Page 1 1 Class XII PHYSICS SQP Marking Scheme 201920 Section – A 1. a , ?= (for one face) 1 2. b , Conductor 1 3. a , 1?. 1 4. c ,12.0kJ 1 5. a , speed 1 6. d, virtual and inverted 1 7. a, straight line 1 8. d, 60 O 1 9. b, work function 1 10. b, third orbit 1 11. 45 O or vertical 1 12. 2 H 1 13. double 1 14. 1.227 A o 1 15. 60° 1 16. Difference in initial mass energy and energy associated with mass of products Or Total Kinetic energy gained in the process 1 17. Increases 1 18. N o/8 1 19. 0.79 eV 1 20. Diodes with band gap energy in the visible spectrum range can function as LED 1 Page 2 1 Class XII PHYSICS SQP Marking Scheme 201920 Section – A 1. a , ?= (for one face) 1 2. b , Conductor 1 3. a , 1?. 1 4. c ,12.0kJ 1 5. a , speed 1 6. d, virtual and inverted 1 7. a, straight line 1 8. d, 60 O 1 9. b, work function 1 10. b, third orbit 1 11. 45 O or vertical 1 12. 2 H 1 13. double 1 14. 1.227 A o 1 15. 60° 1 16. Difference in initial mass energy and energy associated with mass of products Or Total Kinetic energy gained in the process 1 17. Increases 1 18. N o/8 1 19. 0.79 eV 1 20. Diodes with band gap energy in the visible spectrum range can function as LED 1 2 OR Any one use Section – B 21. When electric field E is applied on conductor force acting on free electrons = e m = e = Average thermal velocity of electron in conductor is zero (u t) av= 0 Average velocity of electron in conductors in t (relaxation time) = v d (drift velocity) v d = (u t) av + a t v d = 0 + = 1 1 22. C 2 and C 3 are in series = + = 1 = 1µf & C 4 are in ? C” = 1 + 1 = 2µf C” & c 5 are in series = + ? = 1µf & c 1 are in ? C eq = 1 + 1 = 2µf Energy stored U = cv 2 = ×2×10 6 ×6 2 = 36×10 6 J 1 1 Page 3 1 Class XII PHYSICS SQP Marking Scheme 201920 Section – A 1. a , ?= (for one face) 1 2. b , Conductor 1 3. a , 1?. 1 4. c ,12.0kJ 1 5. a , speed 1 6. d, virtual and inverted 1 7. a, straight line 1 8. d, 60 O 1 9. b, work function 1 10. b, third orbit 1 11. 45 O or vertical 1 12. 2 H 1 13. double 1 14. 1.227 A o 1 15. 60° 1 16. Difference in initial mass energy and energy associated with mass of products Or Total Kinetic energy gained in the process 1 17. Increases 1 18. N o/8 1 19. 0.79 eV 1 20. Diodes with band gap energy in the visible spectrum range can function as LED 1 2 OR Any one use Section – B 21. When electric field E is applied on conductor force acting on free electrons = e m = e = Average thermal velocity of electron in conductor is zero (u t) av= 0 Average velocity of electron in conductors in t (relaxation time) = v d (drift velocity) v d = (u t) av + a t v d = 0 + = 1 1 22. C 2 and C 3 are in series = + = 1 = 1µf & C 4 are in ? C” = 1 + 1 = 2µf C” & c 5 are in series = + ? = 1µf & c 1 are in ? C eq = 1 + 1 = 2µf Energy stored U = cv 2 = ×2×10 6 ×6 2 = 36×10 6 J 1 1 3 23. Gain in KE of particle = Qv = K P = q pV p (i)V p = V ? =V = K ? = q ?V ? (ii) (ii)/(i) = = = = = v ? : v p = 1: 1 1 24. “The angle of incidence at which the reflected light is completely plane polarized, is called as Brewster’s angle (i B) At i = i B, reflected beam 1 to refracted beam ? i B + r = 90 ? r = 90i B Using snell’s law = µ = µ ? = µ µ = tan 1 1 25. wave function ? = 2.14eV (a) Threshold frequency ? = h? 0 ? 0 = = 1 Page 4 1 Class XII PHYSICS SQP Marking Scheme 201920 Section – A 1. a , ?= (for one face) 1 2. b , Conductor 1 3. a , 1?. 1 4. c ,12.0kJ 1 5. a , speed 1 6. d, virtual and inverted 1 7. a, straight line 1 8. d, 60 O 1 9. b, work function 1 10. b, third orbit 1 11. 45 O or vertical 1 12. 2 H 1 13. double 1 14. 1.227 A o 1 15. 60° 1 16. Difference in initial mass energy and energy associated with mass of products Or Total Kinetic energy gained in the process 1 17. Increases 1 18. N o/8 1 19. 0.79 eV 1 20. Diodes with band gap energy in the visible spectrum range can function as LED 1 2 OR Any one use Section – B 21. When electric field E is applied on conductor force acting on free electrons = e m = e = Average thermal velocity of electron in conductor is zero (u t) av= 0 Average velocity of electron in conductors in t (relaxation time) = v d (drift velocity) v d = (u t) av + a t v d = 0 + = 1 1 22. C 2 and C 3 are in series = + = 1 = 1µf & C 4 are in ? C” = 1 + 1 = 2µf C” & c 5 are in series = + ? = 1µf & c 1 are in ? C eq = 1 + 1 = 2µf Energy stored U = cv 2 = ×2×10 6 ×6 2 = 36×10 6 J 1 1 3 23. Gain in KE of particle = Qv = K P = q pV p (i)V p = V ? =V = K ? = q ?V ? (ii) (ii)/(i) = = = = = v ? : v p = 1: 1 1 24. “The angle of incidence at which the reflected light is completely plane polarized, is called as Brewster’s angle (i B) At i = i B, reflected beam 1 to refracted beam ? i B + r = 90 ? r = 90i B Using snell’s law = µ = µ ? = µ µ = tan 1 1 25. wave function ? = 2.14eV (a) Threshold frequency ? = h? 0 ? 0 = = 1 4 = 5.17× H z (b) As k max = eV 0 = 0.6eV Energy of photon E = k max + ? = 0.6eV + 2.14eV = 2.74eV Wave length of photon ? = = = 4530Å 1 26. centripetal force = electrostatic attraction = = (i) as = n . = put in (i) m . = = OR Energy of electron in n = 2 is 3.4eV ? energy in ground state = 13.6eV kE = TE = +13.6eV 1 1 1 E n = ? 3.4eV = ? energy in ground state x =  13.6eV. Page 5 1 Class XII PHYSICS SQP Marking Scheme 201920 Section – A 1. a , ?= (for one face) 1 2. b , Conductor 1 3. a , 1?. 1 4. c ,12.0kJ 1 5. a , speed 1 6. d, virtual and inverted 1 7. a, straight line 1 8. d, 60 O 1 9. b, work function 1 10. b, third orbit 1 11. 45 O or vertical 1 12. 2 H 1 13. double 1 14. 1.227 A o 1 15. 60° 1 16. Difference in initial mass energy and energy associated with mass of products Or Total Kinetic energy gained in the process 1 17. Increases 1 18. N o/8 1 19. 0.79 eV 1 20. Diodes with band gap energy in the visible spectrum range can function as LED 1 2 OR Any one use Section – B 21. When electric field E is applied on conductor force acting on free electrons = e m = e = Average thermal velocity of electron in conductor is zero (u t) av= 0 Average velocity of electron in conductors in t (relaxation time) = v d (drift velocity) v d = (u t) av + a t v d = 0 + = 1 1 22. C 2 and C 3 are in series = + = 1 = 1µf & C 4 are in ? C” = 1 + 1 = 2µf C” & c 5 are in series = + ? = 1µf & c 1 are in ? C eq = 1 + 1 = 2µf Energy stored U = cv 2 = ×2×10 6 ×6 2 = 36×10 6 J 1 1 3 23. Gain in KE of particle = Qv = K P = q pV p (i)V p = V ? =V = K ? = q ?V ? (ii) (ii)/(i) = = = = = v ? : v p = 1: 1 1 24. “The angle of incidence at which the reflected light is completely plane polarized, is called as Brewster’s angle (i B) At i = i B, reflected beam 1 to refracted beam ? i B + r = 90 ? r = 90i B Using snell’s law = µ = µ ? = µ µ = tan 1 1 25. wave function ? = 2.14eV (a) Threshold frequency ? = h? 0 ? 0 = = 1 4 = 5.17× H z (b) As k max = eV 0 = 0.6eV Energy of photon E = k max + ? = 0.6eV + 2.14eV = 2.74eV Wave length of photon ? = = = 4530Å 1 26. centripetal force = electrostatic attraction = = (i) as = n . = put in (i) m . = = OR Energy of electron in n = 2 is 3.4eV ? energy in ground state = 13.6eV kE = TE = +13.6eV 1 1 1 E n = ? 3.4eV = ? energy in ground state x =  13.6eV. 5 PE = 2TE = 2×13.6eV = 27.2eV 1 27. Ptype semiconductor ntype semiconductor 1. Density of holes >> density of electron 2. Formed by doping trivalent impurity 1. density of electron>>density of holes 2. formed by doping pentavalent impurity Energy band diagram for ptype Energy band diagram of ntype semiconductor OR Energy of photon E = = eV =2.06eV As E<E g (2.8eV), so photodiode cannot detect this photon. Any 2x1 =1 1 1 Section – C 28. Principle of potentiometer, when a constant current flows through a wire of uniform area of crosssection, the potential drop across any length of the wire is directly proportional to the length. Let resistance of wire AB be R 1 and its length be ‘l’ then current drawn from driving cell – I = and hence P.D. across the wire AB will be V AB = IR 1 = × Where ‘a’ is area of crosssection of wire AB ? = = constant = k Where R increases, current and potential difference across wire AB will be 1 1 1Read More
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
134 videos388 docs213 tests
