Tiêu đề 2. P2C2 স্থির তড়িৎ-(With Solve).pdf ✅

w ̄’i Zwor  Engineering Question Bank & Practice Book ............................................................................................ 29 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. 4 F aviK‡Z¡i GKwU aviK 200 V G mshy3| aviKwU‡K e ̈vUvwi n‡Z Avjv`v K‡i Gi mv‡_ 2 F aviK‡Z¡i aviK‡K mgvšÍiv‡j hy3 Ki‡j KZ kw3 e ̈q n‡e? wK cwigvY B‡jK‡Uav÷ ̈vwUK kw3 B‡jK‡Uavg ̈vM‡bwUK kw3iƒ‡c wewKwiZ n‡e? [BUET 22-23] mgvavb: Ui = 1 2 CV2 = 1 2 × 4 × 10–6 × 2002  Ui = 0.08 J Uf = Q 2 2Ceq = (8 × 10–4 ) 2 2 × 6 × 10–6  Uf = 0.0533 J  U = Ui – Uf = 0.0267 J (Ans.) Q = CV = 4 × 10–6 × 200 = 8 × 10–4 C 2. GKwU 100 pF aviK‡Z¡i avi‡K mwÂZ kw3i cwigvY wbY©q Ki hLb (i) aviKwU‡Z 4 kV wefe cv_©K ̈ cÖ`vb Kiv nq Ges (ii) avi‡Ki cÖwZwU cv‡Zi PvR© 60 nC| [BUET 20-21] mgvavb: (i) mwÂZ kw3, U1 = 1 2 CV2 = 1 2  100  10–12  (4  103 ) 2  U1 = 8  10–4 J (Ans.) (ii) mwÂZ kw3, U2 = Q 2 2C = (60  10–9 ) 2 2  100  10–12  U2 = 1.8  10–5 J (Ans.) 3. GKwU avi‡Ki cvZ؇qi ga ̈eZ©x `~iZ¡ 5 mm n‡j Gi g‡a ̈ ivLv 10 wU B‡jKUab enbKvix GKwU †Z‡ji †duvUvi fvimvg ̈ iÿv Ki‡Z KZ wefe cÖ‡qvM Ki‡Z n‡e Zv wbY©q Ki| †`Iqv Av‡Q, †Z‡ji †duvUvi fi 3  10–16 kg| [BUET 19-20] mgvavb: mvg ̈ve ̄’vq, FC = FG  qE = mg  q V d = mg Fc + + + + + + – – – – – – – 5 mm FG 10 e E –  V = 3  10–16  9.8  5  10–3 10  1.6  10–19  V = 9.1875 V (Ans.) 4. GKwU mgvšÍivj cvZ avi‡Ki cÖwZwU cv‡Zi †ÿÎdj 0.05 m2 | cvZ؇qi ga ̈eZ©x gva ̈g k~b ̈; G‡`i g‡a ̈ `~iZ¡ 0.0015 m Ges wefe cv_©K ̈ 50 V n‡jÑ (i) avi‡Ki aviKZ¡ (ii) cvZ `ywUi g‡a ̈ mwÂZ kw3 Ges (iii) avi‡Ki GKK AvqZ‡bi mwÂZ kw3 wbY©q Ki| [0 = 8.85  10–12 Fm–1 ] [BUET 18-19] mgvavb: (i) aviKZ¡, C = 0kA d = 8.85  10–12  1  0.05 0.0015  C = 2.95  10–10 F (Ans.) (ii) mwÂZ kw3, U = 1 2 CV2 = 1 2  2.95  10–10  502  U = 3.6875  10–7 J (Ans.) (iii) GKK AvqZ‡b mwÂZ kw3 = U Ad = 3.6875  10–7 0.05  0.0015 = 4.9167  10–3 Jm–3 (Ans.) 5. wP‡Î cÖ`wk©Z •e`y ̈wZK eZ©bxi AskUzKz mvg ̈e ̄’vq i‡q‡Q Ges †iva ̧‡jvi g‡a ̈ wWwm Kv‡i›U cÖevwnZ n‡”Q| aviK C = 4 F Gi g‡a ̈ mwÂZ kw3 wbY©q Ki| [BUET 18-19] 3A 2A 4V 3 3 5 1 3 3 2A 3V 3 4 C 4F 1A mgvavb: 3A 2A 5 1 3 2A 4A 1A 5A 1A 5A
30 ......................................................................................................................................  Physics 2nd Paper Chapter-2 avi‡Ki `yB cÖv‡šÍi wefe cv_©K ̈, V = 5  5 + 5  1 + 3  1  V = 33 V  avi‡K mwÂZ kw3, U = 1 2 CV2  U = 1 2  4  10–6  332  U = 2.178  10–3 J (Ans.) 6. c„w_ex c„‡ôi mwbœK‡U evqyk~b ̈ ̄’v‡b y A‡ÿi, y = 10 m we›`y‡Z GKwU B‡jKUab Aew ̄’Z| y A‡ÿi †Kvb we›`y‡Z cÖ_g B‡jKUa‡bi mv‡c‡ÿ wØZxq B‡jKUab ivL‡j, Zv‡`i ga ̈w ̄’Z w ̄’iwe`y ̈Zxq ej, cÖ_g B‡jKUa‡bi Dci wμqvkxj gva ̈vKl©Y e‡ji fvimvg ̈ iÿv Ki‡e? [g = 9.8 ms–2 ] [BUET 18-19] mgvavb: y 2q e – 1g e – FC FG mvg ̈ve ̄’vq, FC = FG  ke2 y 2 = meg  9  109  (1.6  10–19) 2 y 2 = 9.11  10–31  9.8  y = 5.08 m  2q e – Gi y A‡ÿ Ae ̄’vb = 10 – 5.08 m = 4.92 m (Ans.) 7. GKwU mgvšÍivj cvZ avi‡Ki cÖwZwU cv‡Zi †ÿÎdj 200 cm2 Ges evqy‡Z cvZ؇qi ga ̈eZ©x `~iZ¡ 0.4 cm n‡j GiÑ (i) aviKZ¡ wbY©q Ki| (ii) hw` aviKwU 500 V •e`y ̈wZK Dr‡mi mv‡_ ms‡hvM Kiv nq, Z‡e avi‡K KZ kw3 mwÂZ n‡e? [BUET 17-18] mgvavb: (i) avi‡Ki aviKZ¡, C = 0kA d  C = 8.854  10–12  1  200  10–4 0.4  10–2  C = 4.427  10–11 F (Ans.) (ii) mwÂZ kw3, U = 1 2 CV2  U = 1 2  4.427  10–11  5002  U = 5.533  10–6 J (Ans.) 8. cÖwZwU 220 V G PvwR©Z mgAvKv‡ii AvUwU †QvU †MvjvKvi †duvUv‡K wgwjZ K‡i GKwU eo †duvUvq cwiYZ Kiv nj| eo †duvUvi wefe KZ n‡e? [BUET 14-15, 11-12] mgvavb: G‡ÿ‡Î †gvU AvqZb aaæeK _v‡K| 4 3 R 3 = n  4 3 r 3  R = n 1 3 r †QvU †duvUvi wefe, Vs = 1 40  q r eo †duvUvi wefe, Vb = 1 40 Q R = 1 40  nq 3 nr  Vb = n 2 3 Vs = 8 2 3  220  Vb = 880 V (Ans.) 9. 4.0  10–8 C ÿz`a gv‡bi mgvb I wecixZ RvZxq Avavb 6.0 cm e ̈eav‡b A I B we›`y‡Z Aew ̄’Z| Avavb؇qi ms‡hvM mij‡jLv AB Gi j¤^ mgwØLÛ‡Ki Dci 4.0 cm `~‡i P we›`y‡Z ̄’vwcZ 1.0  10–8 C Avav‡bi Dci wμqvkxj ej wbY©q Ki|     1 40 = 9  109 Nm2C –2 [BUET 13-14] mgvavb: Fnet F F 4cm 3cm 5cm 40nC –40nC  10nC  A B P F = 1 40  q1q2 r 2  F = 9  109  (10  40)  10–18 0.052  F = 1.44  10–3 N Avevi,  = 2 tan–1     3 4   =  – 2 tan–1     3 4 = 106.26  wμqvkxj jwä e‡ji gvb, Fnet = 2F cos     2 = 2  1.44  10–3  cos     106.26 2 = 1.728  10–3 N (Ans.) e‡ji w`K = tan–1     F sin F + F cos = 53.13, PB Gi mv‡_| (Ans.)
w ̄’i Zwor  Engineering Question Bank & Practice Book ............................................................................................. 31 10. 2 F aviKZ¡wewkó GKwU aviK‡K PvwR©Z Kivi ci GKwU cwievnx Zvi Øviv GwU‡K PvR© gy3 Kiv nj| avi‡K mwÂZ mg ̄Í kw3B ZviwU‡K DËß Ki‡Z LiP nj| GB kw3i cwigvY 214.3 K ̈vjwi n‡j, KZ †fv‡ë aviKwU‡K PvwR©Z Kiv n‡qwQj? [BUET 13-14] mgvavb: avi‡K mwÂZ kw3 = Drcbœ Zvckw3  1 2 CV2 = Q  214.3  4.2 = 1 2  2  10–6  V 2  V  30001 V (Ans.) 11. evqy gva ̈‡g 50000 Vm–1 mylg •e`y ̈wZK †ÿ‡Î `ywU e„ËvKvi cvZ 0.002 m `~i‡Z¡ mgvšÍivj Ae ̄’vq Av‡Q| cÖwZwU cv‡Zi e ̈vmva© 0.08 m| MwVZ aviKwU‡Z †gvU mwÂZ kw3 wbY©q Ki| [BUET 10-11] mgvavb: mwÂZ kw3, U = 1 2 CV2  U = 1 2      0kA d  (Ed)2  U = 1 2  8.854  10–12  1    0.082  (5  104 ) 2  0.002  U = 4.45  10–7 J (Ans.) 12. 3  10–10 C Avavbhy3 GKwU †MvjvKvi †Z‡ji †duvUvi Z‡ji wefe 500 V| hw` GiKg `ywU †duvUv wg‡j GKwU †MvjvKvi †duvUvi m„wó nq, Zvn‡j D3 †duvUvi Z‡ji wefe KZ n‡e? [0 = 8.85  10–12 C 2N –1 m –2 ] [BUET 08-09] mgvavb: G‡ÿ‡Î †gvU AvqZb aaæeK _v‡K| 4 3 R 3 = n  4 3 r 3  R = n 1 3 r †QvU †duvUvi wefe, Vs = 1 40  q r eo †duvUvi wefe, Vb = 1 40  Q R = 1 4 0  nq 3 nr  Vb = n 2 3 Vs = 2 2 3  500  Vb = 793.7 V (Ans.) 13. 20 C wewkó GKwU PvR© •e`y ̈wZK †ÿÎ •Zwi K‡i| PvR©wU †_‡K 10 cm Ges 5 cm `~i‡Z¡ `ywU we›`yi Ae ̄’vb| GKwU we›`y n‡Z Aci we›`y‡Z GKwU B‡jKUab wb‡Z Kv‡Ri cwigvY †ei Ki| [BUET 06-07] mgvavb: Avgiv Rvwb, K...ZKvR, W = Qq 40     1 rf – 1 ri  W = 9  109  20  10–6  1.6  10–19     1 0.05 – 1 0.1  W = 2.88  10–13 J (Ans.) 14. wZbwU avi‡Ki aviKZ¡ h_vμ‡g 5 F, 10 F Ges 1 F| G‡`i cÖ_g I Z...ZxqwU‡K †kÖwY‡Z mshy3 K‡i wØZxqwUi mv‡_ mgvšÍiv‡j mshy3 Kiv n‡j Zzj ̈ aviKZ¡ wbY©q Ki|[BUET 04-05; BUTex 00-01] mgvavb: 10F 5 F 1 F Zzj ̈ aviKZ¡, Ceq = 10 + 5  1 5 + 1  Ceq = 10.833 F (Ans.) 15. PviwU aviK, hvi cÖ‡Z ̈KwUi aviKZ¡ 20 F mgvšÍivj mgš^‡q ivLv n‡q‡Q| 2 V e ̈vUvixi m‡1⁄2 G‡K mshy3 K‡i ms‡hvM wew”Qbœ Kiv nj| KZ PvR© GB aviK ̧‡jv‡Z Rgv n‡e? [BUET 01-02] mgvavb: cÖwZwU avi‡Ki wefe cv_©K ̈ = 2 V  Q1 = Q2 = Q3 = Q4 = CV = 20  10–6  2 = 40 C  cÖwZwU avi‡K 40 C PvR© mwÂZ n‡e| (Ans.) 16. 3 F I 6 F aviK‡Z¡i `ywU aviK‡K †kÖwY mgev‡q hy3 K‡i eZ©bxi `yB cÖv‡šÍ 12 V Gi GKwU e ̈vUvwi ms‡hvM †`qv n‡jvÑ (i) eZ©bxi †gvU aviKZ¡ KZ? (ii) cÖ‡Z ̈KwU avi‡Ki wefe cv_©K ̈ KZ? (iii) cÖ‡Z ̈K avi‡K mwÂZ kw3i cwigvY KZ? [BUET 00-01] mgvavb: 3 F 6 F (i) †gvU aviKZ¡, Ceq = 3  6 3 + 6 = 2 F (Ans.) (ii) †gvU cÖevwnZ PvR©, Q = Ceq  V = 2  12 = 24 C  3 F Gi wefe cv_©K ̈, V1 = Q C1 = 24 C 3 F = 8 V (Ans.)  6 F Gi wefe cv_©K ̈, V2 = 12 – 8  V2 = 4 V (Ans.) (iii) 3 F avi‡Ki mwÂZ kw3, U1 = 1 2 C1V 2 1 = 1 2  3  10–6  8 2  U1 = 9.6  10–5 J (Ans.) 6 F avi‡Ki mwÂZ kw3, U2 = 1 2 C2V 2 2 = 1 2  6  10–6  4 2  U2 = 4.8  10–5 J (Ans.)
32 ......................................................................................................................................  Physics 2nd Paper Chapter-2 weMZ mv‡j KUET-G Avmv cÖkœvejx 1. 20 †m.wg. e ̈vmv‡a©i GKwU cwievnx †Mvj‡Ki Zj mylgfv‡e 3 gvB‡μv-Kzj¤^ Avav‡b AvwnZ| †Mvj‡Ki c„‡ô Ges †K›`a †_‡K 30 †m.wg. `~‡i †Kv‡bv we›`y‡Z Zwor cÖvej ̈ wbY©q Ki| [KUET 19-20] mgvavb: †Mvj‡Ki c„‡ô cÖvej ̈, E1 = 1 40  Q R 2  E = 9  109  3  10–6 0.22  E1 = 6.75  105 NC–1 (Ans.) †Mvj‡Ki †K›`a †_‡K 30 cm `~‡i cÖvej ̈, E2 = 1 40  Q r 2  E = 9  109  3  10–6 0.32  E2 = 3  105 NC–1 (Ans.) 2. 12 F Ges 24 F aviK‡Z¡i `yBwU aviK †kÖwYe×fv‡e mshy3 Ki‡j aviKZ¡ KZ n‡e? G‡`i `yB cÖvšÍ 40 V Gi GKwU e ̈vUvixi mv‡_ mshy3 Ki‡j GwU KZ PvR© MÖnY Ki‡e? GKwU 100  †ivaK D3 aviK `yBwUi mv‡_ †kÖwY ms‡hvM Ki‡j M„nxZ Pv‡R©i wK cwieZ©b n‡e? [KUET 03-04] mgvavb: Zzj ̈ aviKZ¡, Ceq = 12  24 12 + 24 = 8 F (Ans.) M„nxZ †gvU PvR©, Q = CV = 8  10–6  40  Q = 3.2  10–4 C (Ans.) 100  †ivaK hy3 Ki‡j M„nxZ Pv‡R©i †Kv‡bv cwieZ©b n‡e bv| †Kbbv e ̈vUvwii wefe cv_©K ̈ I Zzj ̈ aviKZ¡ DfqB aaæeK _vK‡e| (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 1. 1.0 m evûwewkó GKwU eM©‡ÿ‡Îi cÖwZwU †KvYvq 5  10–9 C PvR© ̄’vcb Kiv n‡jv, eM©‡ÿ‡Îi †K‡›`a wefe wbY©q Ki| [RUET 07-08] mgvavb: 1m 1 2 m 5nC 5nC 5nC 5nC e‡M©i †K‡›`a wefe, V = V1 + V2 + V3 + V4  V = 1 40  1 r (q1 + q2 + q3 + q4)  V = 9  109  1 1 2  (4  5  10–9 )  V = 254.558 V (Ans.) 2. PviwU 4 F Gi aviK‡K 100 V e ̈vUvwii mwnZÑ (i) mgvšÍivj ms‡hvM w`‡j wK cwigvY •e`y ̈wZK kw3 mwÂZ n‡e? (ii) wmwiR ms‡hvM w`‡j wK cwigvY •e`y ̈wZK kw3 mwÂZ n‡e? [RUET 03-04] mgvavb: (i) mgvšÍivj ms‡hv‡M, Ceq = 4  4 = 16 F  mwÂZ wefekw3, U = 1 2 CeqV 2  U = 1 2  16  10–6  1002  U = 0.08 J (Ans.) (ii) wmwiR ms‡hv‡M, Ceq = 4 4 = 1 F  mwÂZ wefekw3, U = 1 2 CeqV 2  U = 1 2  1  10–6  1002  U = 5  10–3 J (Ans.) weMZ mv‡j CUET-G Avmv cÖkœvejx 1. 1 bs wP‡Î V1, V2, V3 †fv‡ë‡Ri gvb wbY©q Ki Ges mswkøó K ̈vcvwmU‡ii PvR© wbY©q Ki| [CUET 07-08] V2 V3 C2 C3 6F 12F V1 10 V + – wPÎ-1 C1 6F mgvavb: C1 avi‡Ki †ÿ‡Î, V1 = 10 V (Ans.)  Q1 = C1V1  Q1 = 6  10–6  10  Q1 = 60 C (Ans.) Avevi, Cs = C2C3 C2 + C3 = 6  12 6 + 12 = 4 F  Q2 = Q3 = Cs  10 = 40 C (Ans.)  V2 = Q2 C2 = 40 C 6 F = 6.667 V (Ans.)  V3 = Q3 C3 = 40 C 12 F = 3.33 V (Ans.) 2. 12 C Ges 8 C `ywU we›`y PvR© ci ̄úi †_‡K 10 †m.wg. `~‡i Aew ̄’Z| PvR© `ywU‡K 6 †m.wg. e ̈eav‡b wb‡q Avm‡Z KZUzKz KvR Ki‡Z n‡e| PvR© `ywU k~‡b ̈ Aew ̄’Z| [CUET 05-06] mgvavb: Avgiv Rvwb, K...ZKvR, W = Qq 40     1 rf – 1 ri  W = 12  10–6  8  10–6  9  109      1 0.06 – 1 0.1  W = 5.76 J (Ans.)