Content text 2. OLYMPIAD CURRENT ELECTRICIT KEY SOL VOL-1.pdf
X– Physics (Vol – I) Olympiad Classwork CURRENT ELECTRICITY_KEY & SOLUTIONS WORKSHEET-1_KEY CUQ 1. 1 2. 1 3. 2 4. 1 5. 1 6. 1 7. 1 8. 4 9. 2 10. 3 11. 1 JEE MAINS & ADVANCED 1. 2 2. 3 3. 2 4. 1 5. 1 6. 2 7. 4 8. 3 9. 3 10. 2 11. 4 12. 2 13. 3 14. 3 15. 2 16. 2 17. 4 18. 2 19. 1 20. 2 21. 2 22. 1 23. 1 24. 1 25. 1 26. a-p,t; b-p,s; c-q,r; d-q 27. 1 28. 1 29. 2 30. 4 SOLUTIONS 1) 17 19 ne 6 10 1.6 10 i 1.6 t 60 mA 2) q 900 3 i 1.5 t 10 60 2 A 3) 19 4 1.6 10 q ne i t t n t 19 19 n 4 2.5 10 t 1.6 10 4) 6 i A A 2 2 10 t = 5 minutes = 5 60 sec Q = ? Q i t 6 Q it 2 10 5 60 4 Q 6 10 C. 5) d eE e v V m m l ; V v d 6) (n n )e e p t
Olympiad Classwork X – Physics (Vol – I) 7) i = 1A 2 6 2 A 1mm 1 10 m Atomic weight (A) = 63.5 amu 3 63..5 10 Kg Avogadro number (N) = 23 6.023 10 Density of metal (d)= 3 900Kgm Number of electrons (no) = 1 free electron density (n) = Nnd A 23 3 6.023 10 1 9000 n 63.5 10 = 29 0.853 10 d 29 19 6 e i 1 V n A 0.853 10 1.6 10 1 10 1 V 0.074mms d 8) d i AnqV 2 2 A A A B B B V i r i r v V i r 1 9 10 9 15 . 1.2 4 12 4 8 9) d i Anv q i AV 2 2 1 1 1 2 V i A 2i 1 . 1 V i A i 2 V V V 2 1 10) d e V ( )E m 11) Free electron density (n) = 28 3 8.5 10 m i = 1A A = 2 4 2 1cm 1 10 m d i V neA
X– Physics (Vol – I) Olympiad Classwork 28 19 14 1 8.5 10 1.6 10 10 = 5 0.0735 10 m /s 6 V 0.735 10 m /s 0.735 m /s d 4 6 10 d 0.735 10 l V 10 1.3605 10 60 60 24 365 = 43.1 yrs. 12) d d 1 2 V v,v ? 1 2 r r,r 2r ; 1 2 i i,i 2i 1 2 2 d 2 2 d 1 V i r v 2i r 2 2 d V i 2r v 2i r d2 V v 2 13) 2 1 t t q idt 2 I 3 4 t t 2 3 4 dq t t dt 2 dq t t dt 3 4 3 2 1 dq t t dt 3 4 2 3 3. 4. 2 3 t t 3 4 2 3 3 3 12 3 1 2 3 = 3 4 8 28 2 3 112 12 3 36 112 140 3 3
Olympiad Classwork X – Physics (Vol – I) 14) d 1 V A 15) d i V neA d i v neA 28 19 5 16 4 10 1.6 10 10 d v 16 4 16 10 1 3 10 4 10 4 10 4 4 2.5 10 / d v m s 16) i = 3.6A t = 3h = 3 60 60s Q = ? Q = it Q 3.6 3 60 60 4 Q 3.9 10 c 17) n = 1 19 e 1.6 10 c 6 v 2.2 10 m/s 11 r 5.3 10 m 6 11 v 2.2 10 f 2 r 2 3.142 5.3 10 17 0.066 10 f = 6.6 rps neV i 2 r (or) i = -nef 19 i ( 1) ( 1.6 10 ) (6.6) 3 i 1.06 10 A i 1.06 mA 18. d i v neA 28 19 8 4 10 1.6 10 10