Tiêu đề 3.-P2C3.-Current-Electricity-2024_With-Solve_Ridoy-6.5.24-PDF.pdf ✅

Pj Zwor  Final Revision Batch 1 Z...Zxq Aa ̈vq Pj Zwor Current Electricity Topicwise CQ Trend Analysis UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU †iv‡ai Ici ZvcgvÎvi cÖfve Ñ Ñ Ñ 1 1 2 – 4 †iva I Av‡cwÿK †iva Ñ 1 Ñ Ñ Ñ 6 – 7 Ry‡ji Zvcxq wμqvi m~Î, Zwor ÿgZv, Zwor kw3 Ñ 1 Ñ 3 6 8 5 23 †Kv‡li Af ̈šÍixY †iva Ges Zwo”PvjK ej Ñ Ñ Ñ Ñ 2 Ñ – 2 †iv‡ai †kÖwY I mgvšÍivj mgš^q ms‡hvM, Zzj ̈ †iva 2 6 1 3 1 5 – 18 eZ©bxi cÖevngvÎv 2 6 1 2 9 4 4 28 †Kv‡li †kÖwY I mgvšÍivj ms‡hvM 1 Ñ 1 2 1 Ñ – 5 In‡gi m~Îvbymv‡i †fv‡ëR Kv‡i›U †jLwPÎ Ñ 1 Ñ Ñ Ñ 1 1 3 wKk©‡di m~Î: m~‡Îi aviYv, eZ©bx‡Z e ̈envi 1 1 1 1 2 Ñ 4 10 ûBU‡÷vb eaxR bxwZ Ñ 1 Ñ 1 2 4 2 10 A ̈vwgUvi I †fvëwgUv‡ii cvV Ñ Ñ Ñ Ñ 3 1 – 4 A ̈wgUvi Gi cvjøv e„w× I kv‡›Ui e ̈envi 1 1 Ñ Ñ 3 6 1 12 * we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| DÏxcKwU jÿ ̈ Ki: [Xv. †ev. 23] 8 B 16 32 D A C P Q R S 24 G r = 0.5 E = 12V (K) X = 16 (K) 1 Kzj¤^ Pv‡R©i msÁv `vI| (L) eZ©bx‡Z †Kv‡li ZworPvjK ej m¤ú~Y© Kvh©Ki nq bv †Kb?Ñ e ̈vL ̈v Ki| (M) M ̈vjfv‡bvwgUvi wew”Qbœ Ae ̄’vq Zwor cÖevn wbY©q Ki| mgvavb: 8  16  24  32  12 V 0.5  Req = (24–1 + 56–1 ) –1  Req = 16.8   Zwor cÖevn, I = V Req + r  I = 12 16.8 + 0.5  I = 0.6936 A  M ̈vjfv‡bvwgUvi wewQbœ Ae ̄’vq Zwor cÖevn 0.6936 A| (Ans.) (N) DÏxc‡Ki ‘X’ †ivawU, wea‡R cÖ`Ë †Kv‡bv GKwU †iv‡ai mv‡_ e ̈envi K‡i mvg ̈e ̄’v m„wó m¤¢e wKbv? MvwYwZK e ̈vL ̈v `vI| mgvavb: eZ©bx‡Z S †iva‡K cwieZ©b K‡i mvg ̈ve ̄’v m„wó Kiv m¤¢e| Avgiv Rvwb, mvg ̈ve ̄’vq, P Q = R S  S = 16  24 8 = 48   S †iv‡ai 32  Gi mv‡_ DÏxc‡Ki 16  †iva †kÖwY‡Z mshy3 Ki‡j, S = 32 + 16 = 48  n‡e| (Ans.) 2| R=20 S M T (K) N E = 12V evwZ [iv. †ev. 23]
2  HSC Physics 2nd Paper Chapter-3 M, N Ges S wZbwU †iva hv‡`i gvb h_vμ‡g 18 , 39  Ges 40 | T †ivawU cwieZ©bkxj| cÖv_wgK Ae ̄’vq Pvwe K wU †Lvjv †i‡L evwZwU 2 sec mg‡q Drcbœ Zv‡ci cwieZ©b wbY©q Kiv nq| (K) Zwor w؇giæ Kv‡K e‡j? (L) GKB gv‡bi †Kv‡li mgvšÍivj mgev‡q m‡e©v”P cwigvY ZworcÖevnv cvIqvi †ÿ‡Î Pvwe Af ̈šÍixY †iva Kxiƒc nIqv cÖ‡qvRb? e ̈vL ̈v K‡iv| (M) 1g †ÿ‡Î Pvwe †Lvjv Ae ̄’vq Drcbœ Zv‡ci cwigvY wbY©q K‡iv| mgvavb: 18  40  20  12V 36  Pvwe †Lvjv Ae ̄’vq, Zzj ̈ †iva, Req = 18 + 36 [] 60 36 + 60  Req = 40.5   g~j Zwor cÖevn, I = V Req = 12 40.5 = 0.2963 A  ev‡j¦i ga ̈ w`‡q Zwor cÖevn, I1 = 60–1 60–1 + 36–1  0.2963  I1 = 0.111 A  ev‡j¦i Drcbœ Zvc, H = I2 1Rt = 0.1112  20  2 = 0.493 J (Ans.) (N) PvwewU eÜ Ae ̄’vq cwieZ©bkxj ‘T’ †iva‡K Kx Ae ̄’v Aej¤^b Ki‡j Drm _vKv m‡Ë¡I evwZwU wb‡f hv‡e? mwVK eZ©bx wPÎ AsKbc~e©K MvwYwZKfv‡e we‡kølY K‡iv| mgvavb: Pvwe eÜ Ae ̄’vq, ûBU‡÷vb weaRwU mvg ̈ve ̄’vq Avm‡j evwZwU wb‡f hv‡e, mvg ̈ve ̄’vq, T M = S N  T = 40 36  18 = 20  myZivs T †iv‡ai gvb 20  n‡j evwZwU wb‡f hv‡e| (Ans.) 3| E1 = 12 V – E2 = 14 V r2 = 4  r1 = 4  + – + A B I1 I2 15  R [Kz. †ev. 23] (K) wKk©‡di wØZxq m~ÎwU wee„Z Ki| (L) g ̈v1⁄2vwb‡Ri †iv‡ai DòZv mnM 3  10–5 K –1 ej‡Z †evSvq? (M) DÏxc‡K eZ©bxi I1 wbY©q K‡iv| mgvavb: 12 V 4  15  I 14 V 4  I2 I1 cÖ_g jy‡c KVL cÖ‡qvM K‡i, 12 – 4I1 + 4I2 – 14 = 0  – 4I + 4I = 2 ......... (i) wØZxq jy‡c, – 15 I + 14 – 4I2 = 0  – 15(I1 + I2) – 4I2 = – 14 [∵ I = I1 + I2]  – 15I1 – 19I2 = – 14 ......... (ii) (i) I (ii) mgvavb K‡i, I1 = 0.132 A (Ans.) (N) DÏxc‡K †Kvl `ywU‡K Kxfv‡e hy3 Ki‡j ewnt‡iv‡a AwaK Zvc Drcbœ n‡e? MvwYwZKfv‡e we‡kølY K‡iv| mgvavb: 4  12 V I 4  14 V 15  ÔMÕ n‡Z cvB, I1 = 0.132 A, I2 = 0.632  I = I1 + I2 = 0.764 A Avevi, Zwor †Kvl `ywU †kÖwY‡Z mshy3 _vK‡j, I = E R = 12 + 14 15 + 4 + 4 = 1.130 A Avgiv Rvwb, †iv‡a Drcbœ Zvc, H = I2Rt  H  I 2  H1 H2 = I I = 0.764 1.130 = 0.676 < 1  H1 < H2 myZivs †Kvl `ywU‡K †kÖwY‡Z ms‡hvM w`‡j ewnt‡iv‡a AwaK Zvc Drcbœ n‡e| (Ans.) 4| 90  †iv‡ai GKwU M ̈vjfv‡bvwgUv‡ii mv‡_ 1  †iva mgvšÍiv‡j hy3 Av‡Q| eZ©bxi g~j cÖevn 3 A| GB M ̈vjfv‡bvwgUvi Øviv 3 A Gi AwaK Zwor cÖevn gvcv m¤¢e| [h. †ev. 23] (K) Kvk©‡di cÖ_g m~ÎwU wee„Z Ki| (L) Zvgvi Av‡cwÿK †iva 1.56 × 10–8  ej‡Z Kx eySvq? (M) DÏxc‡K 1  †iv‡ai ga ̈ w`‡q cÖevwnZ Zwo‡Zi gvb wbY©q Ki|

4  HSC Physics 2nd Paper Chapter-3 (M) DÏxc‡Ki ZviwUi •`N© ̈ 12.5 cm I cÖ ̄’‡”Q‡`i †ÿÎdj 0.1 cm2 n‡j Dcv`v‡bi Av‡cwÿK †iva KZ? mgvavb:†iva, R =  L A   = AR L = 0.1  10–4  12.5 12.5  10–2   = 10–3 m myZivs Dcv`v‡bi Av‡cwÿK †iva 10–3 m| (Ans.) (N) DÏxc‡Ki cvwb wK dzU‡e? †Zvgvi DËi MvwYwZK we‡køl‡Y `vI| mgvavb:Zv‡ii AvqZb AcwiewZ©Z _vK‡j, L1A1 = L2A2  A1 A2 = L2 L1 = 2L1 L1 = 2 R =  L A  R  L A  R2 R1 = L2 L1  A1 A2 = 2  2  R2 = 4  12.5 = 50  GLb, H = Q  V 2 R2 t = ms  2202 50  305 = m  4200  (100 – 30)  m = 1.0042 kg myZivs, m‡e©v”P 1.0042 kg cvwb _vK‡j, cvwb dzU‡e| (Ans.) 7| R1 = 10 R2 = 20 R3 = 10 R4 = 20 3V 3V 3V 3V R5 = 10 B C A [e. †ev. 23] (K) aviKZ¡ Kx? (L) †Kv‡bv e ̄‘‡Z Pv‡R©i gvb wbiew”Qbœ n‡Z cv‡i bvÑ e ̈vL ̈v Ki| (M) eZ©bxi AB As‡ki wefe cv_©K ̈ KZ? mgvavb: A 10  10  10  20  20  B 6 V eZ©bxi mijxK...Z wPÎÑ  RP = 30  30 30 + 30 = 15  A 10  B 6 V 15   VAB = 15 15 + 10  6 = 3.6  myZivs, AB As‡ki wefe cv_©K ̈ 3.6 | (Ans.) (N) R5 †iva AcmviY K‡i eZ©bx m¤ú~Y© Ki‡j g~j cÖev‡ni Kxiƒc cwieZ©b n‡e? MvwYwZKfv‡e we‡kølY Ki| mgvavb: 10  6 V 15  R5 AcmviY Kivi c~‡e©, I1 = V Req = 6 15 + 10 = 0.24 A 6 V 15  R5 AcmviY Kivi ci, I2 = V Req = 6 15 = 0.4 A  I2 I1 = 0.4 0.24 = 1.67  I2 = 1.67 I1 myZivs, R5 AcmviY Ki‡j eZ©bxi g~j Zwor g~j Zwor cÖevn c~‡e©i 1.67 ̧Y n‡e| (Ans.) 8| r1 = 0.2  r1 = 0.3  R = 10  R = 10  eZ©bx-1 eZ©bx-2 E1 = 3 V E2 = 6 V E1 = 3 V E2 = 6 V r2 = 0.3  r1 = 0.2  [wm. †ev. 23] (K) kv›U Kv‡K e‡j? (L) wb‡K‡ji †iv‡ai DòZv mnM 6  10–3 C –1 ej‡Z Kx eyS? (M) eZ©bx-1 Gi ewnt‡iva R Gi ga ̈ w`‡q cÖevwnZ Zwor cÖevn wbY©q K‡iv| mgvavb: Zwor cÖevn, I = Eeq Req + r = 3 + 6 10 + (0.2 + 0.3)  I = 0.857 A myZivs, ewnt‡iva R Gi g‡a ̈ w`‡q Zwor cÖevn 0.857 A| (Ans.)