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Content text 2. OLYMPIAD M.F. VOL 2 KEY SOL (77-100).pdf

X – Physics (Vol–II) Olympiad Class Work Book ELECTROMAGNETISM_KEY & SOLUTIONS WORKSHEET-1 CUQ’S 1.4 2.4 3.1 4.4 5.NO 6.1 7.3 8.3 9.3 10.3 11.3 12.4 JEE MAIN & ADVANCED 1.1 2.1 3.2 4.1 5.3 6.2 7.2 8.3 9.4 10.1 11.1 12.1 13.4 14.1 15.2 16.2 17.2 18.3 19.2 20.1,2,3 21.1 22. 4 23.1 24. 3 25.a-q,r;b-p,r:c-p,r;d-q 26.1,2,3,4 27.2 28.1 29.1 30.4 31.1 32.1 33.a-p;b-s;c-r;d-q 34.4 35.4 36. a-q; b-r; c-s; d-p CUQ SOLUTIONS 9. Given q q 1 2  V V 1 2  ; B B 1 2  mv 1 2 R , but K.E mv Bq 2   2 2K.E 2ev v m m   2 2 2 2ev m m v 2mev m R Bq Bq Bq    2 1 2 2 R R m R   10. F V  ; 1 2 F v F 2v  ;F 2F 2  11. 1 2 mv qv 2  2qv r m   and F qvB  12. F qvBsin  
Olympiad Class Work Book X – Physics (Vol–II) JEE MAIN & ADVANCED SOLUTIONS 1. F q V B     6 6 2 i j k F 4 10 2 3 1 10 10 2 5 3                6 6 2 F 4 10 i 4 j 8 k 16 10 10                6 2 F 4 4i 8j 16k 10 10         F 0.16i 0.32j 0.64k N ˆ ˆ ˆ     2.  0 F qvBsin60 15 4.65 10           19 3 3 1.6 10 v 3.5 10 2         15 22 3 4.6 10 1.6 10 3.5 v 2 15 22 4.6 10 10 2 v 1.6 3.5 3         7 9.2k 10 5.6 1.732    V 7   0.947 10 6   9.47 10 m/s 3. F BqV 90       19 5 F 0.8 2 1.6 10 10       14 F 2.56 10 N    4. F q V B     19 6 2 i j k F 1.6 10 2 3 0 10 10 3 15 0                15 F 1.6 10 i 0 j 0 k 39              F 6.24 10 N k   14 ˆ
X – Physics (Vol–II) Olympiad Class Work Book 5. F BqV sin   19 7 F 3 1.6 10 10 2        12 F 2.4 10 N    6. 2mk R Bq  ; k R B   1 1 2 2 2 1 R k B R k B ;  1 2 R k 3B R 2k B 2 R 3 R 2  ; 2 2R R 3  7. 2mk R Bq  ; 2mV 1 R q B   31 3 2 19 2 9.1 10 2 10 1 18 10 1.6 10 B            4 B 8.38 10 T    8. E V B  ; Bqr E m B      3 2 6 1 20 10 5 10 E 20 10 1          E 50 V/m  9. x Vsin T    ; 2 m x vsin Bq     2 mvsin ,0,0 Bq          10.  2mk r Bq (k and B are constants) m m;m 2m;m 4m p duet     q e; q e; q 2e p due t      p m m r ; r q e
Olympiad Class Work Book X – Physics (Vol–II) duet     2m 4m 2m r ; r e 2e 2e p d r r r     11. F q V B          19 6 i j k F 1.6 10 2 2 0 10 0 5 0   13 F 1.6 10 k 10         F 1.6 10 k    13 ˆ 12 F 1.6 10 N     12. F q V B     13 19 6 x y z i j k 1.28 10 k 1.6 10 2 3 0 10 ˆ B B B          1.28 10 i 13ˆ     13 z z y x 1.610 i 3B 2B j k 2B 3B    ˆ       Compare kˆ components both sides 2B 3B 0.8 3B 2B 0.8 y z x y        magneticfield is along ve y direction y B 0.4j   ˆ 13. We know that F B     F B 0 ma B 0   ; a B 0   ˆ ˆ ˆ ˆ 2 xi 2j 3i 4j 10 0               3x 8 0   ; 8 2 x m/s 3   14. F q V B     6 6 1 i j k F 2 10 2 0 0 10 10 0 2 4       

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