l ( ) Þ A A e t 1 2 1 2 1 = ´ l Þ 2 1 2 1 A A e t = l Þ log A A t 2 1 1 2 æ è ç ç ö ø ÷ ÷ = l Þ t A A 1 2 1 1 2 = æ è ç ç ö ø ÷ ÷ l log Þ t T A A 1 2 1 2 2 = æ è ç ç ö ø ÷ ÷ log log Hence Correct option is (c). 16. The given reaction is 1 2 1 2 1 3 1 1 H H H H + ¾® + DM m m m =   [ ( ) ( ) ( )] 2 1 2 1 3 1 1 H H H = ´   ( ) 2 2.014102 3.016049 1.007825 amu = ´  4.33 amu 10 3 DE = ´ ´  4.33 931.5 MeV 10 3 = 4 MeV Hence correct option is (c). 17. Number of fusion required to generate 1 kWh = ´ ´ ´ ´ ´  1 10 3600 4 10 10 3 6 19 1.6 = ´ = ´ » 36 10 10 10 18 18 18 6.4 5.6 Hence correct option is (b). 18. The energy released = 4 MeV This energy produced by two atoms. Hence energy produced per atom = = ´ ´  2 2 10 13 MeV 1.6 J Hence number of atom fused to produced 1 kWJ 91 Page 5 5. Here assertion is wrong since bdecay process is n p e v ® + + – but reason is true hence correct option is (d). 6. Here assertion is true but reason is false. Correct option is (c). 7. Here both assertion and reason are true and reason may or may not be true. Correct option is (a, b) 8. Both assertion and reason are true but reason is not correct explanation of assertion. Hence correct option is (b). 9. Here reason is true but assertion is false Q 1 amu = 931.5 MeV Correct option is (d) 10. Both assertion and reason are true but reason does not correctly explain assertion. Hence correct option is (b). 11. Here both assertion and reason are true and reason may or may not be correct explanation of assertion . Hence correct options are (a, b). Objective Questions (Level 2) ¢ Single option correct 1. Let initially substance have N i nuclei then N N e i t =  l dN dt N e i t =   l l At t t = we get dN dt N e N t t i t æ è ç ö ø ÷ =  = =  l l 0 …(i) At t t = 4 dN dt N e N t t i t æ è ç ö ø ÷ =  = =  4 4 0 16 l l …(ii) Dividing Eq. (i) and Eq. (ii) we get e t 3 16 l = …(iii) Now at t t = æ è ç ö ø ÷ 11 2 dN dt N e t t i t æ è ç ö ø ÷ =  =  11 2 11 2 l l =  ´   l l l N e e i t t 8 2 3 2 =  = ´  l l l N e e N i t t 4 3 0 16 16 From Eqs. (ii) and (iii) = N 0 64 Hence, correct option is (b). 2. We have l l l = + = + 1 2 2 30 2 60 log log Þ l = log 2 20 Now N N e t =  0 l Þ N N e t 0 0 2 20 4 =  log Þ log log 4 20 2 = t Þ 2 2 20 2 log log = t Þ t = 40 yr Hence correct option is (c). 3. From graph it is clear that number of nucleons in X is N 3 and binding energy per nucleon is E 3 for Y nucleon is N 2 and BE per nucleon is E 2 . Hence X Y E N E N + = + 3 3 2 2 Similerly W E N = 1 1 88 The reaction is W X Y ® + The energy released is ( ) E N E N E N 3 3 2 2 1 1 +  Hence Correct option is (b). 4. Energy = ´ + ´  ´ ( ) 110 200 8.2 90 8.2 7.4 = ´  200 (8.2 7.4) = ´ = 200 160 0.8 MeV Hence correct option is (d). 5. The reaction is 1 2 1 2 2 4 H H He energy + ® + Energy = ´  ´ (4 7 2 1.1) MeV =  = (28 4.4) 23.6 MeV Hence Correct option is (b). 6. Total energy released per second = ´ 16 10 6 W = ´ 16 10 6 J/s Energy per fission = 200 MeV = ´ ´ ´  200 10 10 6 19 1.6 = ´ ´  2 10 11 1.6 J Q Efficiency = 50% Hence power (energy converted per second) = ´ ´ ´ = ´   2 10 50 100 10 11 11 1.6 1.6 J Number of fission = ´ ´ =  16 10 10 10 6 11 18 1.6 /s Hence correct option is (d). 7. dN dt A N = l Q After time N become conservation Þ dN dt = 0 Þ N A A T AT = = = l log log 2 2 Hence correct option is (d). 8. By conservation of momentum M v M M H H H = + ¢ ( ) n Þ v v ¢ = 2 Let initial KE of Hatom = K Final KE of each Hatom = K 2 For excitation K E E 2 4 13 6 2 1 =  =  + æ è ç ö ø ÷ 13.6 eV . Þ K 2 = 10.2 eV Þ K = ´ ´ ´  2 10 19 10.2 1.6 Þ 1 2 2 10 2 19 M u H 10.2 1.2 1.6 = ´ ´ ´ ´  Þ u H 1.2 10.2 1.6 1.673 = ´ ´ ´ ´ ´ ´  2 2 10 10 19 27 = ´ 6.25 m/s 10 4 Hence correct option is (c). 9. Let us suppose just before the death no radioactive atoms were present hence original activity A 0 is given as A N 0 0 = l …(i) After death the radioactivity decreases exponentially ie, A dN dt N N e t = = =  l l l 0 …(ii) Dividing eq. (ii) by eq. (i) we get A A e t 0 =  l or lt A A = log 0 or t A A = 1 0 l log Now A 0 15 = decay/min/gram A = 375 200 decay/min/g but l = 0.693 yr 5730 89 t = ´ ´ = ´ 5730 15 200 375 5730 200 25 0.693 0.693 log log Þ t = ´ 5730 8 0.693 log = ´ = ´ 5730 3 2 5730 3 0.693 log Þ t = 17190 yr Hence correct option is (c). 10. N N e P t t =  + 0 1 l( ) …(i) and N N e Q t =  0 l Now A N P P = l and A N Q Q = l Þ A A N N e P Q P Q t = =  l 1 Þ lt A A Q P 1 = æ è ç ç ö ø ÷ ÷ log Þ t A A T A A Q P Q P 1 1 = æ è ç ç ö ø ÷ ÷ = æ è ç ç ö ø ÷ ÷ l log log Hence correct option is (b). 11. The given reactions are 1 2 1 2 1 3 H H H + ® + p 1 2 1 2 4 H H He 3 + ® + n Þ 3 H He 1 2 2 4 ® + + n p Mass defect Dm = ´    ( ) 3 2.014 4.001 1.007 1.008 Dm = 0.026 amu Energy Released = ´ = ´  0.026 MeV 3.87 J 931 10 12 This energy produced by the three deutronatoms. Total energy released by 10 40 deutrons = ´ ´ = ´  10 3 10 10 40 12 28 3.87 J 1.29 J The average power P = = 10 10 16 16 W J/s Therefore total time to exhaust all deutrons of the star will be t = ´ » 1.29 s s. 10 10 10 28 16 12 Hence correct option is (c). 12. N N e N e t t t 1 0 0 2 1 1 = =   l log …(i) N N e N e t t t 2 0 0 2 2 2 = =   l log …(ii) R N 1 1 1 = l …(iii) and R N 2 2 2 2 = = l l …(iv) Let after time t, R R 1 2 = then Þ R R 1 1 1 = Þ l l 1 2 1 2 1 ´ = N N Þ l l 1 2 2 1 2 2 1 1 2 = =  é ë ê ù û ú ´ N N e t t t t t log Þ log log t t t t t t t 2 1 2 1 1 2 2 = ´ ´  æ è ç ç ö ø ÷ ÷ Þ t t t t t t t 1 1 2 2 1 2 1 =  ´ 0.693 ( ) log Hence Correct option is (a). 13. The given reaction is Z A X Y 232 90 ¾® + a Þ Z A X Y 232 90 2 ¾® + He 4 Þ Z = 92 and A = 228 Q Initially X is in rest hence momentum of aparticle after decay will be equal and opposite of Y. Þ M v M v Y Y = a a Þ v M M v Y Y = a a Total kinetic energy K M v M v T Y Y = + 1 2 2 2 ( ) a a Þ K M v M M v M T y Y = + é ë ê ù û ú 1 2 2 2 2 2 a a a a 90 Þ K M v M M T Y = + é ë ê ù û ú 1 2 1 2 a a a Þ K K T = + é ë ê ù û ú a 1 4 228 Þ K K T a = 232 228 Hence Correct option is (b). 14. Energy of emitted photon = 7 MeV = ´ ´ ´  7 10 10 6 19 1.6 J = ´  11.2 J 10 13 Momentum of photon = ´ ´  11.2 J 10 m/s 10 3 13 8 = ´  11.2 kgm/s 3 10 21 Q Initial nucleus is stationary Applying conservation of momentum principle O = + Þ =  ® ® ® ® P P P P nuc photon nuc photon Þ     P P ® ® =  nuc photon Þ P nuc 11.2 kg m/s = ´  3 10 21 Mass of nucleus = 24 amu = ´ ´  24 10 27 1.66 kg But P mK nuc nuc 2 2 = Þ K P m nuc nuc 11.2 11.2 10 1.66 Joule = = ´ ´ ´ ´ ´ ´   2 42 27 2 9 2 24 10 Þ K c nuc 11.2 1.66 1.6 eV = ´ ´ ´ ´ ´ ´ ´    112 10 18 24 10 10 42 27 19 . » 1.1 keV Hence correct option is (b). 15. Let time interval between two instants is t 1 then N N e t t 1 0 1 =  + l( ) and N N e t 2 0 2 = l A N N e t t 1 1 0 1 = =  + l l l( ) A N N e t 2 2 0 2 = =  l l l ( ) Þ A A e t 1 2 1 2 1 = ´ l Þ 2 1 2 1 A A e t = l Þ log A A t 2 1 1 2 æ è ç ç ö ø ÷ ÷ = l Þ t A A 1 2 1 1 2 = æ è ç ç ö ø ÷ ÷ l log Þ t T A A 1 2 1 2 2 = æ è ç ç ö ø ÷ ÷ log log Hence Correct option is (c). 16. The given reaction is 1 2 1 2 1 3 1 1 H H H H + ¾® + DM m m m =   [ ( ) ( ) ( )] 2 1 2 1 3 1 1 H H H = ´   ( ) 2 2.014102 3.016049 1.007825 amu = ´  4.33 amu 10 3 DE = ´ ´  4.33 931.5 MeV 10 3 = 4 MeV Hence correct option is (c). 17. Number of fusion required to generate 1 kWh = ´ ´ ´ ´ ´  1 10 3600 4 10 10 3 6 19 1.6 = ´ = ´ » 36 10 10 10 18 18 18 6.4 5.6 Hence correct option is (b). 18. The energy released = 4 MeV This energy produced by two atoms. Hence energy produced per atom = = ´ ´  2 2 10 13 MeV 1.6 J Hence number of atom fused to produced 1 kWJ 91 = ´ ´ ´ = ´  36 10 2 10 18 10 5 13 18 1.6 1.6 Mass of deutrium which contain 18 10 18 1.6 atom ´ = ´ ´ ´ = ´  18 10 10 10 18 23 5 1.6 6.02 3.7 kg. Hence correct option is (c). ¢ More Than one Option is Correct 1. x N = 0 , y N = l 0 Þ x y N N = = 0 0 1 l l where l is decay constant Hence x y is constant throught. Q x y T T = = = 1 1 l 0.693 0.693 Þ x y T > , xy N = l( ) 0 2 For one half life N N = 0 2 Þ ( ) xy N N xy T = æ è ç ö ø ÷ = = l l 0 2 0 2 2 4 4 Hence correct options are (a), (b) and (d). 2. The correct options are (a), (b), (c) and (d). 3. A nucleus in excited state emits a high energy photon called as gray. The reaction is X X * ¾® + g Hence by gamma radiation atomic number and mass number are not changed. Since after emission of one a atomic number reduced by 2 2 4 ( ) a and after 2b atomic number is increased by (2). Hence correct options are (a), (b) and (c). 4. Here half lives are T and 2T and N N x = 0 , N N y = 0 after 4T for first substance = 4 half lives and after 4T the second substance = 2 Half lines. Þ N N N x = æ è ç ö ø ÷ = 0 4 0 1 2 16 N N N y = æ è ç ö ø ÷ = 0 2 0 1 2 4 Þ x N N N N x y = = = 0 0 16 4 1 4 / / Let their activity are R x and R y . Þ R N x x x = l and R N y y y = l Þ y R R N N T T x y x y x y = = ´ = ´ l l 0.693 0.693 2 1 4 Þ R R y x y = = 1 2 Hence correct options are (b) and (c). 5. Since nuclear forces are vary short range charge independent, no electromagnetic and they exchange (n p ® orp n ® ). Hence the correct options are (a), (b), (c) and (d). 6. Q R R A = 0 1 3 / r p p = = ´ ´  M R A R A 4 3 10 4 3 3 27 0 3 / 1.67 kg Þ r is independent of A. But r = ´ ´ ´ ´   1.67 kg 3.14 1.3 10 4 3 10 27 15 3 ( ) = ´ 1.8 kg/m 10 17 3 Hence correct options are (b) and (c). 92Read More
1. What is the concept of modern physics? 
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3. What are the main topics covered in DC Pandey's book on Modern Physics II? 
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