Title 3. P2C3. চল তড়িৎ (With Solve).pdf ✅

Pj Zwor  Engineering Practice Sheet ............................................................................................................................. 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. eZ©bxi Zwor cÖevn KZ? a I b we›`yi wefe cv_©K ̈ KZ? [BUET 22-23] b 4.4 V 2.1 V 6  1.5  1.2  a mgvavb: Applying KVL, – 4.4 + 2.1 + 1.2I + 6I + 1.5I = 0  8.7I = 2.3  I = 0.264 A (Ans.) Vab = 2.1 + 1.2I + 6I = 2.1 + (7.2 × 0.264) V = 4.003 V (Ans.) 2. wb‡Pi mvwK©UwUi cÖwZwU †iv‡a cÖevwnZ Kv‡i›U wbY©q Ki| A I B we›`yi wefe cv_©K ̈ wbY©q Ki| [BUET 21-22] 2 1 4 3 4V 2V 3V A B mgvavb: 2 1 4 4V 3 2V 3V A C D B O i1 i3 i2 AODA jy‡c, – 4 + i1 + 2 (i1 – i2) = 0  3i1 – 2i2 = 4 ........ (i) DOBD jy‡c, – 3 + 2 (i2 – i1) + 3 (i2 – i3) = 0  – 2i1 + 5i2 – 3i3 = 3 ....... (ii) COBC jy‡c, 2 + 3 (i3 – i2) + 4i3 = 0  – 3i2 + 7i3 = – 2 ......... (iii) (i), (ii) I (iii) n‡Z cvB, i1 = 2.68 A i2 = 2.02 A i3 = 0.58 A  1  †iv‡a cÖevn = i1 = 2.68 A (Ans) 2  †iv‡a cÖevn = i1 – i2 = 0.66 A (Ans) 3  †iv‡a cÖevn = i2 – i3 = 1.44 A (Ans) 4  †iv‡a cÖevn = i3 = 0.58 A (Ans.) VAB = (4 + 3) V = 7 V (Ans.) 3. GKwU Zvgvi Zvi‡K •`N© ̈ eivei †U‡b 0.1% e„w× Ki‡j, Zv‡ii †iva KZ kZvsk e„w× cv‡e? [BUET 21-22] mgvavb: R2 R1 = L2 L1  A1 A2 = 1.001L1 L1  1.001A2 A2 = 1.002001 A1L1= A2L2  A1 A2 = L2 L1  R R1  100% = 0.2001% (Ans.) 4. wP‡Î cÖ`wk©Z eZ©bx‡Z †fvëwgUv‡ii cvV KZ n‡e wbY©q Ki| ai †h, †fvëwgUviwU h_vh_ †cvjvwiwU‡Z ms‡hvM Kiv n‡q‡Q| [BUET 19-20] 5 5 4 a 10 20V 5 5V 30V b I2 I1 V mgvavb: I1 = 20 10 = 2 A I2 = 30 9 = 10 3 A  Vab = 20 – 5I1 + 5 + 0 – 4I2 = 25 – (5 × 2) – 4 × 10 3 = 5 3 V (Ans.)
2 ........................................................................................................................................  Physics 2nd Paper Chapter-3 5. Rbve Avj Avwgb Zvi †kvevi N‡i 1 Ton Gi GKwU GqviKwÛkbvi ̄’vcb Ki‡jb| Gqvi KwÛkbviwU Pvjv‡bv Ae ̄’vq 220 V mvcøvB jvBb †_‡K 6.5 A Kv‡i›U †bq| wZwb M‡o •`wbK 8 hrs K‡i Gqvi KwÛkbviwU e ̈envi K‡ib| hw` we`y ̈‡Zi wej cÖwZ kWh Gi Rb ̈ 5 Tk nq Z‡e H Gqvi KwÛkbviwU GwcÖj gv‡m e ̈env‡ii Rb ̈ we`y ̈r wej KZ n‡e Zv wbY©q Ki| [BUET 18-19] mgvavb: e ̈eüZ we`y ̈r kw3, W = VIt = 220 × 6.5 × 10–3 × 8 × 30 kWh = 343.2 kWh  Bill = 343.2 × 5 UvKv = 1716 UvKv (Ans.) 6. GKwU 1.5 kW B‡jw±aK †KZjx‡Z 2 wjUvi cvwb wb‡q Mig Ki‡j Zv 6 min 20 sec ci dzU‡Z ïiæ K‡i| cÖ_‡g †KZjx‡Z cvwbi ZvcgvÎv KZ wQj? †KZjx‡Z cvwb †dvUv‡Z KZ unit we`y ̈r LiP n‡q‡Q? (Zvcÿq bMY ̈ aiv †h‡Z cv‡i) [BUET 17-18] mgvavb: Pt = mS  1.5 × 103 × (6 × 60 + 20) = 2 × 4200  (100 – )   = 32.14C (Ans.) H = 1.5 ×     6  60 + 20 3600 kWh = 0.1583 kWh = 0.1583 unit (Ans.) 7. wb‡Pi wP‡Î cÖ`wk©Z eZ©bx‡Z voltmeter Gi cvV 5 V, ammeter Gi cvV 2 A Ges we`y ̈r cÖev‡ni w`K Zxi wP‡ýi gva ̈‡g cÖ`wk©Z n‡q‡Q| (a) †ivaK R Gi gvb I (b) E Gi gvb wbY©q Ki| [BUET 16-17] E V A 3 2 10 5V R mgvavb: (a) 3  †iv‡ai g‡a ̈ cÖevn k~b ̈ nIqvq †fvëwgUv‡ii cvV 5 V  R †iv‡ai g‡a ̈ cÖevn 2 A  R = 5 2  = 2.5  (b) – E + 10I + 2I + 2.5I = 0  E = 2 (10 + 2 + 2.5)  E = 29 V (Ans.) 8. 220 V G Kvh©iZ 100 W Gi GKwU evwZi wdjv‡g‡›Ui ga ̈ w`‡q cÖwZ †m‡K‡Û cÖevwnZ B‡jKUa‡bi msL ̈v wbY©q Ki| †`Iqv Av‡Q, B‡jKUa‡bi PvR©, e = 1.6  10–19 C| [BUET 14-15] mgvavb: I = P V  ne t = 100 220  n = 100 220 × 1 1.6 × 10–19 = 2.84 × 1018 wU (Ans.) 9. GKwU bgybvq w؇hvRx avZzi AvqZb 4.0  10–5 m 3 | avZzwUi NbZ¡ 9.0 gcm–3 Ges AvbweK fi 60 gmol–1 | bgybvq KqwU cwievnx B‡jKUab i‡q‡Q? [A ̈v‡fv‡M‡Wav msL ̈v = 6.02  2023 mol–1 ]| [BUET 14-15] mgvavb: m = V = 9 × 106 × 4 × 10–5 g = 360 g B‡jKUab msL ̈v = 2 × w M NA = 2 × 360 60 × 6.02 × 1023 wU = 7.224 × 1024 wU (Ans.) 10. wP‡Î cÖ`wk©Z mvwK©‡U myBP k †Lvjv Ae ̄’vq Kv‡i›U I1, I2 Ges I3 Gi gvb wbY©q Ki| [BUET 14-15] 12V 9V c d b a 7 8 4 I1 I2 I3 k mgvavb: myBP †Lvjv Ae ̄’vq, I3 = 0 A (Ans.) I1 = I2 = 12 – 9 15 A = 0.2 A (Ans.) 11. 2.35 m j¤^v Ges 1.63 mm e ̈vm wewkó Gjywgwbqv‡gi Zv‡ii wfZi w`‡q 1.24 A we`y ̈r cÖevwnZ n‡”Q| GB Zv‡i wK cwigvY kw3 e ̈q n‡”Q? [A ̈vjywgwbqvg Gi †ivav1⁄4,  = 2.80  10–8 m] [BUET 14-15] mgvavb: R =  L A = 2.8 × 10–8 × 2.35  × (0.5 × 1.63 × 10–3 ) 2  = 0.0315  cÖwZ †m‡K‡Û e ̈wqZ kw3, P = I2R = 1.242 × 0.0315 W = 0.0485 W (Ans.)
Pj Zwor  Engineering Practice Sheet ............................................................................................................................ 3 12. we`y ̈‡Zi e ̈envi Kgv‡bvi j‡ÿ ̈ GKwU mvaviY 60 W GSL evwZ‡K GKwU 13 W CFL evwZ w`‡q e`jv‡bv nj| evj¦ `ywUi g~j ̈ h_vμ‡g Tk. 30 Ges Tk. 250| cÖwZ BDwbU we`y ̈‡Zi `vg Tk. 4 n‡j GK eQ‡ii g‡a ̈ evj¦wU e`jv‡bvi LiP DVv‡Z cÖwZw`b M‡o KZ NÈv CFL evj¦wU‡K R¡vjv‡Z n‡e? [BUET 12-13] mgvavb: mvkÖqK...Z kw3 = 60 – 13 = 47 W  mvkÖqK...Z kw3 Øviv LiP DVv‡Z n‡e = 250 – 30 = 220 Tk.  (250 – 30) = (60 – 13) × 10–3 × t × 365 × 4  t = 220 4 × 0.047 × 365 hr = 3.206 hr (Ans.) 13. 6 V-Gi GKwU e ̈vUvixi Af ̈šÍixY †iva 0.25 | Ab ̈ GKwU 0.5  Af ̈šÍixY †ivawewkó 3 V e ̈vUvixi mv‡_ mgvšÍiv‡j ms‡hvM Ki‡j D3 mgev‡qi cÖvšÍ؇qi wefe cv_©K ̈ wbY©q Ki| [BUET 11-12] mgvavb: I I A C B 3V 6V 0.25 0.5 D V ABCDA jy‡c KVL e ̈envi K‡i cvB, – 6 + 3 + 0.5I + 0.25I = 0  0.75I = 3  I = 4 A  VAC = 3 + 0.5I = (3 + 2) V = 5 V (Ans.) 14. GKwU •e`y ̈wZK Bw ̄¿‡Z ‘220 V – 1000 W’ †jLv Av‡Q| Bw ̄¿wU 200 V jvB‡b hy3 n‡q 2 NÈv Pj‡j KZ BDwbU we`y ̈r kw3 LiP Ki‡e? [BUET 09-10] mgvavb: R = V 2 P = 2202 1000  = 48.4  W = V 2 2 R  t = 2002 48.4 × 10–3 × 2 kWh  W = 1.653 Unit (Ans.) 15. 5 ohm †ivawewkó GKwU Zvi‡K †U‡b wZb ̧Y j¤^v Kiv nj| j¤^vK...Z ZviwUi †iva wbY©q Ki| [BUET 08-09, 04-05, 02-03] mgvavb: †U‡b j¤^v Ki‡j AvqZb aaæe _vK‡e|  A1L1 = A2  3L1  A1 = 3A2  R2 R1 = L2 L1 × A1 A2 = 3 × 3 = 9  R2 = 9 × 5  = 45  (Ans.) 16. 27C ZvcgvÎvq 1 kW GKwU B‡jKwUaK †KZwj‡Z 2 litre cvwb Av‡Q| †KZwjwU‡K 10 wgwb‡Ui Rb ̈ myBP Ab Kiv n‡jv| hw` Pvicv‡k Zvc n«v‡mi nvi 160 Js–1 nq Z‡e 10 wgwb‡U †KZwji ZvcgvÎv KZ n‡e? [BUET 06-07] mgvavb: Kvh©Ki ÿgZv, P = 1000 – 160 W = 840 W Pt = mS  840 × 10 × 60 = 2 × 4200 × ( – 27)   = 87C (Ans.) 17. †kÖwY mgev‡q mw3⁄4Z `ywU cwievnxi †iva 40 ohm hv mgvšÍivj mgev‡q 7.5 ohm nq| cÖwZwU cwievnxi †iva †ei Ki| [BUET 05-06] mgvavb: R1 + R2 = 40 ........... (i) R1R2 R1 + R2 = 7.5  R1R2 = 300 ............. (ii)  R1 – R2 = 20 ........... (iii)  (i) I (iii) n‡Z, R1 = 30  (Ans.) R2 = 10  (Ans.) 18. GKwU ûBU‡÷vb wea‡Ri cÖ_g Ges wØZxq evû‡Z h_vμ‡g 10  Ges 12  Gi †iva hy3 Av‡Q| hLb PZz_© evû‡Z 20  Gi `ywU †iva mgvšÍivj ms‡hv‡M hy3 nq ZLb weaRwU mvg ̈ve ̄’vq _v‡K| ARvbv †iv‡ai gvb KZ? [BUET 04-05] mgvavb: P Q = R S  10 12 = R 20 || 20  R = 10 × 10 12  = 8.33  (Ans.) 19. GKwU †cv‡UbwmIwgUvi Zv‡i we`y ̈r cÖevn wbqš¿Y K‡i †Kvb we`y ̈r †Kv‡li Rb ̈ 6 m `~‡i wb ̄ú›` we›`y cvIqv †Mj| †KvlwUi `y-cÖv‡šÍi mv‡_ 3 In‡gi GKwU †iva hy3 Ki‡j 4 m `~‡i wb ̄ú›` we›`y cvIqv hvq| †KvlwUi Af ̈šÍixY †iva wbY©q Ki| [BUET 02-03] mgvavb: r =     l2 l1 – 1 R =     6 4 – 1 × 3 = 1.5  (Ans.) 20. 30  Af ̈šÍixY †iv‡ai GKwU M ̈vjfv‡bvwgUvi 500 A Zwor cÖev‡n c~Y© † ̄‹j we‡ÿc †`q| GB M ̈vjfv‡bvwgUvi‡K 2 mA cÖevngvÎv cwigv‡ci Dc‡hvMx Ki‡Z KZ gv‡bi kv›U e ̈envi Ki‡Z n‡e? [BUET 01-02] mgvavb: IG = S G + S × I  500 × 10–6 = S 30 + S × 2 × 10–3  S = 10  (Ans.)
4 ........................................................................................................................................  Physics 2nd Paper Chapter-3 weMZ mv‡j KUET-G Avmv cÖkœvejx 1. GKwU Zv‡ii •`N© ̈ Aci GKwU Zv‡ii Pvi ̧Y| Zvi `ywUi †iva mgvb n‡j G‡`i e ̈v‡mi AbycvZ †ei Ki| [KUET 05-06] mgvavb: R1 = R2  L1 A1 = L2 A2  d1 d2 = L1 L2 = 4L2 L2 = 2  d1 : d2 = 2 : 1 (Ans.) 2. GKwU wbw`©ó cwigvY Af ̈šÍixY †iva wewkó GKwU †Kv‡li we`y ̈r PvjK ej 1.4 volt| Gi cÖvšÍØq 2.6  †iv‡ai GKwU Zvi w`‡q hy3 Ki‡j cÖvšÍxq wefe cv_©K ̈ 1.3 volt cvIqv hvq| †KvlwUi Af ̈šÍixY †iva wbY©q Ki| [KUET 04-05] mgvavb: VR = R R + r E  1.3 = 2.6 2.6 + r  1.4  r = 0.2  3. GKwU U ̈vb‡R›U M ̈vjfv‡bvwgUv‡ii wfZi w`‡q 10 amp Zwor cÖev‡ni d‡j Gi KuvUvi we‡ÿc 45 nq| KZ Zwor cÖev‡ni d‡j KvUvi we‡ÿc 30 n‡e? [KUET 04-05] mgvavb: I2 I1 = ktan2 ktan1  I2 = I1 tan2 tan1  I2 = 10 × tan30 tan45  I2 = 5.77 A (Ans.) 4. ms‡ÿ‡c DËi `vI: [KUET 03-04] (i) GKwU †iv‡ai Mv‡q wZbwU is Gi e ̈vÛ Av‡Q| †iv‡ai gvb 470  n‡j e ̈vÛ is wK wK n‡e? (ii) 5 A wdDR ej‡Z wK †evS? mgvavb: (i) 1g e ̈vÛ njy`, 2q e ̈vÛ †e ̧bx, 3q e ̈vÛ ev`vgx| (Ans.) (ii) 5 A wdDR ej‡Z Avgiv eywS †mB wdDR m‡e©v”P 5 A Zwor cÖevn mn ̈ Ki‡Z cv‡i| Gi †ewk cÖev‡n wdDRwU M‡j eZ©bx ms‡hvM wew”Qbœ K‡i †`q| (Ans.) 5. GKwU Zvgvi Zv‡ii ga ̈ w`‡q 0.001 sec a‡i 0.001 mA Zwor Pvjbv Ki‡j KZwU B‡jKUab cÖevwnZ n‡e? [KUET 03-04] mgvavb: ne = It  n × 1.6 × 10–19 = 0.001 × 10–3 × 0.001  n = 6.25 × 109 wU (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 1. GKwU M ̈vjfv‡bvwgUv‡ii cvjøv 10 mA – 500 mV| (i) 20 A Ges (ii) 440 V gvc‡Z Kx e ̈e ̄’v wb‡Z n‡e? [RUET 19-20] mgvavb: (i) G = 500 × 10–3 10 × 10–3  = 50  IG = S G + S × I  10 × 10–3 = S 50 + S × 20  S = 0.025  0.025  kv›U M ̈vjfv‡bvwgUv‡ii mv‡_ mgvšÍiv‡j e ̈envi Ki‡Z n‡e| (Ans.) (ii) S = (n – 1) R  S =     440 550  10–3 – 1  50  R = 43950  43950  kv›U M ̈vjfv‡bvwgUv‡ii mv‡_ †kÖwY‡Z hy3 Ki‡Z n‡e| (Ans.) 2. R †iva wewkó GKwU Zwor cwievnx Zv‡ii •`N© ̈‡K †U‡b Gi Avmj •`‡N© ̈i n ̧Y j¤^v Kiv n‡jv| j¤^v Kivi c‡i ZviwUi †iva KZ n‡e? [RUET 18-19] mgvavb: †U‡b j¤^v Ki‡j AvqZb aaæe _vK‡e| A1L1 = A2L2  A1L1 = A2nL1  A1 = nA2  R2 R1 = L2 L1 × A1 A2 = n × n = n2  R2 = n 2R1 (Ans.) 3. 5 , 10  Ges 15  Gi wZbwU †iva †kÖwY I mgvšÍivj mgev‡q mvRv‡bv Av‡Q| Dfq‡ÿ‡Î Zzj ̈ †iva wbY©q Ki| [RUET 18-19] mgvavb: Rp = (5–1 + 10–1 + 15–1 ) –1  = 2.73  (Ans.) Rs = (5 + 10 + 15)  = 30  (Ans.)