Title 2. P2C2.-Static-Electricity For FRB_2024_With Solve_Sha 5.4.24 PDF.pdf ✅

w ̄’i Zwor  Final Revision Batch 1 wØZxq Aa ̈vq w ̄’i Zwor Electrostatics Topicwise CQ Trend Analysis UwcK 2016 2017 2018 2019 2022 2022 2023 †gvU Kzj¤^ m~Î I †ÿÎ Ñ Ñ Ñ Ñ 1 Ñ 3 4 we›`y Pv‡R©i Zwor ej 1 3 Ñ 1 1 Ñ Ñ 6 Zwor †ÿÎ cÖvej ̈ 3 3 1 2 3 6 4 22 Zwor wefe 2 2 1 4 4 7 8 28 avi‡Ki aviYv, aviKZ¡, †kÖwY I mgvšÍivj ms‡hvM, Zzj ̈ aviKZ¡, kw3 I e ̈envi 3 3 Ñ 3 7 5 9 30 * we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb, 2021 mv‡j kU© wm‡jev‡m Aa ̈vqwU AšÍfz©3 wQjbv| weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| DÏxcKwU jÿ ̈ Ki: [Xv. †ev. 23] 1mm + + + + + + – – – – – – M N 1mm 6mm +10C –10C cÖwZwU cv‡Zi †ÿÎdj 2 cm2 ; 0 = 8.854 × 10–12 C 2N –1m –2 | (K) •e`y ̈wZK w؇cv‡ji msÁv `vI| (L)  iwk¥ •e`y ̈wZK †ÿÎ Øviv wewÿß nq bv|Ñ e ̈vL ̈v Ki| (M) DÏxc‡Ki aviKwUi aviKZ¡ wbY©q Ki| mgvavb: C = 0kA d = 8.854  10–12  1  2  10–4 6  10–3 F = 2.95  10–13 F (Ans.) (N) DÏxc‡Ki M we›`y n‡Z N we›`y‡Z + 2C Avavb‡K wb‡Z †Kv‡bv KvR m¤úbœ n‡e wK? MvwYwZK we‡køl‡Yi gva ̈‡g †`LvI| mgvavb: W = Vq = Edq = Q 0A dq = 10 8.854  10–12  2  10–4  (6 – 2)  10–3  2 J = 4.52  10–13 J A_©vr, M n‡Z N we›`y‡Z +2C Avavb‡K wb‡Z 4.52  10–13 J KvR m¤úbœ n‡e| (Ans.) 2| Q2 = – 6 nC Q1 = – 2.25 nC r = 2m r r2=25 cm 1 = 15 cm [iv. †ev. 23] wP‡Î `ywU duvcv †Mvj‡Ki c„‡ô PvR© cÖ`vb Kiv n‡q‡Q| (K) mv›U Kv‡K e‡j? (L) Zwor †ÿ‡Îi aviYv e ̈vL ̈vq Kzj‡¤^i m~‡Îi mxgve×Zv e ̈vL ̈v K‡iv| (M) †MvjK؇qi ms‡hvRK mij‡iLv eivei wbi‡cÿ we›`yi Ae ̄’vb wbY©q K‡iv| mgvavb: A x 2 – x Q1 Q2 awi, Q1 PvR© †_‡K x `~i‡Z¡ A we›`y‡Z wbi‡cÿ we›`y cvIqv hv‡e|  E1 = E2  1 40 . Q1 x 2 = 1 40 . Q2 (2 – x) 2  2.25  10–9 x 2 = 6  10–9 (2 – x) 2  2 – x =1.63 x  x = 0.759 m  Q1 PvR© †_‡K 0.759 m `~i‡Z¡ wbi‡cÿ we›`y cvIqv hv‡e| (Ans.) (N) †MvjKØq GKwU cwievnx Zvi Øviv mshy3 Kiv n‡j †MvjK؇qi P‚ovšÍ Pv‡R©i cwigv‡Yi Zzjbvg~jK we‡kølY K‡iv| mgvavb: V1 = 1 40  Q1 r1 = 9  109  –2.25  10–9 15  10–2 V = – 135 V
2  HSC Physics 2nd Paper Chapter-2 V2 = 1 40  Q2 r2 = 9  109  – 6  0 –9 25  10–2 V = – 216 V  V2 < V1 awi, x cwigvY PvR© wØZxq †MvjK †_‡K cÖ_g †Mvj‡K ̄’vbvšÍwiZ n‡j †MvjK؇qi wefe mgvb n‡e|  V 1 = V 2  1 40 . Q1 + x r1 = 1 40 . Q2 – x r2  – 2.25 + x 0.15 = – 6 – x 0.25  5(x – 2.25) = – 3(x + 6)  5x – 11.25 = – 3x – 18  x = – 0.84 nC  Q 1 = Q1 + x = (– 2.25 – 0.84) nC = – 3.09 nC Q 2 = Q2 – x = (– 6 + 0.84) nC = – 5.16 nC  †MvjKØq GKwU cwievnx Zvi Øviv hy3 Kiv n‡j †MvjK؇q P‚ovšÍ Pv‡R©i cwigvY n‡e – 3.09 nC I – 5.16 nC| (Ans.) 3| q1 = 4nC A q2 = 8nC D B C P [Kz. †ev. 23] wP‡Î GKwU e„ËvKvi c‡_i A Ges B we›`y‡Z `ywU PvR© ̄’vcb Kiv n‡q‡Q| P e„‡Ëi †K›`a, e„‡Ëi e ̈vmva© 15 cm Ges AB = BC = CD = DA| (K) Zwor d¬v· Kv‡K e‡j? (L) †Kv‡bv e ̄‘i Avavb 7.5e n‡Z cv‡i bv †Kb? e ̈vL ̈v K‡iv| (M) DÏxc‡Ki P we›`y‡Z Zwor cÖve‡j ̈i gvb wbY©q K‡iv| mgvavb: EAP = 1 40  q1 AP2 = 9  109  4  10–9 0.152 NC–1 = 1600 NC–1 ; AP eivei EBP = 1 40  q2 BP2 = 9  109  8  10–9 0.152 NC–1 = 3200 NC–1 ; BP eivei EP = E 2 AP + E2 BP = 16002 + 32002 NC–1 = 3577.7 NC–1  P we›`y‡Z Zwor cÖve‡j ̈i gvb 3577.7 NC–1 | (Ans.) (N) C I D we›`yi mv‡_ GKwU cwievnx Zvi hy3 Ki‡j, cwievnxi gy3 B‡jKUab ̧‡jv †Kvb w`‡K MwZkxj n‡e? MvwYwZKfv‡e gZvgZ `vI| mgvavb: VC = VAC + VBC = 1 40 q1 AC + 1 40 q2 BC = 9  109     4  10–9 0.15 2 + 8  10–9 0.15  2 V = 459.4 V VD = VAD + VBD = 1 40 q1 AD + 1 40 q2 BD = 9  109     4  10–9 0.15 2 + 8  10–9 0.15  2 V = 409.7 V  VC > VD  C I D we›`yØq‡K cwievnx Zvi Øviv Ki‡j hy3 Ki‡j, cwievnxi gy3 B‡jKUab ̧‡jv D we›`y †_‡K C we›`yi w`‡K MwZkxj n‡e| (Ans.) 4| C3 = 10F C4 = 20F C1 = 5F C2 = 15F 10 V [Kz. †ev. 23] (K) aviK Kv‡K e‡j? (L) aviK Ges Zwor †Kv‡li Zzjbv Ki| (M) eZ©bxi Zzj ̈ aviKZ¡ †ei Ki| mgvavb: C2 I C3 mgvšÍiv‡j hy3,  CP = C2 + C3 = (15 + 10) F = 25 F C1, CP I C4 †kÖwY mgev‡q hy3  Ceq = (5–1 + 25–1 + 20–1 ) –1 F = 3.45 F  Zzj ̈ aviKZ¡ 3.45 F| (Ans.) (N) eZ©bxi †Kvb aviKwU‡Z mwÂZ kw3 me‡P‡q †ewk? MvwYwZK we‡kølYc~e©K gZvgZ `vI| mgvavb: Q = CeqV = 3.45  10 C = 34.5 C C1 avi‡Ki `yB cÖv‡šÍi wefe cv_©K ̈ V1 = Q C1 = 34.5 5 V = 6.9 V C2 I C3 aviK؇qi `yB cÖv‡šÍi wefe cv_©K ̈, V2 = V3 = Q CP = 34.5 25 = 1.38 V
w ̄’i Zwor  Final Revision Batch 3 C4 avi‡Ki `yB cÖv‡šÍi wefe cv_©K ̈ V4 = Q C4 = 34.5 20 V = 1.725 V  U1 = 1 2 C1V 2 1 = 1 2  5  10–6  6.92 J = 1.19  10–4 J U2 = 1 2 C2V 2 2 = 1 2  15  10–6  1.382 J = 1.43  10–5 J U3 = 1 2 C3V 2 3 = 1 2  10  10–6  1.382 J = 9.52  10–6 J U4 = 1 2 C4V 2 4 = 1 2  10  10–6  1.7252 J = 2.98  10–5 J  U1 > U4 > U2 > U3  C1 avi‡K mwÂZ kw3i cwigvY me‡P‡q †ewk| (Ans.) 5| [h. †ev. 23] 30 4m P A B – 4C O + 4C 0.1 mm 0.1 mm (K) Bwćbi Kg©`ÿZv Kx? (L) PvwR©Z †MvjvKvi cwievnxi †K‡›`a Zwor cÖvej ̈ k~b ̈ nq †Kb? (M) DÏxc‡Ki P we›`y‡Z Zwor wefe †ei Ki| mgvavb: V = 1 40 . P cos r 2 = 9  109  4  10–6  2  10–4  cos30 4 2 V = 0.3897 V (Ans.) (N) OP †iLv w؇giæi ga ̈ we›`y‡Z h_vμ‡g 0 Ges 90 †KvY Drcbœ Ki‡j P we›`y‡Z Zwor cÖve‡j ̈i Kxiƒc cwieZ©b n‡e? MvwYwZK e ̈vL ̈v Ki| mgvavb: P we›`y cÖvej ̈, E = 1 40 . P 1 + 3 cos2  r 3 = 9  109  4  10–6  2  10–4 1 + 3.cos2 30 4 3 NC–1 = 0.2028 NC–1 OP †iLv ga ̈we›`y‡Z 0 †KvY Drcbœ Ki‡j, P we›`y‡Z cÖvej ̈, E1 = 1 40 . P 1 + 3 cos2  r 3 = 9  109  4  10–6  2  10–4 1 + 3.cos2 0 4 3 NC–1 = 0.225 NC–1 OP †iLv ga ̈we›`y‡Z 90 †KvY Drcbœ Ki‡j, P we›`y‡Z cÖvej ̈, E2 = 1 40 . P 1 + 3 cos2  r 3 = 9  109  4  10–6  2  10–4 1 + 3.cos2 90 4 3 NC–1 = 0.1125 NC–1 E1 = E1 – E = 0.222 NC–1 E2 = E – E2 = 0.903 NC–1  OP †iLv w؇giæi ga ̈we›`y‡Z 0 †KvY Drcbœ Ki‡j P we›`y‡Z ZworcÖvej ̈ 0.0222 NC–1 e„w× cvq Ges 90 †KvY Drcbœ Ki‡j ZworcÖvej ̈ 0.0903 NC–1 n«vm cvq| (Ans.) 6| A P(2kg) 2m 2m 1.676C 1.676C B 2m C Øv`k †kÖwYi QvÎx jvweev ejj, A we›`y‡Z ̄’vwcZ P e ̄‘wU Szj‡e| wKš‘ Zvi evÜex jvwgmv ejj, GwU m¤¢e bq| [h. †ev. 23] (K) MvD‡mi m~ÎwU wee„Z Ki| (L) PvwR©Z †MvjvKvi cwievnxi †K›`a †_‡K `~iZ¡ ebvg wefe †jLwP‡Îi cÖK...wZ e ̈vL ̈v Ki| (M) DÏxc‡Ki A we›`yi wef‡ei gvb KZ? mgvavb: V = VB + VC = 1 40 . q1 AB + 1 40 . q2 AC = 9  109     1.676 2 + 1.676 2 V = 1.5084  1010 V A_©vr A we›`yi wef‡ei gvb 1.5084  1010 V| (Ans.) (N) DÏxc‡Ki DwjøwLZ `yB R‡bi g‡a ̈ Kvi Dw3wU mwVK? MvwYwZKfv‡e hvPvB Ki| mgvavb: B C W  A FC Fnet FB   awi, P e ̄‘wUi Avavb q2  FB = FC = 1 40 . q.q2 2 2 = 9  109  1.676 q2 4 N = 3.771  109 q2 N
4  HSC Physics 2nd Paper Chapter-2 Fnet = F 2 B + F2 C + 2FBFC cos60 = 2FB cos  2 = 2  3.771  109 q2  cos30 N = 6.53  109 q2 N e ̄‘wU Szj‡e hw` Fnet = W nq  6.53  109 q2 = 2  9.8  q2 = 3  10–9 C = 3 nC  P e ̄‘i PvR© 3 nC n‡j jvweevi Dw3 mwVK n‡e| Ab ̈_vq jvwgmvi Dw3 mwVK n‡e| (Ans.) 7| 20 cm2 20 cm2 200 V 3cm [P. †ev. 23] 20 cm2 †ÿÎd‡ji `ywU avZe cvZ‡K 3 cm e ̈eav‡b †i‡L wPÎ Abyhvqx aviK •Zwi Kiv nj| cieZ©x‡Z K = 5 Ges 2 mm cyiæZ¡wewkó GKwU eøK cvZ؇qi gv‡S †i‡L aviKZ¡ wbY©q Kiv n‡jv| (K) 1 eV ej‡Z Kx eyS? (L) †Kv‡bv avi‡Ki Mv‡q 0.025 A – 220 V †jLv _vK‡j Kx eySvqÑ e ̈vL ̈v K‡iv| (M) cÖ_g †ÿ‡Î cvZØq m¤ú~Y© PvwR©Z n‡j mwÂZ kw3i cwigvY wbY©q Ki| mgvavb: C = 0kA d = 8.854  10–12  2  10–3 3  10–2 F = 5.902  10–13 F  U = 1 2 CV2 = 1 2  5.902  10–13  2002 J = 1.18  10–8 J  cÖ_g †ÿ‡Î cvZ؇qi mwÂZ kw3i cwigvY 1.18 10–8 J| (Ans.) (N) 1g †ÿ‡Îi Zzjbvq 2q †ÿ‡Î avi‡Ki Kxiƒc cwieZ©b n‡eÑ MvwYwZKfv‡e wbY©q Ki| mgvavb: M n‡Z cvB, C1 = 5.902  10–13 F 2q †ÿ‡Î, C2 = 0A d – t     1 – 1 k = 8.854  10–12  20  10–4 0.03 – 0.002     1 – 1 5 F = 6.235  10–13 F  aviK‡Z¡i cweZ©b = C2 – C1 = (6.235 – 5.902)  10–13 F = 0.333  10–13 F e„w× cv‡e|  1g †ÿ‡Îi Zzjbvq 2q †ÿ‡Îi avi‡Ki aviKZ¡ 0.333  10–13 F e„w× cv‡e| (Ans.) 8| + – C1 C2 C3 9V wPÎvbyhvqx eZ©bx‡Z hy3 aviK ̧‡jvi cÖwZwUi gvb 900 pF| cvZ؇qi e ̈eavb 0.4 cm| 0 = 8.85 × 10–12 C 2N –1m –2 | [e. †ev. 23] (K) B‡jKUa‡bi Zvob †eM Kv‡K e‡j? (L) Kvk©‡di cÖ_g m~Î Pv‡R©i msiÿY bxwZ †g‡b P‡jÑ e ̈vL ̈v Ki| (M) avi‡Ki †h †Kv‡bv GKwU cv‡Zi †ÿÎdj wbY©q Ki| mgvavb: C = 0kA d  900  10–12 = 8.854  10–12  1  A 0.4  10–2  A = 0.407 m2 (Ans.) (N) C3 aviK‡K AcmviY K‡i C1 avi‡Ki g‡a ̈ KvMR (K = 3) Øviv c~Y© Kiv n‡j, eZ©bxi mwÂZ Zwor kw3 c~e©v‡cÿv †ewk n‡e wKbv? MvwYwZKfv‡e hvPvB Ki| mgvavb: C1 = C2 = C3 = C = 900  10–12 F  Ceq = C3 + C1C2 C1 + C2 = C + C 2 = 3 2 C = 3 2  900  10–12 F = 1.35  10–9 F  eZ©bxi mwÂZ kw3, U1 = 1 2 CeqV 2 = 1 2  1.35  10–9  9 2 J = 5.47  10–8 J C 1 = kC = 3  900  10–12 F = 27  10–10 F wØZxq †ÿ‡Î Zzj ̈ aviKZ¡ C eq = C 1C2 C 1 + C2 = 2700  900  10–24 (2700 + 900)  10–12 F = 675  10–12 F