Tiêu đề 1. P2C1. তাপগতিবিদ্যা_(With Solve).pdf ✅

WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. GKwU Kv‡b©v BwÄb hLb 27 C ZvcgvÎvq Zvc MÖvn‡K _v‡K ZLb Gi Kg©`ÿZv 50%| G‡K 60% Kg©`ÿ Ki‡Z n‡j Gi Dr‡mi ZvcgvÎv KZ evov‡Z n‡e? [BUET 21-22; BUTex 18-19; RUET 07-08] mgvavb: 1 = 1 – T2 T1  0.5 = 1 – 300 T1  T1 = 600 K 2 = 1 – T2 T1   0.6 = 1 – 300 T1   T1  = 750 K  T1 = 150 K (Ans.) 2. GKwU B‡jKwUaK †KUwji mvnv‡h ̈ 2 kg cvwbi ZvcgvÎv 25C †_‡K 80C G DbœxZ Ki‡j GbUawci cwieZ©b †ei K‡iv| [BUET 20-21] mgvavb: GbUawci cwieZ©b, S = mS ln T2 T1 = 2 × 4200  ln 80 + 273 25 + 273 = 1422.746392 JK–1 (Ans.) 3. mgvb f‡ii wZbwU wfbœ Zij c`v_© A, B, C Gi ZvcgvÎv h_vμ‡g 12C, 19C Ges 28C| A †K hw` B Gi mv‡_ †gkv‡bv nq Zvn‡j ZvcgvÎv nq 16C| B †K hw` C Gi mv‡_ †gkv‡bv nq Zvn‡j ZvcgvÎv nq 23C| A †K hw` C Gi mv‡_ †gkv‡bv nq Zvn‡j ZvcgvÎv KZ n‡e? [BUET 20-21] mgvavb: QA = QB  mSAA = mSBB  SA (16 – 12) = SB (19 – 16)  SA SB = 3 4  SA = 3 4 SB ....... (i) Avevi, QB = QC  mSBB = mSCC  SB × (23 – 19) = SC × (28 – 23)  SB SC = 5 4  SC = 4 5 SB ....... (ii) QA = QC  mSAA = mSCC  SA ( – 12) = SC (28 – )  SA SC = 28 –   – 12  3 4 SB 4 5 SB = 28 –   – 12  3 4 × 5 4 = 28 –   – 12   = 20.258 C (Ans.) 4. GKwU cÖZ ̈veZ©x BwÄb Zv‡ci 1/6th Ask‡K Kv‡R iƒcvšÍi K‡i| hLb Dr‡mi ZvcgvÎv wVK †i‡L MÖvn‡Ki ZvcgvÎv 62C Kgv‡bv nq, ZLb Bwćbi `ÿZv wØ ̧Y nq| MÖvnK Gi ZvcgvÎv wbY©q K‡iv| [BUET 19-20] mgvavb: W = Q × 1 6  W Q = 1 6 =   = 1 – T2 T1 = 1 6  T2 T1 = 5 6 ..... (i) Avevi, 2 = 2 6 = 1 – T2 – 62 T1  1 – 1 3 = T2 T1 – 62 T1  2 3 = 5 6 – 62 T1  T1 = 372 K  T2 = 310 K = 37C  MÖvn‡Ki ZvcgvÎv 37C (Ans.)
2 ........................................................................................................................................  Physics 2nd Paper Chapter-1 5. T1 Avw` ZvcgvÎvi GKwU Av`k© M ̈v‡mi Avw` AvqZb 2 m3 | iæ×Zvcxq cÖmvi‡Yi d‡j Gi AvqZb 4 m3 nq| Zvici m‡gvò cÖwμqvq cÖmvwiZ Kivq AvqZb 10 m3 nq, cieZ©x av‡c iæ×Zvcxq cÖwμqvq ms‡KvP‡bi d‡j Gi ZvcgvÎv cybivq T1 nq| Gi P‚ovšÍ AvqZb KZ? [BUET 18-19] mgvavb: 1g av‡c, T1V 1  – 1 = T2V 2  – 1  T1 T2 =     4 2  – 1 = 2( – 1) 2q av‡c, m‡gvò cÖmvi‡Y ZvcgvÎvi cwieZ©b nq bv|  V3 = 10 m3 3q av‡c, iæ×Zvcxq ms‡KvP‡b, T2V 3  – 1 = T1V 4  – 1  V4 =     T2 T1 1  – 1 × V3 =     1 2  – 1  – 1 × V3  V4 = 1 2 × 10 = 5 m3 (Ans.) 6. GKwU wmwjÛv‡i 300 K ZvcgvÎvq I 106 Pa Pv‡c 0.001 m3 M ̈vm Av‡Q| M ̈vmwU‡K cÖ_‡g m‡gvò cÖmviY Kiv nj Ges c‡i iæ×Zvcxq cÖwμqvq AveviI cÖmviY Kiv nj, cÖwZ †ÿ‡ÎB cÖmvi‡Yi AbycvZ 1 : 2| cÖmvi‡Y †gvU Kv‡Ri cwigvY †ei K‡iv| [BUET 17-18] mgvavb: 1g av‡c, W1 = nRT ln V2 V1 = P1V1 ln V2 V1 [∵ PV = nRT] = 106 × 0.001 ln2 = 693.147 J iæ×Zvcxq cÖmvi‡Y, T1V 1  – 1 = T2V 2  – 1  T2 =     1 2 1.4 – 1 × 300 = 227.357 K PV = nRT  n = PV RT = 106 × 0.001 8.314 × 300 = 0.4 mole W2 = nR 1 –  (T2 – T1) = 0.4 × 8.314 1 – 1.4 (227.357 – 300) = 603.95 J Wnet = W1 + W2 = 1297.1 J (Ans.) 7. GbUawc ej‡Z Kx eyS? 100C ZvcgvÎvi 4 kg cvwb‡K ev‡®ú cwiYZ Kiv nj| GbUawci e„w× †ei Ki| [BUET 17-18] mgvavb: GbUawc n‡”Q e ̄‘i wek„•LjZvi cwigvc Ges e ̄‘i †fŠZ ag© hv iæ×Zvcxq cÖwμqvq w ̄’i _v‡K| (Ans.) S = mlv T = 4 × 2.26 × 106 373 = 24.2359 × 103 JK–1 (Ans.) 8. 10 g IR‡bi GKwU †jvnvi †c‡iK‡K wKQzÿY GKwU evb©vi wkLvq DËß Kiv nj| DËß †c‡iKwU‡K 10C ZvcgvÎvq 100 g cvwb‡Z Wzev‡bv nj| G‡Z cvwbi ZvcgvÎv e„w× †c‡q 20C nj| cvwb‡Z Wzev‡bvi c~‡e© †c‡i‡Ki ZvcgvÎv wbY©q K‡iv| [†jvnvi Av‡cwÿK Zvc = 0.11 kcal/kgC] [BUET 16-17] mgvavb: †c‡i‡Ki cÖv_wgK ZvcgvÎv  n‡j, M„nxZ Zvc = ewR©Z Zvc  10 × 0.11 ( – 20) = 100 × 1 × (20 – 10)   = 929.09C (Ans.) 9. GK cigvYywewkó GKwU Av`k© M ̈vm‡K 17C ZvcgvÎvq nVvr Gi g~j AvqZ‡bi GK-`kgvsk AvqZ‡b msKzwPZ Kiv nj| ms‡KvP‡bi ci ZvcgvÎv KZ n‡e? [BUET 16-17] mgvavb: T1V 1  – 1 = T2V 2  – 1  T2 =     V1 V2  – 1 × T1 =       1 1 10 1.67 – 1 × (17 + 273) = 1356.43191 K  T2 = 1083.432C (Ans.) 10. 1.2 atm Pvc Ges 310 K ZvcgvÎvq †Kvb M ̈v‡mi AvqZb 4.3 L| iæ×Zvcxq cÖwμqvq M ̈vm‡K msKzwPZ K‡i AvqZb 0.76 L Kiv nj| M ̈vmwUi (i) P‚ovšÍ Pvc Ges (ii) P‚ovšÍ ZvcgvÎv wbY©q Ki| [M ̈vmwU‡K Av`k© M ̈vm wnmv‡e we‡ePbv Kiv hvq hvi  = 1.4] [BUET 14-15] mgvavb: (i) P1V 1  = P2V2   P2 =     V1 V2  × P1 = 1.2 ×     4.3 0.76 1.4  P2 = 13.58 atm (Ans.) (ii) T1V 1  – 1 = T2V 2  – 1  T2 =     V1 V2  – 1 × T1 =     4.3 0.76 1.4 – 1 × 310  T2 = 620.0456 K (Ans.)
ZvcMwZwe` ̈v  Engineering Practice Sheet....................................................................................................................... 3 11. 0C ZvcgvÎvq 1 kg eid‡K 100C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z GbUawci e„w× wbY©q K‡iv| [BUET 13-14] mgvavb: S = S1 + S2 = mlf T + mS ln T2 T1 = 1 × 3.36 × 105 273 + 1 × 4200 ln 373 273  S = 2541.617 JK–1 (Ans.) 12. GKwU cvigvYweK †evgv we‡ùvwiZ n‡j m„ó Av ̧‡bi †Mvj‡Ki e ̈vmva© nq 100 m Ges Gi ZvcgvÎv 105 K| hw` †MvjKwU iƒ×Zvc c×wZ‡Z 1000 m e ̈vmv‡a© ewa©Z nq Z‡e Gi m¤¢ve ̈ ZvcgvÎv KZ n‡e?    Av‡cwÿK Zvc؇qi AbycvZ   CP CV = 1.66 [BUET 13-14; CUET 05-06] mgvavb: T1V 1  – 1 = T2V 2  – 1  T2 =     V1 V2  – 1 × T1 =     r1 r2 3 – 3 × T1    V =  4 3 r 3  V  r 3 = 105 ×     100 1000 3 × 1.66 – 3  T2 = 1047.128548 K (Ans.) 13. 10C ZvcgvÎvi 5 kg cvwb‡K 100C ZvcgvÎvq DbœxZ Ki‡Z GbUawci e„w× wbY©q K‡iv| [BUET 11-12; CUET 08-09] mgvavb: S = mS ln T2 T1 = 5 × 4200 ln 373 283 = 5798.76 JK–1 (Ans.) 14. GKwU Kv‡b©v Bwćbi Zvc Drm I Zvc MÖvn‡Ki ZvcgvÎv h_vμ‡g 500 K I 375 K| hw` BwÄbwU cÖwZ P‡μ 252  104 J Zvc †kvlY K‡i Z‡e, (i) Bwćbi `ÿZv, (ii) cÖwZP‡μ Kv‡Ri cwigvY I (ii) cÖwZP‡μ ewR©Z Zv‡ci cwigvY wbY©q K‡iv| [RUET 08-09; BUET 03-04, 01-02] mgvavb: (i)  = 1 – T2 T1 = 1 – 375 500 = 1 4 = 25% (Ans.) (ii)  = W Q1  0.25 = W 252 × 104  W = 6.3 × 105 J (Ans.) (iii) W = Q1 – Q2  Q2 = Q1 – W = 252 × 104 – 6.3 × 105 = 1.89 × 106 J (Ans.) 15. GKwU Kv‡b©v Pμ cÖv_wgK 327C ZvcgvÎvq KvR m¤úbœ K‡i| cÖwZwU av‡c ms‡KvPb ev cÖmvi‡Yi AbycvZ 1 : 6 n‡j Kv‡b©v P‡μi me©wb¤œ ZvcgvÎv Ges `ÿZv wbY©q K‡iv| [ = 1.4] [BUET 07-08] mgvavb: (P1,V1, T1) A B (P2,V2, T1) (P4,V4, T2) D (P3,V3, T2) P V C wPÎ n‡Z, V2 V1 = 6 I V3 V2 = 6 BC  iæ×Zvcxq cÖmviY myZivs, T1V 2  – 1 = T2V 3  – 1  T2 T1 =     V2 V3  – 1  T2 =     1 6 1.4 – 1  600 = 293.015 K (Ans.)   = 1 – T2 T1 = 1 – 293.015 600 = 51.2% (Ans.) 16. GK †ivMxi †`‡ni ZvcgvÎv GKwU ÎæwUc~Y© _v‡g©vwgUv‡ii mvnv‡h ̈ †g‡c 45C cvIqv †Mj| hw` GB _v‡g©vwgUv‡i eid we›`y Ges ev®úwe›`y h_vμ‡g 3C Ges 107C †Z cvIqv hvq, Zvn‡j †ivMxi †`‡ni cÖK...Z ZvcgvÎv dv‡ibnvBU † ̄‹‡j †ei K‡iv| [BUET 04-05] mgvavb: x – 3 107 – 3 = F – 32 212 – 32  45 – 3 104 = F – 32 180  F = 104.692F (Ans.) 17. GKwU †gvUi Uvqvi‡K 15C ZvcgvÎvq 2 evqygÐjxq Pv‡c cv¤ú Kivq UvqviwU nVvr †d‡U †M‡jv| Gi d‡j, ZvcgvÎv KZ K‡g hv‡e Zv †ei K‡iv| [ = 1.4] [BUET 02-03] mgvavb: †d‡U hvIqvi c‡i Uvqv‡ii Pvc n‡e evqygÛjxq Pv‡ci mgvb|  P2 = 1 atm T1P 1 –   1 = T2P 1 –   2  T2 =     P1 P2 1 –   × T1 =     2 1 1 – 1.4 1.4 × (15 + 273) = 236.26 K = – 36.74C ZvcgvÎv n«vm = 15 – (–36.74)C = 51.74C (Ans.)
4 ........................................................................................................................................  Physics 2nd Paper Chapter-1 18. GKwU Kv‡b©v Bwćbi `ÿZv 1 6 | Zvc MÖvn‡Ki ZvcgvÎv 65C Kgv‡j `ÿZv 1 3 nq| Zvc Drm I Zvc MÖvn‡Ki ZvcgvÎv wbY©q K‡iv| [BUET 02-03] mgvavb: 1 = 1 – T2 T1 = 1 6  T2 T1 = 5 6 ....... (i) 2 = 1 – T2 – 65 T1 = 1 3  1 – T2 T1 + 65 T1 = 1 3  T1 = 390 K (Ans.)  T2 = 5 6 × 390 = 325 K (Ans.) 19. ̄^vfvweK ZvcgvÎv I Pv‡c wnwjqv‡gi GK wK‡jv‡gvj AYyi AvqZb 22.42 L| w ̄’i AvqZ‡b wnwjqv‡gi Av‡cwÿK Zvc hw` 3.0 calmol–1K – 1 nq Z‡e w ̄’i Pv‡c wnwjqv‡gi Av‡cwÿK Zvc wbY©q K‡iv| (†`Iqv Av‡Q, cvi‡`i NbZ¡ = 13.6  103 kgm–3 ; J = 4200 J/kcal) [BUET 01-02] mgvavb: GLv‡b, R = P0V0 nT0 = 0.76  13.6  103  9.8  22.42  10–3 1 × 273 Jmol–1K –1 = 8.31 Jmol–1K –1  CP = CV + R = (3  4.2 + 8.31) Jmol–1K –1 = 20.91 Jmol–1K –1 (Ans.) 20. GKwU Av`k© Kv‡b©v Bwćbi Drm Ges wm‡1⁄4i ZvcgvÎv 450 K Ges 350 K| cÖwZ mvB‡K‡j BwÄbwU hw` Drm n‡Z 1 kcal Zvc MÖnY K‡i Zvn‡jÑ (i) cÖwZ mvB‡K‡j wm‡1⁄4 ewR©Z Zvc (ii) BwÄbwUi `ÿZv Ges (iii) cÖwZ mvB‡K‡j m¤úvw`Z Kv‡Ri cwigvY wbY©q K‡iv| [J = 4.184 kJ/kcal] [BUET 01-02] mgvavb: (i) Q2 Q1 = T2 T1  Q2 = 350 450 × 1 kcal = 0.78 kcal (Ans.) (ii)  = 1 – T2 T1 = 1 – 350 450 = 0.222 = 22.2% (Ans.) (iii) cÖwZ P‡μ m¤úvw`Z KvR, W = Q1 – Q2 = (1 – 0.78) kcal = 0.22 kcal = 0.92048 kJ (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 1. 40 wWwMÖ †mjwmqvm ZvcgvÎvq Ges 76 †m. wg. evqy Pv‡c wKQz cwigvY evqy‡K iæ×Zvcxq cÖwμqvq AvqZb wØ ̧Y Kiv n‡j ZvcgvÎv KZ n‡e? Avevi hw` cwiewZ©Z ZvcgvÎv †_‡K 20 wWwMÖ †mjwmqvm e„w× Kiv nq, Z‡e Gi Pvc KZ n‡e? [ = 1.4] [KUET 19-20] mgvavb: T1V 1  – 1 = T2V 2  – 1  T2 =     V1 V2  – 1 × T1 =     1 2 1.4 – 1 × (40 + 273) = 237.2096 K (Ans.) Avevi, P1V1  = P2V2   P2 =     V1 V2  × P1 =     1 2 1.4 × 76 = 28.7986 cm ZvcgvÎv Av‡iv 20C ev 20 K evov‡j, T2  = 257.2096 K T1  P 1 1 –   = T2  P 2 1 –    P2 =      T1  T2   1 –  × P1 =     237.2096 257.2096 1.4 1 – 1.4 × 28.7986 = 38.23 cm Hg (Ans.) 2. GKwU Kv‡b©v BwÄb AšÍM©vgx Zv‡ci 1 4 Ask Kv‡R iƒcvšÍi K‡i| Gi Zvc MÖvn‡Ki ZvcgvÎv Av‡iv 70C n«vm Ki‡j Zvi `ÿZv wØ ̧Y nq| Dr‡mi ZvcgvÎv I Zvc MÖvn‡Ki ZvcgvÎv †ei K‡iv| [RUET 12-13, 09-10, KUET 03-04] mgvavb: 1 = 1 – T2 T1 = W Q = 1 4  T2 T1 = 3 4 ....... (i) 2 = 21 = 1 2 = 1 – T2 – 70 T1  1 2 = T2 T1 – 70 T1 = 3 4 – 70 T1 ∴ T1 = 280 K (Ans.) ∴ T2 = 3 4 × 280 = 210 K (Ans.)