ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 379 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 06. Rjxq ev‡®úi †ÿ‡Î †KvbwU mwVK? [wm.‡ev 2021] A. Rjxq ev‡®úi cwigv‡Yi Dci †Kv‡bv ¯’v‡bi AvenvIqv wbf©i K‡i bv B. Kzqvkv m„wói Rb¨ Rjxq ev®ú `vqx bq C. Rjxq ev‡®úi NbZ¡ ﮋ Nb‡Z¡i †P‡q †ewk D. Rjxq ev®ú ﮋ evqy A‡cÿv nvjKv Ans D 07. ﮋ I Av`ª© evjyi nvB‡MÖvwgUv‡ii mvnv‡h¨ AvenvIqv c~e©vfv‡mi Rb¨ wb‡Pi †KvbwU mwVK? [wm.‡ev 2021] A. _v‡g©vwgUvi `ywUi cv‡Vi cv_©K¨ bv _vK‡j evZvm m¤úK…Z Av‡Q B. _v‡g©vwgUvi `ywUi cv‡Vi cv_©K¨ Kg n‡j AvenvIqv ﮋ _vK‡e C. _v‡g©vwgUvi `ywUi cv‡Vi cv_©K¨ ax‡i `x‡i Kg‡Z _vK‡j S‡oi m¤¢vebv D. _v‡g©vwgU6vi `ywUi cv‡Vi cv_©K¨ †ewk n‡j AvenvIqv Av`ª© _vK‡e Ans A 08. ‡h ZvcgvÎvq †Kv‡bv wbw`©ó AvqZ‡bi evqy Rjxq ev®ú Øviv m¤ú„³ nq, Zv‡K e‡j- [wm. †ev. 2016] A. wkwkiv¼ B. Av`ª©Zv C. Av‡cwÿK Av`ª©Zv D. cig Av`ª©Zv Ans A 09. GKwU •KwkK bj‡K wMømvwi‡b Wyev‡jÑ [wm. †ev. 2015] i. KvP I wMømvwi‡bi ¯úk© †KvY m~² †KvY nq ii. Zij c„ô AeZ‡j AvKvi aviY K‡i iii. KvP I wMømvwi‡bi ¯úk© †KvY ¯’~j †KvY nq wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans A STEP 2 mg„× †ewmK MvwYwZK cÖ Ö‡qvM (MATH) kU©KvU †UKwbK Concept 1 e‡qj, Pvj©m I Pvcxq m~Î/ Pvc, AvqZb I ZvcgvÎv msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. P1V1 = P2V2 (ZvcgvÎv w¯’i); 02. 2 2 1 1 T V T V (Pvc w¯’i); 03. 2 2 1 1 T P T P (AvqZb w¯’i); 04. P1V1 T1 = P2V2 T2 [mgš^q m~Î] CONCEPTUAL MATH MEx 01 0.755m Pv‡c Ges 15C ZvcgvÎvq †Kvb M¨v‡mi AvqZb 1.25 10–4m 3 | cÖgvY ZvcgvÎvq I Pv‡c H M¨v‡mi AvqZb KZ n‡e? V2 = P1V1T2 T1P2 = 0.755 1.25 10–4 273 288 0.76 = 1.18 10–4m 3 MEx 02 w¯’i DòZvq KZ Pvc cÖ‡qvM Ki‡j GKwU M¨v‡mi AvqZb ¯^vfvweK Pv‡c AvqZ‡bi 4 ¸Y n‡e? Pvc, P2 = P1V1 V2 = (1.013 105 ) 1 4 = 2.53 104Nm–2 MEx 03 GKwU †ejyb‡K 25C ZvcgvÎvq Ge&s 75 10–2m (cvi`) Pv‡c 1 10–3 nvB‡Wªv‡Rb Øviv c~Y© n‡jv, GLb 10C ZvcgvÎvq Ges 75 10–3m Pv‡c †ejybwU‡K Dwo‡q †`Iqv nj| †ejy‡bi AvqZb KZUv e„w× cv‡e? ‡ejy‡bi AvqZb e„w×, V = V2 – V1 = = P1V1T2 T1P2 = 75 10–2 10–3 283 298 75 10–3 × 10–3 = 8.49664 10–3 MEx 04 w¯’i Pv‡c 270C ZvcgvÎvi 2 Litre evZv‡mi AvqZb 4 Litre Ki‡Z n‡j DËß K‡i †h ZvcgvÎvq wb‡Z n‡e| V1 T1 = V2 T2 2 300 = 4 T2 T2 = 4 300 2 = 600K = 327C NOW START PRACTICE 01. T ZvcgvÎvq 1 wjUvi evqy‡K DËß Kiv nj hZ¶Y bv evqyi AvqZb I Pvc wظY nq| evqyi P~ovšÍ ZvcgvÎv KZ? 02. ¯^vfvweK ZvcgvÎv I Pv‡c wKQz cwigvY gy³ evqy‡K aªæe ZvcgvÎvq msbwgZ K‡i AvqZb A‡a©K Kiv nj| P~ovšÍ Pvc KZ n‡e? 03. wQwc AvUv GKwU †evZ‡j ¯^vfvweK Pv‡c 270C ZvcgvÎvq wKQz M¨vm Av‡Q| †evZ‡ji ZvcgvÎv 670C DbœxZ Ki‡j M¨v‡mi Pvc KZ? 04. GKwU k³ cv‡Î 0 0C ZvcgvÎvq wKQz M¨vm iw¶Z Av‡Q KZ ZvcgvÎvq M¨v‡mi Pvc 0 0C ZvcgvÎvq Pv‡ci GK Z…Zxqvsk n‡e? 05. wbw`©ó f‡ii GKwU Av`k© M¨v‡mi AvqZb aªæeK Pv‡c wظY Kiv nj| hw` M¨v‡mi cÖv_wgK ZvcgvÎv 17 0C nq Z‡e P~ovšÍ ZvcgvÎv KZ? 06. †Kvb wbw`©ó f‡ii M¨v‡mi ZvcgvÎv 30°C Pvc w¯’i _vK‡j †Kvb ZvcgvÎvi AvqZb wظb n‡e? 07. GKwU Av`k© M¨v‡mi bgybvi ZvcgvÎv 200C, hw` bgybvwUi Pvc Ges AvqZb wظY Kiv nq, Z‡e cwiewZ©Z ZvcgvÎv KZ?
380 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES NOW PRACTICE SOLVE : 01. P1V1 T1 = P2V2 T2 ev, 1 1 T1 = 2 2 T2 ev, T2 = 4T 02. AvqZb A‡a©K ZvB Pvc wظY| (e‡q‡ji m~Îvbymv‡i) P2 = 2P = 2x1.01 105 = 2.02 105Nm-2 A_ev, P1V1 = P2V2 P2 P1V1 V2 = 1.01 105 1 1 2 = 2.02 105 Nm–2 03. P1 T1 = P2 T2 ev, = 1.012325 105 300 = P2 340 ev, P2 = 1.1485 105 Nm–2 04. Pvc 1 3 ZvB ZvcgvÎv T2 = 273 3 = 91K A_ev P1 T1 = P2 T2 ev, 1 273 = 1 3 T2 ev, T2 = 91K 05. T2 = 2T1 = 2(273 + 17) = 2 290 = 580 K = 307C 06. V1 T1 = V2 T2 ev, 1 303 = 2 T2 ev, T2 = 606k = 333C 07. P1V1 T1 = P2V2 T2 = p.v (20 + 273) = 2p.2v T2 T2 = 899C REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU Av`k© M¨vm Gi bgybvi ZvcgvÎv 20°C, hw` bgybvwUi Pvc Ges AvqZb wظb Kiv nq,Z‡e cwiewZ©Z ZvcgvÎv KZ? [DU. 13-14] A. 20°C B. 80°C C. 900°C D. 1200°C bgybvi Pvc Ges AvqZb wظb Kiv n‡q‡Q| cwiewZ©Z ZvcgvÎv T2 =? T2 = 22T1 = 4(273+20) = 4293 = 1172k = 899°C = 900°C (KvQvKvwQ) 02. wbw`©ó f‡ii GKwU Av`k© M¨v‡mi AvqZb aªæe Pv‡c wظY Kiv n‡jv| hw` M¨v‡mi cÖv_wgK ZvcgvÎv 13C nq Z‡e P‚ovšÍ ZvcgvÎv KZ? [DU. 04-05; JNU. 10-11] A. 299C B. 399C C. 499C D. 199C Pvc w¯’i _vK‡j ZvcgvÎv, T2 = V2 V1 T1 = 2 (13 + 273) = 572 K = 299˚C Ans. 03. mgPv‡c I 170C ZvcgvÎvq 2 wjUvi AvqZb‡K 3 wjUvi Kivi Rb¨ ZvcgvÎv KZ e„w× Ki‡Z n‡e? [DU. 08-09] A. 1000C B. 1520C C. 1620C D. 2550C V1 T1 = V2 T2 ev, 2 290= 3 T2 ev, T2 = 435k ev, T2 = 162C 04. GK evqygÛjxq Pv‡c GKwU Av`k© M¨vm‡K DËß K‡i 0.01m3 AvqZb e„w× Kiv n‡jv| G‡Z m¤úvw`Z Kv‡Ri cwigvY [DU. 08-09] A. 7.610-3 J B. 76 J C. 110-2 J D. 1103 J W = PV = 101325 0.01 = 1.013 103 J 05. ¯^vfvweK ZvcgvÎv I Pv‡c wKQy cwigvY ﮋ evqy‡K m‡gvò cÖwµqvq msbwgZ K‡i AvqZb A‡a©K Kiv n‡jv| P~ovšÍ Pvc KZ n‡e? [DU. 00-01] A. 4.0410–2 Nm–2 B. 2.20 Nm–2 C. 4.04105 Nm–2 D. 2.02105 Nm–2 m‡gvò cÖwµqvq, P2 = V1 V2 P1 = 2 1.01 105 = 2.02 105 Nm–2 06. T ZvcgvÎvq GK wjUvi evqy‡K DËß Kiv n‡jv hZ¶Y bv ch©šÍ evqyi AvqZb Ges Pvc wظY n‡jv| evqyi P‚ovšÍ ZvcgvÎv KZ? [DU. 95-96] A. T/2 B. T/4 C. 2T D. 4T P1V1 T1 = P2V2 T2 T2 = V2 V1 P2 P1 T1 = 2 2 T = 4T 07. w¯’i Pv‡c 270C ZvcgvÎvi 2 Litre evZv‡mi AvqZb 4 Litre Ki‡Z n‡j DËß K‡i †h ZvcgvÎvq wb‡Z n‡e [DU. 02-03; JU. 14-15] A. 540C B. 2370C C. 3000C D. 3270C V1 T1 = V2 T2 ev, 2 300 = 4 T2 ev, T2 = 600 K = 327C STEP 02 ANALYSIS OF JU QUESTION 01. GKwU e¯‘i fi evZv‡m 100 gm I A¨vj‡Kvn‡j 84 gm| A¨vj‡Kvn‡ji NbZ¡ 0.8gm/cc n‡j e¯‘i AvqZb KZ? [JU-A, Set-C. 20-21] A. 8 cc B. 10 cc C. 20 cc D. 8.4 cc S Blank info 1 = m1 V ; 2 = m2 V m1 = V1; m2 = V2 m1 – m2 = V1 – V2 =V(1–2)V = m1–m2 1–2 = 100 – 84 1 – 0.84 = 100cc 02. 0C ZvcgvÎvi †Kvb M¨v‡mi Pvc 3103 Pa n‡j 60C ZvcgvÎvq Gi Pvc KZ n‡e? [JU. 13-14] A. 3.66105 Pa B. 3.85105 Pa C. 3.00106 Pa D. 3.66103 Pa Pvc, P2 = T2 T1 P1 = 333 273 (3 103 ) = 3.66 103 Pa 03. 270C ZvcgvÎvq †Kvb wbw`©ó cwigvb M¨vm nVvr cÖmvwiZ n‡q w`¸b AvqZb jvf K‡i| P~ovšÍ ZvcgvÎv KZ? [JU. 12-13] A. 45 B. 46 C. – 45.84 D. 45.64 T1 V –1 = T2 V –1 2 ev, T2 = V1 V2 –1 T1 = 1 2 1.4–1 300 = 227.357 K = – 45.64C 04. M¨v‡mi AvqZb ¸YvsK 6103 N/m3 M¨v‡mi AvqZb 10% Kgv‡Z n‡j wK cwigvb AwZwi³ Pv‡c cÖ‡qvM Ki‡Z n‡e? [JU. 12-13] A. 300 N/m2 B. 400 N/m2 C. 1000 N/m2 D. 600 N/m2 P=6103 10% = 600N/m2 05. 0˚C ZvcgvÎvq †Kvb M¨v‡mi Pvc 3105 Nm–2 n‡j 60˚C ZvcgvÎvi Pvc KZ n‡e? [JU. 09-10] A. 30.66105 Pa B. 33.6105 Pa C. 3.66105 Pa D. †KvbwUB bq P2 = T2 T1 P1 = 60 + 273 0 + 273 3 105 = 3.66 105 Pa
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 381 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 06. cvi‡`i 0.755m Pv‡c Ges 15C ZvcgvÎvq †Kvb M¨v‡mi AvqZb 1.25 10–4 m 3 | cÖgvY ZvcgvÎvq I Pv‡c H M¨v‡mi AvqZb n‡e- [JU. 09-10] A. 1.210–4 m 3 B. 1.1810–4 m 3 C. 1.1010–4 m 3 D. 1.1710–4 m 3 P1V1 T1 = P2V2 T2 V2 = P1 P2 T2 T1 V1 = 0.755 1.25 10–4 273 288 0.76 = 1.1810–4 m 3 STEP 03 ANALYSIS OF RU QUESTION 01. Pvj©‡mi m~ÎwU wb‡Pi †Kvb †jLwPÎwU w`‡q cÖKvk Kiv nq? [RU-C, Set-3. 19-20] A. B. C. D. V P V V V T T P Pvj©‡mi m~Î : V T [P w¯’i] Avgiv Rvwb, mKj mgvYycvwZK MÖvd g~jwe›`yMvgx mij‡iLv Ges mKj e¨v¯ÍvbycvwZK MÖvd AvqZvKvi Awae„Ë ev eµ‡iLv| 02. mgPv‡c 15C ZvcgvÎvq 200cm3 evqy‡K 65C ZvcgvÎvq DVv‡bv n‡jAvqZb KZ cm3 n‡e? [RU. 16-17] A. 432.72 B. 342.72 C. 234.72 D. 334.72 PV T = Constant V T V2 = T2 T1 V1 = 338 288 200 = 234.72 cm3 03. Av‡cw¶K ¸iæZ¡ 0.8 wewkó GKwU KwVb c`v‡_©i evZv‡m IRb 120 MÖvg cvwb‡Z Dnvi IRb KZ? [RU. 16-17] A. 96 MÖvg B. 99.50 MÖvg C. 98.95 MÖvg D. 98.50 MÖvg evZv‡m KwVb c`v‡_©i IRb = KwVb c`v‡_©i Av‡cw¶K ¸iæZ¡ cvwbi IRb = 0.8 120= 96 gm 04. 5gCO2 M¨v‡mi N.T.P ‡Z AvqZb KZ? [RU.15-16] A. 197.12L B. 2.545L C. 0.255L D. 1.971L CO2 Gi AvYweK fi = 12 + 32 = 44L 44g Gi AvqZb = 22.4 liter 1g Gi AvqZb = 22.4 44 5g Gi AvqZb = 22.4 5 44 = 2.5454L 05. 27C ZvcgvÎvq †Kvb M¨vm bgybvi AvqZb-Pvc ¸Ydj hZ, Zv wظY Ki‡Z KZ ZvcgvÎv cÖ‡qvRb? [RU. 13-14] A. 54C B. 327 K C. 300 K D. 327C P1V1 T1 = P2V2 T2 T2 = V2 V1 P2 P1 T1 = 2 (27 + 273) = 600K = 327C Ans. 06. w¯’i ZvcgvÎvq 105 Nm–2 Pv‡c wbw`©ó f‡ii wKQz M¨v‡mi AvqZb 0.002 m 3 , 4105 Nm–2 Pv‡c M¨vmwUi AvqZb KZ? [RU. 12-13; KU. 03-04] A. 0.0003 m3 B. 0.004 m3 C. 0.0005 m3 D. 0.0007 m3 P1V1 = P2V2 ev, V2 = P1V1 P2 = 105 0.002 4 105 = 0.0005 m3 07. GKwU Aw·‡Rb wmwjÛv‡ii AvqZb 5 105 c.c Ges G‡Z cÖ_‡g 250 evqygÛjxq Pv‡c Aw·‡Rb fwZ© wQj| wKQyUv e¨env‡ii d‡j †`Lv †Mj Pvc 100 evqygÛjxq Pv‡c †b‡g †M‡Q| e¨eüZ Aw·‡R‡bi cwigvb evqygÛjxq Pv‡c KZ? [RU. 10-11] A. 7 105C.C B. 10m3 C. 750m3 D 0.75 108 cc p1v1 = p2v2 v2 = p1v1 p2 = 250 5 105 100 = 12.5 105 cc e¨eüZ M¨v‡m = (12.5 – 5) 105 = 7.5 105 cc (100 atm Pv‡c) 1 evqy gÛjxq Pv‡c AvqZb = 7.5 105 100 1 = 7.5 107 cc = 0.75 108 cc 08. 0˚C ZvcgvÎvq †Kvb M¨v‡mi Pvc 3105 Pa n‡j 91˚C ZvcgvÎvq Dnvi Pvc KZ n‡e? [RU. 08-09] A. 3.66105 Pa B. 4105 Pa C. 2.25105 Pa D. 6105 Pa P2 = T2 T1 P1 = 91 + 273 273 3 105 = 4 105 Pa STEP 04 ANALYSIS OF GST QUESTION 01. 10 wjUvi AvqZ‡bi eÜ cv‡Î 300k ZvcgvÎvq 16g Aw·‡Rb †h Pvc cÖ`k©b K‡i, GKB cv‡Î GKB ZvcgvÎvq KZ MÖvg bvB‡Uªv‡Rb ivL‡j GKB Pvc cÖ`k©b Ki‡e? [GST-A. 20-21] A. 14 B. 16 C. 18 D. 32 S A info P1V1 n1T1 = P2V2 n2T2 n1 = n2 m1 M1 = m2 M2 16 32 = m2 28 m2 = 14 [P1= P2, V1 = V2, T1 = T2] PART A Analysis of General University Question JnU 01. GK evqygÛjxq Pv‡ci mgvb n‡jv- [JnU. 07-08] A. 1 kPa B. 102 kPa C. 300 kPa D. 500 kPa 102 kPa 02. cvi‡`i 0.755 Pv‡c Ges 150C ZvcgvÎvq †Kvb M¨v‡mi AvqZb 1.25104 m 3 cÖgvb ZvcgvÎv I Pv‡c H M¨v‡mi AvqZb n‡e? [JnU. 06-07] A. 1.1104m 3 B. 1.18104m 3 C. 1.10104m 3 D. 1.17 104m 3 P1V1 T1 = P2V2 T2 ev, V2 = P1V1T2 T1P2 = 0.755 1.25 104 288 273 0.76 = 1.18104m 3 KU 01. †Kvb wbw`©ó f‡ii M¨v‡mi ZvcgvÎv 30°C Pvc w¯’i _vK‡j †Kvb ZvcgvÎvq AvqZb wظY n‡e? [KU. 13-14] A. 636°C B. 663°C C. 366°C D. 333°C T2 = 2(273+ 30) = 2303 = 606K = 333°C 02. 27°C ZvcgvÎvi GKwU Uvqvi‡K cv¤ú Ki‡Z Ki‡Z Zvi Pvc 2 evqygÛjxq Pv‡ci mgvb nIqvi mv‡_ mv‡_ †mwU †d‡U †Mj| P~ovšÍ ZvcgvÎv KZ? [ = 1.4] [KU. 13-14] A. 44.3°C B. 22.3°C C. 33.3°C D. 11.3°C S Blank info T1P1 1 – = T2P2 1 – ; T1 = T2 p2 p1 1 – = 300 2 1 1 – 1.4 1.4 = 246 K = –26.9C †d‡U hvIqvi ci Pvc, p1 = 1 atm PART B Analysis of Science & Technology Question SUST 01. GKwU wmwjÛv‡i Ave× M¨v‡mi ZvcgvÎv 30C †_‡K 100C Kiv n‡j Pvc KZ kZvsk †e‡o hv‡e? [SUST. 13-14; g.†ev 2021] A. 19 B. 23 C. 30 D. 33 e) 70 P1 T1 = P2 T2 ev, P 303 = P2 373 ev, P2 = 373 303 =1.231 Pvc e„w× cv‡e = 1.23 – 1.00 = 0.23 = 23% 02. GKwU wmwjÛv‡i 2 atm Pv‡c 27C ZvcgvÎvq 5L evZvm ivLv Av‡Q| Lye ax‡i ax‡i evZv‡mi Pvc wظb Kiv n‡j evZv‡mi ZvcgvÎv Ges AvqZb n‡e- [SUST. 10-11] A. 3L, 95C B. 3.5L, 25C C. 1.5L, 28C D. 2.5L, 27C P1V1 = P2V2 ev, 1 5 = 2V2 V2 = 2.5 Liter T1 = 300K P1V1 T1 = P2V2 T2 ev, 1 5 300 = 2 2.5 T2 C = T2
382 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES JUST 01. ¯^vfvweK ZvcgvÎv I Pv‡ci †Kvb M¨vm‡K Pvc cÖ‡qvM Kivq AvqZb A‡a©K n‡q †Mj| P~ovšÍ Pvc KZ? [JUST. 15-16] A. 2.64 atm B. 1 atm C. 2 atm D. 0.38 atm E. 3.48 atm PyovšÍ Pvc P2 = 1 2 = 2atm MBSTU 01. w¯’i ZvcgvÎvq ‡Kvb M¨v‡mi AvqZb ¯^vfvweK Pv‡ci AvqZ‡bi 4 ¸Y Kivi Rb¨ KZ cwigvb Pvc cÖ‡qvM Ki‡Z n‡e? [MBSTU. 15-16] A. 25.325103Nm–2 B. 253.25103m –2 C. 253.25103Nm–1 D. 2.51010Nm–1 P1V1 = P2V2 ev, P2 = P1V1 v2 = 1.03 105 1 4 = 25.325 103 Nm–2 PSTU 01. w¯’i AvqZ‡bi GKwU Av`k© M¨v‡mi cÖ‡Z¨KwU AYyi fi A‡a©K Ges ‡eM wظY Kiv n‡j Avw` I ‡kl Pv‡ci AbycvZ n‡e- [PSTU. 15-16] A. 2 B. 0.5 C. 0.25 D. 4 P1 P2 = m1 m2 C1 C2 2 ev, P1 P2 = 2 1 1 2 2 = 2 4 = 0.5 STEP 05 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. 30C ZvcgvÎvq wKQz cwigvb ﮋ evqy‡K AvKw¯§Kfv‡e AvqZ‡bi A‡a©‡K msKzwPZ Kiv nj|P~ovšÍ ZvcgvÎv KZ? [=1.4] [KUET. 14-15] A. 122.9C B. 410K C. 126.81C D. 395.6K E. 127C T1V1 –1 T2V2 –1 T2 = T1 v1 v2 –1 = 303 v v 2 –1 = 303 2 0.4 = 399.18K 395.6K Note: AvKw¯§Kfv‡e ZvB ey×Zvcxq| 02. 0 0C ZvcgvÎvq †Kvb wbw`©ó M¨vm‡K nVvr cÖmvwiZ K‡i AvqZ‡b wظY Kiv n‡jv| P~ovšÍ ZvcgvÎv KZ? ( 1 =1.4) [KUET. 11-12] A. –88.25C B. – 166.13C C. –66.104C D. 88.25 T1V 1 –1 = T2V 2 –1 ev, T2 = v1 v2 –1 T1 = 1 2 1.4–1 273 = 206.895 K = – 66.104C 03. 100C ZvcgvÎvq 1 wjUvi evqy‡Z Zvc †`Iqv nj †h ch©šÍ Zvi Pvc I AvqZb wظb bv nq| evqyi P~ovšÍ ZvcgvÎv wbY©q Ki| [KUET. 07-08] A. 283 K B. 566 K C. 8149 K D. 1132K E. 1415K T1 = 273+10 = 283 T2 = 283 2 2 = 1132K = 859°C RUET 01. GKwU cv‡Î 0 0C ZvcgvÎvq wKQz M¨vm iw¶Z Av‡Q| KZ ZvcgvÎvq M¨v‡mi Pvc 0 0C ZvcgvÎvi Pv‡ci GK Z…Zxqvsk n‡e? [RUET. 12-13,JU-A. 20-21] A. 91K B. 81K C. 73K D. 83K T2 = 273 3 = 91K CUET 01. GKwU wmwjÛv‡i iw¶Z Aw·‡Rb M¨vm Gi AvqZb 1 10–2m 3 ZvcgvÎv 300K Ges Pvc 2.5105 Nm-2 . ZvcgvÎv w¯’i †i‡L wKQz Aw·‡Rb †ei K‡i †bqv nj| d‡j Pvc K‡g 1.3105 Nm-2 nq| e¨eüZ Aw·‡R‡bi fi wbY©q Ki? [CUET. 11-12] A. 0.18kg B. 0.015 kg C. 0.018 kg D. None of them e¨eüZ M¨vm = 1 R P1V1 T1 – P2V2 T2 = 1 RT (P1V1 – P2V2) = 1 8.314 300 10–2 (2.5 105 – 1.3 105 ) = 0.4811 m3 e¨eüZ Aw·‡R‡bi fi = 0.4811 32 = 15.39 g = 0.0154 kg STEP 06 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. C 0 20 ZvcgvÎvq 80kPa Pv‡c GKwU wbw`©ó cwigvb M¨v‡mi AvqZb 0.25 3 m | C 0 20 ZvcgvÎvq D³ M¨v‡mi AvqZb 0.50 3 m n‡j M¨vmwUi Pvc KZ n‡e? [MAT. 13-14] A) 20kPa B. 40kPa C. 50kPa D. 60kPa P1V1 = P2V2 , [T1 = T2] ev, 80 × 0.25 = P2 0.5 ev, P2 = 40kPa STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. 100C ZvcgvÎvq 20g Aw·‡Rb GKwU 20 cm •`‡N©¨i NbK‡K c~Y© K‡i| 1 †gvj Aw·‡R‡bi fi 32 gm. Nb‡Ki Af¨šÍ‡i Aw·‡R‡bi Pvc KZ? [Xv. †ev. 2015] A. 7800 kPa B. 242 kPa C. 65 kPa D. 12 kPa PV = m m RT P = m mv RT = 20 × 8.31 × 373 32 × (0.2) 3 = 242 kPa 02. wb‡Pi †Kvb †jLwPÎwU e‡qj Gi m~‡Îi Rb¨ cÖ‡qvR¨? [iv.‡ev 2021] A. B. C. D. PV V V V V PV P P Ans B wb‡Pi DÏxcK Abymv‡i 03 I 04 bs cÖ‡kœi DËi `vI: [iv. †ev. 2016] Dc‡ii †jLwP‡Î wbw`©ó cwigvY Av`k© M¨v‡mi †ÿ‡Î PV ebvg P †jLwPÎ †`Lv‡bv n‡q‡Q| 03. ‡jLwPÎwU †Kvb m~·K mg_©b K‡i? A. e‡q‡ji B. Pvj©‡mi C. Pv‡ci D. †Kjwfb †jLwPÎ †_‡K cvB, PV = aªæeK A_©vr P 1 V hv e‡q‡ji m~Î| myZivs †jLwPÎwU e‡q‡ji m~·K mg_©b K‡i| 04. A I B we›`y‡Z M¨v‡mi AvqZ‡bi AbycvZ A. 1 t 1 B. 1 t 2 C. 1 t 3 D. 2 t 1 A we›`y‡Z, PAVA = 200 1 × 105 × VA = 200 VA = 2 × 10–3 m 3 B we›`y‡Z, PBVB = 200 1 × 105 × VB = 200 VB = 2 × 10–3 m 3 VA VB = 2 × 10–3 m 3 1 × 10–3 m 3 = 2 : 1 05. w¯’i AvqZ‡b 1atm Pv‡ci †Kv‡bv M¨v‡mi ZvcgvÎv 0C evov‡j cwiewZ©Z Pvc n‡e- [w`.‡ev 2021] A. 0.00366atm B. 1atm C. 1.00366atm D. 2atm P = Po + Po 273 = 1 + 1×0 273 P = 1atm 06. 3 2 †gvj M¨v‡mi Rb¨ Av`k© M¨vm mgxKiY n‡e †KvbwU? [Kz.‡ev 2021] A. 3PV = 2RT B. 2PV = 1 3 RT C. 2PV = 3RT D. PV RT= 2 3 PV = nRT PV = 3 2 RT [n = 3 2 ] 2PV = 3RT
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 383 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 07. 1atm Pv‡c I 37C ZvcgvÎvq 32g Aw·‡R‡bi AvqZb KZ? [R = 8.314J mol–1K –1 ] [wm.‡ev 2021] A. 0.003m3 B. 0.025m3 C. 0.097m3 D. 0.814m3 PV = nRT; V = nRT P = 2 8.314 310 1.01 105 = 0.025m3 08. 0C DòZvq †Kv‡bv wbw`©ó AvqZ‡bi M¨v‡mi Pvc 3105Pa n‡j 60C DòZvq Gi Pvc KZ n‡e? [wm. †ev. 2016] A. 3.66105 Pa B. 2.45105 Pa C. 0.27105 Pa D. 0.40105 Pa P1 T1 = P2 T2 P2 = P1 T1 × T2 = 3 × 105 273 × 333 P2 = 3.66 × 105 Pa Concept 2 n«‡`i ev cyKz‡ii MfxiZv wbY©q msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. ey`ey‡`i AvqZ‡bi ¸Y †`Iqv _vK‡j: Rjvk‡qi MfxiZv h = g ( -1)P n GLv‡b, n = AvqZb hZ ¸Y | A_ev h (n 1)10.2 02. ey`ey‡`i c„ôZ‡ji †ÿÎd‡ji ¸Y †`Iqv _vK‡j: h = (n 2 1)P g GLv‡b, n = †ÿÎdj hZ ¸Y nq A_ev h = (n2 1) 10.2 03. ey`ey‡`i e¨v‡mi/e¨vmv‡a©i ¸b †`Iqv _vK‡j: h = g ( -1)P 3 n GLv‡b, n = e¨vm ev e¨vmva© hZ ¸b| A_ev ( 1) 10.2 3 h n CONCEPTUAL MATH Model Example01 ‡Kv‡bv n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmvq GKwU evqy ey`ey` AvqZ‡b cuvP¸Y nq| evqygÛ‡ji Pvc 105Nm–2 n‡j ü‡`i MfxiZv KZ? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] GLv‡b, n«‡`i Zj‡`‡ki ey`ey‡`i AvqZb, V1 = V n«‡`i c„‡ô ey`ey‡`i AvqZb, V2 = 5V cvwbi NbZ¡, = 103 kgm3 wfKl©R Z¡iY, g = 9.8ms2 n«‡`i MfxiZv, h =? aiv hvK, n«‡`i Zj‡`‡ki Pvc, p1 ü‡`i c„ô‡`‡k Pvc, p2 = evqygÛ‡ji Pvc = 105Nm 2 p1 = evqygÛ‡ji Pvc + h MfxiZvq cvwbi Pvc = p2 + hpg GLb Avgiv Rvwb, p1V1 = p2V2 (p2 + hg) V = p2 5V hg = 4p2 h = 4p2 g = 4 105Nm2 103 kgm3 9.8ms2 = 40.82m h = (n – 1) 10.2 h = (5 – 1) 10.2 = 40.82 Model Example02 ‡Kvb n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmvq GKwU ey`ey‡`i e¨vm wظY nq| n«‡`i c„‡ô evqygÛ‡ji Pvc 105Nm-2 n‡j n«‡`i MfxiZv KZ? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] GLv‡b, n«‡`i Zj‡`‡k ey`ey‡`i AvqZb, v1 cvwbi DcwiZ‡j ey`ey‡`i AvqZb, v2 cvwbi NbZ¡, = 1 103 kgm3 AwfKl©R Z¡iY, g = 9.8ms2 n«‡`i MfxiZv, h =? evqygÛ‡ji ¯^vfvweK Pvc, P2 = 105 Nm2 Avgiv Rvwb, V= 4 3 r 3 = 4 3 d 2 3 = 4 24d 3 = 6 d 3 = Kd3 V d 3 V2 = 8V1 myZivs e¨vm wظY n‡j AvqZb 8 ¸Y n‡e| g‡bKwi, n«‡`i Zj‡`‡k Pvc= P1 Ges n«‡`i c„‡ô Pvc – P2 P1 = P2 + hg Avgiv Rvwb, P1V1 = P2V2 ev, (P2 + hg) V = P2V2 = P2 8V ev, hg = 8P2 – P2 = 7P2 h = 7P2 g = 7 105 1 103 9.8 = 72.01m = 71.4 m h = (n3 1) 10.2 = (23 – 1) 10.2 = 7 10.2 = 71.4
384 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES NOW START PRACTICE 01. ‡Kvb n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmvq GKwU evqy ey`ey‡`i e¨vm 1.5 ¸Y nq| n«‡`i c„‡ô evqygÛ‡ji Pvc ¯^vfvweK evqygÛjxq Pv‡ci mgvb Ges n«‡`i DòZv aªæeK n‡j n«‡`i MfxiZv KZ? 02. ‡Kvb n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmvi mgq evZv‡mi ey`ey` AvqZ‡b wظY nq| evqygÛ‡ji Pvc 105Nm-2 n‡j, n«‡`i MfxiZv KZ? 03. †Kvb n«‡`i Zj‡`k †_‡K c„‡ô Avmvi d‡j GKwU evZv‡mi ey`ey‡`i AvqZ‡b mvZ¸b n‡q hvq| n«‡`i c„‡ô evqygÛ‡ji Pvc 105Nm-2 n‡j n«‡`i MfxiZv KZ? NOW PRACTICE SOLVE : 01. v d 3 n = 1.53 n«‡`i MfxiZv, h = (n – 1) P 9800 = (1.53 – 1) (1.013 105 ) 9800 = 24.55m ASPECT SPECIAL: h = (n 3 – 1) 10.2= (1.53 – 1) 10.2 = 24.23m 02. h = ( 2 –1) 10.2 = 10.2; 03. h (71)10.2 61.2m REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. ‡Kvb GKwU n«‡`i Zj‡`k †_‡K cvwbi Dcwi Z‡j Avmvq GKwU ey`ey` AvqZ‡b cuvP¸b nq| evqy gÛ‡ji Pvc Ges cvwbi NbZ¡ h_vµ‡g 105 N/m2 Ges 103 kg – m –3 n‡j n«‡`i MfxiZv KZ? g = 9.8ms–2 [DU. 10-11, KU. 10-11, 05-06, JUST. 15-16] A. 40.8m B. 10.2m C. 51m D. 49m MfxiZv, h = (n 1) 10.2 = (5 1) 10.2 (n = 5) = 4 10.2 = 40.8m STEP 02 ANALYSIS OF JU QUESTION 01. 4200 m DuPz GKwU RjcÖcv‡Zi Zj‡`k I kxl©‡`‡ki g‡a¨ ZvcgvÎvi e¨eavb KZ n‡e hw` cZbkxj cvwbi mg¯Í kw³B ZvcgvÎv e„wׇZ e¨q nq| [JU.15-16] A. 20C B. 9.8C C. 15.6C D. †KvbwUB bq = h 428.6 = 4200 428.6 = 9.8°C STEP 03 ANALYSIS OF GST QUESTION JUST 01. nª‡`i Zj‡`‡k †_‡K c„ó‡`‡k Avmvq GKwU evqy ey`ey`‡`i e¨vm wظY nq| n«v‡`i c„‡ô evqygÛjxq Pvc 105Nm2- n‡j n«‡`i MfxiZv KZ? [JUST. 15-16] A. 63.59m B. 54.89m C. 48.37 m D. 71.43m E. 81.43m MfxiZv h = (n3 –1)10.2 = (8–1)10.2 = 71.4 STEP 04 ANALYSIS OF HSC BOARD QUESTION 01. GKwU ey`ey` n«‡`i Zj‡`k †_‡K Dcwic„‡ô D‡V Avmvq Gi AvqZb AvU¸Y nq| evqygÛ‡ji Pvc H wgUvi D”PZvq cvwb¯Í‡¤¢i Pv‡ci mgvb n‡j n«‡`i MfxiZv KZ? [h. †ev. 2016] A. H B. 3 H C. 5 H D. 7 H h = (n – 1) H = 7H Concept 3 Av`k© M¨vm msµvšÍ FORMULA 01. Av`k© M¨vm mgxKiY: PV = nRT = m M RT hLb n = m M 02. f¨vbWviIqvjm Gi ms‡kvab P + n 2 a V 2 (V – nb) = nRT [a I b aªæeK] 03. †evëRg¨vb aªæeK: k = R NA [NA = A¨v‡fv‡M‡Wªvi msL¨v] CONCEPTUAL MATH MEx 01 18gm wnwjqvg M¨vm c~Y© †ejy‡bi AvqZb 0.10m3 | †ejy‡bi wfZi M¨v‡mi Pvc 1.2105Nm–2 †ejy‡bi ga¨eZx© M¨v‡mi ZvcgvÎvPV = m M RT ev, T = PVM mR = 1.2 105 0.1 4 18 8.316 = 320.9K NOW START PRACTICE 01. 20°C DòZvq 740mm Pv‡c 400ml AvqZ‡bi †Kvb M¨v‡mi fi 0.842g| M¨vmwUi AvYweK fiNOW PRACTICE SOLVE : 01. pv = m M RT ev, M = mRT pv = 0.842 0.0821 293 760 1000 740 400 ev, M = 52
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 385 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF RU QUESTION 01. GKwU Av`k© M¨v‡mi ZvcgvÎv T n‡Z e„w× K‡i 2T Kiv n‡j †Kvb ivwkwU wظY n‡e? [RU. 17-18] A. AYy¸wji Mo eM©‡e‡Mi eM©g~j B. AYy¸wji Mo †e‡Mi eM© C. AYy¸wji MwZkw³ D. AYy¸wji Mo eM©‡eM M¨v‡mi MwZ kw³, E = 2 3 nRT; E T 02. cÖgvb Ae¯’vq 2.2 g CO2 M¨v‡mi AvqZb †KvbwU? [RU.15-16] A. 1.12 L B. 0.12 L C. 2.12L D. 3.12 L CO2 Gi AvYweK fi = 44 MÖvg 44 MÖvg CO2 Gi AvqZb = 22.4 L 1 MÖvg CO2 Gi AvqZb = 22.4 44 L 2.2 MÖvg CO2 Gi AvqZb = 22.4 2.2 44 = 1.12L 03. GKwU †Lvjv wjUvi d¬v‡¯‹ 27C ZvcgvÎvq 1.32103 kg evqy Av‡Q| 97C ZvcgvÎvq d¬v¯‹ n‡Z Kx cwigvY evqy †ei n‡q hv‡e? [RU. 13-14] A. 3.310–3 kg B. 3.310–4 kg C. 2.510-4 kg D. 2.510–3 kg PV = m M RT ; mα 1 T m2 = T1 T2 m1 = 300 97 + 273 1.32 10–3 = 1.07 10–3 kg d¬v¯‹ n‡Z AcmvwiZ evqy = 1.32 10–3 kg – 1.07 10–3 kg = 2.5 10-4 kg STEP 02 ANALYSIS OF GST QUESTION PART A Analysis of Science & Technology Question SUST 01. 18gm wnwjqvg M¨vm c~Y© †ejy‡bi AvqZb 0.10m3 | †ejy‡bi wfZi M¨v‡mi Pvc 1.2105Nm-2 ‡ejy‡bi ga¨eZx© M¨v‡mi ZvcgvÎv- [SUST. 03-04] A. 320.9 K B. 321.9 K C. 319.5 K D. 322.5K PV = m M RT ev T = PVM mR = 1.2 105 0.1 4 18 8.316 = 320.9K BSMRSTU 01. 20°C DòZvq 740mm Pv‡c 400ml AvqZ‡bi †Kvb M¨v‡mi fi 0.842g| M¨vmwUi AvYweK fi- [BSMRSTU. 14-15] A. 50 B. 52 C. 60 D. 80 pv = m M RT ev, M = mRT pv = 0.842 0.0821 293 760 1000 740 400 ev, M = 52 DU Technology 01. Av`k© M¨v‡mi AvqZb †Kvb ZvcgvÎvq k~b¨ nq? [DU-Tech.19-20] A. –273K B. 0C C. 30F D. 0K – 273C ev 0 K ev cigk~b¨ ZvcgvÎvq M¨v‡mi AvqZb MwZkw³ k~b¨ nq| STEP 03 ANALYSIS OF ENGINEERING & BUTex QUESTION CKRuet. Combind 01. 30C ZvcgvÎvq 150m3 AvqZ‡bi K‡ÿ GKwU cvwbi cvÎ ivLv Av‡Q| KZUzKz cvwb ev®ú nIqvi ci Aewkó cvwb I ev®ú mvg¨ve¯’vq _vK‡e? [30C ZvcgvÎvq m¤ú„³ ev®úPvc 31.83 mm Hg Pvc] [CKRUET. 2020-21] A. 1.1kg B. 3.03 kg C. 4.32 kg D. 4.55 kg E. 4.55 gm cvwb I ev‡®úi mvg¨ve¯’vq, ev‡®úi AvqZb, V = 150m3 , ev‡®úi Pvc, P = 31.83 mm Hg = 4243.65 Pa, ZvcgvÎv, T = 30C = 303K ev‡®úi †gvj msL¨v, n = PV RT = 4243.65 × 150 8.314 × 303 = 252.68 = cvwbi †gvj msL¨v ev®úxf~Z cvwbi fi, W = nM = (252.68 × 18)g = 4548.32g = 4.55 kg [1 †gvj cvwbi fi = 18g] STEP 04 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. cig k~b¨ ZvcgvÎv n‡”Q FYvZ¥K- [MAT; 19-20] A. 273.45 B. 273.25C C. 273.15C D. 273.35C †h ZvcgvÎvq w¯’i Pv‡c †Kv‡bv wbw`©ó f‡ii M¨v‡mi AvqZb k~b¨ nq Ges MwZkw³ m¤ú~Y©iƒ‡c †jvc cvq Zv‡K cigk~b¨ ZvcgvÎv e‡j| Gi gvb – 273.15C ev 0 K| STEP 05 ANALYSIS OF HSC BOARD QUESTION 01. 27˚C ZvcgvÎvq 4g Aw·‡Rb M¨v‡mi †gvU MwZkw³- [iv. †ev. 2015] A. 116.86 J B. 207.75 J C. 467.44 J D. 1498 J Ek = 3 2 nRT Ek = 3 2 × 4 32 ×8.31×300 = 467.44 J 02. GK evqygÐjxq Pvc mgvb- [Kz.‡ev 2021] i. 1.013105Nm–2 ii. 1.013 105 Pa iii. 760mmHg wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii 1 atm = 1.01 × 105 Nm–2 = 1.01× 105 Pa = 760 mm Hg = 76 cm Hg
386 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 4 ZvcgvÎv I g~j MoeM© †eM msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. M¨v‡mi g~j Mo eM©‡eM I ZvcgvÎvi m¤úK©: Crms = 3RT M ; R = †gvjvi M¨vm aªæeK = 8.314 J mol–1K –1 ; M = AvYweK fi; R = 0.0821 Latm mol–1 k –1 Avevi, Crms = 3KT m ; K = AbycÖwZ M¨vmaªæe‡Ki gvb = R NA = 1.38 10–23JK–1 ; m = cÖwZwU AYyi fi Crms T C1 C2 = T1 T2 02. M¨v‡mi g~j Mo eM©‡eM I Pv‡ci m¤úK©: Crms = 3P ; Crms = 3PV m ; P = M¨v‡mi Pvc; = M¨v‡mi NbZ¡ ASPECT SPECIAL: (i) g~jMo eM©‡eM e„w×, crms = T2 T1 – 1 (ii) T2 = n2 .T1 CONCEPTUAL MATH MEx 01 0C ZvcgvÎvq O2 AYyi g~j Mo eM©‡eM wbY©q Ki? 27C ZvcgvÎvq Gi gvb KZ? C0 = 3RT M = 3 8.31 273 32 10–3 = 461.18m/s; C27 T27 T0 C0 = 300 273 461.18 = 483.44m/s. MEx 02 GKwU wiG±‡i 1.675 10–27kg f‡ii wbDUªb KYvi 27C ZvcgvÎvq ¯^vfvweKfv‡e wePiYiZ Ae¯’vq †Kvb Aÿ eivei G‡`i g~j-Mo eM© †eM KZ? Crms = 3kT m = 3 1.38 10–23 300 1.675 10–27 = 2723.7 m/s MEx 03 ‡Kvb ZvcgvÎvq nvB‡Wªv‡R‡bi g~j MoeM© †eM mvaviY Pvc I ZvcgvGvi g~j MoeM© †e‡Mi wظY| GLv‡b, n = hZ ¸b; ‡Kjwf‡b ZvcgvGv †ei Ki‡Z → T2= n2 273K (22 1) × 273 = 819C = 22 273K = 1092K ASPECT SPECIAL: wWwMÖ‡Z ZvcgvGv †ei Ki‡Z → T = (n2 – 1) 273C NOW START PRACTICE 01. STP-‡Z nvB‡Wªv‡Rb AYy¸‡jvi †e‡Mi Mo e‡M©i eM©g~j wbY©q Ki| S. T. P.- †Z nvB‡Wªv‡R‡bi NbZ¡ 0.09kgm–3 | 02. nvB‡Wªv‡Rb‡K Av`k© a‡i Ges AYy¸wji g~j Mo eM©‡eM 2000 wg./‡m. Ges NbZ¡ 0.09Km/m3 g‡b K‡i M¨v‡mi Pvc †ei Ki| 03. †Kvb ZvcgvÎvq nvB‡Wªv‡R‡bi g~j Mo eM©‡eM mvaviY Pv‡c I ZvcgvÎvq g~j Mo eM©‡e‡Mi wظY n‡e? NOW PRACTICE SOLVE : 01. C = 3P0 = 3 (1.013 105 ) 0.09 = 1.84kms–1 02. P = C 2 3 = (2000) 2 0.09 3 = 1.2 105 N–m –2 03. CT = T C0 2 T0 = 22 273 = 1092 K REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. †Kv‡bv Av`k© M¨v‡mi ZvcgvÎv †Kjwfb †¯‹‡j 4 ¸Y e„w× †c‡j Zvi AYy¸‡jvi g~j Mo eM©‡eM KZ ¸Y e„w× cvq? [DU. 20-21] A. 4 B. 1 2 C. 2 D. 1 †gvjvi M¨vm aªæeK, R Ges †gvjvi fi M aªæeK n‡j, C = 3RT M C T C2 C1 = T2 T1 = 4T1 T1 C2 = 2C1 02. †Kvb Av`k© M¨v‡mi ZvcgvÎv †Kjwfb †¯‹‡j wظY Kiv n‡j, Zvi AYy¸‡jvi rms †eM KZ ¸Y e„w× cvq? [DU.15-16] A. 4 B. 2 C. 1.41 D. 0.5 eM©g–j Mo eM©‡eM, C = 3RT M Cα T; C2= 2C1=1.41¸Y 03. GKwU wmwjÛv‡i ivLv GKwU Av`k© M¨v‡mi AYy¸‡jvi eM©g–j-Mo-eM©‡eM u| M¨v‡m Zvc cÖ‡qv‡Mi d‡j Pvc 9 ¸Y e„w× †cj| wmwjÛv‡ii AvqZb AcwiewZ©Z _vK‡j M¨v‡mi AYy¸‡jvi cwiewZ©Z eM©g~j-Mo-eM©‡eM KZ? [DU. 14-15] A. 9u B. 6u C. 3u/2 D. 3u eM©g~j-Mo-eM©‡eM, C = 3P ; C P C2 = P2 P1 C1 = 9 u = 3u 04. GK †gvj nvB‡Wªv‡Rb Ges GK †gvj Aw·‡R‡bi fi h_vµ‡g 2g Ges 32g n‡j †Kvb GK wbw`©ó ZvcgvÎvq AbycvZ nvB‡Wªv‡Rb AYyi g~j Mo eM©‡eM Aw·‡Rb AYyi g~j Mo eM©‡eM Gi gvb n‡e- [DU. 04-05, 03-04] A. 5 B. 4 C. 6 D. 7 C = 3RT M C 1 M nvB‡Wªv‡Rb AYyi g~j Mo eM©‡eM Aw·‡Rb AYyi g~j Mo eM©‡eM = CH2 CO2 = MO2 MH2 = 32 2 = 4
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 387 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 02 ANALYSIS OF JU QUESTION 01. 27C ZvcgvÎvi M¨vm‡K KZ ZvcgvÎvq †bqv n‡j Mo †eM wظY n‡e? [JU-H, Set-A. 20-21] A. 1200K B. 1050K C. 1300K D. 1350K Crms = 3RT M C T C 2 T C2 2 C1 2 = T2 T1 T2 = 22 300 = 1200 K [C2 =2C1 C2 C1 = 2 Ges T1 = 300K] kU© †UKwbK: T2 = n2 T1 = 22 300 = 1200 K 02. w¯’i Pv‡c †Kvb ZvcgvÎvq †Kv‡bv M¨v‡mi AYyi g~j Mo eM©‡eM cÖgvY Pvc I ZvcgvÎvi g~j Mo eM©‡e‡Mi A‡a©K n‡e? [JU-A, Set-A. 19-20] A. 58.20 K B. 68.25 K C. 20.25 K D. 100.20 K Avgiv Rvwb g~j Mo eM©‡eM, Crms = 3RT M 3RT M = 3 R 273 M 1 2 3RT M = 3R 273 M 1 4 T = 273 4 k = 68.25K 03. 0C ZvcgvÎvq Aw·‡R‡bi g~j Mo eM©‡eM wbY©q Ki| [JU. 17-18] A. 560 m/s B. 461 m/s C. 861 m/s D. 961 m/s eM©g~j Mo eM©‡eM,C = 3RT M = 3 8.314 273 32 10–23 = 461ms–1 04. ¯^vfvweK ZvcgvÎv I Pv‡c nvB‡Wªv‡R‡bi NbZ¡ 0.09 kg m–3 nvB‡Wªv‡Rb AYyi Mo eM© ‡e‡Mi eM©g~j KZ? [JU. 11-12] A. 18.38102ms-1 B. 87.3610-3ms-1 C. 29.38102ms-1 D. 37.38102ms-1 C = 3P = 3 1.01325 105 0.09 = 18.37 102 ms–1 STEP 03 ANALYSIS OF RU QUESTION 01. n msL¨K M¨v‡mi AYyi cÖ‡Z¨KwUi `ªæwZ 2ms–1 | AYy¸‡jvi r.m.s. `ªæwZ KZ ms–1 ? [RU. 18-19] A. 2 n B. n 2 C. 2 n D. 2 Crms = n 2 2 n = 2 02. 32˚K ZvcgvÎvq Aw·‡Rb M¨v‡mi R.M.S. †eM KZ? [RU. 17-18] A. R B. 6R C. 2R D. 3R eM©g~j Mo eM©‡eM, C = 3RT M = 3 R 32 32 = 3R 03. cuvPwU AYyi †eM , 2, 3, 4, 5| AYyM‡jvi Mo eM©‡e‡Mi eM©g~j KZ? [RU. 17-18] A. 11 B. 11 C. / 11 D. 11 eM©g–j Mo eM©‡eM, C = C 2 1 + C2 2 + C2 3 + ........ + C2 n n 1 2 + 22 + 32 + 42 + 52 5 v = 55 5 v = 11v 04. RMS †eM- [RU. 17-18] A. 3 2 RT B. 8RT M C. 3RT M D. w M RT Ans C 05. `ywU K…ò e¯‘i wbM©Z Zvckw³i AbycvZ 16:1| wØZxq e¯‘i ZvcgvÎv 300 k n‡j, cÖ_g e¯‘i ZvcgvÎv KZ? [RU.15-16] A. 1200 K B. 1600 K C. 600 K D. 300 K E1 E2 = T1 T2 4 ev, 16 1 = T1 300 4 ev, 2 = T1 300 ev, T1 = 600K 06. w¯’i Pv‡c KZ ZvcgvÎvq †Kvb M¨v‡mi AYyi Mo eM©‡e‡Mi eM©g~j cÖgvY Pvc I ZvcgvÎviMo eM©‡e‡Mi eM©g~‡ji wظY n‡e? [RU. 14-15, 15-16, 16-17] A. 546K B. 1092K C. 273K D. 2184K eM©g~j Mo eM©‡eM, C = 3RT M C T; T C 2 T1= C1 C2 2 T2 = 22 273 = 1092 K STEP 04 ANALYSIS OF CU QUESTION 01. †Kvb Av`k© M¨v‡mi ZvcgvÎv 120K †_‡K 480K G DbœxZ Kiv nj| hw` 120K-G g~j Mo eM© †eM v nq Z‡e 480K-G Zv n‡e- [CU. 18-19] A. 4v B. 2v C. v 2 D. v 4 Crms T Crms2 Crms1 = 480 120 Crms2 = 2v 02. dzUšÍ cvwb ev‡®ú cwiYZ n‡”Q G Ae¯’vq cvwbi Av‡cw¶K Zvc n‡e- [CU. 15-16] A. k~b¨ B. GK C. GK Gi †P‡q †QvU D. Amxg E. †KvbwUB bq Ans D 03. 0C ZvcgvÎvq 1kg eid 30C ZvcgvÎv 5 wjUvi cvwbi mv‡_ †gkv‡bv n‡jv| wgkÖ‡Yi †kl ZvcgvÎv KZ n‡e? [CU. 15-16] A. 11.67C B. 11.68C C. 11.69C D. 11.66C E. 11.65C = m22 – 80m1 m1 + m2 ev, = 5 30 – 80 1 5 + 1 = 150 – 80 6 = 70 6 = 11.68C 04. 0 0C ZvcgvÎvi Aw·‡R‡bi eM©g~j Mo eM©‡eM KZ n‡e? [CU. 13-14] A. 461ms-1 B. 641ms-1 C. 146ms-1 D. 416 m s-1 E. 164 m s-1 c = 3RT M = 3 8.316 273 103 32 = 461ms–1 STEP 05 ANALYSIS OF DU-7 Clg QUESTION 01. GKwU M¨v‡mi cig ZvcgvÎv Pvi¸Y Kiv n‡j, Gi AYy¸‡jvi g~j Mo eM©‡eM KZ n‡e? [DU-7Clg.A.19-20] A. AcwiewZ©Z _vK‡e B. A‡a©K n‡e C. Pvi¸Y n‡e D. wظY n‡e Crms = 3RT M Crms T, C2 = 4T T C1 = 2C1 STEP 06 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. N.T.P †Z nvB‡Wªv‡Rb AYy¸‡jvi g~jMo eM©‡eM wbY©q Ki N.T.P †Z nvB‡Wªv‡Rb NbZ¡ 0.088 kgm-3 [JnU. 09-10] A. 1.90 Kms-1 B. 1.84 Kms-1 C. 1.88 ms-1 D. 1.86 ms-1 C = 3P = 3 1.01325 105 0.088 = 1.86ms–1 02. cÖgvY ZvcgvÎv I Pv‡c †Kvb M¨v‡mi AYy¸‡jvi Mo †e‡Mi eM©g~j wbY©q Ki| (cÖgvb ZvcgvÎvq Ges Pv‡c M¨v‡mi NbZ¡ 1.4kg/m3 [JnU. 08-09] A. 469.1 ms-1 B. 421.2 ms-1 C. 465.9 ms-1 D. 591.2 ms-1 C = 3P = 3 1.01325 105 1.4 = 465.9ms–1
388 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 03. GKwU weKv‡i 60 cm3 cvwb Av‡Q| H cvwb‡Z GKwU 128 g f‡ii avZe e¯‘ LÛ wbgw¾Z Ki‡j cvwbi AvqZb `vovq 78 cm3 | avZe H e¯‘ L‡Ûi NbZ¡ wbY©q Ki? [JnU. 07-08] A. 1.6 g/cm3 B. 2.1 g/cm3 C. 7.11 g/cm3 D. 18.0 g/cm3 e¯‘i AvqZb v = (78–60) = 18 cm3 e¯‘i NbZ¡ = m v = 128 18 = 7.11 g/cm3 04. wZbwU M¨vm AYyi †eM n‡”Q 25m/s, 30m/s Ges 35m/s| Zv‡`i g~j Mo eM©‡eM n‡e- [JnU. 06-07] A. 17.5 m/s B. 30.3 m/s C. 30 m/s D. 916.7 m/s 252 + 302 + 352 3 = 30. 28ms–1 CoU 01. ZvcgvÎv e¨eavb Kg n‡j †Kvb e¯‘ KZ…©K Zvc nviv‡bvi nvi e¯‘i I cwicv‡k¦©i ZvcgvÎv e¨eav‡bi- [CoU.15-16] A. mgvb B. mgvbycvwZK C. A‡a©K D. e¨v¯ÍvbycvwZK Ans B IU 01. 1 kg cvwbi ZvcgvÎv 1 k e„w× Ki‡Z cÖ‡qvRbxq Zvc- [IU. 15-16] A. 4.2 J B. 42 J C. 3.36×105 J D. 4200 J Ans D PART B Analysis of Science & Technology Question SUST 01. †Kvb M¨v‡mi ZvcgvÎv 1000 C ‡_‡K evwo‡q 2000 C Kiv nj| M¨v‡mi Mo †eM KZ ¸Y evo‡e? [SUST. 10-11, MBSTU-A, Set-2 19-20] A. 1.13 ¸Y B. 1.31 ¸Y C. 2 ¸Y D. 4 ¸Y C2 C1 = T2 T1 ev, C2 = 473 373 C1 = 1.126 = 1.13 ¸b 02. 0 0C ZvcgvÎvq Aw·‡R‡bi gyjMo eM©‡eM 461 m/s n‡j 1000C ZvcgvÎvq GB †eM KZ n‡e? [SUST. 07-08] A. 539 m/s B. 458 m/s C. 798 m/s D. 4610 m/s C2 C1 = T2 T1 ev C2 461 = 373 273 ev C2 = 1.168885 461 = 539 ms–1 03. ‡Kv‡bv Ave× M¨v‡mi ZvcgvÎv 0C †_‡K evwo‡q 273C Kiv n‡j M¨v‡mi AYy¸‡jvi Mo †eM KZUzKz e„w× cv‡e? [SUST. 06-07] A. 40% B. 50% C. 60% D. 100% C = 3RT M Cα T C = C2 – C1= T2 T1 C1 – C1 = 273 + 273 273 C1 – C1 = 0.41C1 40% 04. 230C ZvcgvÎvq bvB‡Uªv‡Rb AYyi eM©g~j Mo eM© MwZ‡eM: [SUST. 05-06] A. 1.92x105 cms-1 B. 102 ms-1 C. 5.192x104 cms-1 D. 1.92x105 ms-1 C = 3RT M = 3 8.316 296 28 10–3 = 514ms–1 = 5.14x104 cms1 05. ¯^vfvweK ZvcgvÎv I Pv‡c †Kvb Ave× M¨v‡mi NbZ¡ 0.0892 kg/m3 , H M¨v‡mi AYy¸‡jvi g~j Mo eM©‡eM KZ? [SUST. 05-06] A. 1.846x104 m/s B. 0.1846x104m/s C. 0.846x106m/s D. 0.1846x106 m/s C = 3P = 3 1.01325 105 0.0892 = 1.846x103 m/s = 0.1846x104 m/s 06. 15 gm f‡ii Ges 30 gm f‡ii `yBwU e¯‘ Gi AvqZ‡bi 2 7 Ask cvwbi Dc‡ii w`‡K fv‡m| e¯‘ `yBwUi NbZ¡ KZ? [SUST. 05-06] A. 0.714 gm/cc Ges 0.714 gm/cc B. 0.714 gm/cc Ges 1.428 gm/cc C. 1.428 gm/cc Ges 1.428 gm/cc D. †KvbwU bq| cvwbi wb‡P wbgw¾Z As‡ki AvqZb = 1 – 2 7 = 5 7 cÖ_g e¯‘i NbZ¡ = 5 7 = 0.7143 gm/cc e¯‘ ؇qi f‡ii AbycvZ = 15 : 30 = 1 : 2 2q e¯‘i NbZ¡ = 0.714 2 = 1.428 gm/cc JUST 01. 30˚C ZvcgvÎvq wnwjqvg Ges †Rbb M¨v‡mi eM©g~j Mo eM©‡eM Gi AbycvZ KZ? [JUST-B, 19-20] A. 1.80 B. 0.16 C. 0.17 D. 0.18 Cx e CHe = MHe Mxe = 4 131 = 0.17 BSMRSTU 01. 20°C ZvcgvÎvq nvB‡Wªv‡Rb AYyi eM©g~j Mo eM© †eM KZ? [BSMRSTU. 14-15] A. 1.87103ms–1 B. 7.81103ms–1 C. 1.91103ms–1 D. 9.11103ms–1 C = 3RT M = 3 8.31 293 1000 2 = 1.911 103ms–1 STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. P~ovšÍ ZvcgvÎv Avw` ZvcgvÎvq KZ ¸Y n‡j †Kv‡bv wbw`©ó M¨v‡mi g~jMo eM©‡eM wظY n‡e? [Xv.‡ev 2021] A. 2 ¸Y B. 3 ¸Y C. 4 ¸Y D. 9 ¸Y Ans C 02. ¯^vfvweK ZvcgvÎv I Pv‡c bvB‡Uªv‡R‡bi NbZ¡ n‡jv 1.25 kgm–3 | Gi g~j Mo eM© †eM (Crms) n‡jv- [Kz. †ev. 2015] A. 491.07 ms–1 B. 492.07 ms–1 C. 493.07 ms–1 D. 495.07 ms–1 Crms = 3P = 3 × 1.01 × 105 1.25 = 493.07 ms–1
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 389 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 5 M¨v‡mi MwZkw³ msµvšÍ MvwYwZK cÖ‡qvM FORMULA n †gvj M¨v‡mi MwZkw³, Ek = 3 2 nRT; n = †gvj msL¨v (GK cvigvYweK M¨v‡mi Rb¨) 1 †gvj M¨v‡mi MwZkw³, Ek = 3 2 RT; n =1 †gvj| (GK cvigvYweK M¨v‡mi Rb¨) M¨v‡mi cÖwZ AYyi Mo MwZkw³, Ek = 3 2 KT; K = †evjRg¨vb aªæeK| (GK cvigvYweK M¨v‡mi Rb¨) Av`k© M¨v‡mi Pv‡ci mgxKiY, PV = 1 3 mn c –2 cÖwZ GKK AvqZ‡b Mo MwZkw³, E = 1 3 c 2 GKK AvqZ‡b AYy¸‡jvi M¨v‡mi Pvc, PV = 2 3 E wØ cvigvYweK M¨vm n‡j: n †gvj M¨v‡mi MwZkw³, Ek = 5 2 nRT cÖwZ AYyi Mo MwZkw³, Ek = 5 2 KT CONCEPTUAL MATH MEx 01 : 270C ZvcgvÎvq 2gm bvB‡Uªv‡R‡bi MwZkw³ wbY©q Ki| Ek=5/2nRT 5 2 2 28 8.314 300 445.4J MEx 02 : 270C ZvcgvÎvq cÖwZ †gvj wnwjqv‡gi MwZkw³ KZ? Ek=3/2 RT =1.5 RT 1.5 8.314 300 3741.3J MEx 03 : 27C ZvcgvÎvq 1gm-mol nvB‡Wªv‡Rb Ges 1gm nvB‡Wªv‡R‡bi MwZkw³ KZ? Ek = 5 2 nRT = 5 2 1 8.314 300 = 6235.5 J l Ek = 5 2 m M RT = 3 2 1 2 8.314 300 = 3117.75J NOW START PRACTICE 01. GKwU cv‡Î 27C ZvcgvÎvq wnwjqvg M¨vm Av‡Q| wnwjqvg AYyi Mo MwZkw³ KZ? (†evjRg¨vb aªæeK k = 1.38 10–23J/K) 02. Avgiv k^vm cÖk^v‡m 1.0 wjUvi evqy †meb Ki‡j †gvU KZ¸‡jv AYy †meb K‡i _vwK? 03. ¯^vfvweK Pv‡c 1 wjUvi M¨v‡mi MwZkw³ KZ? NOW PRACTICE SOLVE : 01. EK = 3 2 KT = 3 2 1.38 10–23 300 = 6.21 10–21J 03. PV = 2 3 E E = 3 2 PV = 3 × 0.76 × 1.6 × 103 × 9.81 × 10–3 2 = 1.52 × 102 J 02. †gvU AYy, N = 6.023 1023 22.4 = 2.7 1022 wU REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU KYvi ¯^vaxbZvi gvÎvi msL¨v 5 n‡j kw³i mgwefvRb bxwZ Abyhvqx KYvwUi †gvU kw³ KZ? [DU. 15-16] A. kT/2 B. kT C. 3kT/2 D. 5kT/2 H2, N2, CO2 BZ¨vw` wØ-cigvYyK M¨vm AYyi ¯^vaxbZvi gvÎv 5| wØ-cigvYyK M¨vm AYyi kw³i cwigvY = 5 2 kT 02. GKwU cv‡Î 270C ZvcgvÎvq wnwjqvg M¨vm Av‡Q| wnwjqvg AYyi Mo kw³ KZ? (†evëRg¨vb aªæeK K=1.3810-23J/K) [DU. 11-12] A. 6.2110-21j B. 5.610-22j C. 1.910-21j D. 210-21j T = 27 + 273 = 300k Ek = 3 2 kT = 3 2 1.38 10–23 300 = 6.21 10–21J STEP 02 ANALYSIS OF JU QUESTION 01. T ZvcgvÎvq Av`k© M¨v‡mi †¶‡Î GKwU AYyi Mo MwZkw³- [JU. 18-19; DU. 10-11; CU. 15-16] A. 2 3 KT B. 3 2 KT2 C. 3 2 KT4 D. 3 2 KT Ans D 02. 270C ZvcgvÎvq 3gm bvB‡Uªv‡R‡bi MwZkw³ wbb©q Ki| bvB‡Uªv‡R‡bi MÖvg AvbweK fi = 28 gm| [JU. 10-11] A. 420.66 J B. 440.66 J C. 220.66 J D. 400.66 J EK = 3 2 m M RT = 3 2 3 28 8.316 300 = 400.66J STEP 03 ANALYSIS OF RU QUESTION 01. S.T.P †Z 2 mol Av`k© M¨v‡mi MwZkw³ KZ? (R = 8.31 J mol–1K –1 ) [RU. 17-18] A. 1300 J B. 6806 J C. 2700 J D. 3403 J M¨v‡mi MwZ kw³, E = 2 3 nRT = 2 3 2 8.31 273 = 6806 J
390 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. 250C ZvcgvÎvq 22g CO2 M¨v‡mi AYymg~‡ni MwZkw³ KZ? [RU. 12-13] A. 84.46 j B. 1858.12 j C. 1558.12 k j D. 84.46 K j EK = 3 2 m M RT = 3 2 22 44 8.316 298 = 1858.6 J 03. 25˚C G 4.4 g CO2 AYyi MwZkw³ n‡e- [RU. 11-12] A. 84.46 J B. 371.64 J C. 371.64 K D. 371.64 kJ CO2 AYyi MwZkw³, E = 3 2 m M RT = 3 2 4.4 44 8.314 298 = 371.64 J Ans. STEP 04 ANALYSIS OF CU QUESTION 01. 300 K ZvcgvÎvq GKwU Av`k© M¨v‡mi MwZkw³ KZ? [CU-A, Set-2. 20-21] A. 3500J B. 3741J C. 3650J D. 3845J S B info Kinetic Energy: Ek = 3 2 RT = 3 2 8.31 300 = 3740J 3741J 02. 27C ZvcgvÎvq cÖwZ MÖvg AYy wnwjqvg M¨v‡mi MwZkw³ KZ? [CU. 15-16] [R = 8.3 Jk-1 mol-1 ] A. 3600 J B. 3635 J C. 3700 J D. 3735 J Ex = 3 2 nR = 3 2 × 8.30 × 300 = 3735J STEP 05 ANALYSIS OF DU-7 Clg QUESTION 01. P1 Pv‡c †Kv‡bv M¨vm‡K GKwU cv‡Î ivLv n‡jv| hw` †mB M¨vm AYyi fi‡K A‡a©K Ges `ªæwZ‡K wظY Kiv nq, Z‡e PyovšÍ Pvc KZ n‡e? [DU. 7Clg-A: 20-21] A. P1/2 B. P1 C. 2 P1 D. 4 P1 P1 = 1 3 mNc2 V ; A_ev P2 = 1 3 M 2 N(2c) 2 v P2 P1 = 1 2 2 2 = 2 P2 = 2P1 STEP 06 ANALYSIS OF GST QUESTION PART A Analysis of Science & Technology Question JUST 01. 127˚C ZvcgvÎvq 22 MÖvg CO2 M¨v‡mi Abymg~‡ni MwZkw³ KZ n‡e? R = 8.314 Jmol–1K –1 [JUST-B, 19-20] A. 791.53 J B. 2493 J C. 249.3 J D. 1246.5 J Ek = 3 2 nRT = 3 2 × 22 44 × 8.31 × 400 = 2493J 02. 29˚C ZvcgvÎvq 3g bvB‡Uªv‡R‡bi †gvU MwZkw³ KZ Ryj? [JUST-A, Set-Ka 19-20] A. 323 B. 383 C. 403 D. 423 †gvU MwZkw³ = 3 2 nRT = 3 2 × 3 28 × 8.31 × 302 = 403 J 03. 27C ZvcgvÎvq 3g bvB‡Uªv‡R‡bi †gvU MwZ kw³ KZ? [JUST. 15-16] A. 300 J B. 400 J C. 500 J D. 600 J E. 700 J Ek = 3 2 m M RT = 3 2 3 24 8.316 300 = 467.78J STEP 07 ANALYSIS OF ENGINEERING & BUTex QUESTION 01. 270C ZvcgvÎvq GKwU M¨vm AYyi Mo MwZkw³ 6.2110-21J 2270C ZvcgvÎvq Mo MwZkw³ n‡e [BUET. 04-09] A. 11.3510-21j B. 9.3510-21j C. 12.3510-21j D. 10.3510-21j E2 E1 = T2 T1 ev, E2 = 500 6.21 10–21 300 = 10.35 10–21J STEP 08 ANALYSIS OF HSC BOARD QUESTION wb‡Pi DÏxc‡Ki Av‡jv‡K 01 I 02 bs cÖ‡kœi DËi `vI : GKwU wmwjÐv‡i 27C ZvcgvÎvq wnwjqvg M¨vm Av‡Q| M¨v‡mi `ywU AYyi †eM h_vµ‡g 9ms–1 I 10ms–1 | 01. wnwjqvg AYyi Mo MwZkw³ KZ? [P.‡ev 2021] A. 6232.5J B. 3739.5J C. 6.21 × 10–21J D. 10.3510–21J Ans C 02. AYy؇qi Mo eM©‡eM KZ? [P.‡ev 2021] A. 9.5ms–1 B. 9.51ms–1 C. 90.5 m2 s –2 D. 181 m2 s –2 c = c1 2 + c2 2 2 = 9 2 + 102 2 = 90.5 m2 s –2 03. Aw·‡Rb M¨v‡mi cÖwZwU AYyi Mo MwZkw³ KZ? [P. †ev., iv. †ev. 2021] A. 1 2 KT B. 3 2 KT C. 5 2 KT D. 7 2 KT Ans C 04. 30C ZvcgvÎvq 7gm bvB‡Uªv‡Rb M¨v‡mi †gvU MwZkw³ wnmve Ki| [bvB‡Uªv‡R‡bi MÖvg AvYweK fi = 28gm ] [P.‡ev. 2019] A. 125.55 J B. 128.62 J C. 944.22 J D. 14958.00 J Ek = 3 2 nRT = 3 2 × 7 28 × 8.31 × 300 = 944.22 J 05. 30˚C ZvcgvÎvq cÖwZ MÖvg AYy wnwjqvg M¨v‡mi MwZkw³- [h. †ev. 2016] A. 7544.7 Jmol–1 B. 3772.35 Jmol–1 C. 1676.6 Jmol–1 D. 373.5 Jmol–1 E = 3 2 × 8.314 × 300 = 3772.35 J mol–1 06. ZvcgvÎvq Av`k© M¨v‡mi †ÿ‡Î GKwU AYyi Mo MwZkw³- [wm. †ev. 2017, 2016; w`. †ev. 2015; iv. †ev. 2015] A. 1 2 KT B. 3 2 KT C. 2 3 KT D. 3 2 KT2 Ans B 07. 15˚C ZvcgvÎvq cÖwZ MÖvg AYy wnwjqvg M¨v‡mi MwZkw³ KZ? (R = 8.31 JK–1 mol–1 ) [wm. †ev. 2015] A. 12.47 J B. 1196.64 J C. 3589.92 J D. 7179.84 J E = 3 2 nRT = 3 2 × 8.314 × 298 = 3589.92 08. 12 ¯^vaxbZv gvÎvm¤úbœ †Kv‡bv AYyi †gvU kw³ n‡e- [e. †ev. 2017] A. 6 KT B. 1 2 KT C. 3 2 KT D. 12 KT Ans A 09. ‘h’ D”PZvwewkó Nb‡Ki g‡a¨ m f‡ii Av`k© M¨vm Av‡Q| Zvi wefekw³Ñ [e. †ev. 2016] A. mgh B. 1 2 mc2 C. 3 2 KT D. k~b¨ M¨v‡mi wefe kw³ †bB ïay MwZkw³ Av‡Q|
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 391 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 6 Mogy³ c_ msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. Mo gy³ c_ = 1 2 2 n (g¨v·I‡qj) 02. Mo gy³ c_ = 1 2 n (K¬wmqvm); n = GKK AvqZ‡b AYy msL¨v| 03. Mo gy³ c_ B = 3 4 2 n (†evëRg¨vb) 04. cÖwZ †m‡K‡Ð av°vi msL¨v, N = 2 ln 05. Mo gy³ c_ c = l n = l 2 l n = 1 2 n Note: = d CONCEPTUAL MATH MEx 01 : †Kvb M¨vm AYyi e¨vm 7 10-10m. Ges cÖwZ NbwgUv‡i M¨vm AYyi msL¨v 2.791025| M¨vmwUi Mo gy³ c_ KZ n‡e| = 1 2d 2 n = 1 2 3.14 (7 10–10) 2 2.79 1025 = 1.65 x 10-8m NOW START PRACTICE 01. †Kvb M¨v‡mi AYyi Kvh©Ki e¨vm 210-10m Ges Mo gy³ c_ 2.410-8m. D³ M¨v‡m GKK AvqZ‡b AYyi msL¨v wbY©q Ki| hw` AYy¸‡jvi g~jMo eM©‡eM 103ms-1 nq Z‡e cÖwZ †m‡K‡Û av°vi msL¨v KZ? 02. GKwU M¨vm AYy cici 20wU av°vi ga¨ w`‡q 3 × 10–5 cm `~iZ¡ AwZµg K‡i| M¨v‡mi AYyi Mo gy³c_ wbY©q K‡iv| NOW PRACTICE SOLVE : 01. = 1 2d 2 n ev, n = 1 2d 2 = 1 2 3.14 (2 10–10) 2 2.4 10–8 = 2.35 1026 molecule m–3 cÖwZ †m‡K‡Û av°vi msL¨v, Crms = 103 2.4 10–8 = 4.167 1010 02. Avgiv Rvwb, Mo gy³ c_ = AwZµvšÍ `~iZ¡ av°vi msL¨v = 3 × 10–5 cm 20 = 1.5 × 10–6 cm REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU M¨vm AYyi e¨vm 21010m Ges cÖwZ Nb †mw›UwgUv‡i AYyi msL¨v 31019 n‡j M¨vm AYyi Mo gy³c_ n‡e- [DU. 16-17] A. 3103 cm B. 3104 cm C. 3105 cm D. 6106 cm AYyi Mo gy³c_, = 1 d 2N = 1 3.14 (2 10–10) 2 (3 1019) = 3105 cm 02. ¯^vfvweK ZvcgvÎv I Pv‡c nvB‡Wªv‡Rb AYyi Mo gy³ c_ cÖvq- [DU. 05-06] A. 10-9m B. 10-7m C. 10-5m D. 10-4m Ans B STEP 02 ANALYSIS OF JU QUESTION 01. ‡Kvb GKwU M¨v‡mi Mo gy³ c_ m 8 6 10 I AYyi e¨vm m 10 2.5 10 | cÖwZ Nb wgUv‡i AYyi msL¨v KZ? [JU. 09-10] A. 3 1025 m –3 B. 6 1025 m –3 C. 3 1020 m –3 D. 36 1020 m –3 = 1 2 d 2N ev, 6 10–8 = 1 2 (2.5 10–10) 2 N ev, N = 610 25 m –3 STEP 03 ANALYSIS OF CU QUESTION 01. M¨v‡mi Mo gy³c_ Ges ZvcgvÎvi g‡a¨ m¤úK© Kx? [CU-A, Set-2. 20-21, MBSTU-C, Set-2 19-20] A. T B. 1 T C. 1 T 2 D. 1 T Avgiv Rvwb, ZvcgvÎv e„w× †c‡j M¨v‡mi MwZkxjZv †e‡o hvq| d‡j Mo gy³c_I e„w× cvq mgvbycvwZK nv‡i| STEP 04 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. ‡Kvb GKwU M¨v‡mi AYyi Mo gy³ c_ 2.410-6 cm I AvbweK e¨vm 2.010-8 cm cÖwZ Nb †mw›UwgUvi AYyi msL¨v KZ? [KUET. 14-15] A. 2.3451026 B. 3.21022 C. 2.342 1028 D. 3.510-20 E. 2.3441020 = 1 2d 2N ev, N = 1 2d 2 = 1 2 (2.0 10–8 ) 2 2.4 10–6 ev, N = 2.344 1020
392 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 7 Av‡cwÿK Av`ª©Zv I wkwkiv¼ msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. Av‡cwÿK Av`ª©Zv, R = f F 100% [f = wkwkivs‡K m¤ú„³ ev®úPvc; F = evqyi ZvcgvÎvq m¤ú„³ ev®úPvc] 02. wkwkivsK = 1 – G (1 2); [1 2] [1 = ﮋ ev‡j¦i ZvcgvÎv; 2 = Av`ª© ev‡j¦i ZvcgvÎv; G = †MøBmv‡ii aªæeK] CONCEPTUAL MATH MEx 01 ‡Kvb ¯’v‡b GKwU wbw`©ó mg‡q evqyi ZvcgvÎvq m¤ú„³ Rjxq ev‡®úi Pvc 16 mmHg Ges wkwkivs‡K m¤ú„³ Rjxq ev‡®úi Pvc 12 mmHg n‡j H ¯’v‡b evZv‡mi Av‡cw¶K Av`ª©Zv KZ? R= f F 100% 12 16 100% = 75% MEx 02 †Kvb ¯’v‡b evZv‡mi ZvcgvÎv 25C Ges wkwkivsK 16C| 16C, 24C Ges 26C ZvcgvÎvq m¤ú„³ Rjxq ev‡®úi Pvc h_vµ‡g 13.6 10–3m, 22.3 10–3 m Ges 25.1 10–3m cvi` n‡j H ¯’v‡b Av‡cwÿK Av`ª©Zv KZ? 25C ZvcgvÎvq m¤ú„³ Rjxq ev‡®úi Pvc = 22.3 10–3 + 25.1 10–3 2 = 23.7 10–3m Av‡cwÿK Av`ª©Zv, R = f F 100% = 13.6 10–3 23.7 10–3 100% = 57.38% MEx 03 †Kvb GKw`b wm³ I ﮋ evj¦ _v‡g©vwgUv‡i cvV h_vµ‡g 28C I 30C cvIqv †Mj| 30C ZvcgvÎvq G Gi gvb 1.65| H w`b wkwkivsK KZ n‡e? = 1 – G (1 – 2) = 30 – 1.65 (30 – 28) = 26.7C NOW START PRACTICE 01. †Kvb mg‡q evqygÛ‡ji ZvcgvÎv 170C I wkwkivsK 120C; 170C I 120C ZvcgvÎvq m¤ú„³ ev®úPvc h_vµ‡g 14.42x10-3m I 10.46x10-3m cvi`| G mg‡qi evqyi Av‡cw¶K Av`ª©Zv wbY©q Ki| 02. †Kvb ¯’v‡b †Kvb GKw`b evq~i ZvcgvÎv 180C I wkwkivsK 120C| 180C I 120C ZvcgvÎvq m¤ú„³ ev®úPvc h_vµ‡g 15.48x10-3 m I 10.52x10-3m cvi`| H w`‡bi Av‡cw¶K Av`ª©Zv KZ? 03. †Kvb GKw`b wm³ I ﮋ evj¦ _v‡g©vwgUv‡i cvV h_vµ‡g 27C I 30C cvIqv †Mj| 30C ZvcgvÎvq G Gi gvb 1.65| H w`b wkwkivsK KZ n‡e? NOW PRACTICE SOLVE : 01. R = f F 100% So, R = 10.46 10–3 14.42 10–3 100% = 72.54% 02. R = f F 100% So, R = 10.52 10–3 15.4810–3 100% = 67.96% 03. = 1 – G (1 – 2) = 30 – 1.65 (30 – 27) = 25.05C REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF JU QUESTION 01. †Kvb GKw`‡b evqygÛ‡ji ZvcgvÎv 20C Ges wkwkiv¼ 15C| 20C I 15C ZvcgvÎvq m¤ú„³ Rjxq ev‡®úi Pvc h_vµ‡g 4103m I 2103m cvi` n‡j, H w`‡bi Av‡cw¶K Av`ªZv KZ? [JU. 14-15] A. 20% B. 40% C. 50% D. 60% Av‡cw¶K Av`ª©Zv, R = f F 100% = 2 10–3 4 10–3 100% = 50% STEP 02 ANALYSIS OF RU QUESTION 01. ‡Kvb GKw`‡bi wkwkivsK 10 ‡mjwmqvm, Av‡cw¶K Av`ª©Zv 70%| H w`‡bi evqyi m¤ú„³ ev®úPvc KZ? (10 †mjwmqvm ZvcgvÎvq m¤ú„³ Rjxq ev‡®úi Pvc 14 mmHg) [RU.15-16] A. 20 mmHg B. 25 mmHg C. 14 mmHg D. 35 mmHg R = f F 100% ev, 70% = 14 F 100% ev, F = 14 70 100 = 20 02. 200C ZvcgvÎvq †Kvb GK mgq evqy‡Z cÖK…Z ev®ú NbZ¡ 17.5 gm/m3 Ges m¤ú„³ ev®ú NbZ¡ 20 gm/m3 n‡j Av‡cw¶K Av`ªZv KZ? [RU.15-16] A. 67.5% B. 87.5% C. 77.5% D. 47.5% R = 17.5 20 100% = 87.5% STEP 03 ANALYSIS OF GST QUESTION PART A Analysis of Science & Technology Question JUST 01. †Kvb mgq evqyi wkwkivsK 10˚C Ges Av‡cwÿK Av`ª©Zv 75%| evqyi ZvcgvÎvq m¤ú„³ ev®úPvc KZ? 10˚C G m¤ú„³ ev®úPvc 9.21 mmHg P. [JUST-B, 19-20] A. 6.91 mmHg P B. 12.28 mmHg P C. 18.42 mmHg P D. 13.63 mmHg P R = f F × 100% F = f R × 100 = 9.21 75 × 100 = 12.28 mm HgP.
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 393 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 04 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. cvwb c~Y© GKwU cvÎ cÖwZ N›Uvq 150mg IRb nvivq| cÖwZ †m‡K‡Û cvÎwU †_‡K KZwU Rjxq ev‡®úi AYy m„wó nq? [KUET. 18-19] A. 1.395 1015 B. 1.395 1016 C. 1.39 1017 D. 1.393 1018 E. 1.395 1019 N = 150 10–3 18 3600 6.023 1023 = 1.394 1018 02. ‡Kvb GKw`‡bi wkwkivsK 20C I Av‡cw¶K Av`ªZv 75% | H w`‡bi m¤ú„³ evqyi m¤ú„³ ev®úPvc KZ? [20C ZvcgvÎvq m¤ú„³ Rjxq ev‡®úi Pvc 17.7x10–3 m] [KUET. 14-15] A. 17.7mm B. 17.710–4m C. 23.610–5m D.23.610–4m E. 23.610–3m R = f F 100% 75 100 = 17.7 10–3 F F = 2.36 10–2 = 23.6 10–3 m SAQ Short Answer Questions WRITTEN PART BAQ Broad Answer Questions 01. n«‡`i Zj‡`k †_‡K c„‡ô Avmvi d‡j GKwU evZv‡mi ey`ey‡`i e¨vmva© wZb¸Y n‡q hvq| e¨v‡ivwgUv‡i cvi` ¯Í‡¤¢i D”PZv 75 †mw›UwgUvi n‡j n«‡`i MfxiZv KZ? (cvi‡`i NbZ¡ 13596Kgm–3 ) Solve GLv‡b, V1 = V ey`ey‡`i e¨vmva© wZb¸Y nIqvq, V2 = 27V1 = 27V P = 75cm cvi` Pvc = 0.75m cvi` Pvc n«‡`i MfxiZv = hm hm cvwbi Mfx‡i cvwbi Pvc = hg n«‡`i Zj‡`‡k Pvc P1 = cvwbi Pvc + evqygÛjxq Pvc = hg + P n«‡`i DcwiZ‡j Pvc P2 = evqygÛjxq Pvc P Avgiv Rvwb, P1V1 = P2V2 (hg + P) V = P 27V h = 26P g = 265.122m (Ans.) 02. w¯’i Pv‡c †Kvb ZvcgvÎvq †Kvb M¨v‡mi AYyi g~j Mo eM©‡eM cÖgvY Pvc I ZvcgvÎvi g~j Mo eM©‡e‡Mi A‡a©K n‡e? Solve GLv‡b, T1 = 273K C2 = C1/2 T2 = ? AZGe, C2 C1 = T2 T1 T2 = T1/4 = 68.25K (Ans.) 03. ‡Kvb GKw`b evqyi ZvcgvÎv 30C, Pvc 0.756m cvi` Ges Av‡cwÿK Av`ªZv 60%| H w`‡bi Rjxqev‡¯úi I ﮋ evqyi Pvc wbY©q Ki| 30C ZvcgvÎvq m¤ú„³ Rjxq ev‡®úi Pvc 0.0316mHg| Solve GLv‡b, T1 = 30C P = 0.756m R = 60% F = 0.0316mHg f = ? Avgiv Rvwb, R = f F 100% f = RF/100% = 0.01896m Avevi, ﮋ evqyi Pvc = P – f = 0.73704 mHg 04. 20C ZvcgvÎv Ges 101 kPa Pv‡c evqyi 1.5 1029 msL¨K AYy GKwU AwW‡Uvwiqvg‡K Kvbvq Kvbvq c~Y© K‡i| AwW‡Uvwiqv‡gi †g‡Si †ÿÎdj 600 m2 n‡j D”PZv KZ? Solve Avgiv Rvwb, PV = nRT ev, PV = N NA RT; [N n‡jv †gvU M¨vmxq AYyi msL¨v Ges NA A¨v‡fv‡M‡Wªv msL¨v] ev, V = NRT PNA ev, V = 1.5 1029 8.31 J/K. mol (20 + 273) K 101 103 Pa 6.02 10 23 m0l = 6006.78 m3 Avevi, Ah = V h = V A 6006.78 m3 600 m 2 = 10.01 m 10.01 m 05. †Kvb M¨v‡mi cÖwZ NbwgUv‡i AYyi msL¨v 2.79 1025 Ges AYyi e¨vm 7.2 m n‡j, H M¨v‡mi Mo gy³ c_ KZ n‡e? Solve Avgiv Rvwb, Mo gy³ c_, = 1 2d 2N = 1 2 3.14 (7.2 1010) 2 (2.79 10) 25 = 1020 642.26 1025 = 0.001557 105 = 1.557 108m 1.557 108m GLv‡b, N = GKK AvqZ‡b AYyi msL¨v = 2.79 1025 molecules/m3 d = e¨vm = 7.2 1010m 06. w¯’iPv‡c 4 10 3 m 3 AvqZ‡bi †Kvb M¨vm‡K 0C n‡Z 68.25 ch©šÍ DËß Kivi d‡j Gi AvqZb 1 103 m 3 e„w× †c‡j cig k~b¨ ZvcgvÎvi gvb KZ? Solve awi, cig k~b¨ ZvcgvÎv = xC = 0 K 0C = x K †h †Kv‡bv ZvcgvÎv, y C = (x + y) K Pvj©m-Gi m~Î †_‡K, w¯’i Pv‡c, V2 V1 = T2 T1 5 103 4 103 = x + 68.25 x ev, 5 4 = x + 68.25 x ev, 5x = 4x + 273 ev, x = 273 GLv‡b, V1 = Avw` AvqZb = 4 103 m 3 V2 = †kl AvqZb = (4 103 + 1 103 ) m3 = 5 103 m 3 T1 = Avw` ZvcgvÎv = x K T2 = †kl ZvcgvÎv = (x + 68.25) K x = ? cig k~b¨ ZvcgvÎv = 273C 07. w¯’i Pv‡c KZ ZvcgvÎvq †Kv‡bv M¨vm AYyi Mo eM©‡e‡Mi eM©g~j cÖgvY Pvc I ZvcgvÎvi Mo eM©‡e‡Mi eM©g~‡ji wظY n‡e? Solve aiv hvK, cÖgvY ZvcgvÎv T1-G †Kv‡bv M¨vm AYyi g~j Mo eM©‡eM C1 Ges T2 ZvcgvÎvq C2| cÖkœg‡Z, C2 = 2C1 ev, 3 kT2 m = 3 kT1 m ev, T2 = 2 T1 ev, T2 = 4T1 = 4 273 K = 1092 K GLv‡b, cÖgvY ZvcgvÎv, T1 = 273 K wb‡Y©q ZvcgvÎv, T2 = ?
394 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 08. 29C ZvcgvÎvq 3g bvB‡Uªv‡R‡bi †gvU MwZkw³ wbY©q Ki| [bvB‡Uªv‡R‡bi MÖvg AvYweK fi 28g] Solve EK = 3 2 nRT = 3 2 3 28 8.316 (273 + 29) = 403.623J 09. wbw`©ó †Kvb w`‡b wkwkivsK 8.5°C Ges evqyi ZvcgvÎv 18.4C| Av‡cwÿvK Av`ª©Zv wbY©q Ki| 8C, C, 18C I 19C ZvcgvÎvq me©vwaK ev®úPvc h_vµ‡g 8.04 10–3m, 8.61 10–3m, 15.46 10– 3m I 16.46 10–3m cvi`| Solve Av‡cwÿK Av`ª©Zv R = f F 100% (9 – 8)C ZvcgvÎv e„wׇZ ev®úPv‡ci cwieZ©b = (8.61 – 8.04) 10– 3m = 0.57 10–3m wkwkivsK ev 8.5C G ev®úPvc, f = (8.04 + 0.57 0.5) 10–3m = 8.325 10–3m 18.4C G ev®úPvc, f = {15.46+(16.46–15.46)0.4}10–3m=15.8610–3m R = 8.32510–3 15.8610–3 100% = 52.49% (Ans.) 10. GKRb e¨w³ k¦vm-cÖk¦v‡m 1.12 litre evqy †meb Ki‡j (i) †m †gvU KZ¸‡jv AYy †meb K‡i? (ii) 27C ZvcgvÎvq H AYy¸‡jvi Mo MwZkw³ KZ? [mve©Rbxb M¨vm aªæeK = 8.314Jmole–1K –1 ] Solve (i) AYyi msL¨v = 1.12 22.4 6.022 1023 = 3.01 1022 (Ans.) (ii) Mo MwZkw³, K = 3 2 RT N = 3 2 8.314 300 6.0221023 = 6.21 10–21 Joule/molecule. 11. hw` R = 8.31Jk–1 mol–1 nq Z‡e 72cm cvi` Pv‡c Ges 27C ZvcgvÎvq 20gm Aw·‡R‡bi AvqZb wbY©q Ki| Solve PV = m M RT V = m PM RT = 20 10–3 (0.7213.6103 9.8)3210–38.31(27+273) = 16.2410–3m 3 12. 18gm wnwjqvg M¨vm c~Y© GKwU †ejy‡bi AvqZb 0.10m3 | †ejy‡bi wfZ‡i M¨v‡mi Pvc 1.2105Nm–2 | †ejy‡bi ga¨eZ©x M¨v‡mi ZvcgvÎv KZ? Solve PV = m M RT ZvcgvÎv, T = PV R M m = 1.2105 0.1 8.31 410–3 1810–3 = 320.9K 13. GKwU U¨vs‡K 27C ZvcgvÎvq I 2 evqygÛjxq Pv‡ci 1660L Aw·‡Rb Av‡Q| U¨vs‡K Aw·‡R‡bi fi wbY©q Ki| [Aw·‡R‡bi fi = 32kg kmol–1 , 1 evqygÛjxq Pvc = 1.013 105 PaR = 8.314 J mol–1K –1 ] Solve PV = m M RT fi, m = M PV RT = 3210–3 21.013105 166010–3 8.314300 = 4.3 kg 14. KZ Pvc cÖ‡qvM Ki‡j – 20C ZvcgvÎvq Ges 1 evqygÛjxq Pv‡c 1L M¨vm 40C ZvcgvÎvq 1 2 L M¨v‡mi AvqZb `Lj Ki‡e? Solve P1V1 T1 = P2V2 T2 P1 = V2 V1 T1 T2 P2 = 1 2 1 253 313 1 = 0.404 atm 15. GKwU †ejyb‡K 25C ZvcgvÎvq Ges 75 10–2m (cvi`) Pv‡c 1 10–3 m 3 nvB‡Wªv‡Rb Øviv c~Y© Kiv n‡jv, GLb, 10C ZvcgvÎvq Ges 75 10–3m Pv‡c †ejybwU‡K Dwo‡q †`Iqv n‡jv| †ejy‡bi AvqZb KZUv e„w× cv‡e? Solve P1V1 T1 = P2V2 T2 V2 = P1 P2 T2 T1 V1 = 7510–2 10–3 283 2987510–3 = 9.5 10–3m 3 16. †Kvb n«‡`i Zj‡`k †_‡K cvwbi DcwiZ‡j Avmvq GKwU evqy ey`ey‡`i e¨vm 1.5 ¸Y nq| n«‡`i c„‡ô evqygÛ‡ji Pvc ¯^vfvweK evqygÛjx Pv‡ci mgvb Ges ü‡`i DòZv aªæeK n‡j ü‡`i MfxiZv KZ? Solve v d 3 n = 1.53 ü‡`i MfxiZv, h = (n – 1) P 9800 = (1.53 –1) (1.013105 ) 9800 = 24.55m 17. †Kvb ZvcgvÎvq nvB‡Wªv‡R‡bi eM©g~j Mo eM©‡eM, mvaviY Pv‡c I ZvcgvÎvq eM©g~j Mo eM©‡e‡Mi wظY n‡e? Solve eM©g~j Mo eM©‡eM, C = 3RT M C T; TC 2 T1 = C1 C2 2 T2 = 22 273 = 1092K = 819C 18. ¯^vfvweK ZvcgvÎv I Pv‡c Aw·‡R‡bi NbZ¡ nvB‡Wªv‡R‡bi Nb‡Z¡i 16 ¸Y n‡j Aw·‡Rb AYyi Mo eM©‡e‡Mi eM©g~jxq gvb wbY©q Ki| Solve C = 3P C 1 Co 2 = H2 O2 C H2 = H2 O2 3RT MH2 = 1 16 38.134273 210–3 = 461.2ms–1 19. †Kvb M¨vm AYyi e¨vm 7.2 10–10m Ges cÖwZ NbwgUv‡i M¨vm AYyi msL¨v 2 1025| M¨vmwUi Mo gy³ c_ KZ n‡e? Solve Mo gy³c_, = 1 d 2N = 1 3.14(7.210–10) 2 (21025) = 310–8m 20. †Kvb M¨v‡mi AYyi Kvh©Ki e¨vm 210–10m Ges Mo gy³c_ 2.410–8 m| D³ M¨v‡mi GKK AvqZ‡b AYyi msL¨v KZ? Solve Mo gy³c_, = 1 2d 2N N = 1 2d 2 = 1 23.14 (210–10) 2 (2.410–8 ) = 2.3461026 21. Av`k© M¨v‡mi (Ideal gas) †¶‡Î cÖgvY Ki †h, PV = nRT. GLv‡b, cÖZxK¸‡jv cÖPwjZ A_© enb K‡i| [JnU-A, Set-1. 19-20] Solve Av`k© M¨v‡mi mgxKiY: e‡q‡ji m~Îvbymv‡i, V 1 p Ges Pvj©‡mi m~Îvbymv‡i VT m~Î `ywU‡K GK‡Î wjL‡Z cvwi| V T P V = K T P PV T = K PV = KT G‡ÿ‡Î K Gi cwie‡Z© R †jLv nq| R = me©Rbxb M¨vm aªæeK PV = RT GKMÖvg AYy M¨v‡mi AvqZb n‡e M m V P. M m .V = RT PV = m M RT PV = nRT GUvB Av`k© M¨v‡mi mgxKiY| 22. 27˚C ZvcgvÎvi M¨vm‡K KZ ZvcgvÎvq †bIqv n‡j Mo‡eM wظY n‡e? [RUET: 19-20] Solve M¨v‡mi Mo`ªæwZ, C = 8RT M C T (M w¯’i _vK‡j); C2 C1 = T2 T1 T2 T1 = C2 C1 2 T2 27 + 273 = 22 T2 = 4 300 = 1200 K 1200 K ev 927˚C ZvcgvÎvq Mo †eM wظY n‡e| Note: M¨v‡mi AYymg~‡ni Random MwZi Kvi‡Y Gi Mo †eM k~b¨|
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 395 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 3 mKj cvV¨eB‡qi NCTB QUESTIONS ANALYSIS cÖkœ e¨vLvmn mgvavb 01. cig †¯‹‡j Pv‡ci m~Î n‡jv- [Bm&nvK, Avwgi] cÖkœ A. P T 2 B. P 1 T C. P T D. P T Ans C 02. wbw`©ó cwigvY †Kv‡bv M¨vm‡K 27°C †_‡K †h ZvcgvÎvq DbœxZ Ki‡j AYy¸‡jvi rms †eM wظY nq Zv n‡jv- [Bm&nvK, Avwgi] A. 327°C B. 600°C C. 927°C D. 1200°C T2 = n2 × T1 = (2)2 × 300 k = 1200 k = 927°C 03. AvYweK MwZkw³ †Kvb ivwki Ici wbf©ikxj? [Bm&nvK, Avwgi] A. Nl©Y B. ZvcgvÎv C. AšÍ¯’kw³ D. Zvc Ans B 04. Rjxq ev‡®úi Pvc †e‡o hvq- [Bm&nvK, Avwgi] i. ZvcgvÎv n«vm †c‡j ii. ZvcgvÎv e„w× †c‡j iii. ZvcgvÎv w¯’i _vK‡j wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans C 05. m¤ú„³ ev®ú Pv‡ci •ewkó¨ n‡jv- [Bm&nvK, Avwgi] i. GwU GKwU Ave× ¯’v‡b •Zwi Kiv hvq ii. m¤ú„³ ev®ú e‡qj I Pvj©m-Gi m~Î †g‡b P‡j iii. ZvcgvÎv e„w× K‡i GKwU wbw`©ó cwigvY m¤ú„³ ev®ú‡K Am¤ú„³ ev‡®ú cwiYZ Kiv hvq wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans D 06. †Kv‡bv GKw`‡bi wkwkiv¼ 10°C I Av‡cwÿK Av`ª©Zv 67.30%| H w`‡bi evqyi m¤ú„³ ev®ú Pvc KZ? (10°C ZvcgvÎvq m¤ú„³ Rjxq ev®úPvc 13.64 × 10–3 m) [Bm&nvK, Avwgi] A. 1.01 × 104 m B. 2.02 × 104 m C. 2.02 × 10–4 m D. 1.01 × 10–4 m R = F 67.30% = 13.64 × 10–3 m F × 100% F = 13.64 × 10–3 0.673 = 2.02 × 10–4 m 07. `ywU M¨vm A Ges B-Gi Nb‡Z¡i AbycvZ 1 : 2| Zv‡`i Pvc P, ZvcgvÎv T| G‡ÿ‡Î Zv‡`i rms †e‡Mi AbycvZ KZ? [Bm&nvK, Avwgi] A. 1 : 1 B. 1 : 4 C. 1 : 2 D. 2 : 1 C = 3P C 1 A : B = 1 : 2 CA : CB = 2 : 1 08. †Kvb e³e¨wU mwVK? [Bm&nvK, Avwgi] A. Mo †eM gvb, rms †eM A‡cÿv Kg nq B. rms †eM I me©v‡cÿv m¤¢ve¨ †eM GKB C. rms †eM I rms `ªæwZ GKB D. rms †eM I Mo †e‡Mi gvb GKB nq Ans A 09. wbw`©ó f‡ii †Kv‡bv M¨v‡mi AYy¸wji Mo MwZkw³ K‡g †M‡j- [Bm&nvK, Avwgi] A. M¨vm Mig nq B. M¨vmwU VvÛv nq Ans B C. M¨vmwU cÖmvwiZ nq D. M¨vmwU msKzwPZ nq 10. wbw`©ó f‡ii GKwU M¨vm‡K w¯’i ZvcgvÎvq cÖmvwiZ Ki‡j †h ivwkwUi gvb cwiewZ©Z nq Zv n‡jv- [Bm&nvK, Avwgi] A. M¨v‡mi Pvc B. M¨v‡mi Af¨šÍixY kw³ C. M¨vm AYy¸wji rms †eM D. M¨vm AYy¸wji MwZkw³ Ans A 11. Mo gy³c_ †KvbwUi Ici wbf©i K‡i bv? [Bm&nvK, Avwgi] A. M¨v‡mi NbZ¡ B. GKwU AYyi e¨vm C. GKwU AYyi fi D. rms †eM Ans D 12. e‡q‡ji m~Îvbymv‡i wbw`©ó f‡ii †Kv‡bv M¨v‡mi AvqZb- [Zcb, AvwRR] A. Gi Pv‡ci mgvbycvwZK B. Gi Pv‡ci mgvbycvwZK e¨¯ÍvbycvwZK C. w¯’i ZvcgvÎvq Gi Pv‡ci e¨¯ÍvbycvwZK D. †Kv‡bvwUB bq Ans C 13. w¯’i Pv‡c wbw`©ó f‡ii M¨v‡mi AvqZb Gi- [Zcb, AvwRR] A. †Kjwfb ZvcgvÎvi e¨¯ÍvbycvwZK B. †Kjwfb ZvcgvÎvi mgvbycvwZK C. †Kjwfb ZvcgvÎvi eM©g~‡ji e¨¯ÍvbycvwZK D. †Kjwfb ZvcgvÎvi eM©g~‡ji mgvbycvwZK Ans B 14. w¯’i AvqZ‡b wbw`©ó f‡ii M¨v‡mi Pvc, Gi- [Zcb, AvwRR] A. †Kjwfb ZvcgvÎvi eM©g~‡ji mgvbycvwZK B. †Kjwfb ZvcgvÎvi mgvbycvwZK C. †Kjwfb ZvcgvÎvi eM©g~‡ji e¨¯ÍvbycvwZK D. †Kjwfb ZvcgvÎvi e¨¯ÍvbycvwZK Ans B 15. w¯’i Pv‡c M¨v‡mi AvqZb cÖmviY mn‡Mi gvb [Zcb, AvwRR] A. 0.0366o C -1 B. 0.366o C -1 C. 1 273 C -1 D. 273o C -1 mKj M¨v‡mi Rb¨ AvqZb cÖmviY mn‡Mi gvb 1 273 C -1 0.00366°C–1 | AvqZb cÖmviY mnM‡K p v Øviv m~wPZ Kiv nq| 16. †h ZvcgvÎvq †Kv‡bv wbw`©ó AvqZ‡bi evqy, Gi g‡a¨ Aew¯’Z Rjxq ev®ú Øviv m¤ú„³ nq †mB ZvcgvÎv‡K Kx ejv nq? [Zcb, AvwRR] A. cig Av`ª©Zv B. Av‡cw¶K Av`ª©Zv C. wkwkiv¼ D. †Kv‡bvwUB bq Ans C 17. M¨v‡mi MwZZË¡ Abymv‡i 0K ZvcgvÎvq M¨v‡mi MwZkw³ n‡e- [Zcb, AvwRR] A. mev©waK B. k~Y¨ C. Lye †ewkI bv Avevi KgI bv D. †Kv‡bvwUB bq T = 0 K n‡j , E = 3 2 KT 3 2 K 0 = 0 18. GKwU Av`k© M¨v‡mi Pvc I AvqZ‡bi ¸Ydj [Zcb, AvwRR] A. aªæeK B. mve©Rbxb M¨vm aªæe‡Ki cÖvq mgvb C. Gi ZvcgvÎvi mgvbycvwZK D. Gi ZvcgvÎvi e¨¯ÍvbycvwZK Ans C 19. Av`k© M¨v‡mi Pv‡ci ivwkgvjv n‡e [Zcb, AvwRR] (i) P = 1 3 c 2 (ii) P = 1 3 c 2 (iii) P = 2 3 RT A. i I iii B. ii I iii C. ii D. i, ii I iii P = 1 3 c 2 20. Av`k© M¨v‡mi †¶‡Î Pvc, AvqZb I ZvcgvÎvi g‡a¨ m¤úK© n‡”Q [Zcb, AvwRR] (i) pV = KT (ii) pV = nRT (iii) pV = RT A. i I ii B. i I ii C. i I iii D. i, ii I iii Ans D 21. Av`k© M¨vm mgxKiY n‡”Q: PV= nRT; GLv‡b n Kx? [Zdv¾j, gwnDwÏb] A. †gvjvi msL¨v B. †gvj msL¨v C. AYy msL¨v D. K¤úv¼ Ans B 22. cig k~b¨ ZvcgvÎv nj †h ZvcgvÎvq- [Zdv¾j, gwnDwÏb] (i) M¨v‡mi Pvc k~b¨ nq (ii) M¨v‡mi AvqZb k~b¨ nq (iii) †h ZvcgvÎvq Zij wnwjqvg KwV‡b iƒcvšÍwiZ nq wb‡Pi †KvbwU mwVK? A. i B. i I ii C. ii D. ii I iii Ans B 23. ms–1 GK‡K `kwU KYvi †eM h_vµ‡g 0, 1, 2, 3, 3, 3, 4, 4, 5, 6| G‡`i g‡a¨ †KvbwU m‡e©vËg m¤¢ve¨ †eM? [Mwb, mykvšÍ] A. 3.1 ms–1 B. 3 ms–1 C. 3.5 ms–1 D. GKwUI bv Ans B
396 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 24. cig ZvcgvÎv I Mo eM©‡e‡Mi eM©g~‡ji m¤úK© †KvbwU? [Mwb, mykvšÍ] A. Cr.m.s t B. Cr.m.s T C. Cr.m.s T 2 D. Cr.m.s T –1 Ans B 25. 27°C ZvcgvÎvq GKwU M¨vm AYyi Mo MwZkw³ 6.2110–21J . 27°C ZvcgvÎvq Mo MwZkw³ n‡e- [Mwb, mykvšÍ] A. 11.3510–21J B. 9.3510–21 J C. 12.3510–21 J D. 10.3510–21 J E2 = T2 T1 E1 = 227+273 27+273 6.2110-21= 10.3510-21 J 26. ¯^vfvweK Pv‡ci gvb- [†gvK‡Q` m¨vi] A. 1.013×103 Nm–2 B. 1.013×104 Nm–2 C. 1.013×105 Nm–2 D. 1.013×100 Nm–2 Ans C wb‡Pi Aby‡”Q` co Ges 27 I 28 bs cÖ‡kœi DËi wjL| [†gvK‡Q` m¨vi] GKwU cv‡Î iw¶Z M¨v‡mi `yÕwU AYyi †eM h_vµ‡g 4 ms–1 I 5ms–1 27. AYy؇qi Mo eM©‡eM n‡e- [†gvK‡Q` m¨vi] A. 20.5 m2 s –2 B. 20.5 ms–1 C. 16 m2 s –2 D. 25 ms–1 Mo eM©‡eM = 4 2 +52 2 = 20.5m2 s -2 28. AYy؇qi eM©g~j Mo eM©‡eM n‡e- [†gvK‡Q` m¨vi] A. 4.53 ms–2 B. 4.53 ms–1 C. 20.5 ms–1 D. 9 m2 s –2 eM©g~j Mo eM©‡eM = 4 2 + 52 2 = 4.5 ms-1 29. wb‡Pi †Kvb m~Îvbymv‡i PjšÍ Mvwoi PvKvi ZvcgvÎv e„w× cvq- [†gvK‡Q` m¨vi] A. e‡q‡ji m~Î B. Pv‡ci m~Î C. Pvj©‡mi m~Î D. A¨v‡fv‡M‡Wªvi m~Î MvwowU Pjvi mgq PvKvq Pvc cÖ‡qvM nq| Pvcxq m~Îvbymv‡i P T Pvc cÖ‡qvM Ki‡j ZvcgvÎv e„w× cv‡e| 30. evqyi Av‡cw¶K Av`ª©Zv Kg n‡j- [†gvK‡Q` m¨vi] i) evqy‡Z Rjxq ev®ú Kg _v‡K ii) wfRv Kvc‡o ev®úvqb `ªæZ nq iii) Av`ª© I ﮋ evj¦ nvB‡MÖvwgUv‡ii `yB _v‡g©vwgUv‡ii e¨eavb †ewk nq wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. iii I i D. i, ii I iii Ans D 31. Av`k© Pv‡c Ges 27°C ZvcgvÎvq 32g Aw·‡R‡bi AvqZb n‡jv- [†gvK‡Q` m¨vi] A. 0.0246 cm3 B. 0.0246 m3 C. 0.246 cm3 D. 0.246 m3 PV = m M RT 1 V = 3 32 0.0821(27+273) V = 24.6 L = 0.0246 m3 wb‡Pi Aby‡”Q`wU co Ges 32 I 33 bs cÖ‡kœi DËi wjL| [†gvK‡Q` m¨vi] GKwU cv‡Î 0°C ZvcgvÎvq wKQz M¨vm iw¶Z Av‡Q| ZvcgvÎv n«vm Ki‡j M¨v‡mi Pvc 0°C ZvcgvÎvi Pv‡ci GK-Z…Zxqvsk nq| 32. M¨vm cv‡Îi †kl ZvcgvÎv KZ? A. 91°C B. 182°C C. 91 K D. –182 K T2 = P2 P1 T1 = P p 3 1 273 = 91 K 33. DÏxc‡Ki M¨v‡mi cwieZ©bkxj PjK¸‡jvi Rb¨ wb‡Pi †Kvb †jLwPÎwU mwVK? A. V O · °C –273°C V0 B. P O · °C –273°C P0 C. V O T D. P O T S AB info A bs †jLwUi mgxKiY,V = V0 1 + 273 Ges B bs †jLwUi mgxKiY P = P0 1 + 273 34. M¨v‡mi GKK AvqZ‡b Pvc (P) Gi MwZ kw³i (E) KZ Ask? [†Mvjvg m¨vi] A 3 2 Ask B. 1 3 Ask C. 2 3 Ask D. 1 2 Ask PV = 1 3 mNC 2= 2 3 1 2 C 2 = 2 3 E GKK AvqZ‡b, P = 2 3 E 35. evqy‡Z Av`ª©Zv K‡g ‡M‡j- [†Mvjvg m¨vi] i. `ªæZ ev®úvqb N‡U ii. nvB‡MÖvwgUv‡ii `yB _v‡g©vwgUv‡ii e¨eavb K‡g hvq iii. ev®úvqb eÜ n‡q hv‡e wb‡Pi ‡KvbwU mwVK? A. i B. ii C. iii D. i I ii Ans A 36. M¨v‡mi Pjivwk wZbwU Kx Kx? [cÖvgvwbK m¨vi] A. AvqZb, fi I NbZ¡ B. AvqZb, ZvcgvÎv I NbZ¡ C. AvqZb, fi I ZvcgvÎv D. AvqZb, ZvcgvÎv I Pvc Ans D 37. GKwU cv‡Î P1 Pv‡c wKQz M¨vm Ave× Av‡Q| hw` M¨vm AYy¸‡jvi cÖ‡Z¨KwUi fi A‡a©K nq Ges Zv‡`i ‡eM wظY nq Z‡e Pvc KZ n‡e? [igv m¨vi] A. GKB _vK‡e B. A‡a©K n‡e C. wظY n‡e D. Pvi¸Y n‡e P2 = m2 m1 c 2 2 c 2 1 P1 = m1 2m1 4c2 1 c 2 1 P1 = 2P1 38. e‡q‡ji m~Îvbymv‡i w¯’i ZvcgvÎvq wbw`©ó f‡ii †Kvb M¨v‡mi AvqZb- [BKivg m¨vi] A. Gi Pv‡ci mgvbycvwZK B. Gi Pv‡ci e¨¯ÍvbycvwZK C. Gi Pv‡ci e‡M©i e¨¯ÍvbycvwZK D. Gi Pv‡ci e‡M©i mgvbycvwZK Ans B 39. w¯’i ZvcgvÎvq 7105Nm–2 Pv‡c wbw`©ó f‡ii wKQz M¨v‡mi AvqZb 0.002m3 | 14105Nm–2 Pvc M¨vmwUi AvqZb KZ n‡e? [BKivg m¨vi] A. 1.010–4m 3 B. 1.010–3m 3 C. 1.010–2m 3 D. 1.010–5m 3 V2 = P1 P2 V1 = 7105 10105 0.002 =0.001 cm3 =110–3 m 3 40. ZvcgvÎv e„w× ‡c‡j AYy¸‡jvi Mo eM©‡eM- [kvgmyi m¨vi] A. n«vm cvq B. e„w× cvq C. AcwiewZ©Z _v‡K D. k~b¨ nq C 2 T , eM©g~j Mo eM©‡eM ZvcgvÎvi eM©g~‡ji mgvbycvwZK| A_©vr ZvcgvÎv e„w× †c‡j †eM e„w× cvq| 41. M¨v‡mi Pjivwk KqwU? [kvgmyi m¨vi] A. 2 B. 3 C. 4 D. 5 Ans B 42. Mig e¯‘ VvÛv Ki‡Z ‡KvbwU AwaK Dc‡hvMx? [kvgmyi m¨vi] A. kxZj evZvm B. kxZj cvwb C. eid D. me¸‡jv mgvb Ans B 43. cvwb‡K 7°C ‡_‡K 1°C G VvÛv Ki‡j Kx N‡U? [kvgmyi m¨vi] A. GwU ïaygvÎ msKzwPZ nq B. GwU ïaygvÎ cÖmvwiZ nq C. GwU cÖ_‡g msKzwPZ nq Ges c‡i cÖmvwiZ nq D. GwU cÖ_‡g cÖmvwiZ nq, Zvici msKzwPZ nq Ges c‡i Avevi cÖmvwiZ nq Ans C 44. 1 wjUv‡i KZ NbwgUvi? [CU. 14-15] [kvgmyi m¨vi] A. 1.0 × 10–3 B. 1.0 × 10–2 C. 1.0 × 10–1 D. 1.0 × 102 E. 1.0 × 10 3 Ans A
ASPECT PHYSICS cÖ_g cÎ Av`k© M¨vm I M¨v‡mi MwZZË¡ 397 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 4 MCQ CONCEPT TEST WRITTEN cÖkœ 01. ¯^vfvweK ZvcgvÎv I Pv‡c †Kv‡bv M¨v‡mi NbZ¡ 1.25 kg m–3 | AYyi g~j Mo eM©‡eM KZ? A. 480 ms–1 B. 490 ms–1 C. 493 ms–1 D. 393 ms–1 02. wbw`©ó f‡ii GKwU Av`k© M¨v‡mi AvqZb aªæe Pv‡c wظY Kiv n‡jv| hw` M¨v‡mi cÖv_wgK ZvcgvÎv 13˚C nq Z‡e P~ovšÍ ZvcgvÎv KZ? A. 299˚C B. 339˚C C. 499˚C D. 199˚C 03. GK †gvj nvB‡Wªv‡Rb Ges GK †gvj Aw·‡R‡bi fi h_vµ‡g 2g Ges 32g n‡j †Kv‡bv wbw`©ó ZvcgvÎvq AbycvZ nvB‡Wªv‡Rb AYyi g~j Mo eM©‡eM/ Aw·‡Rb AYyi g~j Mo eM©‡eM Gi gvb KZ n‡e? A. 4 B. 5 C. 6 D. 7 04. 10˚C DòZvq Ges 750 mm cvi` Pv‡c evZv‡mi Nb‡Z¡i mv‡_ 15˚C DòZvq Ges 760 mm cvi` Pv‡c evZv‡mi Nb‡Z¡i Zzjbv Ki| A. 1 B. 0.5 C. 2 D. 2.5 05. ¯^vfvweK ZvcgvÎv I Pv‡c wKQz cwigvY ﮋ evqy‡K m‡gvò cÖwµqvq msbwgZ K‡i AvqZb A‡a©K Kiv n‡jv| P~ovšÍ Pvc KZ? A. 4.04 × 10–2 Nm–2 B. 2.20 Nm–2 C. 4.04 × 105 Nm–2 D. 2.02 × 105 Nm–2 06. GKwU Av`k© M¨v‡mi †¶‡Î Cp/Cv = x n‡j, wb‡Pi †Kvb m¤úK©wU GK †gv‡ji Rb¨ mwVK? A. Cv = (x–1)R B. Cv = R/(x–1) C. Cv = R/(1–x) D. CV = R/(1+x) 07. GK †gvj nvB‡Wªv‡Rb Ges GK ‡gvj Aw·‡R‡bi fi h_vµ‡g 2g I 32g n‡j †Kvb GK wbw`©ó ZvcgvÎvq AbycvZ,nvB‡Wªv‡Rb Abyi g~j Mo eM©‡eM Aw·‡Rb Abyi g~j Mo eM©‡eM Gi gvb n‡eA. 1 8 B. 1 4 C. 4 D. 8 08. M¨v‡mi MwZZË¡ Abymv‡i 0K ZvcgvÎvq M¨v‡mi MwZkw³ KZ? A. Amxg B. Mo MwZkw³ C. k~b¨ D. ‡KvbwUB bv 09. Av‡cwÿK Av`ª©Zv KZ n‡j wkwkiv¼ evqyi ZvcgvÎvi mgvb n‡e? A. k~b¨ B. 50% C. 75% D. 100% 10. `ywU K…ò e¯‘i wbM©Z Zvckw³i AbycvZ 16:1| wØZxq e¯‘i ZvcgvÎv 300 k n‡j, cÖ_g e¯‘i ZvcgvÎv KZ? A. 1200 K B. 1600 K C. 600 K D. 300 K 11. GKRb Wyeyix n«‡`i Zj‡`‡k KvR Kivi mgq 2cm3 AvqZ‡bi ey`ey` Dc‡ii w`‡K cÖevwnZ n‡”Q| cvwbi DcwiZ‡j ey`ye‡`i AvqZb 4cm3 nq; wKš‘ ZvcgvÎv AcwiewZ©Z _v‡K| hw` evqygÛjxq Pvc 10m cvwbi Pv‡ci mgvb nq, n«‡`i MfxiZv KZ? A. 10m B. 20m C. 30m D. 40m 12. Rjxq ev‡®úi Nb‡Z¡i mv‡_ evqyi Pv‡ci m¤úK© n‡jvA. P 2 B. P P C. P D. 12cm 13. wØ-cigvYyK M¨vm AYyi ¯^vaxbZvi gvÎv KqwU? A. 2 B. 3 C. 4 D. 5 14. Av‡cw¶K Av`ª©Zv 100% n‡j wkwkiv¼ evqyi ZvcgvÎvA. A‡a©K n‡e B. wظb n‡e C. mgvb n‡e D. Kg n‡e 15. ‡Kvb k‡Z© ev¯Íe M¨vm mg~n Av`k© AvPiY K‡i? A. wbgœ ZvcgvÎvq I D”P Pv‡c B. wbgœ ZvcgvÎvq I wbgœ Pv‡c C. D”P ZvcgvÎvq I D”P Pv‡c D. wbgœ Pv‡c I D”P ZvcgvÎvq 16. PV=k GB mgxKiYwU mvaviY fv‡e †Kvb m~‡Îi cÖKvkA. Pvj©‡mi m~Î B. e‡q‡ji m~Î C. Pv‡ci m~Î D. Av`k© M¨vm mgxKiY 17. 0˚C ZvcgvÎvq †Kvb M¨v‡mi Pvc 3105 Nm–2 n‡j 60˚C ZvcgvÎvi Pvc KZ n‡e? A. 30.66105 Pa B. 33.6105 Pa C. 3.66105 Pa D. †KvbwUB bq 18. T ZvcgvÎvq 1 wjUvi evqy‡K DËvß Kiv n‡jv hZ¶b bv evqyi Pvc I AvqZb DfqB wظY nq| P~ovšÍ ZvcgvÎv KZ? A. T 2 B. T 4 C. 2T D. 4T E. 8T 19. evqygÛ‡j Rjxqev®ú Nbxf~Z nIqvi dj bq †KvbwU? A. wkwki B. Kyqvkv C.So D. e„wó 20. cÖgvY ev Av`k© ZvcgvÎvi gvb n‡jvA. 0K B. –273°C C. 273°C D. 273 K OMR SHEET 01. A B C D 06. A B C D 11. A B C D 02. A B C D 07. A B C D 12. A B C D 03. A B C D 08. A B C D 13. A B C D 04. A B C D 09. A B C D 14. A B C D 05. A B C D 10. A B C D 15. A B C D WRITTEN PART 21. GKB ZvcgvÎvq wfbœ wfbœ GK †gvj M¨v‡mi †ÿ‡Î Mo MwZkw³ wK mgvb bv wfbœ e¨vL¨v K‡iv| DËi:....................................................................................... 22. †Kv‡bv ¯’v‡b evZv‡mi Av‡cwÿK Av`ª©Zv 70% ej‡Z Kx eySvq? DËi:....................................................................................... 23. wm³ I ﮋ evi nvB‡MÖvwgUv‡i wm³ I ﮋ evj¦ _v‡g©vwgUv‡ii ZvcgvÎvi cv_©K¨ Lye †ewk n‡j †m ¯’v‡b AvenvIqv e¨vL¨v K‡iv| DËi:....................................................................................... 24. el©vKvj A‡cÿv kxZKv‡j wfRv Kvco `ªæZ ïKvq †KbÑ e¨vL¨v K‡iv| DËi:....................................................................................... 25. kxZKv‡j †VuvU gyL †d‡U hvq †Kb? e¨vL¨v Ki| DËi:....................................................................................... 26. kw³i mgwefvRb bxwZ e¨vL¨v Ki| DËi:....................................................................................... 27. – 273C Gi wb‡P ZvcgvÎv _vK‡Z cv‡i bv †Kb? e¨vL¨v Ki| DËi:....................................................................................... 28. †Kv‡bv M¨v‡mi ZvcgvÎvi mv‡_ Nb‡Z¡i m¤úK© Kxiƒc? DËi:....................................................................................... 29. AvswkK Zij c~Y© e× cv‡Îi ev®ú Pvc m¤ú„³ bv Am¤ú„³? †Kb? DËi:....................................................................................... 30. Av‡cwÿK Av`ª©Zv ej‡Z Kx eyS? DËi:.......................................................................................
398 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES MCQ ANSWER ANALYSIS WRITTEN cÖkœ bs DËi e¨vL¨v cÖkœ bs DËi e¨vL¨v 01 C Crms = 3P = 3 × 1.013 × 105 1.25 = 493.07 ms–1 07 C CH CO = MO MH ev CH CO = 32 2 = 4 02 A V1 V2 = T1 T2 T2 = V2T1 V1 = 2V × (273 + 13) V = 2 × 286 = 572K = 299˚C 08 C Ek = 3 2 nRT = 3 2 nR 0 = 0 03 A H2 Gi Rb¨ C1 = 3RT 2 O2 Gi Rb¨ C2 = 3RT 32 = 3RT 16 × 2 = 1 4 3RT 2 C2 = C1 4 C1 = 4 × C2 09 D R = f F 100% = 1 1 100% R = 100% 04 A 1 2 = T2 T1 = 288 283 = 1 10 C E1 E2 = T1 T2 4 ev, 16 1 = T1 300 4 2 = T1 300 T1 = 600k 05 D P2 = P1V1 V2 = 1.01 × 105 × V V/2 = 2.02 × 105 Nm–2 11 A n«‡`i MfxiZv, h = (n – 1) 10.2 = 4 2 – 1 10.2 = (2 – 1) 10.2 = 10.2m 06 B Cp Cv = x ev, Cp = Cvx Ges Cp – Cv = R ev, Cvx – Cv = R ev, Cv(x – 1) = R ev, Cv = R/(x – 1) 12. C 13. C 14. C 15. D 16. B 17. C 18. D 19. C 20. D 21 GKB ZvcgvÎvq wfbœ wfbœ GK †gvj M¨v‡mi †ÿ‡Î Mo MwZkw³ mgvb _v‡K| †Kbbv wbw`©ó T ZvcgvÎvq GK †gvj †h‡Kv‡bv M¨v‡mi Mo MwZkw³, E = 3 2 RT mgxKiY Abyhvqx, Mo MwZkw³ M¨v‡mi AvYweKfi, NbZ¡ ev Pv‡ci Ici wbf©i K‡i bv, †Kej cig ZvcgvÎvi Ici wbf©i K‡i| G Kvi‡Y GKB ZvcgvÎvi wfbœ wfbœ GK †gvj M¨v‡mi †ÿ‡Î Mo MwZkw³ aªæe _v‡K| 22 †Kv‡bv ¯’v‡b evZv‡mi Av‡cwÿK Av`ª©Zv 70% ej‡Z eySvq, H ¯’v‡bi evZvm‡K m¤ú„³ Ki‡Z †h cwigvY Rjxq ev®ú `iKvi Zvi kZKiv 70 fvM Rjxq ev®ú H gyn~‡Z© H ¯’v‡bi evqy‡Z i‡q‡Q| 23 wm³ I ﮋ Øviv evj¦ _v‡g©vwgUv‡ii ZvcgvÎv GKB n‡j eyS‡Z n‡e evZvm bZzb K‡i Rjxq ev®ú †kvlY Ki‡Q bv| †m‡ÿ‡Î evZvm Rjxq ev®ú Øviv m¤ú„³- Giƒc eyS‡Z n‡e| Z‡e evZvm hZ `ªæZ nv‡i Rjxq ev®ú †kvlY Ki‡e wm³ ev‡j¦i ZvcgvÎv ZZ n«vm cv‡e| myZivs wm³ I ﮋ evj¦ _v‡g©vwgUv‡ii ZvcgvÎvi cv_©K¨ Lye †ewk n‡j eyR‡Z n‡e, H ¯’v‡bi evZvm Lye `ªæZ nv‡i Rjxq ev®ú †kvlY Ki‡Q A_©vr evZv‡mi Av‡cwÿK Av`ª©Zv LyeB Kg| myZivs AvenvIqv fv‡jv Ae¯’vq i‡q‡Q| e„wó nIqvi m¤¢vebv †bB| 24 el©vi w`‡b evqygÛj Rjxq ev®ú Øviv cÖvq m¤ú„³ _v‡K, A_©vr Av‡cwÿK Av`ª©Zv 100% Gi KvQvKvwQ _v‡K| d‡j evZvm AwaK cwigv‡Y Rjxq ev®ú †kvlY Ki‡Z cv‡i bv| G Kvi‡Y el©vKv‡j †fRv Kvco mn‡R ïKv‡Z Pvq bv| †m Zzjbvq kxZKv‡ji evZvm ïK‡bv _v‡K| ïK‡bv evZv‡m Rjxq ev®ú AZ¨šÍ Kg _v‡K| GB evZvm wfRv Kvco †_‡K `ªæZ Rjxq ev®ú †kvlY K‡i wb‡Z cv‡i| d‡j kx‡Zi w`‡b †fRv Kvco ZvovZvwo ïKvq| 25 kxZKv‡j evZv‡m Av`ª©Zv AZ¨šÍ K‡g hvq| ZLb †VuvU n‡Z cvwb Rjxq ev®ú AvKv‡i evZv‡m P‡j hvq| †Vuv‡Ui Z¡‡Ki cvwb¯Í‡ii wewfbœ Ask n‡Z cvwb P‡j hvIqvq wew”Qbœ Ask¸‡jv cvwbi c„ôUv‡bi `iæb †ÿÎdj msKzwPZ n‡Z Pvq d‡j †VuvU †d‡U hvq| GKB Kvi‡b kxZKv‡j gyL †d‡U hvq| 26 28vbyhvqx GKwU AYyi Mo kw³ = 3 2 kT| wØcvigvYweK M¨v‡mi GKwU AYyi ¯^vaxbZvi gvÎv 5, AZGe cÖwZwU AYyi Mokw³ = 5 2 kT| 27 Pvj©‡mi m~Îvbymv‡i; v = v0 1 + 273 ; GLv‡b v = 0 emv‡j, = – 273C cvIqv hvq| Avevi Pv‡ci (†i‡bvi) m~Îvbymv‡i; P = P0 1 + 273 GLv‡b P = 0 emv‡j = – 273C cvIqv hvq| A_©vr, – 273C ZvcgvÎvq ZvwË¡Kfv‡e †h‡Kv‡bv M¨v‡mi AvqZb I Pvc k~b¨ nq| e¯‘Z ZvcgvÎv Kwg‡q – 273C ZvcgvÎvq c`v‡_©i AYy-cigvYymg~‡ni mKj K¤úb †_‡g hvq| †h‡nZz c`v‡_©i Acy-cigvYymg~‡ni mKj K¤úb †_‡g hvq| †h‡nZz c`v‡_©i AYy-civgvYymg~‡ni K¤ú‡bi gvÎv w`‡qB Avgiv ZvcgvÎv Abyaveb Kwi, ZvB – 273C –Gi wb‡P †Kv‡bv ZvcgvÎv _vK‡Z cv‡i bv| 28 A_©vr, 1 T ; hLb Pvc I fi w¯’i _v‡K| 29 AvswkK Zijc~Y© e× cv‡Îi ev®úPvc n‡jv m¤ú„³ ev®úPvc| Gi KviY n‡jv Zij †_‡K cÖ_g w`‡K hLb ev®úxfeb ïiæ n‡e, ZLb ev®úPvc Am¤ú„³ _vK‡jI µgvMZ Zij AYymg~n hLb AviI ev®úxf~Z n‡e, ZLb ev®úPvc GK mgq m¤ú„³ gv‡b †cuŠQv‡Z eva¨ n‡e 30 †Kv‡bv wbw`©ó ZvcgvÎvq GKwU wbw`©ó AvqZ‡bi evqy‡Z †h cwigvY Rjxq ev®ú _v‡K Ges H ZvcgvÎvq H AvqZ‡bi evqy‡K m¤ú„³ Ki‡Z †h cwigvY Rjxq ev‡®úi cÖ‡qvRb nq Zv‡`i AbycvZ‡K Av‡cwÿK Av`ª©Zv e‡j| Avgiv Rvwb, w¯’i ZvcgvÎvq wbw`©ó AvqZ‡bi evqyi Rjxq ev‡®úi Pvc Zvi f‡ii mgvbycvwZK Ges †h‡Kv‡bv ZvcgvÎvq GB Rjxq ev®ú Pvc wkwkiv‡¼ D³ evqyi m¤ú„³ Rjxq ev®ú Pv‡ci mgvb| myZivs, Av‡cwÿK Av`ª©Zv, R = wkwkiv‡¼i m¤ú„³ Rjxq ev®úPvc evqyi ZvcgvÎvq m¤ú„³ Rjxq ev®úPvc ev, R = f F 100%
400 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES SURVEY TABLE Kx coe? †Kb coe? †Kv_v n‡Z coe? KZUzKz coe? TOPICS MAGNETIC DECISION [hv co‡e] MAKING DECISION [†h Kvi‡Y co‡e] VVI For This Year DU JU RU CU GST Engr. HSC Written MCQ THEORY CONCEPT-01 ZvcgvÎv cwigv‡ci bxwZ I ZvcMwZwe`¨vi k~b¨Zg m~Î 35% 30% 20% 25% 30% 50% 70% CONCEPT-02 ZvcMwZwe`¨vi cÖ_g m~Î (Ryj I K¬wmqv‡mi gZev`, 1g m~‡Îi Zvrch©) 40% 45% 35% 30% 40% 15% 45% CONCEPT-03 ZvcMZxq cwieZ©b (m‡gvò, iæ×Zvcxq, mgPvc Ges mgAvqZb) 50% 45% 40% 35% 40% 40% 25% CONCEPT-04 CP, CV Ges R Gi m¤úK© 75% 70% 65% 30% 70% 50% 20% CONCEPT-05 ZvcMwZwe`¨vi wØZxq m~Î I Zvc BwÄb 70% 75% 70% 60% 75% 15% 70% CONCEPT-06 †iwd«Rv‡iUi I Kvh©K…Z mnM 20% 10% 20% 25% 30% 10% 30% CONCEPT-07 GbUªwc msµvšÍ 30% 40% 35% 40% 30% 60% 70% MvwYwZK cÖ‡qvM CONCEPT-01 ZvcgvÎv †¯‥j I ÎæwUc~Y© _v‡g©vwgUvi msµvšÍ MvwYwZK cÖ‡qvM 80% 85% 85% 80% 80% 60% 80% CONCEPT-02 Zvc Øviv K…ZKvR msµvšÍ 75% 80% 70% 75% 80% 70% 85% CONCEPT-03 kw³i iƒcvšÍi msµvšÍ 60% 70% 70% 75% 70% 40% 50% CONCEPT-04 ZvcMZxq cÖwµqv msµvšÍ 40% 50% 60% 65% 70% 50% 75% CONCEPT-05 Bwćbi `ÿZv msµvšÍ MvwYwZK cÖ‡qvM 90% 85% 80% 85% 80% 70% 80% CONCEPT-06 GbUªwci cwieZ©b msµvšÍ MvwYwZK cÖ‡qvM 70% 75% 70% 75% 70% 60% 90% DU. = Dhaka University, JU. = Jahangirnagar University, RU. = Rajshahi University, CU = Chittagong University, GST = General, Science & Technology, Engr. = Engineering. STEP 1 mvRv‡bv me Z_¨ THEORY fwZ© †ivMxi c_¨ Concept 1 ZvcgvÎv cwigv‡ci bxwZ I ZvcMwZwe`¨vi k~b¨Zg m~Î ✅ ZvcgvÎv: ZvcgvÎv e¯‘i GKwU Zvcxq Ae¯’v, hv H e¯‘ n‡Z Ab¨ e¯‘‡Z Zv‡ci cÖevn wbqš¿Y K‡i Ges Zvc cÖev‡ni AwfgyL wba©viY K‡i| ✅ ZvcgvÎv cwigv‡ci bxwZ: _v‡g©vwgUvi: †h hš¿ Øviv ZvcgvÎv wbfy©j fv‡e cwigvc Kiv hvq ZvcgvÎv cwigv‡ci hš¿ _v‡g©vwgUvi (Thermometer) Zvc cwigv‡ci hš¿ K¨vjwiwgUvi (Calorimeter) DòZvwgwZK c`v_©: †h me c`v‡_©i DòZvwgwZK ag© e¨envi K‡i _v‡g©vwgUvi •Zix Kiv nq DòZvwgwZK ag©: †Kvb c`v‡_©i †hme ag© ZvcgvÎv cwieZ©‡bi mv‡_ mgvbycv‡Z cwiewZ©Z nq c`v‡_©i †mme ag©‡K DòZvwgwZK ag© e‡j| †hgb- hw` X T nq Zvn‡j X GKwU DòZvwgwZK ag© Ges T nj ZvcgvÎv| ✅ wb¤œ w¯’i we›`y: †h ZvcgvÎvq cÖgvY Pv‡c weï× eid cvwbi mv‡_ mvg¨e¯’vq _v‡K Zv‡K wb¤œ w¯’i we›`y/Mjbv¼/ Melting point e‡j| ✅ EaŸ© w¯’i we›`y: †h ZvcgvÎvq cÖgvY Pv‡c weï× cvwb Rjxq ev‡®úi mv‡_ mvg¨e¯’vq _v‡K Zv‡K EaŸ© w¯’i we›`y/ùyUbv¼/Boiling Point e‡j| ✅ †g․wjK e¨eavb: (EaŸ© w¯’i we›`y – wb¤œ w¯’i we›`y)| ✅ •Îawe›`y: †h ZvcgvÎvq wbw`©ó Pv‡c cvwb KwVb, Zij I ev®úxq Ae¯’vq GKB mvg¨ve¯’vq Ae¯’vb K‡i| •Îa we›`yi ZvcgvÎv Ttr = 273.16K Ges •Îa we›`y‡Z cvwbi ev®ú Pvc 4.58mm cvi` ¯Í‡¤¢i D”PZvi mgvb| ✅ †÷dvb-Gi m~Î: E = T 4 ✅ fx‡bi/DB‡bi m~Î: mT = aªæeK (b) [fx‡bi ev DB‡bi aªæeK, b = 2.89 10–3 mK = 2898 mK] ✅ ZvcMwZwe`¨vi k~b¨Zg m~Î: m~‡Îi wee„wZ: `ywU e¯‘ hw` Z…Zxq †Kvb e¯‘i mv‡_ Zvcxq mvg¨ve¯’vq _v‡K, Z‡e cÖ_‡gv³ e¯‘ `ywU ci®ú‡ii mv‡_ Zvcxq mvg¨ve¯’vq _vK‡e| cÖ‡qvM: GB m~‡Îi Dci wfwË K‡i _v‡g©vwgUvi •Zix Kiv n‡q‡Q| Avwe®‥viK: R.H. dvIjvi| ✅ ZvcgvÎvi aviYv: ZvcgvÎv ej‡Z e¯‘i DòZvi cwigvY eySvq| A_©vr Zvc n‡jv KviY Ges ZvcgvÎv Gi dj| ZvcgvÎv e¯‘i GKwU Zvcxq Ae¯’v, hv H e¯‘ n‡Z Ab¨ e¯‘‡Z Zv‡ci cÖevn wbqš¿Y K‡i Ges Zvc cÖev‡ni AwfgyL wba©viY K‡i| ‡iva _v‡g©vwgUv‡i cøvwUbvg †iva e¨envi K‡i Ges Zwor †iv‡ai DòZv wgwZK a‡g©i Ici j¶ †i‡L ZvcgvÎv cwigvc Kiv nq| ZvcMwZwe`¨v THERMODYNAMICS wØZxq cÎ 01 Aa¨vq c„ôv 400
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 401 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ✅ D”PZi ZvcgvÎv wewKiY cvB‡ivwgUv‡i DËß e¯‘i wewKiY ag© Kv‡R jvwM‡q 500°C Gi D‡aŸ© ZvcgvÎv cwigvc Kiv nq| ZvcgvÎv †¯‥j wb¤œ w¯’i we›`y EaŸ© w¯’i we›`y †g․wjK e¨eavb †mjwmqvm (mve©Rbxb) 0C 100C 100 †Kjwfb 273K 373K 100 dv‡ibnvBU (Wv³vwi _v‡g©vwgUvi) 32F 212F 180 i¨vswKb †¯‥j 492 672 180 †ivgvi †¯‥j 0 80 80 ✅ wewfbœ †¯‥‡j ZvcgvÎvi cwieZ©b: 1C = 9 5 F = 1K _v‡g©vwgUv‡ii bvg DòZvwgwZK c`v_© DòZvwgwZK ag© cvjøv Zij _v‡g©vwgUvi ‣KwkKb‡j Zij ¯Í¤¢| †hgb: cvi`, A¨vj‡Kvnj Zij ¯Í‡¤¢i •`N©¨ – 39° †_‡K 1500°C w¯’i AvqZb M¨vm _v‡g©vwgUvi w¯’i AvqZ‡b M¨vm (†hgb-evqy) M¨v‡mi Pvc – 270° †_‡K 1500°C w¯’i Pvc M¨vm _v‡g©vwgUvi w¯’i Pv‡c M¨vm M¨v‡mi AvqZb – 270° †_‡K 1500°C ‡iva _v‡g©vwgUvi cøvwUbvg †iva Zvi cwievnxi †iva – 2000° †_‡K 1200°C _v‡g©vKvcj/ZvchyMj `ywU avZe c`v‡_©i hyMj Zvcxq Zwo”PvjK ej – 250° †_‡K 1500°C _v‡g©vwgUvi/_vwg©÷i Aa©cwievnK c`v_© Zwor †iva 70° †_‡K 300°C wewKiY cvB‡ivwgUvi K…òKvq cvZ DËß e¯‘i wewKiY 500°C Gi D‡aŸ© REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. A¨vjywgwbqvg cvZ †_‡K †K‡U GKwU ejqvKvi A¨vjywgwbqvg wis •Zix Kiv n‡q‡Q| GwU Mig Ki‡j wK N‡U? [DU: 2018-19] A. A¨vjywgwbqvg evB‡ii w`‡K ewa©Z nq I wQ`ª GKB AvKv‡ii _v‡K B. wQ‡`ªi e¨vm K‡g hvq C. wQ‡`ªi †ÿÎdj A¨vjywgwbqv‡gi †h‡Kv‡bv As‡ki †ÿÎd‡ji mgvb Abycv‡Z e„w× cvq D. wQ‡`ªi †ÿÎdj A¨vjywgwbqv‡gi †h‡Kv‡bv As‡ki †ÿÎd‡ji †P‡q †ewk Abycv‡Z e„w× cvq Ans C 02. GKwU Av`k© M¨vm GKwU Zvc AšÍi‡Ki AveiYhy³ `„p cv‡Î k~b¨ gva¨‡g cÖmvwiZ n‡jv| d‡j wb‡¤œi †KvbwU N‡U? [DU: 2018-19] A. AšÍ¯’ kw³i †Kv‡bv cwieZ©b nq bv B. ZvcgvÎv n«vm cvq C. Pv‡ci †Kv‡bv cwieZ©b nq bv D. `kvi cwieZ©b nq Ans B 03. †Kvb e¨w³ ce©‡Z cvwb dzUv‡Z PvB‡j cvwbi cv·K †h ZvcgvÎvq DËß Ki‡Z n‡e Zv- [DU: 2009-10, RU: 2018-19] A. Higher than 100°C B. Lower than 100°C C. To 100°c D. Cannot be determine Ans B 04. kx‡Zi †`‡k iv¯Ívq eid Mjv‡bvi Rb¨ jeY e¨envi Kiv nq KviY- [DU: 2008-09; RU: 13-14] A. jeY ei‡di Mjbv¼ evwo‡q †`q B. jeY ei‡di Mjbv¼ Kwg‡q †`q C. jeY I eid wg‡j GKwU bZzb Zij ivmvqwbK †h․M •Zwi nq D. cÖK…Zc‡¶ G c×wZ KvR K‡i bv Gi †Kvb •eÁvwbK wfwË †bB Ans B 05. cvwb, eid I Rjxq ev®ú †h ZvcgvÎvq GK m‡½ _vK‡Z cv‡i Zv n‡jv [DU: 2003-04] A. 0K B. 273.16K C. 100K D. 4K Ans B STEP 02 ANALYSIS OF JU QUESTION 01. DòZvwgwZK ag© m¤úbœ e¯‘i D`vnib- [JU: 2017-18] A. •KwkK b‡j iw¶Z cvi` B. cøvwUbvg Zvi C. ZvchyM‡ji cwievnx Zvi D. mK‡jB Ans D 02. GKwU RjšÍ Pzjvi cv‡k `uvov‡j Mig Abyf‚Z nIqvi KviY? [JU: 2017-18] A. cwienb B. cwiPvjb C. wewKiY D. mK‡jB Ans C STEP 03 ANALYSIS OF RU QUESTION 01. Pvc e„w× †c‡j ùyUbvsKÑ [RU. Astrazeneca, Set-1. 20-21] A. K‡g B. ev‡o C. AcwiewZ©Z _v‡K D. †KvbwUB bq S B info Pvc e„w× †c‡j ùzUbv¼ ev‡o Ges Pvc n«vm Ki‡j ùzUbv¼ K‡g| 02. ‡h e¯‘ Zvi Dci AvcwZZ mKj wewKiY †kvlY K‡i †bq Zv‡K Kx e‡j? [RU: 2018-19] A. `~e©j wewKiY B. DËß e¯‘ C. Av`k© K…ò e¯‘ D. †KvbwUB bq Ans C 03. ‡h h‡š¿i mvnv‡h¨ Zvc cwigvc Kiv nq- [RU: 2017-18] A. _vwg©÷vi B. cvB‡ivwgUvi C. _v‡g©vwgUvi D. K¨v‡jvwiwgUvi Ans D 04. M¨v‡mi Af¨šÍixY kw³ wbf©i K‡i †Kvb ivwki Ici| [RU: 17-18, 16–17] A. Pvc B. ZvcgvÎv C. AvqZb D. GbUªwc Ans B 05. †h ZvcgvÎvq eid cvwb I Rjxq ev®ú mnve¯’v‡b _v‡K Zv‡K e‡j- [RU: 16-17] A. •Îa we›`y B. wkwkivsK C. Kzix we›`y D. cig ZvcgvÎv Ans A 06. m~h© n‡Z c„w_exi evq~gÛ‡ji DcwifvM ch©šÍ Zvc ¯’vbvšÍ‡ii †¶‡Î †Kvb cÖwµqvwU mwVK? [RU: 2014–15] A. wewKiY I cwienb B. cwienb I Av‡ek C. cwiPjb D. wewKiY Ans D 07. ‡Kvb ¯’v‡bi ZvcgvÎv †mjwmqvm †¯‥‡j k~b¨ wWMÖx n‡j H ¯’v‡bi ZvcgvÎv †Kjwfb †¯‥‡j KZ? [RU: 2011-12] A. 173 K B. 273 K C. 173 K D. –273 K S B info †Kjwfb †¯‥‡j ZvcgvÎv, K = 273 + 0 = 273 K 08. DuPz ce©‡Zi Dci †Lvjv cv‡Î ivbœv Kiv KwVb nIqvi KviY wK? [RU: 11-12] A.D”PZv B. evqy Pv‡ci e„w× C. ùzUbvs‡Ki n«vm D. ùzUbvs‡Ki e„w× Ans C 09. cvwbi ùzUbvsK me‡P‡q †ewk n‡e- [RU: 2007-08] A. bxPz mgZjf‚wg‡Z B. DuPz cvnv‡oi P‚ovq C. mgy‡`ªi av‡i D. me RvqMvq GKB _vK‡e Ans A STEP 04 ANALYSIS OF CU QUESTION 01. ZvcMwZwe`¨vi †Kvb m~·K wfwË K‡i _v‡g©vwgUvi •Zwi Kiv nq? [CU-A, 19-20; P. †ev. 2017, e. †ev. 2017, h.†ev. 2015] A. k~b¨Zg B. cÖ_g C. wØZxq D. Z…Zxq ZvcMwZwe`¨vi k~b¨Zg m~‡Îi Dci wfwË K‡i _v‡g©vwgUvi •Zwi Kiv nq| k~b¨Zg m~ÎwU n‡jv- `ywU e¯‘ hw` Z…Zxq †Kv‡bv e¯‘i mv‡_ Zvcxq mgZvq _v‡K Z‡e cÖ_‡gv³ e¯‘ `ywU ci¯ú‡ii mv‡_ Zvcxq mgZvq _vK‡e|
402 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. cvwbi wngv¼ KZ? [CU: 2015-16] A. 0F B. 32F C. 80F D. 100F E. 212F cvwbi wngv¼ 0 0C ev, 32F 03. dzUšÍ cvwb ev‡®ú cwiYZ n‡”Q G Ae¯’vq cvwbi Av‡cw¶K Zvc n‡e- [CU: 15-16] A. k~b¨ B. GK C. GK Gi †P‡q †QvU D. Amxg E. †KvbwUB bq Ans D 04. cig k~b¨ (absolute zero) mgvb KZ? [CU: 2015-16, IU: 2015-16] A. 0°C B. 0°F C. 0°R D. 0K E. –273°F Ans D 05. ZvcgvÎvi cwieZ©b bv K‡i Zij Ae¯’v †_‡K KwVb Ae¯’vq cwiYZ n‡Z GKK f‡ii ewR©Z Zvc‡K ejv nq- [CU: 2012–13] A. myßZvc B. Mj‡bi Av‡cw¶K myßZvc C. KwVb fe‡bi Av‡cw¶K myßZvc D. ev®úxfe‡bi Av‡cw¶K myßZvc E. NYxfe‡bi Av‡cw¶K myßZvc Ans C 06. ‡mjwmqvm I dv‡ibnvBU †¯‥‡j ZvcgvÎv h_vµ‡g C I F n‡j, G‡`i gv‡S m¤úK© n‡”Q- [CU: 2008-09] A. C = 5 9 F B. C = 5 9 (F – 32) C. F = 5 9 (F – 32) D. F = 5 9 (C – 32) Ans B STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. ‡÷dv‡bi m~Î (Stefan law) †KvbwU? [JnU: 2013-14] A. ET B. ET 2 C. ET 3 D. ET 4 Ans D 02. _vwg©÷i †Kvb ai‡Yi c`v_© w`‡q •Zix nq? [JnU: 2011-12] A. cwievnx B. Kzcwievnx C. Aa©cwievnx D. †KvbwUB bq Ans C KU 01. KZ †Kjwfb ZvcgvÎvq Aa©cwievnx AšÍiK wnmv‡e KvR K‡i? [KU: 2018-19] A. –273 B. 0 C. 100 D. 273 Ans B 02. †Kvb e¯‘‡K Gi cvwicvwk¦©K kxZjZg Ask n‡Z AwaKZi kxZj K‡i kw³i weivg mieivn cvIqv m¤¢e bq| GB wee„wZwU Kvi? [KU: 2010-11] A. †Kjwf‡bi B. cø¨v‡¼i C. K¬wmqv‡mi D. Kv‡iviB bq Ans A 03. GKK f‡ii †Kvb KwVb c`v‡_©i ZvcgvÎvi †Kvbiƒc cwieZ©b bv K‡i ïaygvÎ KwVb n‡Z Zij Ae¯’vq cwiYZ Ki‡Z GKwU wbw`©ó cwigvY Zv‡ci cÖ‡qvRb nq| GB Zvc‡K D³ c`v‡_©i Mj‡bi- [KU: 2004-05] A. MjbvsK e‡j B. myßZvc e‡j C. wngvsK e‡j D. ùzUbvsK e‡j Ans B IU 01. wewKY© Zvc P‡j- [IU: 16–17] A. k‡ãi †e‡M B. we`y¨‡Zi †e‡M C. Av‡jvi †e‡M D. evZv‡mi †e‡M Ans C 02. ZvcgvÎv cwigv‡ci Rb¨ Aa©cwievnx Øviv •Zix Zvc my‡e`x †ivaK‡K ejv nq- [IU: 2015-16] A. _v‡g©vwgUvi B. _vwg©÷i C. cvB‡ivwgUvi D. †iva _v‡g©vwgUvi Ans B 03. wb‡Pi †KvbwU MÖxbnvDR M¨vm bq? [IU: 2013–14] A. CO2 B. CH4 C. CO D. †KvbwUB bq Ans D BRUR 01. 120C ZvcgvÎvq cvwb KZ Pv‡c ev‡®ú cwiYZ nq? [BRUR: 2015-16] A. 0.7534 atm B. 1.7354 atm C. 0.7354 atm D. 1.07354 atm Ans A JKKNIU 01. A¨v‡gvwbqvi wngv¼ KZ? [JKKNIU-B, Set-3, 19-20] A. 0˚C B. 77.5˚C C. –77.73˚C D. 87.1˚C Ans C PART B Analysis of Science & Technology Question SUST 01. ei‡di Dc‡i mvB‡Kj Pvjv‡Z Amyweav nIqvi KviY- [SUST: 2006-07] A. ei‡di c„ôPvc Kg B. ei‡di Dci evqyi Pvc Kg C. Zvc K‡g hvq D. ei‡di Nl©YRwbZ ej AwZwi³ Kg Ans D STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. Mig e¯‘ VvÛv Ki‡Z †KvbwU AwaK Dc‡hvMx? [BUET: 2012-13] A. kxZj evZvm B. kxZj cvwb C. eid D. me¸‡jv mgvb Ans B STEP 07 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. e¯‘i †Kvb •ewkó¨wU cvB‡ivwgUv‡i e¨eüZ nq? [MAT. 2020-21] A. •`N©¨ m¤úªmviY •ewkó¨ B. •e`y¨wZK •ewkó¨ C. wewKiY •ewkó¨ D. AcwUK¨vj •ewkó¨ S C info wewKiY cvB‡ivwgUv‡i DËß e¯‘i wewKiY ag© Kv‡R jvwM‡q 500C Gi D‡aŸ©i ZvcgvÎv cwigvc Kiv nq| 02. cvwb ei‡d iƒcvšÍi Kiv n‡j Zvi NbZ¡- [MAT; 19-20] A. K‡g B. ev‡o C. k~b¨ nq D. GKB _v‡K S A info 4C ZvcgvÎvq cvwbi NbZ¡ me‡P‡q †ewk| 4C †_‡K ZvcgvÎv evov‡bv ev Kgv‡bv hvB †nvK bv †Kb cvwbi NbZ¡ K‡g| 03. †KvbwU K¶ ZvcgvÎv? [MAT: 2014-15] A. 298K B. 310K C. 313K D. 288K K¶ ZvcgvÎv = 25°C = 298k 04. cvwb‡K 0°C ZvcgvÎv n‡Z 10°C ZvcgvÎvq DËß Ki‡j Dnvi AvqZb- [MAT: 2014-15] A. e„w× cvq B. K‡g C. cwiewZ©Z _v‡K D. cÖ_‡g ev‡o Zvici K‡g Ans D 05. MÖxb nvDR wµqv wb‡gœ †Kvb m~Î Øviv e¨vL¨v Kiv hvq? [MAT: 2007-08, BRUR 13-14] A. fx‡bi m~Î B. w÷‡d‡bi m~Î C. wbDU‡bi m~Î D. †KvbwUB bq Ans A DAT 01. cvwbi •Îawe›`y (Triple point) ZvcgvÎv- [DAT. 2020-21, JUST-B, 19-20] A. 263.16K B. 253.16K C. 373.15K D. 273.16K S D info cvwbi •Îa we›`yi ZvcgvÎv Tr = 273.16 K GKwU wbw`©ó Pv‡c †h ZvcgvÎvq †Kv‡bv c`v_© KwVb, Zij I evqexq iƒ‡c mvg¨e¯’vq _v‡K Zv‡K IB c`v‡_©i •Îawe›`y e‡j| 02. Kv‡b©v P‡µi cÖ_g av‡ci †ÿ‡Î wb‡Pi †KvbwU mwVK? [DAT: 2016-17; P. †ev. 2015] A. ZvcgvÎv e„w× cvq B. AšÍt¯’ kw³ n«vm cvq C. ZvcgvÎv w¯’i _v‡K D. Zv‡ci wewKiY nq Ans C AFMC 01. gvbe‡`‡ni ¯^vfvweK ZvcgvÎvÑ [AFMC. 2020-21] A. 36.6C B. 90F C. 40C D. 41C S A info gvby‡li kix‡ii ¯^vfvweK ZvcgvÎv cÖvq 37 wWMÖx †mjwmqvm (98.6 dv‡ibnvBU)|
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 403 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 08 ANALYSIS OF HSC BOARD QUESTION 01. Zvcxq PjK n‡jvÑ [Xv. †ev. 2015, P. †ev. 2017] i. Pvc ii. AvqZb iii. AšÍt¯’ kw³ wb‡Pi †KvbwU mwVK? A. i, ii I iii B. i I ii C. ii I iii D. i I iii Ans B 02. GKwU w¯’i we›`y c×wZ‡Z ZvcgvÎv cwigv‡ci g~jbxwZ e¨eüZ nq wb‡¤œi †Kvb †¯‥‡j? [iv.‡ev. 2016] A. †mjwmqvm B. †ivgvi C. †Kjwfb D. dv‡ibnvBU Ans C 03. wb‡Pi †Kvb kw³ Ab¨ kw³‡Z mn‡R iƒcvwšÍZ n‡Z Pvq bv? [P. †ev. 2016] A. Zvc B. Av‡jv C. kã D. Zwor Ans A 04. wbw`©ó cwigvY M¨v‡mi Af¨šÍixY kw³ wb‡¤œv³ †KvbwUi Ici wbf©i K‡i bv? [P. †ev. 2016] A. Pvc B. AvqZb C. ZvcgvÎv D. fi Ans D 05. M¨vm KZ©„K K…ZKvR m¤úbœ n‡j wb‡Pi †KvbwU cÖ‡hvR¨ n‡e? [P. †ev. 2015] A. AvqZb e„w× cvq B. AvqZb n«vm cvq C. fi e„w× cvq D. fi n«vm cvq Ans A 06. wZbwU wm‡÷g Zvcxq mvg¨ve¯’vq _vK‡j Zv‡`i wb‡Pi †Kvb ivwkwU GKB n‡e? [P. †ev. 2015] A. fi B. ZvcgvÎv C. AšÍt¯’ kw³ D. wefe kw³ Ans B 07. dv‡ibnvBU †¯‥‡j cvwbi •Îawe›`yi ZvcgvÎv n‡jvÑ [w`. †ev. 2017] A. 0 F B. 32 F C. 273 F D. 273.16 F Ans B 08. Ò†Kv‡bv wbw`©ó cwigvY Zvckw³ m¤ú~Y©fv‡e hvwš¿K kw³‡Z iƒcvšÍi Kivi g‡Zv hš¿ wbg©vY m¤¢e bq|ÓÑ Dw³wU †Kvb weÁvbx cÖ`vb K‡ib? [w`. †ev. 2021] A. cøvsK B. †Kjwfb C. K¬wmqvm D. Kv‡Y©v Ans D 09. GKwU c`v‡_©i ZvcwgwZK ag©Ñ [w`. †ev. 2015] i. Pv‡ci mgvbycvwZK ii. AvqZ‡bi mgvbycvwZK iii. ZvcgvÎvi mgvbycvwZK wb‡Pi †KvbwU mwVK? A. i B. ii C. iii D. i I iii Ans C 10. wP‡Îi wZbwU eø‡Ki ZvcgvÎv h_vµ‡g 1C, 2C I 3C hviv ci¯ú‡ii mv‡_ Zvcxq ms¯úk© Av‡Q| [wm. †ev. 2016] 1C 2C 3C eøK-1 eøK-2 eøK-3 †Kvb ZvcgvÎv Zvcxq mvg¨e¯’v wb‡`©k K‡i? 1C 2C 3C A. 5 10 5 B. 10 5 10 C. 15 15 15 D. 20 15 15 Ans C 11. †Kv‡bv M¨v‡mi `ywU †gvjvi Av‡cwÿK Zv‡ci AbycvZ GKwU aªæe ivwk| GB aªæe ivwk‡K †h cÖZxK Øviv cKvk Kiv nq Zv n‡jv- [mKj †evW© 2018] A. B. R C. D. K Ans A Concept 2 ZvcMwZwe`¨vi cÖ_g m~Î (Ryj I K¬wmqv‡mi gZev`, 1g m~‡Îi Zvrch©) ✅ ZvcMwZwe`¨vi cÖ_g m~Î: cÖ`vb K‡ib: weÁvbx †Rgm †cÖmKU Ryj mvaviYfv‡e cÖKvk K‡ib: K¬wmqvm ✅ Ry‡ji gZev`: weÁvbx Ryj me© cÖ_g Zvc I Kv‡Ri g‡a¨ m¤úK© ¯’vcb K‡ib| G‡K Ry‡ji gZev` e‡j| Ry‡ji gZev`: hLb KvR m¤ú~Y© fv‡e Zv‡c ev Zvc m¤ú~Y© fv‡e Kv‡R iƒcvšÍwiZ nq ZLb KvR I Zvc ci®ú‡ii mgvbycvwZK nq| A_©vr W H W = JH Zv‡ci hvwš¿K ÿgZv/Ryj ZzjvsK mgZv Zzj¨vsK J = W H J Gi GKK Jcal–1 Gm AvB GK‡K J Gi GKK bvB; W I H Df‡qiB GKK Joule ✅ K¬wmqv‡mi gZev`: hLb †Kvb e¨e¯’vq Zvc mieivn Kiv nq ev e¨e¯’v KZ…©K Zvc M„nxZ nq, ZLb Zvi wKq`sk Af¨šÍixY kw³ e„w× Ki‡Z A_©vr ZvcgvÎv e„w× Ki‡Z Ges Aewkó Ask evwn¨K KvR m¤úv`‡b e¨eüZ nq| m~‡Îi cÖKvk: dQ = dU + dW [dQ = M„nxZ ev ewR©Z Zvc kw³; dU = Af¨šÍixY kw³i cwieZ©b; dW = PV K…Z KvR] wm‡÷g Zvckw³ eR©‡b (–) dQ wm‡÷g Zvckw³ MÖn‡Y (+) Af¨šÍixY kw³ n«vm †c‡j (–) dU Af¨šÍixY kw³ e„wׇZ (+) wm‡÷‡gi Dci evwni †_‡K KvR Kiv n‡j/ AvqZb n«vm †c‡j (–) dW wm‡÷g wb‡R KvR Ki‡j/AvqZb e„w× †c‡j (+) ✅ ZvcMwZ we`¨vi 1g m~‡Îi Zvrch©: †Kvb wKQz e¨q bv K‡i KvR/kw³ cvIqv Am¤¢e| KvR I Zvc G‡K Ac‡ii mgZyj¨| GwU kw³i msi¶bkxjZv m~‡Îi we‡kliƒc R¡vjvwb kw³ e¨wZZ †Kvb hš¿B KvR Ki‡Z m¶g bq| ✅ 1847 mv‡j †Rgm †cÖmKU Ryj Zvc I hvwš¿K kw³i g‡a¨ m¤úK© wbY©q K‡ib; ✅ ZvcgvÎv w¯’i _vK‡j AšÍt¯’ kw³ cwiewZ©Z nq bv; ✅ gvby‡li kix‡ii mvavib ZvcgvÎv 98.4°F Ges †mjwmqvm †¯‥‡j GB ZvcgvÎv 36.9°C/37°C| ✅ wm‡÷g (System): cix¶v-wbix¶vi mgq Avgiv Ro RM‡Zi LvwbKUv wbw`©ó Ask we‡ePbv Kwi| Ro RM‡Zi GB wbw`©ó Ask‡K wm‡÷g ev e¨e¯’v e‡j| ✅ Zvcxq wm‡÷‡gi cÖKvi‡f`: cÖKvi •ewkó¨ D`vniY i. Db¥y³ wm‡÷g fi I kw³ DfqB wewbgq K‡i| cwievnx c`v_©| ii. e× wm‡÷g ïay kw³ wewbgq K‡i| – iii. wew”Qbœ wm‡÷g ev Kzcwievnx wm‡÷g fi I kw³ wKQzB wewbgq K‡i bv| Kzcwievnx c`v_©|
404 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF RU QUESTION 01. kw³i wbZ¨Zv m~ÎwU ZvcMwZwe`¨vi †Kvb m~‡Îi mvnv‡h¨ e¨vL¨v Kiv hvq? [RU. Astrazeneca, Set-1. 20-21, MBSTU-B, Set-2 19-20] A. k~b¨Zg m~Î B. cÖ_g m~Î C. wØZxq m~Î D. cÖ_g I wØZxq m~Î S B info kw³i wbZ¨Zv m~ÎwU ZvcMwZwe`¨vi cÖ_g m~Î, dQ = dU + dW Gi mvnv‡h¨ e¨vL¨v Kiv hvq| 02. ZvcMwZwe`¨vi cÖ_g m~Î cÖK…Zc‡¶- [RU: 2017-18 ; P. †ev. 2016] A. fi‡e‡Mi wbZ¨Zvi m~Î B. kw³i wbZ¨Zvi m~Î C. f‡ii wbZ¨Zvi m~Î D. ZvcgvÎv wba©vi‡Yi m~Î Ans B 03. ZvcMwZwe`¨vi cÖ_g m~Î bx‡Pi †Kvb `ywUi g‡a¨ m¤úK© ¯’vcb K‡i? [RU: 2016–17; KU: 2017-18; wm. †ev. 2015] A. Zvc I KvR B. ej I kw³ C. Zvc I ej D. KvR I ¶gZv Ans A STEP 02 ANALYSIS OF CU QUESTION 01. ZvcMwZwe`¨vi cÖ_g m~Î †KvbwU? [CU-A, Set-1. 20-21; ; Xv. †ev. 2017] A. Q = U + W B. W = Q + U C. Q =W – U D. W = Q – U Ans A STEP 03 ANALYSIS OF GST QUESTION PART A Analysis of General University Question CoU 01. KvR †c‡Z n‡j Aek¨B Zvc mieivn Ki‡Z n‡e-GwU cvIqv hvq- [CoU: 16-17] A. ZvcMwZwe`¨vi 1g m~Î n‡Z B. ZvcMwZwe`¨vi 2q m~Î n‡Z C. ZvcMwZwe`¨vi 3q m~Î n‡Z D. Ry‡ji m~Î n‡Z Ans A STEP 04 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. evB‡ii kw³i mvnvh¨ Qvov †Kvb ¯^qswµq h‡š¿i c‡¶ wbgœ DòZvi e¯‘ n‡Z D”PZi DòZvi e¯‘‡Z Zvc ¯’vbvšÍi m¤¢e bq| GwU †Kvb wee„wZ? [MAT: 2006-07] A. K¬vwmqv‡mi B. cøvs‡Ki C. †Kjwf‡bi D. Kv‡bv© Ans B STEP 05 ANALYSIS OF HSC BOARD QUESTION 01. Zvc AšÍi‡Ki AveiYhy³ `„p cv‡Î GKwU Av`k© M¨vm k~b¨ gva¨‡g cÖmviY Kiv n‡jv| d‡j wb‡¤œi †KvbwU NU‡e? [P. †ev. 2021] A. AšÍt¯’ kw³i cwieZ©b n‡e bv B. ZvcgvÎv n«vm cv‡e C. Pv‡ci †Kv‡bv cwieZ©b n‡e bv D. `kvi cwieZ©b n‡e Ans B 02. T1 I T2 ZvcgvÎvi `ywU e¯‘i Avf¨šÍixY kw³ h_vµ‡g U1 I U2 †hLv‡b T1 > T2| wew”Qbœ wm‡÷‡g e¯‘Øq Zvcxq mvg¨e¯’vq Avm‡j wb‡Pi †KvbwU N‡U? [wm. †ev. 2021] A. U1 e„w× U2 n«vm B. U1 n«vm U2 e„w× C. U1 e„w× U2 e„w× D. U1 n«vm U2 n«vm Ans B Concept 3 ZvcMZxq cwieZ©b (m‡gvò, iæ×Zvcxq, mgPvc Ges mgAvqZb) ✅ ZvcMZxq cwieZ©b Pvi cÖKvi: mgPvcxq cwieZ©b (Isobaric) Pvc w¯’i, P = 0 mgAvqZb cwieZ©b (Isochoric) AvqZb w¯’i, V = 0 m‡gvò cwieZ©b (Isothermal) DòZv/ZvcgvÎv w¯’i, T = 0 iæ×Zvcxq cwieZ©b (Adiabetic) Zvckw³ w¯’i, Q = 0 ◈ m‡gvò cÖwµqvi †¶‡Î, wm‡÷g ev e¨e¯’v KZ…©K KvR, wm‡÷‡g mieivnK…Z ev M„nxZ Zvckw³i mgvb; ◈ iƒ×Zvcxq cÖmviY cÖwµqvi †¶‡Î, Af¨šÍixb kw³ n«vm cvq + ZvcgvÎv n«vm cvq (dU = – dW); ◈ aªæe AvqZb cÖwµqvi †¶‡Î, AšÍt¯’ kw³i e„w× mieivnK…Z Zvckw³i mgvb (dQ = dU); ✅ m‡gvò cwieZ©b: ZvcgvÎv w¯’i _vK‡j V 1 P A_©vr e‡q‡ji m~Î †g‡b P‡j| V Gi mv‡_ P Gi e¨¯ÍvbycvwZK m¤úK© myZivs †jLwPÎ hyperbola/Awae„ËvKvi n‡e| P [V 1 P ] ; [PV = aªæeK] AvB‡mv_vg©vj †iLv PV V 1 P AvB‡mv_vg©vj †iLv Xvj = – P V AvB‡mv_vg©vj †iLv [AvqZvKvi Awae„Ë] P V ✅ m‡gvò cwieZ©‡bi kZ©: M¨vm, mycwievnx cv‡Î _vK‡e| PZzc©vk©¦¯’ gva¨‡gi ZvcMÖvnxZv D”P n‡Z n‡e| cÖ‡qvRbxq ZvcMÖnY ev eR©‡bi Øviv ZvcgvÎv w¯’i _vK‡e| Pv‡ci cwieZ©b ax‡i ax‡i msNwUZ Ki‡Z n‡e| ✅ iƒ×Zvcxq cwieZ©b: †h cÖwµqvq wm‡÷g Zvckw³ MÖnY ev eR©b K‡i bv| ✅ iƒ×Zvcxq cwieZ©‡bi kZ©: M¨vm Kzcwievnx cv‡Î _vK‡e| PZzc©vk©¦¯’ gva¨‡gi ZvcMÖnxZv Kg n‡e| Pvc cwieZ©b `ªæZ Ki‡Z n‡e| P iæ×Zvcxq †iLv V m‡gvò †iLv
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 405 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. iæ× Zvcxq cÖwµqvq wb‡gœi †Kvb mgxKiYwU ï×? [DU: 01-02, JU: 2011-12, 2012-13; RU: 2009-10, 2017-18; CU: 2010-11; IU: 2011-12] A. PV1- = Constant B. PV = Constant C. PV = Constant D. T = Constant Ans B STEP 02 ANALYSIS OF JU QUESTION 01. `ywU e¯Íyi Nl©‡Yi d‡j Zvc Drcbœ nq, GwU GKwU- [JU: 2018-19] A. cÖZ¨veZ©x cÖwµqv B. AcÖZ¨veZ©x cÖwµqv C. iæ× ZvcxqcÖwµqv D. m‡gvòcÖwµqv Ans B STEP 03 ANALYSIS OF RU QUESTION 01. iæ×Zvcxq cwieZ©‡bi †ÿ‡Î mZ¨ bqÑ [RU. Sinovac, Set-1. 20-21] A. M¨v‡mi AšÍwb©wnZ kw³ w¯’i _v‡K B. Zv‡ci Av`vb-cÖ`vb nq bv C. ZvcgvÎvi cwieZ©b nq D. PV = aªæeK S A info M¨v‡mi iæ×Zvcxq cwieZ©‡bi †ÿ‡Î AšÍ©bxwnZ kw³ w¯’i _v‡K bv| 02. iæ×Zvcxq cwieZ©‡bi •ewkó¨ n‡jv- [RU: 2015-16] i. GwU GKwU AwZ `ªæZ cÖwµqv ii. GB cwieZ©‡b ZvcgvÎv w¯’i _v‡K iii. GB cwieZ©‡b cvÎwU Zvc Kzcwievnx nIqv `iKvi wb‡Pi †KvbwU mwVK? Ans B A. i I ii B. i I iii C. ii I iii D. i, ii I iii 03. cÖZ¨vMvgx cÖwµqvq †h †Kvb e¨e¯’vi GbUªwc- [RU: 2006-07] A. n«vm cvq B. e„w× cvq C. w¯’i _v‡K D. mvgvb¨ cwiewZ©Z nq Ans C STEP 04 ANALYSIS OF CU QUESTION 01. †Kvb e¯‘i KwVb Ae¯’v †_‡K Zij Ae¯’vq bv †h‡q mivmwi evqexq Ae¯’vb iƒcvšÍi c×wZ‡K wK e‡j? [CU-A, Set-1. 20-21] A. Nbxfeb B. ev®úxKiY C. DaŸ©cvZb D. GKxfeb S C info †Kv‡bv KwVb e¯‘‡K DËß Ki‡j Zv Zi‡j iƒcvšÍwiZ bv n‡q mivmwi ev‡®ú ev evqexq `kvq cwiYZ nIqv‡K DaŸ©cvZb e‡j| STEP 05 ANALYSIS OF DU-7 Clg QUESTION 01. GKwU Av`k© M¨vm m¼zwPZ n‡q wb‡P D‡jøwLZ wewfbœ Zvcxq cÖwµqvq Zvi cÖK…Z AvqZ‡bi A‡a©K AvqZ‡b n«vm cvq| bx‡Pi †KvbwU‡Z me‡P‡q †ewk KvR m¤úbœ nq? [DU. 7Clg-A: 20-21] A. m‡gvò B. iæ×Zvcxq C. mgAvqZb D. mgPvcxq S A info m‡gvò cÖwµqvq ms‡KvPb I cÖmvi‡Yi Rb¨ K…ZKv‡Ri gvb iæ×Zvcxq cÖwµqvq ms‡KvPb I cÖmvi‡Yi Rb¨ K…ZKvR| STEP 06 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. PV = aªæeK mgxKi‡Y wØcigvYyK (diatomic) M¨v‡mi †¶‡Î Gi gvb KZ? [JnU: 2015-16] A. 1.20 B. 1.40 C. 1.33 D. †KvbwUB bq Ans B 02. iæ×Zvcxq (Adiabatic) cÖwµqvq GbUªwc (Entropy) [JnU: 2014–15] A. e„w× cvq B. K‡g hvq C. †Kvb cwieZ©b nq bv D. †KvbwUB bq Ans C 03. GKwU iæ×Zvcxq cÖwµqvq cwi‡e‡ki ZvcgvÎv e„w× †c‡j wm‡÷‡gi AYyi MwZkw³- [JnU: 2011-12] A. e„w× cvq B. n«vm cvq C. mgvb _v‡K D. k~Y¨ nq Ans A KU 01. PjšÍ Mvwoi Uvqv‡ii †fZi †Kvb ZvcMZxq cÖwµqv P‡j? [KU:2018-19] A. m‡gvò B. iæ×Zvcxq C. mgAvqZb D. mgPvc Ans C 02. •e`y¨wZK †iv‡ai ga¨ w`‡q we`y¨r cÖevwnZ n‡j Zvc m„wó nq| GwU GKwU- [KU: 2009-10] A. AcÖZ¨vMvgx cÖwµqv B. iƒ×Zvcxq cÖwµqv C. m‡gvò cÖwµqv D. cÖZ¨vMvgx cÖwµqv Ans A 03. GKwU Mvwo Pj‡Z _vK‡j Zvi Uvqv‡ii wfZi wKQz ZvcMZxq cÖwµqv P‡j| GB cÖwµqvwU nj- [KU: 2007-08] A. m‡gvò cÖwµqv (Isothermal process) B. iæ× Zvcxq cÖwµqv (Adiabatic process) C. aªæe-AvqZb cÖwµqv (Isochoric process) D. aªæe Pvc cÖwµqv (Isobaric process) Ans C 04. cÖZ¨vMvgx cÖwµqvi †¶‡Î †KvbwU mZ¨ bq? [KU: 2001-02] A. GB cÖwµqvq wecixZgyLx n‡q cÖZ¨veZ©b K‡i B. GwU Lye axi cÖwµqv C. GwU ¯^Ztd‚Z© cÖwµqv D. f‚-c„ô †_‡K Af¨šÍ‡i Ans D CoU 01. iæ×Zvcxq cÖwµqvq Pvc cÖ‡qvM Ki‡j wb‡gœi †Kvb mgxKiYwU ï×- [CoU: 2012-13] A. Q =0 B. T =0 C. A I B D. †KvbwUB bq Ans A IU 01. †Kvb wm‡÷‡gi Ae¯’vi cwieZ©‡bi †¶‡Î Reversible Process GKwU- [IU: 2015-16] A. GKgyLx cÖwµqv B. axi cÖwµqv C. `ªæZ cÖwµqv D. ¯^ZtùzZ© cÖwµqv Ans B 02. m‡gvò cwieZ©‡bi Rb¨ †Kvb wm‡÷‡gi Pvc cwieZ©b Ki‡Z n‡e- [IU: 2011-12] A. Lye `ªæZ B. gvSvwifv‡e C. Lye Av‡¯Í Av‡¯Í D. †h †Kvb fv‡e Ans C BRUR 01. iæ×Zvcxq cÖwµqvq †Kvb †f․Z ivwk w¯’i _v‡K?[BRRU-F, Set-2, 19-20;Xv. †ev. 2018, e.‡ev. 2015,17] A. ZvcgvÎv B. Pvc C. GbUªwc D. Af¨šÍixY kw³ iæ×Zvcxq cÖwµqvq GbUªwc w¯’i _v‡K Ges AcÖZ¨vMvgx cÖwµqvq GbUªwc e„w× cvq| PART B Analysis of Science & Technology Question SUST 01. w¯’i ZvcgvÎvi Av`k© M¨v‡mi P-V †jLwPÎwU n‡e GKwU: [SUST:2018-19] A. AvqZvKvi Awae„Ë B. P-A‡ÿi mgšÍivj mij‡iLv C. Dce„Ë D. cive„Ë E. V-A‡ÿi mgvšÍivj mij‡iLv Ans A MBSTU 01. iæ×Zvcxq cwieZ©‡b AvqZb I ZvcgvÎvi g‡a¨ m¤úK© nj- [MBSTU-B, Set-2 19-20;IU. 12-3,BUET. 13-14] A. TV1 B. TV C. TV D. T1V (i) PV = aªæeK (ii) TV – 1 = aªæeK (iii) TP 1 – = aªæeK NSTU 01. ‡h cÖwµqvq wm‡÷g †_‡K Zvc evB‡i hvq bv ev evB‡i †_‡K Zvc wm‡÷‡g Av‡m bv Zv‡K †Kvb cÖwµqv e‡j? [NSTU: 2014-15] A. m‡gvò cÖwµqv B. iƒ×Zvcxq cÖwµqv C. cÖmviY cÖwµqv D. mgPvc cÖwµqv Ans B STEP 07 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. †Kvb Zvc-hyM‡ji Rb¨ wb‡Pi gšÍe¨ ¸‡jvi g‡a¨ †KvbwU mwVK bq? [BUET: 2011-12] A. †Kvb GKwU wbw`©ó Zvc-hyMj †m‡Ui Rb¨ wbi‡c¶ ZvcgvÎv w¯’i _v‡K| B. wbi‡c¶ ZvcgvÎv kxZj ms‡hv‡Mi ZvcgvÎvi Dci wbf©i K‡i bv| C. Drµg ZvcgvÎv kxZj ms‡hv‡Mi ZvcgvÎvi Dci wbf©i K‡i bv| D. wbi‡c¶ ZvcgvÎvq m‡e©v”P Zvcxq ZworPvjK kw³ cvIqv hvq| Drµg ZvcgvÎv kxZj ms‡hv‡Mi ZvcgvÎvi Dci wbf©i K‡i bv|
406 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 08 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. iæ× Zvc cÖwµqvi †¶‡Î wb‡¤œi †KvbwU mwVK bq? [MAT: 2011-12] A. ZvcgvÎv aªæe _v‡K bv wKš‘ Zv‡ci cwieZ©b nq bv| A_©vr dQ = 0 B. GwU GKwU axi cÖwµqv C. GB cÖwµqvq Zvc †kvlY ev eR©b Kiv nq bv| D. GB cÖwµqvq M¨v‡mi Pvc I AvqZ‡bi mm¤úK© pv = aªæeK iæ×Zvcxq cÖwµqv GKwU AwZ `ªæZ cÖwµqv| 02. wb‡Pi †KvbwU iƒ× Zvcxq cÖwµqvi •ewkó¨;- [MAT: 2010-11] A. e‡q‡ji m~Î Abymib K‡i B. iƒ× Zvcxq †jL A‡c¶vK…Z Lvov C. GB cÖwµqvq wm‡÷‡g ZvcgvÎv w¯’i _v‡K bv D. AvšÍwiZ Ki‡Z nq Ans B 03. †h cÖwµqvq †Kvb wm‡÷‡gi ZvcgvÎv w¯’i †i‡L c`v‡_©i Pvc I AvqZ‡bi cwieZ©b NUv‡bv nq, Zvnv bx‡P D‡jøwLZ †Kvb cÖwµqv? (MAT: 2007-08] A. ZvcRvZxq B. mgPvc C. m‡gvò D. iƒ×Zvcxq Ans C 04. †Kvb Dw³wU mZ¨ bq? [MAT: 2006-07] A. fi †e‡Mi GKK nj f‡ii GKK X †e‡Mi GKK (Kgms–1 ) B. †K․wbK †eM‡K e„‡Ëi e¨vmva© Øviv fvM Ki‡j ‣iwLK †eM cvIqv hvq C. k‡ãi ZxeªZv Dr‡mi K¤úvs‡Ki e‡M©i mgvbycvwZK Ans B D. iƒ× Zvcxq †iLv m‡gvò †iLvi †P‡q AwaKZi Lvov 05. wb‡Pi †KvbwU ZvcRvZxq cwieZ©b bq? [MAT: 2003-04] A. m‡gvò cwieZ©b B. mgAvqZb cwieZ©b C. mgPvc cwieZ©b D. mgag©x cwieZ©b Ans D 06. wb‡¤œi †KvbwU AcÖZ¨vMvgx cÖwµqv bq? [MAT: 2003-04] A. e¨vcb B. cwiPvjb C. wewKiY D. cÖwZmiY Ans D STEP 09 ANALYSIS OF HSC BOARD QUESTION 01. m‡gvò †iLvi Xvj iƒ×Zvcxq †iLvi Xvj A‡cÿv KZ¸Y Lvov? [Xv. †ev. 2021] A. + 1 B. + C. 1 D. S A info iæ×Zvcxq = m‡gvò m‡gvò †iLv = 1 iæ×Zvcxq| 02. evqy gva¨‡g kã mÂvjb †Kvb ai‡bi cÖwµqv? [Xv. †ev. 2021] A. m‡gvò B. mgPvcxq C. mgAvqZb D. iƒ×Zvcxq Ans C 03. iƒ×Zvcxq cÖmvi‡Y †KvbwU mwVK? [Xv. †ev. 2021] [GLv‡b W = evwn¨K KvR, U = AšÍt¯’ kw³i cwieZ©b, Q = cÖhy³ Zvc] A. W = U B. w = U C. Q = U D. Q = U S B info Q = U + W U + w = 0 [Q = 0, iæ×Zvcxq] w = U 04. wb‡Pi †Kvb †jLwPÎwU P V Gi cwieZ©b wb‡`©k K‡i? [Xv. †ev. 2016] A. B. C. D. O P V O P V O P V O P V Ans D 05. m‡gvò †iLv †KvbwU? [Xv. †ev. 2015] A. B. C. D. P V P V P V P V Ans A 06. wb‡Pi wee„wZ¸‡jv jÿ¨ Ki: [P. †ev. 2021; g. †ev. 2021] i. †h ZvcgvÎvq †Kv‡bv c`v_© KwVb, Zij I evqexqiƒ‡c mvg¨ve¯’vq _v‡K Zv‡K H c`v‡_©i •Îa we›`y e‡j ii. †h cwieZ©‡bi Kvi‡Y ZvcMZxq ¯’vbvs‡Ki gv‡bi cwieZ©b nq †mB cwieZ©b‡K ZvcMZxq cÖwµqv e‡j iii. †Kv‡bv wm‡÷‡gi kw³ iƒcvšÍ‡ii Rb¨ kw³i AcÖvc¨Zv‡K GbUªwc e‡j wb‡Pi †KvbwU mwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans D 07. ZvcMwZwe`¨vi †Kvb cÖwµqvq M¨v‡mi Dci †Kv‡bv KvR nq bv? [iv. †ev. 2021] A. m‡gvò B. mg-AvqZb C. mgPvcxq D. iæ× Zvcxq Ans B 08. iæ×Zvcxq cwieZ©‡bi mgq Av`k© M¨v‡mi †ÿ‡Î Pvc I ZvcgvÎvi m¤úK© n‡jvÑ [P. †ev. 2021, g.‡ev. 2021] A. TP 1 – = aªæeK B. PT 1 – = aªæeK C. TP 1 – = aªæeK D. PT 1 – = aªæeK Ans A 09. GKwU Kv‡b©v P‡µ iæ×Zvcxq cÖmviY KqwU? [P. †ev. 2017] A. 1 wU B. 2 wU C. 3 wU D. 4 wU S A info Kv‡b©vi P‡µ GKwU m‡gvò I GKwU iæ×Zvcxq cÖwµqvq cÖmviY N‡U Ges GKwU m‡gvò I GKwU iæ×Zvcxq ms‡KvPb N‡U| 10. †Kvb M¨v‡mi Rb¨ iæ×Zvcxq †jL AwaK Lvov? [w`. †ev. 2021] A. wbqb B. Aw·‡Rb C. I‡Rvb D. Kve©b-WvB-A·vBW Ans A 11. m‡gvò cÖwµqvi †ÿ‡ÎÑ [w`. †ev. 2021] i. P V ii. 1 P V iii. PV V wb‡Pi †KvbwUmwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans B 12. iæ× Zvcxq cwieZ©‡bi †ÿ‡ÎÑ [w`. †ev. 2021] i. nVvr msNwVZ nq ii. ZvcgvÎv w¯’i _v‡K iii. GbUªwci cwieZ©b k~b¨ wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans C 13. m‡gvò cÖwµqvq wm‡÷‡gi †Kvb ivwkwU w¯’i _v‡K? [w`. †ev. 2016] A. AvqZb B. Zvc C. ZvcgvÎv D. Pvc Ans C 14. cÖZ¨veZx© cÖwµqv GKwU- [h. †ev. 2016] A. ¯^Ztù’Z© cÖwµqv B. `ªæZ cÖwµqv C. GKgyLx cÖwµqv D. ZvcMZxq cÖwµqv Ans D 15. Kv‡b©vi P‡µi PZz_© av‡c wK N‡U? [Kz. †ev. 2017] A. m‡gvò cÖmviY B. m‡gvò ms‡KvPb C. iæ×Zvcxq ms‡KvPb D. iæ×Zvcxq cÖmviY Ans C 16. †KvbwU iæ×Zvcxq cwieZ©‡bi •ewkó¨ wb‡`©k K‡i? [Kz. †ev. 2016] A. GB cwieZ©‡b ZvcgvÎvi cwieZ©b N‡U B. iæ×Zvcxq †jL m‡gvò †jL A‡cÿv Lvov C. GwU GKwU `ªæZ cÖwµqv D. GB cwieZ©‡b cvÎ Zvc mycwievnx Ans D 17. wb‡Pi †Kvb¸‡jv ZvcMZxq PjK wb‡`©k K‡i? [Kz. †ev. 2016] A. P. V. T. M B. P. T. V. U C. P. V. T. S D. P. V. T. Q Ans C 18. iæ×Zvcxq cÖmvi‡Yi †ÿ‡Î †KvbwU mwVK? [wm. †ev. 2017] A. wm‡÷‡gi Ici KvR m¤úbœ nq B. ZvcgvÎv w¯’i _v‡K C. AšÍt¯’ kw³ n«vm cvq D. Zvc ewR©Z nq Ans C 19. iæ×Zvcxq cwieZ©‡bÑ [wm. †ev. 2016] i. ZvcgvÎvi cwieZ©b N‡U bv ii. cvÎ Zvc Kzcwievnx nIqv cÖ‡qvRb iii. Av`k© M¨v‡mi mgxKiY n‡jv, P1V1 = P2V2 wb‡Pi †KvbwU mwVK? Ans B A. i I ii B. ii I iii C. i I iii D. i, ii I iii wb‡Pi Pvc ebvg AvqZb †jLwP‡Îi Av‡jv‡K 20, 21 I 22 bs cÖ‡kœi DËi `vI: [wm. †ev. 2015] AvqZb, V B A Pvc, P O Y X C
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 407 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 20. AB †jLwP‡Îi †ÿ‡Î †Kvb m¤úK©wU mwVK? A. PV–1 = aªæeK B. PV = aªæeK C. PV–1 = aªæeK D. PV = aªæeK Ans B 21. AC †jLwP‡Îi †ÿ‡Î †Kvb m¤úK©wU mwVK? A. PV = aªæeK B. PV–1 = aªæeK C. PV+1 = aªæeK D. PV = aªæeK Ans D 22. wØcvigvYweK M¨v‡mi †ÿ‡Î AC †jL AB †jL A‡cÿv KZ ¸Y Lvov? A. 1.33 B. 1.41 C. 1.50 D. 1.67 Ans B 23. hw` evqyc~Y© GKwU †ejyb dz‡U hvq, cÖwµqvwU‡ZÑ [wm. †ev. 2016] i. KvR m¤úbœ n‡q‡Q ii. Af¨šÍixY kw³ I ZvcgvÎv K‡g †M‡Q iii. GbUªwci cwieZ©b n‡q‡Q wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans D 24. Zvc AšÍi‡Ki AveiYhy³ `„p cv‡Î GKwU Av`k© M¨vmk~b¨ gva¨‡g cÖmviY Kiv n‡jv| d‡j wb‡¤œi †KvbwU NU‡e? [g. †ev. 2021] A. AšÍt¯’ kw³i cwieZ©b n‡e bv B. ZvcgvÎv n«vm cv‡e C. Pv‡ci †Kv‡bv cwieZ©b n‡e bv D. `kvi cwieZ©b n‡e Ans D 25. m‡gvò cÖwµqvq w¯’i AvqZ‡b †Kvb ivwkwU cwiewZ©Z nq? [gv. †ev. 2018] A. Pvc B. GbUªwc C. ZvcgvÎv D. AšÍt¯’ kw³ Ans A 26. GK Kvc Mig `y‡a GKwU VvÛv PvgP Wzev‡bv n‡j- [gv. †ev. 2017, Kz. †ev. 2016] i. `y‡ai AšÍt¯’ kw³ e„w× cvq ii. Pvg‡Pi AšÍt¯’ kw³ e„w× cvq iii. `y‡ai AšÍt¯’ kw³ GKB _v‡K wb‡Pi †KvbwU mwVK? A. i B. ii C. i I iii D. ii I iii Ans B 27. PV = aªæeK mgxKiYwU †Kvb cÖwµqv‡K mg_©b K‡i? [gv. †ev. 2017] A. m‡gvò B. mgAvqZb C. mgPvc D. iæ×Zvcxq Ans A Concept 4 CP, CV Ges R Gi m¤úK© ✅ GKwU Av`k© M¨v‡mi Cp Ges Cv Gi g‡a¨ m¤úK©: Cp – Cv = R †h‡nZz mve©Rbxb ev †gvjvi M¨vm aªæeK R me©`v GKwU abvZ¥K ivwk, myZivs, Cp > Cv Cp me©`vB Cv Gi †P‡q eo| †gvjvi M¨vm aªæeK : R = 8.314 Jmole–1K –1 A_ev, R = 0.0821 Latm mol–1K -1 ✅ w¯’i Pv‡c M¨v‡mi †gvjvi Av‡cwÿK Zvc Cp Ges w¯’i AvqZ‡b M¨v‡mi †gvjvi Av‡cwÿK Zvc CV Gi AbycvZ‡K Øviv m~wPZ Kiv nq; A_©vr = Cp Cv M¨v‡mi MwZZ‡Ë¡i mvnv‡h¨ GK cvigvYweK M¨v‡mi †ÿ‡Î cvIqv hvq, Cv = 3 2 R Cp = Cv + R = 3 2 R + R = 5 2 R ✅ wewfbœ M¨v‡mi Gi gvb: M¨v‡mi MV‡b cigvYy msL¨v D`vniY Cp Gi gvb Cv Gi gvb Gi gvb GK cvigvYweK M¨vm He, Ne, Ar 5 2 R 3 2 R 1.67 wØ-cvigvYweK M¨vm H2, O2, N2, Cl2 7 2 R 5 2 R 1.41 wÎ-cigvYyK/eû cvigvYweK M¨vm CO2, C2H6, NH3 4R 3R 1.33 ◈ wKQz ejv bv _vK‡j Gi gvb 1.4 e¨envi Ki‡Z n‡e| ◈ M¨vm Øviv KvR m¤úvw`Z n‡j, AšÍt¯’ kw³ n«vm cvq; ◈ iƒ×Zvcxq †iLv m‡gvò †iLv A‡c¶v ¸Y Lvov, iæ×Zvcxq †iLvi Xvj m‡gvò †iLvi Xvj = ◈ ev¯Íe M¨vm mg~n ïaygvÎ D”P ZvcgvÎvq I wbgœ Pv‡c Av`k© M¨v‡mi b¨vq AvPiY K‡i ◈ M¨v‡mi †gvjvi Av‡cw¶K Zvc: †Kvb e¯‘i GK †gv‡ji ZvcgvÎv GK †Kjwfb e„w× Ki‡Z †h cwigvY Zv‡ci cÖ‡qvRb nq, Zv‡K M¨v‡mi †gvjvi Av‡cwÿK Zvc e‡j| C = dQ ndT| ◈ M¨v‡mi †gvjvi Av‡cwÿK Zvc 2 cÖKvi: (1) w¯’i Pv‡c †gvjvi Av‡cw¶K Zvc; (2) w¯’i AvqZ‡b †gvjvi Av‡cw¶K Zvc| ◈ †gqv‡ii cÖKí: Ò‡Kvb wbw`©ó cwigvY M¨v‡mi Af¨šÍixY kw³ ïaygvÎ Gi ZvcgvÎvi Dci wbf©i K‡i, Pvc ev AvqZ‡bi Dci bq|Ó REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU Av`k© M¨v‡mi †¶‡Î Cp/Cv = x n‡j, wb‡Pi †Kvb m¤úK©wU GK †gv‡ji Rb¨ mwVK? [DU: 2015-16] A. Cv = (x – 1)R B. Cv = R/(x – 1) C. Cv = R/(1 – x) D. Cv = R/(1 + x) iæ× Zvcxq M¨vm aªæeK, = V P C C = x CP= xCv ; Avevi, †gvjvi M¨vm aªæeK, R= Cp – Cv = xCv – Cv = Cv(x–1) Cv = R/(x – 1) 02. GKwU wØ cigvby wewkó M¨v‡mi †¶‡Î Cp Cv nj- [DU: 2010-11] A. 1.67 B. 1.4 C. 1.33 D. 1.11 Ans B STEP 02 ANALYSIS OF JU QUESTION 01. GK cigvYweK Av`k© M¨v‡mi †¶‡Î CP = ? [JU: 2009-10] A. 2.8Jmol-1 k -1 B. 20.8Jmol-1 k -1 C.12.5Jmol-1 k -1 D.8315Jmol-1 k -1 2 5 R CP 8.314 2 5 = 20.8 STEP 03 ANALYSIS OF RU QUESTION 01. †Kv‡bv M¨v‡mi Av‡cwÿK Zvc؇qi AbycvZ = 1.4 n‡j, M¨vmwUi AYyÑ [RU. Sinovac, Set-1. 20-21,IU. 11-12;e. †ev. 2015] A. GK cvigvYweK B. wØ-cvigvYweK C. wÎ-cvigvYweK D. eû cvigvYweK S B info * GK-cvigvYweK M¨v‡mi Rb¨, = 1.67 * wØ-cvigvYweK M¨v‡mi Rb¨, = 1.41 * wÎ/ eû cvigvYweK M¨v‡mi Rb¨, = 1.33 STEP 04 ANALYSIS OF CU QUESTION 01. = 1.41 †Kvb& M¨v‡mi †¶‡Î cÖ‡hvR¨? [CU: 2006-07] A. O2 B. H2 C. N2 D. me¸wj Ans D 02. Cp Cv = R m¤ú‡K© ejv nq- [CU: 2006-07] A. e‡qjm m¤úK© B. Pvj©‡mi m¤úK© C. †gvjvi m¤úK© D. †evëm&g¨vbm m¤úK© Ans C
408 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. = 1.67 n‡j M¨vmwUi AYy KZ cvigvYweK n‡e? [KU: 2017-18] A. GK B. wØ C. wÎ D. eû Ans A IU 01. GKwU Av`k© M¨v‡m Cp Ges Cv Gi g‡a¨ m¤úK©- [IU: 16–17] A. Cp+ Cv = R B. CP + R = Cv C. R – Cv = CP D. CP – Cv = R Ans D 02. mKj wØ-cigvYyK M¨v‡mi †¶‡Î Gi gvb KZ ? [IU: 2012-13] A. 1.33 B. 1.4 C. 2.44 D. 1.66 Ans B STEP 06 ANALYSIS OF HSC BOARD QUESTION 01. = 5 3 Gi Rb¨ †KvbwU mwVK? [iv. †ev. 2016] A. Cp = 5 3 R B. CV = 5 3 R C. CV = 3 2 R D. CV = 2R S C info Cv = R –1 = R 5 3 – 1 = 3 2 R 02. wnwjqvg M¨v‡mi †ÿÎ Cp/Cv Gi AbycvZ wb‡Pi †KvbwU? [w`. †ev. 2021] A. 1.67 B. 1.41 C. 1.33 D. k~b¨ Ans A 03. Ne M¨v‡mi †ÿ‡Î -Gi gvb- [e. †ev. 2016] A. 1.33 B. 1.40 C. 1.67 D. 1.76 Ans C Concept 5 ZvcMwZwe`¨vi wØZxq m~Î I Zvc BwÄb ✅ 1854 mv‡j K¬wmqvm ZvcMwZ we`¨vi wØZxq m~‡Îi cÖ¯Ívebv K‡ib| ✅ 2q m~‡Îi wfwË (mvw` Kv‡b©vi wm×všÍ): Zvckw³‡K KL‡bvB m¤ú~Y©iƒ‡c Kv‡R cwiYZ Kiv hvq bv| ✅ K¬wmqv‡mi g‡Z, ÒevB‡ii †Kvb kw³i mvnvh¨ Qvov, †Kvb ¯^qswµq h‡š¿i c‡¶ wbgœ ZvcgvÎvq †Kvb e¯‘ n‡Z, D”P ZvcgvÎvi e¯‘‡Z Zv‡ci ¯’vbvšÍi m¤¢e bq|Ó ✅ ‡Kjwf‡bi g‡Z, Ò‡Kvb e¯‘‡K Zvi cvwicv‡k¦©i kxZjZg Ask n‡Z AwaKZi kxZj K‡i kw³i Aweivg mieivn cvIqv m¤¢e bq|Ó ✅ cøvsK Gi g‡Z, Ò‡Kvb Zvc Drm n‡Z Aweivgfv‡e Zvc †kvlY Ki‡e Ges Zv m¤ú~Y©iƒ‡c Kv‡R cwiYZ n‡e, Giƒc GKwU Zvc BwÄb •Zix Kiv m¤¢e bq|Ó ✅ Kv‡b©vi g‡Z, Ò‡Kvb wbw`©ó cwigvY Zvckw³ m¤ú~b©fv‡e hvwš¿K kw³‡Z iƒcvšÍi Ki‡Z mÿg hš¿ •Zix m¤¢e bq|Ó ASPECT SPECIAL: wewfbœ weÁvbxi g‡Z ZvcMwZwe`¨vi wØZxq m~‡Îi wee„wZ: K¬v‡mi evB‡i A‡K Ab¨ †Kvb RvqMvq cvB? K¬v‡mi evB‡i A‡K †Kvb RvqMvq cv B K¬wmqvm evB‡ii kw³ †Kjwfb Aweivg mieivn Kv‡b©v hvwš¿K kw³ cø¨v¼ Zvc BwÄb ✅ cÖZ¨veZ©x cÖwµqv (Reversible Process): †h cÖwµqv wecixZ gyLx n‡q cÖZ¨veZ©b K‡i Ges m¤§yLeZ©x I wecixZ gyLx cÖwµqvi cÖwZ ¯Í‡i Zvc I Kv‡Ri djvdj mgvb I wecixZ nq| D`vniY: cvwb eid ✅ AcÖZ¨veZ©x cÖwµqv (Irreversible process): †h cÖwµqv m¤§yLMvgx nIqvi ci wecixZgyLx n‡q cÖZ¨veZ©b Ki‡Z cv‡i bv, Zv‡K AcÖZ¨vMvgx cÖwµqv e‡j| D`vniY: Nl©‡Yi d‡j †h Zvc Drcbœ nq| ✅ cÖZ¨veZ©x I AcÖZ¨veZ©x cÖwµqvi g‡a¨ cv_©K¨: cÖZ¨veZ©x cÖwµqv AcÖZ¨veZ©x cÖwµqv 1. GwU AwZ axi cÖwµqv 1. GwU `ªæZ cÖwµqv 2. ¯^ZtùzZ© cÖwµqv bq 2. ¯^ZtùzZ© cÖwµqv 3. ms¯’v ZvcMZxq mvg¨ Ae¯’v eRvq iv‡L| 3. ms¯’v ZvcMZxq mvg¨ Ae¯’v eRvq iv‡L bv| 4. GB cÖwµqvq Ae¶xq djvdj `„ó nq bv| 4. GB cÖwµqvq Ae¶xq djvdj `„ó nq| 5. D`vnviY: ax‡i ax‡i msNwUZ m‡gvò Ges iæ×Zvcxq cÖwµqv, Dci n‡Z w¯’wZ ¯’vcK ej‡K w¯’wZ¯’vcK B¯úv‡Zi Dci †djv, w¯úªs Gi ms‡KvPb I cÖmvi‡Y K…ZKvR Ges eid cvwb‡Z A_ev cvwb ei‡d iƒcvšÍi| 5. D`vnviY: •e`y¨wZK †iv‡ai ga¨ w`‡q we`y¨r cÖev‡ni d‡j m„ó Zvc, Ryj _gmb cÖmviY, `ywU e¯‘i Nl©‡Y m„ó Zvc, †Kvb w¯cÖs‡K Lye `ªæZ m¤cÖmviY Ki‡j G‡Z m„ó K¤ú‡bi Øviv K…Z KvR AwaK ZvcgvÎvi e¯‘ †_‡K Kg ZvcgvÎvi e¯‘‡Z hvq| wKš‘ Kg ZvcgvÎvi e¯‘ †_‡K AwaK ZvcgvÎvi e¯‘‡Z hvevi †Kvb cÖebZv bvB| e›`yK n‡Z ¸wj †Qvo‡j eviæ‡`i we‡ùviY, e¨vcb, cwiPvjb Ges wewKiY| ✅ Zvcxq/Zvc BwÄb I Bwćbi `ÿZv: Zvc BwÄb: †h hvwš¿K e¨e¯’v Zvc kw³‡K hvwš¿K kw³‡Z cwiYZ K‡i Zv‡K Zvc BwÄb ejv nq| Zvc Drm BwÄb Zvc‡K hvwš¿K kw³‡Z iƒcvšÍi K‡i Zvc wbM©gb ewnt`©n BwÄb: ev®ú BwÄb AšÍt`©n BwÄb: †c‡Uªvj BwÄb Kg©`ÿZv = output input 100% hw` wm‡÷g Q1 cwigvY kw³ †kvlY K‡i, Q2 cwigvY kw³ Z¨vM K‡i Zvn‡j KvR W = Q1 – Q2 = Q1 – Q2 Q1 100% = 1 – Q2 Q1 100% [ev¯Í‡e mKj Zvc BwÄb AcÖZ¨vMvgx cÖwµqvq P‡j Ges GUvB mKj AcÖZ¨vMvgx Zvc Bwćbi Kg©`ÿZvi mgxKiY] Kv‡b©vi Av`k© Zvc BwÄb cÖZ¨vMvgx cÖwµqvq P‡j a‡i wb‡j, Q T = T1 – T2 T1 100% = 1 – T2 T1 100% ; Q1 Q2 = T1 T2 ✅ Bwćbi `ÿZv wnmve †_‡K jÿ Kiv hvq †h, Bnv †Kej Zvc Drm I Zvc MÖvn‡Ki ZvcgvÎv T1 – T2 Gi Dci wbf©i K‡i- Kvh©wbe©vnK e¯‘i cÖK…wZi Ici wbf©i K‡i bv|
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 409 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ✅ †h †Kv‡bv `ywU wbw`©ó ZvcgvÎvi g‡a¨ Kvh©iZ mKj cÖZ¨vMvgx Bwćbi Kg©`ÿZv mgvb nq| ✅ †h‡nZz T1 (T1 – T2), Kv‡RB Bwćbi `ÿZv KLbB 100% n‡Z cv‡i bv| ✅ Zvc Drm I ZvcMÖvn‡Ki ga¨eZ©x ZvcgvÎvi g‡a¨ cv_©K¨ hZ †ewk n‡e Bwćbi `ÿZvI ZZ †ewk n‡e| ✅ Kv‡b©v Pµ I Kv‡b©v BwÄb: Kv‡b©v Pµ Av‡jvPbv Kivi c~‡e© Kv‡b©v BwÄb m¤^‡Ü wKQzUv aviYv _vKv `iKvi| divwm weÁvbx mvw` Kv‡b©v (1892) mKj †`vl-ÎæwU gy³ GKwU Bwćbi cwiKíbv K‡ib| GwU GKwU Av`k© BwÄb hvi Kg©`ÿZv m‡e©v”P| ✅ Kv‡b©v P‡µi g~jbxwZ: Kv‡b©v P‡µ cÖZ¨vMvgx cÖwµqvi gva¨‡g Kvh©wbe©vnK e¯‘ Drm †_‡K Zvc MÖnY K‡i GKwU wbw`©ó Pvc, AvqZb I ZvcgvÎvq n‡Z Avi¤¢ K‡i GKwU m‡gvò cÖmviY I GKwU iæ×Zvcxq cÖmviY Ges GKwU m‡gvò ms‡KvPb I GKwU iæ×Zvcxq ms‡KvP‡bi gva¨‡g Zv‡ci wKQz Ask Kv‡R iƒcvšÍwiZ K‡i Ges evwK Ask Zvc MÖvn‡K eR©b K‡i Avw` Ae¯’vq wd‡i Av‡m| ◈ •ewkó¨: Kv‡b©v Pµ GKwU cÖZ¨vMvgx Pµ| wc÷b I †PvO ev wmwjÛv‡ii g‡a¨ †Kv‡bv Nl©Y †bB| Kvh©Kix c`v_© (M¨vm)-Gi Dci cÖhy³ cÖwµqv¸‡jv Lye ax‡i ax‡i msNwUZ nq| wc÷b I wmwjÛvi wbg©v‡Y Av`k© Zvc wb‡ivaK I AšÍiK I Av`k© Zvc cwievnx e¨envi Kiv nq Ges Zvc Drm I Zvc MÖvn‡Ki Dcv`v‡b Ggb AwZ D”P Zvc MÖnxZvhy³ Kiv nq †h m‡gvò cÖwµqv¸wj w¯’i ZvcgvÎvq msNwUZ nq| REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU Zvcxq BwÄb cÖwZwU P‡µi abvZ¥K KvR K‡i Ges Zvc nvivq wKš‘ BwÄbwU †Kvb Zvc MÖnY K‡i bv| BwÄbwU ZvcMwZwe`¨vi †Kvb m~·K j•Nb K‡i? [DU: 2018-19] A. k~b¨Zg m~Î B. cÖ_g m~Î C. wØZxq m~Î D. Z…Zxq m~Î Ans C STEP 02 ANALYSIS OF JU QUESTION 01. Zvc MwZwe`¨vi wØZxq m~‡Îi MvwYwZK iƒc n‡jvÑ [JU-A, Set-G. 20-21,19-20] A. dQ = ds T B. dS = dQ T C. dS = T dQ D. †Kv‡bvwUB bq S B info Zvc MwZwe`¨vi wØZxq m~Î: dS = dQ T = mlf T = mlv T STEP 03 ANALYSIS OF RU QUESTION 01. ZvcMwZwe`¨vi wØZxq m~Î n‡Z cvIqv hvq- [RU: 2016–17] A. kw³i msi¶YkxjZv B. RoZvi aviYv C. `kv cv_©K¨ D. GbUªwci aviYv Ans D STEP 04 ANALYSIS OF GST QUESTION PART A Analysis of General University Question IU 01. ‡KvbwU †c‡Uªvj Bwćbi •ewkó¨ bq ? [IU: 2006-07, 02-03] A. R¡vjvbx n‡jv †c‡Uªvj B. Pµ `yB hv‡Z m¤úbœ nq C. evqy Øviv KvR m¤úbœ K‡i D. Zvcxq `¶Zv †ekx Ans C PART B Analysis of Science & Technology Question SUST 01. GKwU Zvc Bwćbi `¶Zv wK‡mi Dci wbf©ikxj? [SUST: 2012-13] A. Dr‡mi ZvcgvÎv B. Zvc MÖn‡Ki ZvcgvÎv C. Drm I ZvcMÖvn‡Ki ZvcgvÎv D. Drm I ZvcMÖvn‡Ki ZvcgvÎvi cv_©K¨ E. gva¨‡gi cÖK…wZ Ans A STEP 05 ANALYSIS OF MEDICAL & DENTAL QUESTION AFMC 01. ZvcMwZwe`¨vi wØZxq m~Î cÖ`vb K‡ib †K? [AFMC. 2020-21] A. mvw` Kv‡b©v B. weÁvbx K¬wmqvm C. weÁvbx †Kjwfb D. g¨v· cøvsK S B info ÒevB‡ii †Kv‡bv kw³i mvnvh¨ Qvov, †Kv‡bv ¯^qswµq h‡š¿i c‡ÿ wb¤œ ZvcgvÎvi †Kv‡bv e¯‘ n‡Z, D”P ZvcgvÎvi †Kv‡bv e¯‘‡Z Zv‡ci ¯’vbvšÍi m¤¢e bq| A_ev Zvc ¯^Ztù~Z©fv‡e kxZjZi e¯‘ n‡Z D”P ZvcgvÎvi e¯‘‡Z cÖevwnZ n‡Z cv‡i bv|Ó (K¬wmqv‡mi gZB wbLyuZ I DbœZ)| wZwb GB wee„wZ 1854 mv‡j cÖKvk K‡ib| STEP 06 ANALYSIS OF HSC BOARD QUESTION 01. †Kvb m~·K Kv‡R jvwM‡q Zvcxq BwÄb I †iwd«Rv‡iUi •Zwi Kiv nq? [iv. †ev. 2016] A. ZvcMwZwe`¨vi k~b¨Zg m~Î B. ZvcMwZwe`¨vi 1g m~Î C. ZvcMwZwe`¨vi 2q m~Î D. ZvcMwZwe`¨vi 3q m~Î Ans C 02. Kv‡Y©v P‡µi wØZxq av‡c Kvh©wbe©vnK e¯‘iÑ [wm. †ev. 2021] i. Zv‡ci †kvlY N‡U ii. Pvc n«vm cvq iii. ZvcgvÎv n«vm cvq wb‡Pi †KvbwUmwVK? A. i I ii B. i I iii C. ii I iii D. i, ii I iii Ans C 03. Kv‡b©v Bwćbi †Kvb av‡c Zvc eR©b nq? [gv. †ev. 2018] A. cÖ_g B. wØZxq C. Z…Zxq D. PZz_© Ans C Concept 6 †iwd«Rv‡iUi I Kvh©K…Z mnM ✅ ‡h h‡š¿i mvnv‡h¨ cwi‡ek A‡c¶v Kg ZvcgvÎv m„wó Kiv hvq Ges GB ZvcgvÎv me©`v w¯’i Ae¯’vq ivLv hvq Zv‡K †iwd«Rv‡iUi ejv nq| ✅ g~jbxwZ: †iwd«Rv‡iUi wbgœ ZvcgvÎvi Drm n‡Z Zvc MÖnb ev Acmvib K‡i I D”P ZvcgvÎvi Avav‡i eR©b K‡i| K‡¤cÖmi GB hvwš¿K KvR K‡i| GLb, †iwd«Rv‡iU‡ii ev®úxfeb KzÛjx n‡Z AcmvwiZ Zvc Q1, K‡¤úªmi KZ…©K mieivnK…Z KvR W Ges Nbxfeb KzÛjx‡Z ewR©Z Zvc Q2 n‡j, kw³i wbZ¨Zv m~Î Abymv‡i cvIqv hvq, Q2 = Q1 + W W = Q2 – Q1 ✅ Kvh©K…Z mnM (Co-efficient of Performance ): †iwd«Rv‡iUi n‡Z AcmvwiZ Zvc I K‡¤úªmi KZ…©K mieivnK…Z hvwš¿K Kv‡Ri AbycvZ‡K Kvh©K…Z mnM e‡j| G‡K K Øviv cÖKvk Kiv nq| GLv‡b, K = AcmvwiZ Zvc mieivnK…Z Zvc W Q Q Q Q 1 2 1 1 K = Kvh©wbe©vnx mnM A_ev C.O.P. (Co-efficient of Performance ) Gi gvb 2 †_‡K 6 Gi g‡a¨ nq| Q1 = AcmvwiZ Zvc; Q2 = ewR©Z Zvc
410 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF HSC BOARD QUESTION 01. GKwU †iwd«Rv‡iU‡ii Kvh©K…Z mnM K = 2.5| GwU VvÐv cÖ‡Kvô n‡Z cÖwZ P‡µ 500 J Zvc AcmviY Ki‡j, cÖwZ P‡µ mieivnK…Z KvR KZ n‡e? [Xv. †ev. 2016] A. 2450 J B. 502.5 J C. 500 J D. 200 J S D info W = Q2 K = 500 2.5 = 200J 02. †iwd«Rv‡iUi cÖ‡Kv‡ô iwÿZ Lv`¨`ªe¨ n‡Z M„nxZ Zvc Q2 Ges cwi‡e‡k ewR©Z Zvc Q1 n‡j Kvh©mnM K Gi gvb n‡jvÑ [wm. †ev. 2021] A. R = Q1 Q2 – Q1 B. K = Q1 Q1 – Q2 C. K = Q2 Q1 – Q2 D. R = Q2 Q2 – Q1 Ans C Concept 7 GbUªwc msµvšÍ ✅ Avwe®‥viK: K¬wmqvm, 1854 mv‡j| ✅ GbUªwc: GKK JK–1 iƒ×Zvcxq cÖwµqvq e¯‘i †h Zvcxq ag© w¯’i _v‡K Zv‡K GbUªwc e‡j| mvg¨ve¯’vq GbUªwci gvb m‡e©v”P| GbUªwc n‡”Q kw³i †mB iƒc hv‡K Avi KL‡bv Kv‡R iƒcvšÍi Kiv hv‡e bv| AcÖZ¨vMvgx cÖwµqvq GbUªwc e„w× cvq| cÖZ¨vMvgx cÖwµqvq †h †Kvb ms¯’vi GbUªwc w¯’i _v‡K| ✅ Zvrch©: GwU e¯‘i ZvcMZxq Ae¯’v wba©vi‡b mnvqZv K‡i| GbUªwc GKwU cÖvK…wZK ivwk hvi gvb Zvc I cig ZvcgvÎvi Abycv‡Zi mgvb| ZvcgvÎv I Pv‡ci b¨vq G‡K Abyfe Kiv hvq bv| GwU e¯‘i GKwU Zvcxq ag© hv Zvc mÂvj‡bi w`K wb‡`©k K‡i| AcÖZ¨vMvgx cÖwµqvq GbUªwc e„w× cvq| AšÍ¯’ kw³i cig gvb A‡c¶v, AšÍ¯’ kw³i cwieZ©bB AwaK Zvrch© c~Y©| G›Uªwc †ewk M¨vwm Ae¯’vq, me‡P‡q Kg KwVb Ae¯’vq| kw³ iƒcvšÍ‡ii A¶gZvB GbUªwc Ges GbUªwc e„w× †c‡j e¯‘i w¯’wZkxjZv n«vm cvq| GwU ZvcgvÎv, Pvc, AvqZb, AšÍwb©wnZ kw³, Pz¤^Kxq Ae¯’vi b¨vq †Kvb e¯‘i Ae¯’v cÖKvk K‡i| GbUªwc e„w× †c‡j e¯‘ k„sLj Ae¯’v (ordered state) n‡Z wek„sLjv Ae¯’vq (disordered state) cwiYZ nq| GbUªwci cwieZ©b me©`v abvZ¡K Ges mvg¨ve¯’vq GbUªwc m‡ev©”P nq| GbUªwc n‡”Q wm‡÷‡gi wek„sLjZvi gvcKvwV| Zvcxq g„Zy¨: RM‡Z GbUªwc µgvMZ ev‡o‡Q| RM‡Zi GbUªwc hLb m‡ev©‡”P †c․Qv‡e ZLb me wKQyi ZvcgvÎv GK n‡q hv‡e| d‡j Zvckw³‡K Avi hvwš¿K kw³‡Z iæcvšÍwiZ Kiv hv‡e bv| GB Ae¯’v‡K †Kjwfb RM‡Zi Zvcxq g„Zz¨ bv‡g AwfwnZ K‡ib| ✅ w¯’i ZvcgvÎvq GbUªwci cwieZ©b: ◈ ds = mLf T = mLv T ◈ L = Av‡cwÿK myß Zvc [Latent Heat] ◈ eid Mj‡bi Av‡cwÿK myß Zvc, Lf = 336000 J/Kg = 3.36 105 J/Kg [Latent Heat of Fusion] ◈ cvwbi ev®úxfe‡bi Av‡cwÿK myß Zvc Lv = 2268000 J/Kg = 2.268 106 J/Kg [Latent Heat of Vaporization] ✅ cwiewZ©Z ZvcgvÎvq GbUªwci cwieZ©b, ◈ ds = ms ln T2 T1 , ◈ T2 = †kl ZvcgvÎv; ◈ T1 = Avw` ZvcgvÎv| ◈ eid = 2100 Jkg1 k 1 S = Av‡cwÿK Zvc cvwb = 4200 Jkg1 k 1 ev®ú = 2000 Jkg1 k 1 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. mycviKÛvKUi mvaviY KÛvKU‡ii †P‡q †ewk myk„sLj| hw` mycviKÛvKUi Ges mvaviY KÛvKUi Ae¯’vq GbUªwc h_vµ‡g Ss Ges Sn nq Z‡e wb‡¤œi †KvbwU mwVK? [DU: 2013-14] A. Ss = Sn B. Ss > Sn C. Ss < Sn D. Ss Sn Ans C 02. GbUªwc wK‡mi cwigvc wb‡`©k K‡i? [DU: 2011-12; CU: 2016-17, 2009-10, CU: 2013-14, BR: 2011-12, BRUR: 2012-13] A. †gvU Zvc (total heat) B. myk„•LjZv (order) C. wek„•LjZv (disorder) D. ZvcgvÎv (temperature) Ans C STEP 02 ANALYSIS OF RU QUESTION 01. G›Uªwci †ÿ‡Î †KvbwU mwVK bq? [RU. Moderna, Set-2. 20-21] A. wm‡÷‡gi wek„•Ljvi cwigvcK B. Zvc I Pv‡ci b¨vq Abyfe Kiv hvq C. wek¦RM‡Zi G›Uªwc µgvMZ e„w× cv‡”Q D. cÖZ¨vMvgx cÖwµqvi G›Uªwc w¯’i _v‡K S B info G›Uªwc Zvc I Pv‡ci b¨vq Abyfe Kiv hvq bv| GwU n‡jv wek„•Ljvi cwigvcK| 02. GbUªwc K‡g GiKg D`vniY †KvbwU? [RU: 2015-16] A. KvV cyov‡j B. eid‡K cvwb‡Z cwiYZ Ki‡j C. †iwd«Rv‡iU‡ii wfZ‡ii evZv‡m D. cvwb‡Z jeY ¸jv‡j KvV cyov‡j GbUªwc e„w× cvq; eid cvwb‡Z cwiYZ n‡j GbUªwc e„w× cvq 03. kw³i AcP‡qi aviYv †`b †K? [RU: 2015-16] A. jW© †Kjwfb B. IqvUmb C. †g·I‡qj D. cøv¼ Ans A STEP 03 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. ax‡i ax‡i msNwUZ m‡gv I iæ×Zvcxq (adiabatic) cwieZ©‡b GbUªwci cwieZ©b (s) KZ? [JnU: 2017-18] A. s = 0 B. s > 0 C. s < 0 D. Dc‡ii me¸‡jv ax‡i ax‡i msNwUZ m‡gv I iæ×Zvcxq cwieZ©b cÖZ¨vMvgx cÖwµqv | ZvB GbUªwci cwieZ©b ds = 0|
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 411 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. GbUªwc (Entropy) Gi GKK †KvbwU? [JnU: 13-14, PbSTU: 2015-16; Xv. †ev. 2016, wm. †ev. 2016, e. †ev. 2017] A. wgUvi/wK‡jvMÖvg B. KT–1 C. JK–1 D. molK–1 Ans C 03. hw` GKwU wm‡÷g T cig ZvcgvÎvq dQ cwigvb Zvc MÖnY ev eR©b K‡i Z‡e GbUªwci cwieZ©b dS Øviv cÖKvk Kiv nq- [JnU: 2012-13] A. dS = dQ H B. dS = dQ dT C. dS = dQ T D. dS = T dQ Ans C KU 01. †Kvb c×wZi wek„sLj Ae¯’v e„w× †c‡j Dnvi †Kvb ivwkwU e„w× cvq? [KU: 2012-13] A. GbUªwc B. KvR C. kw³ D. †d‡iv‡P․¤^K c`v_© Ans A 02. G›Uwci cwieZ©b dS cÖKvk Kiv nq- [KU: 2011-12, 2009-10] A. dS = dQ T B. dS = dS T C. dS = T dQ D. dS = dQ dT Ans A 03. GbUªwc n‡jv- [KU: 2008-09] A. kw³i iƒcvšÍ‡ii ¶gZv B. f‡ii iƒcvšÍ†ii ¶gZv C. kw³i iƒcvšÍ†ii A¶gZv D. e‡ji iƒcvšÍ†ii ¶gZv Ans C 04. ‡Kvb wm‡÷‡gi ZvcgvÎv mv‡c‡¶ Zvc cwieZ©‡bi nvi Øviv wK‡mi cwieZ©b cwigvc Kiv nq? [KU: 2003-04] A. Zvcxq `¶Zv B. GbUªwc C. myßZvc D. Av‡cw¶K Zvc Ans B IU 01. KwVb Ae¯’vq GbUªwci gvb- [IU: 2016–17; w`. †ev. 2017, Xv.‡ev. 2015] A. †ewk B. Kg C. mgvb D. me©”Qv Ans B 02. ‡Kvb e¨e¯’vi djcÖmy Kvh©cÖvwßi m¤¢ebv GbUªwci gv‡bi- [IU: 2013-14] A. mgvbycvwZK B. e¨¯ÍvbycvwZK C. Kv‡h©i m‡½ m¤ú©‡K †bB D. †KvbwUB bq Ans A 03. GbUªwc e„w× †c‡Z †c‡Z hLb m‡e©v”P gv‡b †c․Qv‡e ZLb we‡k¦i mKj e¯‘- [IU: 2009-10] A. Zij cwibZ n‡e B. Zvcxq mvg¨ve¯’vq DcbxZ n‡e C. M¨vm cwibZ n‡e D. †KvbwUB bq Ans B PART B Analysis of Science & Technology Question MBSTU 01. GbUªwc me‡P‡q †ekx- [MBSTU: 14-15; RU: 2017-18] A. KwVb Ae¯’vq B. Zij Ae¯’vq C. M¨vmxq Ae¯’vq D. †KvbwUB bq Ans C BSMRSTU 01. GKgyLx cÖwµqvq GbUªwc- [BSMSTU: 14–15] A. e„w× cvq B. n«vm cvq C. gvb k~b¨ nq D. †Kvb cwieZ©b N‡U bv GKgyLx ev AcÖZ¨vMvgx cÖwµqvq GbUªwc e„w× cvq| HSTU 01. GbUªwci cwieZ©b me©`vÑ [HSTU: 2015-16] A. abvZ¥K B. FYvZ¥K C. k~b¨ D. †KvbwUB bq Ans A STEP 04 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. GKwU Kv‡bv©P‡µ †gvU GbUªwci cwieZ©b n‡jv- [BUET: 2011-12] A. Zero B. 1 2 1 2 T T C. less than zero D. greater than zero Ans A STEP 05 ANALYSIS OF MEDICAL & DENTAL QUESTION MAT 01. ‡Kvb Dw³wU mZ¨? [MAT: 2006-07] A. †Kvb wm‡÷‡gi kw³ iƒcvšÍ‡ii A¶gZv‡K GbUªwc e‡j| B. †h mKj †f±‡ii gvb k~b¨ bq Zv‡`i bvj †f±i e‡j| C. ‣iwLK †e‡Mi gvÎv LT–2 D. †Kvb e¯‘i Z¡ib e¯‘i Dci cÖhy³ wbUe‡ji e¨v¯ÍvbycvwZK| Ans A AFMC 01. gnvwe‡k¦i GbUªwc w`b w`bÑ [AFMC. 2020-21, R.U 09-10] A. evo‡Q B. Kg‡Q C. AcwiewZ©Z D. †Kv‡bvwUB bq Ans A STEP 06 ANALYSIS OF HSC BOARD QUESTION 01. Kv‡b©v P‡µi 1g av‡ci †ÿ‡Î wb‡Pi †KvbwU mwVK? [iv. †ev. 2016; P. †ev. 2015] A. ZvcgvÎv e„w× cvq B. ZvcgvÎv w¯’i _v‡K C. Af¨YÍixY kw³ e„w× cvq D. Zvc ewR©Z nq Ans B 02. GbUªwc n‡jv- [h.‡ev. 2016] A. k„•Ljvi cwigvc B. kw³i iƒcvšÍi ÿgZvi cwigvc C. iƒcvšÍ‡ii Rb¨ kw³ cvIqvi cwigvc D. Zvcxq g„Zz¨i m¤¢vebvi cwigvc Ans D 03. ¯^Ztù~Z© cwieZ©‡b- [wm. †ev. 2017] A. G›Uªwc I wek„•Ljv n«vm cvq B. G›Uªwc I k„•Ljv e„w× cvq C. G›Uªwc I k„•Ljv n«vm cvq D. G›Uªwc I wek„•Ljv e„w× cvq Ans D STEP 2 mg„× †ewmK MvwYwZK cÖ Ö‡qvM (MATH) kU©KvU †UKwbK Concept 1 ZvcgvÎv †¯‹j I ÎæwUc~Y© _v‡g©vwgUvi msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. C 5 = F-32 9 = K-273 5 = Rn-492 9 = Rm 4 02. cv_©‡K¨i †ÿ‡Î ZvcgvÎvi wewfbœ †¯‥‡ji g‡a¨ m¤úK©: 5 9 5 C F K = 9 R = 4 R m 03. 100 C ice x steam x ice x x c 04. †¯‥‡ji mv‡_ µwUc~Y© _v‡g©vwgUvi Gi m¤úK©: C 100 = F-32 180 = k-273 100 = R-492 180 = Xt Xice Xsteam Xice Note: ◈ dv‡ibnvBU I †Kjwfb †¯‥‡j GKB cvV 574.25F Or 574.25 K ◈ †mw›U‡MÖW I dv‡ibnvBU †¯‥‡j GKB cvV 40C or 40F ◈ †mw›U‡MÖW I †Kjwfb †¯‥j KLbB mgvb n‡e bv| Model Example01 †Kvb ZvcgvÎvq †Kjwfb Ges dv‡ibnvBU †¯‥‡j GKB cvV cvIqv hv‡e? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] Avgiv Rvwb, C 5 = F–32 9 = K – 273 5 = Rn – 492 9 †h‡nZz †Kjwfb I dv‡ibnvBU †¯‥‡ji Zzjbv Kiv n‡q‡Q| K – 273 5 = F–32 9 [K = F = x] K – 273 5 = F–32 9 9x – 5x = 2457 – 160 4x = 2297 x = 574.25F dv‡ibnvBU I †Kjwfb †¯‥‡ji GKB cvV cvIqv hvq 574.25K ev 574.25F (wbw`©ógvb)
412 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Model Example02 †h ZvcgvÎvq †mjwmqvm I dv‡ibnvBU †¯‥‡j GKB cvV cvV`vb Ki‡e, †m ZvcgvÎvwU n‡”QGeneral Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] Avgiv Rvwb, C 5 = F–32 9 = K – 273 5 = Rn – 492 9 †h‡nZz †Kjwfb I dv‡ibnvBU †¯‥‡ji Zzjbv Kiv n‡q‡Q| C 5 = F–32 9 x 5 = x – 32 9 [C = F = x] 95x = 5x – 160 4x = –160 x = – 40 †mw›U‡MÖW I dv‡ibnvBU †¯‥j GKB cvV †`q –40C ev –40F (wbw`©ó gvb) CONCEPTUAL MATH MEx 01 20C ZvcgvÎv dv‡ibnvBU †¯‥‡j KZ n‡e? 5 C = 9 F - 32 5 20 = 9 F - 32 F = 68° F MEx 02 †Kvb GKwU †iva _v‡g©vwgUv‡ii †iva 0 0C Ges 1000C ZvcgvÎvq h_vµ‡g 12 Ges 24 | _v‡g©vwgUviwU‡K GKwU Mig †Z‡ji mv‡_ ¯’vcb Ki‡j †iva 36 nq| †Z‡ji ZvcgvÎv wbY©q Ki? 100 1 0 R oo Ro R R 100 200 C 24 12 36 12 0 NOW START PRACTICE 01. GKwU µzwUc~Y© _v‡g©vwgUvi mvavib evq~ Pv‡c MwjZ ei‡d 4°C Ges ï®‥ ev‡®ú 98°C cvV †`q| _v‡g©vwgUviwU 51°C cvV w`‡j cÖK…Z cvV KZ? 02. dv‡ibnvBU †¯‥‡j †Kvb e¯‘i ZvcgvÎv 500F n‡j †Kjwfb †¯‥‡j H ZvcgvÎv n‡e03. gqgbwmsn kn‡ii kxZ I MÖx®§Kvjxb ZvcgvÎvi cv_©K¨ 150C n‡j dv‡ibnvBU †¯‥‡j GB cv_©K¨ KZ n‡e? 04. †Kvb ZvcgvÎvq dv‡ibnvBU †¯‥‡ji cvV †mjwmqvm †¯‥‡ji cv‡Vi wظb n‡e? 05. dv‡ibnvBU †¯‥‡ji ‡Kvb ZvcgvÎv †mjwmqvm †¯‥‡ji ZvcgvÎvi cuvP¸b n‡e? NOW PRACTICE SOLVE : 01. B M S M 100 0 C 0 ev, 98 4 51 4 100 C ev, 94 47 100 C ev, C 50C 02. 9 32 5 273 T F ev, 9 50 32 5 273 T ev 9 18 5 273 T ev T 283K 03. C 5 = F 9 15 5 = F 9 F = 27F 04. 9 2 32 5 x x ev, x F o C o 160 320 . 05. 9 5x 32 5 x 25x 9x = 160 x = 10C 5x = 500 F REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. †Kvb ZvcgvÎv †mjwmqvm I dv‡ibnvBU Dfq †¯‥‡j GKB msL¨v w`‡q cÖKvk Kiv hvq? [DU. 2017-18, 2000-01, 1995-96; CU: 2012-13; 10-11; RU:2013-14;06-07] A. 40 B. 32 C. 40 D. 12 5 C = 9 F32 Let, C = F= x 5 x = 9 x 32 x = – 40 – 40˚C Ges – 40˚F 02. GKwU ÎæwUc~Y© _v‡g©vwgUv‡ii eid we›`y 5C Ges ÷xg we›`y 99C| hLb G _v‡g©vwgUv‡i 52C cÖ`k©b K‡i ZLb dv‡ibnvBU †¯‥‡j ZvcgvÎv KZ? [2007-08] A. 132F B. 122F C. 302F D. 322F 180 F32 = 100 0 0 X X X X = 99 5 52 5 ZvcgvÎv F =122˚F STEP 02 ANALYSIS OF JU QUESTION 01. 0C I 100C ZvcgvÎvq GKwU †iva _v‡g©vwgUv‡ii †iva h_vµ‡g 11 I 16 | _v‡g©vwgUviwU GKwU Pzjvq Zi‡ji ùzUbv‡¼ ivL‡j †iva cvIqv hvq 36 | Zi‡ji ùzUbv¼ wbY©q Ki| [JU-H, Set-A. 20-21] A. 20C B. 500C C. 250C D. 10C S D info = R – R0 R100 – R0 100 = 36 – 11 16 – 11 100 = 500C 02. ZvcgvÎvq †¯‥‡ji m¤úK©wU n‡jv [JU. 2013-14] A. C 5 = F 9 = K 5 B. C 5 = F – 32 9 = K – 273 5 C. C 9 = F – 32 9 = 273 – K 5 D. C 5 = F 32 = R 9 Ans B 03. 27C ZvcgvÎvq †Kvb wbw`©ó cwigvY M¨vm nVvr cÖmvwiZ n‡q wظY AvqZb jvf K‡i| P~ovšÍ ZvcgvÎv KZ? ( =1.4) [JU. 11-12, 09-10; KU: 06-07; RU: 06-07] A. 45.65C B. – 45.65C C. – 35.05C D. –15.65C T1V1 –1 = T2V2 –1 P~ovšÍ ZvcgvÎv, T2 = V1 V2 –1 T1 = (0.5)0.4300 = 227.357K = – 45.65˚C 04. GKwU †iva _v‡g©vwgUv‡ii †iva 0˚C ZvcgvÎvq 8 Ges 100˚C ZvcgvÎvq 20| _v‡g©vwgUviwU‡K GKwU Pzjøx‡Z ¯’vcb Ki‡j †iva 32 nq| Pzjøxi ZvcgvÎv KZ ? [JU. 2011-12, RU. 18-19] A. 300˚C B. 100˚C C. 200˚C D. 400˚C 100 C = 100 0 0 R R R R 100 C = 20 8 32 8 _v‡g©vwgUv‡i ZvcgvÎv †mw›Uª‡MÖW †¯‥‡j, C = 200˚C
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 413 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 05. wbw`©ó f‡ii GKwU Av`k© M¨v‡mi AvqZb aªæe Pv‡c wظY Kiv n‡jv| hw` M¨v‡mi cÖv_wgK ZvcgvÎv 130C nq Z‡e P~ovšÍ ZvcgvÎv KZ? [JU. 2004-05] A. 7.50C B. 2990C C. 130C D. 260C T2= 2 x (273+13) = 572K = 2990C STEP 03 ANALYSIS OF RU QUESTION 01. GKLÛ B‡Ui NbZ¡ 2gm/cc Dnvi IRb 6kg. A‡a©K cvwb‡Z wbgw¾Z ivL‡j IRb KZ n‡e? [RU. 2015-16] A. 2.5kg B. 3.5kg C. 4.5 kg D. 5 kg B‡Ui NbZ¡ = 2gm/cc; B‡Ui AvqZb = 6 1000 2 = 3000 cc BU KZ…©K AcmvwiZ cvwbi AvqZb = 3000 cc A‡a©K BU KZ…©K AcmvwiZ cvwbi AvqZb = 3000 2 = 1500cc A‡a©K BU KZ…©K AcmvwiZ cvwbi IRb = 1500gm = 1.5kg A‡a©K BU cvwb‡Z wbgw¾Z ivL‡j IRb = 6 – 1.5 = 4.5 kg 02. †Kvb ZvcgvÎvq d‡ibnvBU I †Kjwfb †¯‥‡j GKB gvb cvIqv hvq- [RU. 2014-15, CUET: 2012-13, KUET: 2010-11; w`. †ev. 2015] A. 322o B. 273o C. 650o D. †KvbwUB bq x – 273 5 = x – 32 9 ev, 9x–2457 =5x–160 ev, 4x = 2297 ev, x =574.25 03. 15˚C ZvcgvÎv KZ wWwMÖ dv‡ibnvB‡Ui mgvb? [RU. 2009-10] A. 47˚C B. 59˚ C. 64˚C D. †KvbwUB bq 5 C = 9 F32 dv‡ibnvBU †¯‥‡j ZvcgvÎv, F = 5 C 9 + 32 = 5 15 9 + 32 = 59˚F 04. GKwU ÎæwUc~Y© _v‡g©vwgUvi ¯^vfvweK Pv‡c ei‡di 1C Ges ev‡®ú 98C cvV †`q| D³ _v‡g©vwgUvi 40C cvV w`‡j cÖK…Z ZvcgvÎv KZ? [RU. 08-09, JUST. 15-16] A. 32.3˚C B. 40.2˚C C. 48.4˚C D. 52.6˚C C 100 = X – X0 X100 – X0 C 100 = 40 – 1 98–1 _v‡g©vwgUv‡i ZvcgvÎv †mw›U‡MÖW †¯‥‡j, C = 40.2˚C STEP 04 ANALYSIS OF CU QUESTION 01. dv‡ibnvBU †¯‥‡j ZvcgvÎv 212 n‡j †mjwmqvm †¯‥‡j Gi gvb KZ? [CU. 15-16] A. 50 B. 100 C.150 D. 200 E. 250 C 5 = F – 32 9 ev, C 5 = 212 – 32 9 ev, C 5 = 180 9 ev, C = 100 02. GKwU K‡¶ Mig cvwbc~Y© GKwU cvÎ ivLv nj| GwU 80°C †_‡K 75°C G kxZj n‡Z T1 †m‡KÛ, 75°C †_‡K 70°C G kxZj n‡Z T2 †m‡KÛ Ges 70°C †_‡K 65°C G kxZj n‡Z T3 †m‡KÛ jvM‡j, wb‡Pi †Kvb ivwkwU mwVK? [CU. 2015-16] A. T1 <T2<T3 B. T1 > T2 > T3 C. T1 < T2 > T3 D. T1 + T2 = T3 E. T1 = T2 = T3 Ans A 03. †mjwmqvm †¯‥‡j ZvcgvÎv 0 wWwMÖ n‡j d‡ibnvBU †¯‥‡j GB gvb KZ? [CU. 14-15] A. 0 B. 32 C. 80 D. 100 E. 212 C 5 = F 32 9 0 5 = F 32 9 F = 32 STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. †mjwmqvm †¯‥‡j ZvcgvÎvi cwieZ©b 35C n‡j dv‡ibnvBU †¯‥‡j Gi cwieZ©b KZ n‡e? [2009-10] A. 63F B. 53F C. 73F D. 83F dv‡ibnvBU †¯‥‡j ZvcgvÎvi cv_©K¨, F = 1.8 C = 1.8 35 = 63˚F IU 01. GKwU †iva _v‡g©vwgUv‡ii †iva 0C I 100C ZvcgvÎvq h_vµ‡g 10I 20| _v‡g©vwgUviwU GKwU Pzwjø†Z ¯’vcb Kivq †iva 35 nq| Pzwjøi ZvcgvÎv KZ? [IU: 2016–17] A. 200C B. 225C C. 250C D. 275C 100 C = 100 0 0 R R R R 100 C = 20 10 35 10 C = 250C BRUR 01. dv‡ibnvBU †¯‥‡j †Kvb e¯‘i ZvcgvÎv 50°F| †Kjwfb †¯‥‡j D³ e¯‘i ZvcgvÎv- [BRUR: 2014-15] A. 273K B. 293K C. 283K D. 298K F – 32 9 = T – 273 5 ev, 50 – 32 9 = T – 273 5 ev,T = 283K PART B Analysis of Science & Technology Question SUST 01. GKwU ÎæwUc~Y© †mjwmqvm _v‡g©vwgUvi MwjZ ei‡d 10˚ Ges dzUšÍ cvwb‡Z 90˚ cvV †`q| †Kvb ZvcgvÎvq GwU mwVK cvV w`‡e? [SUST: 2009-10] A. 0˚ B. 25˚ C. 50˚ D. 80˚ 100 C = 100 0 0 X X X X = 90 10 C 10 ZvcgvÎv C = 50˚C 02. ‡Kvb& ZvcgvÎvq dv‡ibnvB‡Ui gvb †mjwmqv‡mi gv‡bi wظY? [SUST: 2006-07, JUST: 2015-16] A. – 40C B. 100C C. 160C D. 320C Let, C = x F = 2x; Now, 5 C = 9 F32 5 x = 9 2x 32 x = 160˚ x = 160˚C 2x = 320C MBSTU 01. GKwU ÎæwUc~Y© _v‡g©vwgUvi mvaviY evqy Pv‡c MwjZ ei‡d 4°C Ges ï®‥ ev‡®ú 98°C cvV †`q| _v‡g©vwgUviwU 51°C cvV w`‡j cÖK…Z cvV KZ? [2015-16] A. 51°C B. 50°C C. 52°C D. 49°C cvV-wbgœ w¯’ivsK DaŸ© w¯’ivsK-wbgœw¯’ivsK = C – 0 100 – 0 ev, 51 – 4 98 – 4 = C 100 ev, 47 92 = C 100 ev, C = 50 02. dv‡ibnvBU †¯‥‡j †Kvb e¯‘i ZvcgvÎv 98°F n‡j, †mjwmqvm †¯‥‡j e¯‘wUi ZvcgvÎv KZ? [2014-15] A. 20.67°C B. 30.67°C C. 36.67°C D. 40.67°C C 5 = F – 32 9 ev, C 5 = 98 – 32 9 ev, C = 36.67°C PSTU 01. gvbe‡`‡ni ¯^vfvweK ZvcgvÎv 98.4°F| †mjwmqvm †¯‥‡j GB ZvcgvÎv KZ n‡e? [PSTU-2014-15] A. 36.90C B. 360C C. 370C D. 36.80C C 5 = F – 32 9 ev, C 5 = 98.4 – 32 9 ev, C = 36.9°C STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION CUET 01. ‡Kvb Zvcgvb hš¿ 0°C ZvcgvÎvq 0.5°C cvV †`q Ges 100°C ZvcgvÎvq 100.8°C cvV †`q| 26°C ZvcgvÎvq hš¿wU KZ cvV w`‡e? [CUET: 12-13] A. 25°C B. 27°C C. 26.57°C D. None 26 100 = T – 0.5 100.8 – 0.5 T = 26.578C
414 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES BUTex 01. †mw›U‡MÖW †¯‥‡j 95.5 dv‡ibnvBU ZvcgvÎvi gvb KZ? [MAT; 19-20] A. 45.456C B. 35.944C C. 36.544C D. 36.944C S D info C 5 = F 32 9 C = (F 32) 5 9 C = (98.5 32) 5 9 C = 36.944C STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. 98.6F ZvcgvÎvi mgZzj¨ _v‡g©vWvqbvwgK ZvcgvÎv KZ? [P. †ev. 2021; g. †ev. 2021] A. 310.16 K B. 345.16 K C. 393.16 K D. 408.16 S A info K 273.16 5 = F 32 9 K = 5 9 (F 32) + 273.16 = 5 9 (98.6 32) + 273.16 = 310.16K 02. dv‡ibnvBU †¯‥‡ji EaŸ© w¯’ivsK I wb¤œ w¯’ivs‡Ki e¨eavb KZ? [gv. †ev. 2017] A. 32 B. 100 C. 180 D. 212 Ans C Concept 2 Zvc Øviv K…ZKvR msµvšÍ FORMULA 01. dw = PdV 02. W = JH; ZvcMwZ we`¨vi 1g m~Î: (kw³i wbZ¨Zv m~Î) 03. dQ = dU + dW: Note: i. Zv‡ci †kvlY n‡j dQ ( ve) ii. Zv‡ci eR©b dQ ( ve) iii. e¨e¯’v Øviv KvR dw(+ve) iv.e¨e¯’vi Dci KvR dw(–ve) hw` wm×všÍ †bqv bv hvq A_©vr e¨e¯’v Øviv ev e¨e¯’vi Dci D‡jøL bv _vK‡j dw(+ve); AšÍt¯’ kw³ e„w× †c‡j dU(+)ve; AšÍt¯’ kw³ n«vm †c‡j dU(–)ve CONCEPTUAL MATH MEx 01 †Kvb ms¯’v cwi‡ek †_‡K 800J Zvckw³ †kvlb Kivq Gi AšÍ¯’ kw³ 500J e„w× †cj| ms¯’v KZ…©K cwi‡e‡ki Dci m¤úvw`Z Kv‡Ri cwigvb wbY©q Ki| dQ = dU + dW ev, 800 = dU + 500 ev, dW = 300J MEx 02 wcóbhy³ GKwU wmwjÛv‡i wKQz M¨vm Ave× Av‡Q| M¨v‡mi Pvc 400 Pa w¯’i †i‡L wm‡÷‡g ax‡i ax‡i 800 J Zvckw³ mieivn Kiv nj Ges 1200 J KvR m¤úvw`Z nq| M¨v‡mi AvqZb I AšÍt¯’ kw³i cwieZ©b wbY©q Ki| dQ=dW + dU or, dU=dQ – dW = 800-1200 = -400J; dW = PdV ev, dV = dW P = 400 1200 = 3m3 MEx 03 †Kvb wm‡÷g 1500 J Zvc †kvlY K‡i Ges 82J KvR m¤úv`b K‡i| wm‡÷‡gi AšÍt¯’ kw³i cwieZ©b n‡e dU = dQ – dW = 1500 – 82 = 1418 J NOW START PRACTICE 01. †Kvb e¨e¯’v 1200 J Zvc ‡kvlb K‡i Ges e¨e¯’vi Dc‡i 400 J KvR m¤úvw`Z nq| e¨e¯’vi Dci AšÍt¯’ kw³i cwieZ©b wbY©q Ki| 02. 1.0 105 Nm-2 w¯’i Pv‡c †Kvb Av`k© M¨v‡mi AvqZb 0.04 m3 ‡_‡K cÖmvwiZ n‡q 0.08 m 3 n‡jv| ewnt¯’ Kv‡Ri cwigvY wbY©q Ki| 03. GK evqygÛjxq Pv‡c GKwU Av`k© M¨vm‡K DËß K‡i 0.01m3 AvqZb e„w× Kiv nj| G‡Z m¤úvw`Z Kv‡Ri cwigvY KZ? NOW PRACTICE SOLVE : 01. dU = dQ – dW = 1200 – (400) =1600 J 02. ewnt¯’ Kv‡Ri cwigvY, dW = PdV = 1.0 105 (0.08 0.04) = 4000 J 03. dW = PdV dW = 105 0.01 = 1 103 J REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GK evqygÛxq Pv‡c GKwU Av`k© M¨vm‡K DËß K‡i 0.01 m3 AvqZb e„w× Kiv nj| G‡Z m¤úvw`Z Kv‡Ri cwigvY- [DU. 2008-09; CU: 2013-14; RU: 2017-18, JU. 07-08] A. 7.6 103 J B. 76 J C. 1 102 J D. 1 103 J m¤úvw`Z Kv‡Ri cwigvY, W=PV= 105 0.01 =1103 J 02. †Kv‡bv e¯‘i Zvc aviY ¶gZv 150 J/0C n‡j Gi ZvcgvÎv 800C n‡Z 200C-G bvwg‡q Avb‡Z wK cwigvY Zvckw³ †ei Ki‡Z n‡e?[DU. 2001-02] A. 3000 J B. 90000J C. 12000 J D. 15000 J †`Iqv Av‡Q, C = 150 J0C –1 , Zvckw³i cwigvY, Q = C = 150(800–200) = 90000 J STEP 02 ANALYSIS OF JU QUESTION 01. eid Mj‡bi myß Zvc n‡jv- [JU. 2014-15] A. 180103 J kg1 B. 336103 J kg1 C. 310103 J kg1 D. 370103 J kg1 Ans B 02. 1 kg cvwb‡K 1 evqygÛjxq Pv‡c ev‡®ú cwiYZ Ki‡Z Af¨šÍixY kw³i cwieZ©b wbY©q Ki| (Rjxq ev‡®úi myß Zvc 2.268106 Jkg–1 I 1kg Rjxq ev‡®úi AvqZb = 1.671 m3 ) [JU. 2009-10] A. 2.099106 J B. 2.300106 J C. 2.190106 J D. 2.400106 J U = Q – W = mLv – PV = 12.268106 – 1.013105 1.671 = 2.099106 J 03. ‡Kvb wm‡÷g 1500 J Zvc †kvlY K‡i Ges 82J KvR m¤úv`b K‡i| wm‡÷‡gi AšÍt¯’ kw³i cwieZ©b n‡e- [JU. 2006-07] A. 1418J B. 1232J C. 1582J D. 1237J dU= dQ – dW = 1500–82 = 1418 J STEP 03 ANALYSIS OF RU QUESTION 01. hw` 2 K¨v‡jvwi Zvc m¤ú~Y©fv‡e Kv‡R iƒcvšÍwiZ nq Z‡e Kv‡Ri cwigvY KZ? [RU: 2017-18 ; Kz.‡ev. 2016] A. 4.2 J B. 4.8 J C. 8.2 J D. 8.4 J W= JH= 4.2 2 = 8.4
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 415 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 02. 10 mole M¨v‡mi iæ×Zvcxq cÖmvi‡Yi mgq 100j KvR m¤úvw`Z nq| D³ e¨e¯’vi AšÍt¯’ kw³i cwieZ©b KZ? [RU.2013-14] A. 1000J B. 100J C. -100J D. 0J iæ×Zvcxq cÖmvi‡Yi †¶‡Î, dQ =0 du = –dW = – 100J STEP 04 ANALYSIS OF CU QUESTION 01. 1 cal Zvc m¤ú~Y©fv‡e Kv‡R iƒcvšÍwiZ n‡Z Kx cwigvY KvR m¤úbœ nq? [CU-A, Set-2. 20-21 KUET: 09-10] A. 1J B. 2.4J C. 4.2J D. 4.8J S C info 1cal = 4.2J 02. 1.01105 N/m2 w¯’i Pv‡ci †Kvb Av`k© M¨v‡mi AvqZb 0.04m3 †_‡K cÖmvwiZ n‡q 0.05m3 n‡jv| ewnt¯’ Kv‡Ri cwigvb- [CU. 02-03; SUST. 02-03] A. 1 J B. 30 J C. 100 J D. 1000 J W = PV = 1105 (0.05–0.04) =1105 0.01=1000 J STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. ¯^vfvweK Pv‡c 100 m3 AvqZ‡bi GKwU M¨v‡m 5103 J Zvc w`‡j M¨v‡mi AvqZb 100.2 m3 nq Z‡e H M¨v‡mi K…Z KvR n‡e- [ KU. 2007-2008] A. 20260 J B. 38340 J C. 22384 J D. 25380 J W = P v = 1.013 105 (100.2 – 100) = 20260J 02. 1kg eid hLb 0 0C ZvcgvÎvq cvwb‡Z cwibZ nq ZLb GbUªwci e„w× nq- [KU. 2006-07] A. 1231 KJ-1 B. 3112 JK-1 C. 1213 JK-1 D. 2113 JK-1 ds = mL T = 1 336 103 273 = 1230.769 = 1231JK–1 PART B Analysis of Science & Technology Question MBSTU 01. †Kvb e¨e¯’v aªæe AvqZ‡b 300J Zvc eR©b K‡i| e¨e¯’vi AšÍ:¯’ kw³i cwieZ©b n‡e- [MBSTU. 2014-15] A. – 150J B. – 300J C. 300J D. – 450J dQ = dU + dW; GLv‡b dQ = –300 Kvib Zvc eR©b K‡i| ev, – 300 = dU ev, dU = – 300J STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION KUET 01. GKwU †gvUi Mvwoi Uvqvi 270C ZvcgvÎv I 2 evqy gÛjxq Pv‡c Av‡Q| hw` UvqviwU nVvr †d‡U hvq Zvi P~ovšÍ ZvcgvÎv wbY©q Ki? [KUET: 2008-09] A. 270C B. 3000C C. 246.10C D. 2730C Ans A CUET 01. ‡Kvb system cwi‡ek †_‡K 800 J Zvckw³ †kvlb Kivq Gi AšÍ¯’kw³ 500J e„w× †cj| system KZ…©K cwi‡e‡ki Dci m¤úvw`Z Kv‡Ri cwigvb n‡jv| [CUET: 2012-13 JU. 18-19] A. 300 J B. 600 J C. 800 J D. 500 J E. 100J dQ = dU + dW ev, dW = dQ–dU=800–500 = 300J STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. †kvwlZ Zvc Q = 700 J Ges m¤úvw`Z KvR W = 200 J n‡j †Kv‡bv wm‡÷‡gi Af¨šÍixY kw³ KZ e„w× cv‡e? [w`. †ev. 2016] A. 900 J B. 700 J C. 600 J D. 500 J S D info U = Q – W = (700 – 200) J = 500 J DÏxc‡Ki Av‡jv‡K wb‡Pi 02 I 03 bs cÖ‡kœi DËi `vI: [h. †ev. 2016] 20C dW=2J V,Q V,Q+dQ V+dV Q+dQ 20C 20C A B C 02. dQ = 5J n‡j A n‡j B †Z AšÍt¯’ kw³i cwieZ©b KZ? A. –3 J B. 0 J C. 3 J D. 7 J S C info dU = dQ – dW = 5J – 2J = 3J 03. hw` wZb Ae¯’vq wm‡÷gwUi AšÍt¯’kw³ h_vµ‡g UA, UB, UC nq Z‡e †KvbwU mwVK? A. UA = UB = UC B. UC = UB > UA C. UB < UC = UA D. UA = UB < UC Ans B 04. ax‡i ax‡i Pvc e„w× Kivq †Kv‡bv wm‡÷‡gi Pvc 2Pa n‡Z 4Pa n‡jv| G‡ÿ‡Î mg AvqZb cÖwµqvq wm‡÷‡gi Af¨šÍixY kw³ 200 J e„w× †c‡jv| wm‡÷‡giÑ [Kz. †ev. 2021] i. mieivnK…Z Zvc 200J ii. K…ZKvR k~b¨ iii. ZvcgvÎv e„w× cv‡e wb‡Pi †KvbwUmwVK? A. i I ii B. ii I iii C. i I iii D. i, ii I iii Ans D 05. †Kvb wm‡÷‡gi Dci evwn¨K ej Øviv 500 J KvR m¤úv`b Kivq wm‡÷g n‡Z 300 J Zvc kw³ †ewi‡q †Mj| wm‡÷‡gi AšÍt¯’ kw³i cwieZ©b KZ? [Kz. †ev. 2017] A. – 800 J B. – 200 J C. + 200 J D. + 800J S C info dQ = – 300 J . dW = – 500 J dU = dQ – dW = + 200 J DÏxcKwU jÿ Ki Ges 06 I 07 bs cÖ‡kœi DËi `vI : P = 3.5 105 Nm–2 dQ X Y wP‡Î wmwjÛv‡i iwÿZ 1 mole M¨v‡m dQ Zvc mieivn Kivq wc÷b X Ae¯’vb n‡Z Y Ae¯’v‡b Av‡m| G‡Z AšÍ¯’tkw³ 207 J n«vm cvq| wc÷‡bi cÖ¯’‡”Q‡`i †ÿÎdj = 0.1 m2 ; X I Y Gi `~iZ¡ = 5 10–2m. 06. m¤úbœ K…ZKvR KZ? [wm. †ev. 2021] A. 1.75 103 J B. 1.75 105 J C. 7 105 J D. 7 107 J S A info dw = Pdv = 3.5 105 0.1 5 102 [v = A l ] = 1750 J = 1.75 103 J 07. wmwjÛv‡i mieivnK…Z Zvckw³ dQ Gi cwigvY njÑ [wm. †ev. 2021] A. 7.002 105 JB. 6.998 105 J C. 1.957 103 J D. 1.543103 J S D info dQ = du + dw = 207 + 1.75 103 [dQ = n«vm cvq] = 1.54 103 J 08. 40C ZvcgvÎvq 1 mole O2 M¨vm‡K ax‡i ax‡i cÖmvwiZ K‡i AvqZb wظY Ki‡j m¤úbœ K…ZKvR n‡jvÑ [wm. †ev. 2017] A. 230.4 J B. 664.8 J C. 1802.9 J D. 5202.1 J S C info W = RT ln V2 V1 = 8.31 313 ln 2V V = 1802.9 J Concept 3 kw³i iƒcvšÍi msµvšÍ FORMULA 01. mv ms 2 2 1 ; v = 2s; = v 2 2s 02. mgh = ms = gh s = h 428.6 [∵ cvwbi Av‡cwÿK Zvc 4200 J kg1K 1 ] 03. Zvc, H = mLf ; H = mLv; H = ms ; Lf = eid Mj‡bi Av‡cwÿK myß Zvc = 3.36 105 Jkg1 ; Lv = ev®úxfe‡bi Av‡cwÿK myß Zvc = 2.26 106 Jkg1 04. h D”PZv n‡Z eid c‡o m¤ú~b© M‡j †M‡j D”PZv, h = Lf g ; h D”PZv n‡Z eid c‡o x% M‡j †M‡j D”PZv, h = Lf g x 100 05. †÷dv‡bi m~Î: E = T 4 06. fx‡bi m~Î: mT = b [fx‡bi aªæeK, b = 2.89 × 103 mk]
416 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES CONCEPTUAL MATH MEx 01 GKwU RjcÖcv‡Zi cvwb 4200m Dci n‡Z wb‡P cwZZ nq| Rjvk‡qi kxl©‡`k I Zj‡`‡k cvwbi ZvcgvÎvi cv_©K¨ KZ? = h 428.6 = 4200 428.6 = 9.8 K MEx 02 KZ D”PZv n‡Z co‡j ei‡di 1% M‡j hv‡e? [g = 10 ms–2 ] h = Lf g x 100 = 3.36 105 10 100 = 360 m MEx 03 †Kvb GKwU K¨vjwiwgUv‡ii cvwbmg 7 kg| Gi ZvcgvÎv 5C n‡Z 25C-Gi DbœxZ Ki‡Z Zv‡ci cwigvY wbY©q Ki| M„nxZ Zvc, Q = msw =7(25–5) =140 J MEx 04 GKwU Zvgvi K¨vjwiwgUv‡ii fi 0.3kg Ges Gi Dcv`v‡bi Av‡cw¶K Zvc 400 Jkg–1K –1 | Gi ZvcgvÎvi cwieZ©b KZ n‡j 6000J Zv‡ci cÖ‡qvRb n‡e? cÖ‡qvRbxq Zvc, Q = ms ZvcgvÎvi cwieZ©b, = 0.3 400 6000 ms Q = 50K MEx 05 GKwU K…òe¯‘ 327C ZvcgvÎvq ivLv Av‡Q| KZ Zi½‣`‡N©¨i kw³ K…òe¯‘wU wewKiY Ki‡e? GLv‡b, fx‡bi m~Î: mT = b m T = 2.898 × 103 [fx‡bi aªæeK, b = 2.89 × 103 mk] m = 2.898 × 103 600 = 4.83 × 106m NOW START PRACTICE 01. 150m DuPz GKwU Rj cÖcv‡Zi Zj‡`k I kxl©‡`‡ki ZvcgvÎvi cv_©K¨ †ei Ki| hw` cZbkxj cvwbi mg¯Í kw³B Zv‡c cwibZ nq| 02. 126 ms-1 ‡eM cÖvß GKwU mxmvi ey‡jU †Kv_vI _vwg‡q †`qvi d‡j mg¯Í MwZ kw³ Zv‡c cwibZ nj| ey‡jUwUi ZvcgvÎv KZ e„w× cv‡e? (mxmvi Av‡cw¶K Zvc 126 J kg-1 k -1 ) 03. GKwU MwZ m¤úbœ mxmvi ey‡jU‡K nVvr _vwg‡q †`qvi d‡j Gi mg¯Í MwZkw³ Zv‡c cwiYZ n‡jv Ges ey‡j‡Ui ZvcgvÎv 160°C e„w× †cj| ey‡j‡Ui †eM KZ wQj? (mxmvi Av‡cw¶K Zvc 125Jkg1 k 1 ) 04. 2kg eid Mwj‡q cvwb‡Z cwiYZ Ki‡Z Kx cwigvY Zv‡ci cÖ‡qvRb? 05. 0C ZvcgvÎvi 1kg eid‡K 0˚C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z KZ Zv‡ci cÖ‡qvRb? NOW PRACTICE SOLVE : 01. = h 428.6 = 150 428.6 = 0.35 K/ C 02. mv ms 2 1 2 ev 63k 2 126 (126) 2S v 2 2 03. 2s v 2 ev, v 2s 2 v 2s v 2125160 = 200 ms-1 04. H = mLf = 2 3.36 105 = 6.72 105 J 05. †gvU cÖ‡qvRbxq Zvc, Q = mLf =13.36105 = 33.6104 J REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GK Kvc Mig Kwd‡K 80C ZvcgvÎv †_‡K 30C ZvcgvÎvq VvÛv Kiv nj| KvcwUi Zvc aviKZ¡ 2.0 kJK1 n‡j kxZjxKiY cÖwµqvq KZ Zvc wbM©Z nj? [DU. 2008-09] A. 0.04 kJ B. 60 kJ C. 100 kJ D. 160 kJ Q = ms = C=2000(80 – 30) = 100,000J=100 kJ STEP 02 ANALYSIS OF JU QUESTION 01. 4200 m DuPz GKwU RjcÖcv‡Zi Zj‡`k I kxl©‡`‡ki g‡a¨ ZvcgvÎvi e¨eavb KZ n‡e hw` cZbkxj cvwbi mg¯Í kw³B ZvcgvÎv e„wׇZ e¨q nq| [JU. 2015-16] A. 20C B. 9.8C C. 15.6C D. †KvbwUB bq w¯’wZ kw³ Zvc kw³‡Z iƒcvšÍwiZ n‡j, mgh= ms = s gh = 4200 9.8 4200 = 9.8K = 9.8˚C 02. 100 wgUvi DuPz GKwU RjcÖcvZ n‡Z cvwb wb‡P cwZZ nq| Dc‡ii I wb‡Pi cvwbi ZvcgvÎvi cv_©K¨ n‡e- [JU. 2012-13] A. 0.822C B. 0.234C C. 1.220C D. 0.880C w¯’wZ kw³ Zvc kw³‡Z iƒcvšÍwiZ n‡j, = 428 h = 428 100 = 0.234K= 0.234C STEP 03 ANALYSIS OF RU QUESTION 01. GKwU K…ò e¯‘i ZvcgvÎv 1000 K n‡j Zv n‡Z KZ kWm2 nv‡i Zvckw³ wewKiY n‡e? [ = 5.67 108 Wm2K 4 ] [RU-C, Set-1. 19-20] A. 0.567 B. 5.67 C. 56.7 D. 567 E = T 4 = 5.67 10–8 (1000)4 = 56700 Wm–2 = 56.7 kWm–2 02. 5g f‡ii GKwU ey‡jU †Kvb †`qv‡j evav cÖvß n‡j Zvi ZvcgvÎv 160K e„w× cvq| Ab¨ †Kvb fv‡e Zvc bó bv n‡j ey‡j‡Ui †eM KZ wQj? (mxmvi AvtZvc = 125 j kg-1 K -1 ) [RU. 2010-11] A. 137 ms-1 B. 141.42 ms-1 C. 200 ms-1 D. 500 ms-1 1 2 mv2 = ms ev, v = 2s = 2 125 160 = 200ms–1 03. 0˚C ZvcgvÎvi 1kg eid‡K 0˚C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z KZ Zv‡ci cÖ‡qvRb? [RU. 2010-11] A. 33.6104 J B. 9.17106 J C. 336 J D. †KvbwUB bq †gvU cÖ‡qvRbxq Zvc, Q = mLf =13.36105 = 33.6104 J
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 417 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 04 ANALYSIS OF CU QUESTION 01. K…òe¯‘i ZvcgvÎv wظY e„w× Ki‡j wewKiY nvi KZ¸Y e„w× n‡e? [2015-16] A. 16 B. 2 C. 4 D. 10 E. 32 E2 E1 = T2 T1 4 ev, E2 = 2 1 4 ev, E2 = 16 E 02. 1kg cvwbi ZvcgvÎv 1k e„w× Ki‡Z cÖ‡qvRbxq Zvc- [2014–15] A. 4.2J B. 4.5J C. 3.36105 J D. 4200J E. 450J H = ms = 1 4200 1 = 4200 J STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question KU 01. GKLÐ eid Dci †_‡K f~wg‡Z cwZZ n‡jv| G‡Z cZb kw³i 50% Zv‡c iƒcvšÍwiZ nIqvq eid LÐwUi GK-PZz_©vsk M‡j †Mj| LÐwU KZ km D”PZv n‡Z cwZZ n‡qwQj? [KU. 2018-19] A. 1.714 B. 8.57 C.17.14 D. 34.28 0.5 mgh = 1 4 mLf 0.5 9.8 h = 1 4 336000 h = 17.14km 02. 200ms–1 †eM cÖvß GKwU mxmvi ey‡jU †Kv_vI _vwg‡q †`qvi d‡j mg¯Í MwZkw³ Zv‡c iƒcvšÍwiZ nj| ey‡j‡Ui ZvcgvÎv KZ e„w× cv‡e? (mxmvi Av‡cw¶K Zvc 126Jkg–1 k –1 ) [KU. 2013-14] A. 158.73K B. 158.73 J/kg C. 108.73 J/kg D. 108.73K 1 2 mv2 = ms ev, = v 2 2s = (200) 2 2 126 = 158.73K 03. 5gm weï× cvwbi ZvcgvÎv 5˚C e„w× Ki‡Z Zv‡ci cÖ‡qvRb-[KU. 2011-12] A. 1 Cal B. 5 Cal C. 25 Cal D. 25 J M„nxZ Zvc, Q= msw = 510-3 42005=105 J =25Cal PART B Analysis of Science & Technology Question JUST 01. m~h© cÖwZ †m‡K‡Û KZ kw³ wewKiY K‡i? [JUST-A, Set-Ka 19-20] A. 6 1026 J B. 8 1026 J C. 3 1026 J D. 4 1026 J m~h© cÖwZ †m‡K‡Û 4×1026J kw³ wewKiY K‡i| STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION CKRuet. Combind 01. Avnbvd 250 gm f‡ii 0C ZvcgvÎvq GKwU eid LÐ GKwU wbw`©ó D”PZv †_‡K †d‡j w`j| gvwU‡Z covi ci kw³i msiÿYkxjZv bxwZi Kvi‡Y Drcvw`Z Zv‡ci Rb¨ eid LÐwU 10% M‡j †Mj| Avnbvd KZ D”PZv †_‡K eid LÐwU †d‡jwQj? [CKRUET. 2020-21] A. 4428.57 m B. 3428.57 m C. 3227.60 m D. 3957.57 m E. 3528.9m S B info eid L‡Ûi fi m ; M‡j †M‡Q = m Gi 10% = m 10 GLb, mgh = m 10 × lf h = lf 10g = 3.30 × 105 10 98 m = 3428.57m [ei‡di Mjb myßZvc (latent heat of fusion) lf = 334 Jg–1 = 334 103 Jkg–1 = 3.34 105 Jkg–1 ] STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. 500 m DuPz RjcÖcv‡Zi Zj‡`k I kxl© †`‡ki cvwbi ZvcgvÎvi cv_©K¨ KZ n‡e? [g = 10 ms2 , cvwbi Av‡cwÿK Zvc = 4200 Jkg1K 1 [e. †ev. 2016] A. 0.50C B. 1.19C C. 5.0C D. 50C S B info ms = mgh m 4200 = m 10 500 = 1.19C Concept 4 ZvcMZxq cÖwµqv msµvšÍ FORMULA 01. m‡gvò cÖwµqv, P1V1 = P2V2 †Kvb As‡K nVvr kãwU D‡jøL _vK‡j m~‡Î (Mvgv) e¨envi Ki‡Z n‡e, KviY cÖwµqvwU iy×Zvcxq| †Kvb As‡K ax‡i ax‡i kãwU D‡jøL Kiv _vK‡j H As‡Ki m~‡Î e¨envi Kiv hv‡e bv| KviY cÖwµqvwU m‡gvò| 02. iæ× Zvcxq cÖwµqv, P1V1 = P2V2 ; T1P1 1–/ = T2P2 1–/ ; T1V1 –1 = T2V2 –1 [GK cigvYyK M¨v‡mi Rb¨ = 1.66; wØcigvYyK M¨v‡mi Rb¨ = 1.41; eû cigvYyK M¨v‡mi Rb¨ = 1.33] 03. iæ× Zvcxq M¨vm aªæeK, = V P C C ; †gvjvi M¨vm aªæeK, R = Cp – Cv ; Cp = R 1 ; Cv = R –1 04. m‡gvò cÖwµqvq KvR, W = nRT ln V2 V1 05. iæ×Zvcxq cÖwµqvq KvR, W = nR 1 (T1 T2) = Cv(T1 – T2) [GK †gvj M¨v‡mi †ÿ‡Î] CONCEPTUAL MATH MEx 01 ¯^vfvweK ZvcgvÎv I Pv‡c wKQy ï®‥ evqy‡K m‡gvò cÖwµqvq wظY AvqZ‡b cÖmvwiZ Kiv n‡jv| P‚ovšÍ Pvc KZ? P1V1 = P2V2 P‚ovšÍ Pvc, P2 = 2 1 V V P1 = 1 1 2V V (1.013105 ) = 5.065104 Nm–2 MEx 02 Av`k© ZvcgvÎv I Pv‡c wbw`©ó AvqZ‡bi ï®‥ M¨vm‡K (i) m‡gvò Ae¯’vq Ges (ii) iæ× Zvc Ae¯’vq wZb¸Y AvqZ‡b cÖmvwiZ n‡Z †`Iqv nj| cÖwZ‡¶‡Î P‚ovšÍ Pvc KZ n‡e wbY©q Ki| [ = 1.4] (i) P1V1 = P2V2 P‚ovšÍ Pvc, P2 = 2 1 V V P1= 1 1 3V V (1.013105 )= 3.377104 Nm–2 (ii) P1V1 =P2V2 P‚ovšÍ Pvc, P2 = 2 1 V V P1= 1.4 1 1 3V V (1.013105 )=2.17104Nm–2
418 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES MEx 03 27C ZvcgvÎvq 0.02 kg nvB‡Wªv‡Rb M¨vm‡K m‡gvò cÖwµqvq msbwgZ K‡i cÖv_wgK AvqZ‡bi GK PZz_©vsk Kiv n‡jv| K…ZKv‡Ri gvb KZ? W = nRT ln V2 V1 = m M RT ln V2 V1 = 0.02 8.314 300 ln 1 4 2 10–3 = – 34576.95 J FYvZ¥K wPý Øviv cwi‡ek KZ…©K K…ZKvR wb‡`©k K‡i| MEx 04 CO2 M¨v‡mi Rb¨ w¯’i AvqZ‡b I w¯’i Pv‡c †gvjvi Av‡cwÿK Zvc wbY©q Ki| ( = 1.33 Ges R = 8.31 Jmol–1 K –1 ) Cv = R – 1 = 8.31 1.33 – 1 = 25.18 Jmol–1 K –1 Cp = Cv + R Cp = 25.18 + 8.31 = 33.49 Jmol–1 K –1 NOW START PRACTICE 01. 25°C ZvcgvÎvq I evqygÛjxq Pv‡c Ave× ï®‥ evqy‡K nVvr ev iƒ×Zv‡c msbwgZ K‡i AvqZb A‡a©K Kiv nj| Pvc wbY©q Ki| 02. evqy†K iy×Zv‡c cÖmvwiZ K‡i Gi AvqZb wZb¸Y Kiv n‡jv| hw` cÖv_wgK Pvc 1 evqygÛjxq Pvc nq Zvn‡j P~ovšÍ Pvc KZ n‡e? NOW PRACTICE SOLVE : 01. P1V1 = P2V2 ev, P2 = 1 2.64 1 2 . 1.4 1 2 1 P v v evqygÛjxq Pvc 02. P1V1 P2V2 1 2 1 2 .P V V P 5 1.4 1.01 10 3 1 = 2.176104 Nm–2 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. ¯^vfvweK ZvcgvÎv I Pv‡c wKQy cwigvY ï®‥ evqy‡K m‡gvò cÖwµqvq msbwgZ K‡i AvqZb A‡a©K Kiv n‡jv| P‚ovšÍ Pvc KZ n‡e? [DU. 2000-01] A. 4.0410–2 Nm–2 B. 2.20 Nm–2 C. 4.04105 Nm–2 D. 2.02105 Nm–2 P1V1 = P2V2 P~ovšÍ Pvc, P2 = 2 1 V V P1 = V1 1 2V1 (1.013105 ) = 2.02105 Nm2 STEP 02 ANALYSIS OF JU QUESTION 01. ¯^vfvweK ZvcgvÎv I Pv‡c wKQz evqy‡K m‡gvò cÖwµqvq wظb AvqZ‡b cÖmvwiZ Kiv n‡j, PzovšÍ Pvc- [JU. 2013-14] A. 500×105Nm–2 B. 505×102Nm–2 C. 505×103Nm–2 D. 502×104Nm–2 P2 = P1 2 ev, P2 = 1.01325 105 2 = 505 102 Nm–2 02. ¯^vfvweK ZvcgvÎv I Pv‡ci †Kvb Av`k© M¨vm‡K iæ×Zvcxq cÖwµqvq msKzwPZ K‡i AvqZb A‡a©K Kiv n‡j, P~ovšÍ Pvc KZ n‡e? [JU. 2009-10] A. 20.057 m cvi` Pvc B. 2.0057m cvi` Pvc C. 40.057m cvi` Pvc D. 4.0057m cvi` Pvc P1V1 =P2V2 P~ovšÍ Pvc, P2 = V1 V2 P1 = (2)1.4 0.76 = 2.0057 m cvi` Pvc 03. Av`k© ZvcgvÎv I Pv‡c wKQy cwigvY ï®‥ evZvm‡K iæ×Zvc cÖwµqv‡Z Dnvi Avw` AvqZ‡bi GK cÂgvs‡k msKzwPZ Kiv n‡jv| ZvcgvÎv e„w× wbY©q Ki| ( = 1.4) [JU. 2009-10] A. 246.69˚C B. 446.69˚C C. 146.89˚C D. 246.89˚C T1V1 –1 = T2V2 –1 P~ovšÍ ZvcgvÎv, T2 = V1 V2 1.41–1 T1 = (5)0.4 273 = 519.69K ZvcgvÎv e„w×, T = T2 – T1 = 519.69– 273 = 246.69K = 246.69C STEP 03 ANALYSIS OF RU QUESTION 01. GKwU Kv‡b©v BwÄb 700C ZvcgvÎvq Zvc MÖnY K‡i Ges 350C ZvcgvÎvq Zvc eR©b K‡i| BwÄb cÖwZ P‡µ 1 kcal Zvc MÖnY Ki‡j, cÖwZ P‡µ m¤úvw`Z Kv‡Ri cwigvY KZ? [RU: 2018-19] A. 2.1kJ B. 4.2 kJ C.1 kJ D. 0.5kJ 700–350 700 = W 42 W = 2.1 02. 400C ZvcgvÎvq †Kvb wbw`©ó cwigvb M¨vm‡K iæ×Zvcxq cÖwµqvq msKzwPZ K‡i AvqZb A‡a©K Kiv n‡j P~ovšÍ ZvcgvÎv KZ? [RU. 2006-07] A. 140 0C B. 1500C C. 1600C D. 1800C T2 = v1 v2 –1 T1 = 2 1 0.4 313 = 413K = 140C STEP 04 ANALYSIS OF CU QUESTION 01. m‡gvò cÖwµqvq GK MÖvg †gvj †Kvb M¨vm 127C G Gi AvqZb wظY nIqv ch©šÍ cÖmvwiZ nq| G‡ÿ‡Î †gvU K…Z KvR: [CU. 2018-19] A. 239 Cal B. 239 Joule C. 549 Joule D. 549 Cal W = nRTlnv2 v1 = 1 8.314 400ln2=2305.13J=548.84Cal 02. ¯^vfvweK ZvcgvÎv I Pv‡c wKQz cwigvb M¨vm‡K nVvr msKzwPZ K‡i AvqZb‡K GK PZz_©vsk Kiv n‡j ZvcgvÎv KZ¸b e„w× cv‡e? [CU. 2006-07, SUST. 09-10] A. 1.74 B. 4 C. 6.96 D. 16 T2 = v1 v2 –1 T1 = 4 1 1.4–1 T2 = 1.74T1 STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of General University Question JnU 01. 30C ZvcgvÎvq †Kvb M¨v‡mi Dci iæ× Zvc cÖwµqvq Pvc wظY Kiv nj| ZvcgvÎv e„w× wbY©q Ki| ( 1.4 ) [JnU. 2008-09] A. 66.36 K B. 369.36 K C. 69.36 K D. 96.36 K S A info T1P1 1 = T2P2 1 P~ovšÍ ZvcgvÎv, T2 = 1 2 1 P P T1 = 1.4 0.4 2 1 (30+273) = 369.36 K ZvcgvÎv e„w×, T = T2 – T1 = 369.36 K – 303 K = 66.36 K
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 419 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES KU 01. 27C ZvcgvÎvi GKwU Uvqvi‡K cv¤ú Ki‡Z Ki‡Z Zvi Pvc 2 evqygÛjxq Pv‡ci mgvb nIqvi mv‡_ mv‡_ †mwU †d‡U †Mj| P‚ovšÍ ZvcgvÎv KZ? ( = 1.4) [KU. 2013-14] A. 44.3C B. 92.3C C. 33.3C D. 11.3C T1 P1 1– = T2 P2 1– P~ovšÍ ZvcgvÎv, T2 = 1 2 1 P P T1 = 1.4 0.4 2 1 (27+273) = 365.3K = 92.3C 02. 270C ZvcgvÎvq †Kvb wbw`©ó cwigvb M¨vm nVvr cÖmvwiZ n‡q wظb AvqZb jvf K‡i, P~ovšÍ ZvcgvÎv n‡e- [KU. 2006-07] A. 59.92 K B. 597.2 K C. 69.72 K D. †KvbwUB bq T2 = v1 v2 –1 T1 = 1 2 14–1 300 = 227.35k BRUR 01. w¯’i DòZvq KZ Pvc cÖ‡qvM Ki‡j GKwU M¨v‡mi AvqZ‡bi ¯^vfvweK Pv‡ci AvqZ‡bi 4 ¸Y nq| [BRUR:2011-12] A. 1.5 atm B. 1.42 atm C. 0.142 atm D. 2.56 atm P2 = v1 v2 P1 = 1 4 1.4 1 = 0.143atm PART B Analysis of Science & Technology Question SUST 01. m‡gvò cÖwµqvq (T = 400K) 4 †gvj Av`k© M¨v‡mi AvqZb V1 †_‡K e„w× †c‡q V2 = 2V1 n‡jv| M¨vmwU KZ©„K K…Z KvRKZ? [SUST: 18-19; KU:2012-13] A. 9216 J B. 2304 J C. –2304J D. –1329J E. 4000 J m‡gvò cÖwµqvq K…Z KvR nRTlnv2 v1 = 4 8.31 400ln 2 1 = 9213 9216 J 02. evqy‡Z iæ×Zv‡c cÖmvwiZ K‡i Gi AvqZb 5 ¸Y Kiv nj| hw` cÖv_wgK Pvc 1 evqygÛjxq Pvc nq Zvn‡j P~ovšÍ Pvc KZ N/m2 n‡e? ( = 1.4) [SUST: 2016-17, JUST-C, 19-20] A. 1.06104 B. 3.36104 C. 4.13104 D. 5.36104 E. 7.56104 P1V1 =P2V2 P~ovšÍ Pvc, P2 = 2 1 V V P1 = 1.4 5 1 1105 = 1.06104 N/m2 03. GKwU wmwjÛv‡ii 1Atm Pv‡c Ave× 10 wjUvi M¨vm‡K nVvr K‡i evowZ Pvc w`‡q 5 Liter Ki‡Z n‡j KZ Pvc w`‡Z n‡e? =1.4 [SUST: 11-12] A. 1.4Atm B. 2.0Atm C. 2.64.4Atm D. 7.0Atm E. 10.0Atm P1V1 = P2V2 ev, P2 = V1 V2 P1 = 10 5 1.4 1 = 2.639 = 2.64Atm 04. iƒØ Zv‡c 1 evqy gÛjxq Pv‡c ivLv M¨vm‡K cÖmvwiZ K‡i wظb Kiv n‡j †h P~ovšÍ Pvc nq, m‡gvò cÖwµqvq †mB GKB Pvc †c‡Z n‡j M¨vm‡K KZ ¸Y cÖmvwiZ Ki‡Z n‡e| [SUST: 08-09, P. †ev. 2021, g. †ev. 2021] A. 1.4 B. 2.63 C. 5.2 D. 7.8 P2 = V1 V2 P1 = 1 2 1.4 1 = 0.378 V2 = P1 P2 V1 = 1 0.378 V1 = 2.63V1 BSMRSTU 01. ¯^vfvweK ZvcgvÎv I Pv‡c †Kvb Av`k© GK cvigvYweK M¨vm‡K iæ×Zvcxq cÖwµqvq msKzwPZ K‡i AvqZb GK Aógvsk Kiv n‡j, P~ovšÍ Pvc KZ n‡e? [BSMRSTU-C, 19-20] A. 32 atm B. 33 atm C. 34 atm D. 35 atm P1V1 = P2V2 P2 P1 = V1 V2 P2 P1 = V V 8 P2= 81.67 1 = 32.2 atm STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. iæ×Zvc cÖwµqvq (=1.4) wØ-cigvYy M¨v‡mi Rb¨ Pvc 0.5% e„w× Kiv n‡j M¨v‡mi AvqZb Kg‡e ? [BUET. 2010-11] A. 0.5% B. 0.70% C. 1.0% D. 0.36% P1V1 = P2V2 ev, V2 V1 = P1 P2 ev, V2 = P1 P2 1 .V = 100 100 + 100 0.5 100 1 1.4 . v = 0.9964v v = (1 – 0.9964) 100% = 0.36% 02. 87.230C ZvcgvÎvq †Kvb wbw`©ó cwigvb M¨vm nVvr cÖmvwiZ K‡i Gi AvqZb wظb Kiv n‡jv| P~ovšÍ ZvcgvÎv n‡e- [BUET. 2009-10] A. 00C B. 450C C. 100C D. 250C T2 = v1 v2 –1 T1 = 1 2 1.4–1 (273 + 87.23) = (0.5)0.4 360.23 = 273k = 0C KUET 01. 137C ZvcgvÎvq †Kvb wbw`©ó cwigvb M¨vm nVvr cÖmvwiZ n‡q 5 ¸Y AvqZb jvf Kij| PzovšÍ ZvcgvÎv KZ n‡e? [ = 1.4] [KUET:2018-19] A. –215C B. – 137C C.–58C D. 58C E. 137C T1V1 –1 = T2V2 –1 T2= T1 ( ) V1 V2 –1 = 215.375K –58C (cÖvq) 02. iæ×Zvcxq cÖwµqvq evqyi AvqZb e„w× †c‡q wظY n‡jv| cÖvi‡¤¢i Pvc GK evqyPvc, P~ovšÍ Pvc KZ? ( = 1.4) [KUET: 2012-13] A. 0.49 B. 0.93 C. 0.38 D. 0.83 E. 0.15 S C info P1V1 = P1V2 ev, P2 = ( ) V1 V2 P1 = ( ) 1 2 1.4 1 = 0.38atm 03. 0°C ZvcgvÎvq †Kvb wbw`©ó M¨vm‡K nVvr cÖmvwiZ K‡i AvqZ‡b wظY Kiv n‡jv| P~ovšÍ ZvcgvÎv KZ? ( = 1.4) [KUET: 2011-12] A. – 88.25°C B. – 166.13°C C. 88.25°C D. – 66.10°C E. 166.13°C S D info T1V1 -1 = T2V2 -1 T2 = T1V1 –1 V2 –1 = 273 (V) –1 (2V) –1 = 273 2 1.4–1 = 206.895K = – 66.105°C – 66.10°C 04. 270C ZvcgvÎvq I 1 evqy gÛjxq Pv‡c wKQz cwigvb ï®‥ evqy nVvr msKzwPZ K‡i A‡a©K AvqZ‡b bvwg‡q Avbv n‡jv| ZvcgvÎv n‡e- (-1.4) [KUET: 2010-11] A. 270C B. 2730C C. 00C D. 122.90C T2 = v1 v2 –1 T1 = 2 1 1.41 300 k C 0.4 0 (2) 300 395.85 122.85
420 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES Concept 5 Bwćbi `ÿZv msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. 100 % 1 1 2 T T T ; 100 % 1 1 2 Q Q Q W = Q1 100% 02. ZvcgvÎvi cwieZ©b = `¶Zvi cv_©K¨ MÖvn‡Ki ZvcgvÎv (1-1g `¶Zv) (1-2q `¶Zv) ev, T = (1 )(1 ) T 1 2 2 03. Dr‡mi ZvcgvÎv 1 T T 2 1 ; MÖvn‡Ki ZvcgvÎv T T (1 ) 2 1 , = `¶Zv 04. Bwćbi †ÿ‡Î, W = Q1 – Q2; Q1 > Q2 ; †iwd«Rv‡iU‡ii †ÿ‡Î, W = Q2 – Q1; Q2 > Q1 , Kvh© m¤úv`b mnM, k = Q1 W = Q1 Q2 Q1 Dr‡mi ZvcgvÎv Dr‡mi M„nxZ Zvc T1 T2 → MÖvn‡Ki ZvcgvÎv Q1 Q2 → MÖvn‡K ewR©Z Zvc Kv‡b©v Bwćbi `¶Zv wbY©‡qi †¶‡Î T1>T2; Q1> Q2 ; 2 1 2 1 T T Q Q Model Example01 GKwU Kv‡b©v Bwćbi `ÿZv 60%| hw` Zvc Dr‡mi ZvcgvÎv 450K Z‡e Zvc MÖvn‡Ki ZvcgvÎv KZ? General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] GLv‡b Dr‡mi ZvcgvÎv T1 = 450K = 1 – T2 450 100% 60 100 = 1 – T2 450 0.6 = 1 – T2 450 T2 450 = 1 – 0.6 T2 = 0.4 450 T2 = 180K MÖvn‡Ki ZvcgvÎv 180K T2 = (1 – )T1 T2 = (1 – 0.6) 450 = 180K Model Example02 GKwU Zvc Bwćbi Kvh©Ki c`v_© 600K, ZvcgvÎvi Drm †_‡K 1200J Zvc MÖnY Ges 300K ZvcgvÎvi MÖvn‡K 600J Zvc eR©b K‡i| Zvc Bwćbi `ÿZv wbY©q Ki| General Rules [Written] 3 in 1 Shortcut Tricks & Tips [MCQ] = 1 – Q2 Q1 100% = 1 – 600 1200 100% = (1 – 0.5) 100% = 0.5 100% = 50% Zvc Bwćbi `ÿZv 50% Q2 = (1 – )Q1 600 1200 = 1 – = 50% CONCEPTUAL MATH MEx 01 GKwU Kv‡b©v BwÄb cvwbi wngv¼ I ùzUbvs‡Ki g‡a¨ Kvh©iZ Av‡Q| BwÄbwUi `¶Zv KZ? 100% 1 1 2 T T T 100% 373 100 0 26.8% [gvbwU gyL¯’ ivLv fv‡jv] MEx 02 GKwU †iwd«Rv‡iUi kxZj Zvcvavi †_‡K 450J MÖnY K‡i Dò Zvcvav‡i 600J Zvckw³ eR©b K‡i| †iwd«Rv‡iUiwUi Kvh© m¤úv`b mnM KZ n‡e? Kvh© m¤úv`b mnM, k = Q1 Q2 Q1 = 450 (600 – 450) = 3 NOW START PRACTICE 01. GKwU Kv‡b©v BwÄb 500 K ZvcgvÎvi Zvc Drm n‡Z 1250 J Zvc MÖnY K‡i Ges Zvc MÖvn‡K 700 J Zvc eR©b K‡i| Bwćbi `¶Zv wbY©q Ki| 02. GKwU Zvc BwÄb Drm ‡_‡K 600k ZvcgvÎvq 1 kcal Zvckw³ avib K‡i Zvc MÖvn‡K 500 cal Zvc kw³ eR©b K‡i| Zvc MÖvn‡Ki ZvcgvÎv KZ? 03. GKwU Kv‡b©v BwÄb hLb 270C DòZvi ZvcMÖvn‡K _v‡K ZLb Kg©`¶Zv 50%| G‡K 60% `¶ Ki‡Z n‡j Gi D³ ZvcgvÎvi wK cwieZ©b Avb‡Z n‡e| 04. GKwU Kv‡b©v BwÄb 3000C I 1000C Ges Av‡iKwU Kv‡b©v BwÄb 5000C I 3000C Gi g‡a¨ KvR Ki‡Q| cÖ_gwU Zzjbvq wØZxqwUi `¶ZvNOW PRACTICE SOLVE : 01. Bwćbi `¶Zv, 100% 44% 1250 700 100% 1 Q Q 1 1 2 02. 2 1 2 1 T T Q Q 600 T2 = 1000 500 T2 = 300 K 03. T = T (1 – 1)(1 – 2) = 0.10 300 (1 – 0.5) (1 – 0.6) = 150K 04. 1 2 = 2q Bwćbi Dr‡mi ZvcgvÎv 1g Bwćbi Dr‡mi ZvcgvÎv = 773 573 100% = 135%
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 421 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. GKwU K‡b©v BwÄb 500 K Ges 250 K ZvcgvÎvi `yBwU Avav‡ii gva¨‡g cwiPvwjZ nq| cÖ‡Z¨K P‡µ BwÄb hw` Drm †_‡K 1 kcal Zvc MÖnY K‡i Zvn‡j cÖ‡Z¨K P‡µ Zvc MÖvn‡K Zvc eR©b Kivi cwigvY KZ? [DU.A 19-20] A. 500 cal B. 1000 cal C. 500 kcal D. 10 kcal S A info †`Iqv Av‡Q, T1 = 500 k , T2 = 250 k ,Q1 = 1 k cal ; Q2 = ? Q2 = T2 T1 Q1 = 250 500 1 = 0.5 Kcal = 0.5 1000 = 500 cal Q2 = 500 cal 02. 127C Ges 27C ZvcgvÎvi g‡a¨ Kg©iZ GKwU Kv‡b©v Bwćbi Kg©`¶Zv- [DU. 2016–17, CU. 07-08, BU. 15-16,] A. 15% B. 25% C. 35% D. 50% Bwćbi `¶Zv, 100 T T 1 1 2 % = 100 127 273 27 273 1 % = 25% STEP 02 ANALYSIS OF JU QUESTION 01. 27 C Ges 230 C ZvcgvÎv؇qi g‡a¨ Kvh©iZ GKwU Kv‡b©v Bwćbi Kg©`ÿZv n‡eÑ [JU-A, Set-B. 20-21] A. 20% B. 40% C. 60% D. †Kv‡bvwUB bq S B info Avgiv Rvwb, = 1 – T2 T1 100% = 1 – 300 503 100% = 40.35% 40% 02. GKwU Kv‡b©v BwÄb ev®úwe›`y I eid we›`y؇qi g‡a¨ Kvh©iZ n‡j GwUi `ÿZv KZ? [JU-A, Set-E. 20-21, RUET. 14-15, RU. 14-15, MBSTU. 19-20] A. 26.81% B. 30.81% C. 36.81% D. †KvbwUB bq S A info = 1 – T2 T1 100% = 1 – 273 373 100% = 26.81% †UKwbK: T2 T1 1 n‡e memgq| d‡j, T1 T2 n‡e| 03. hw` †Kvb Zvc BwÄb †_‡K Zvc ewR©Z bv nq, Z‡e Bwćbi ÿgZv KZ n‡e? [JU-A, Set-A. 19-20] A. 0% B. 1% C. 30% D. 100% Bwćbi `ÿZv, = 1 T2 T1 100% T2 = 0 = 100% 04. GKwU Kv‡b©v Bwćbi Rb¨ hw` Zvc Dr‡mi ZvcgvÎv AcwiewZ©Z †i‡L Zvc MÖvn‡Ki ZvcgvÎv ax‡i ax‡i Kgv‡bv nj Zvn‡j Bwćbi Kg©`¶Zv †Kgb fv‡e cwiewZ©Z n‡e? [JU. 2015-16] A. e„w× cvq B. AcwiewZ©Z _v‡K C. Kg‡Z _vK‡e D. ejv m¤¢e bq = 1 – T2 T1 100% T2 Gi gvb Kgv‡bv _v‡K| `¶Zv T2 T1 e„w× †c‡Z _v‡K| 05. 100W ¶gZv I 60% `¶Zv wewkó GKwU †gvU‡ii cÖwZ †m‡K‡Û m¤úvw`Z KvR wbY©q Ki| [JU. 2015-16] A. 70 J B. 60 J C. 80 J D. 40 J Ans B 06. 33% Kg©`¶Zv m¤úbœ GKwU Zvc BwÄb 9.0107 J Zvckw³ mieivn Kiv n‡jv| BwÄbwU KZUzKz Zvckw³‡K Kv‡R iƒcvšÍwiZ Ki‡Z cvi‡e? [JU. 14-15] A. 3000J B. 8400J C. 30000J D. 10000J = out put in put 100% ev, 33% = output 9 107 100% ev, output = 29700 30000J 07. GKwU BwÄb ZvcDrm n‡Z 3000 J Zvc MÖnY K‡i| BwÄbwUi `¶Zv 40% n‡j,BwÄbwU n‡Z wK cwigvY Zvc wbM©Z n‡e? [JU. 2014-15] A. 1200 J B. 3000 J C. 1800 J D. 1400 J BwÄbwUi `¶Zv, = 1 – Q2 Q1 100% Q2 = (1 – ) Q1 = (1 – 0.4) 3000 = 1800J 08. GKwU KvY©wU BwÄb 800K I 400K ZvcgvÎvq †h `¶Zvi KvR K‡i, wVK mg `¶Zvq KvR K‡i T I 900K Zvc gvÎvq| ZvcgvÎv T Gi gvb KZ? [JU. 2011-12, JnU: 10-11] A. 900K B. 450 C. 1800K D. 500K T = cÖv_wgK ZvcgvÎv؇qi AbycvZ wØZxq †¶‡Îi ZvcgvÎv ev, 900 400 800 T ev, T=1800K 09. GKwU Zvc BwÄb ÷xg we›`y I 270 C ZvcgvÎvi g‡a¨ Kvh©iZ| Bwćbi `¶Zv KZ? [JU. 2010-11] A. 29.5% B. 21.75% C. 15.25% D. 19.57% = T1 – T2 T1 100% = 100 – 27 373 100 = 73 373 100% = 19.57% 10. gnvwe‡k¦i me‡P‡q †ewk cwigvY c`v_© †Kvb Ae¯’vq Av‡Q? [JU. 2009-10] A. KwVb B. Zij C. evqexq D. cøvRgv Ans D 11. GKwU Kv‡b©v BwÄb 3000C I 1000C Ges Av‡iKwU Kv‡b©v BwÄb 5000C I 3000C Gi g‡a¨ KvR Ki‡Q| cÖ_gwU Zzjbvq wØZxqwUi `¶Zv- [JU. 07-08] A. Same B. 75% C. 135% D. 167% 1 2 = 2q Bwćbi Dr‡mi ZvcgvÎv 1g Bwćbi Dr‡mi ZvcgvÎv = 773 573 100% = 135% 12. GKwU Kv‡b©v BwÄb 2270C ZvcgvÎvq Zvc MÖnY K‡i Ges 770C ZvcgvÎvq Zvc eR©b K‡i| Bwćbi `¶Zv n‡jv- [JU. 2005-06, JUST: 2015-16] A. 70% B. 35% C. 30% D. 66% = T1 – T2 T1 100% = 227 – 77 500 100% = 30% STEP 03 ANALYSIS OF RU QUESTION 01. GKwU †iwd«Rv‡iU‡ii Kg©m¤úv`b mnM 2| GwU kxZj Zvcvavi n‡Z cÖwZ P‡µ 250J Zvc MÖnY K‡i| †iwd«Rv‡iUiwU cÖwZ P‡µ Kx cwigvY Zvc Dò Zvcvav‡i eR©b Ki‡e? [RU-C, Set-1. 19-20] A. 500 ] B. 125 ] C. 252 ] D. 375 ] Kg©m¤úv`b mnM, K = Q1 Q2 – Q1 2 = 250 Q2 – 250 2Q2 – 500 = 250 Q2 = 750 2 = 375 J 02. GKwU Kv‡b©v Bwćbi Drm ZvcgvÎv 500K| Drm †_‡K 1000 Ryj Zvc MÖnY K‡i `yBwU wm‡¼ h_vµ‡g 200 Ryj I 50 Ryj Zvc eR©b K‡i| BwÄbwUi Kg©`¶Zv KZ? [RU. 2017-18] A. 75% B. 80% C. 95% D. †KvbwUB bq †gvU ewR©Z Zvc = 200 + 50 = 250 J Bwćbi `¶Zv, = 1 – Q2 Q1 100% = 1 – 250 1000 100% = 75% 03. GKwU Kv‡b©v BwÄb 227C ZvcgvÎvq Zvc MÖnY K‡i I 167C ZvcgvÎvq Zvc eR©b K‡i| BwÄbwUi `¶Zv KZ? [RU. 2015-16] A. 20% B. 25% C. 50% D. †KvbwUB bq = 227 – 167 227 + 273 100% = 12%
422 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 04. GKwU Zvc Bwćbi Kvh©Ki e¯‘ 400K ZvcgvÎvi Drm n‡Z 840J Zvc MÖnY K‡i kxZj Avav‡i 420J Zvc eR©b K‡i| kxZj Avav‡ii ZvcgvÎv- [RU. 2015-16] A. 200 K B. 420 K C. 300 K D. 100 K Q1 Q2 = T1 T2 ev, 840 420 = 400 T2 ev, 2 1 = 400 T2 ev, T2 = 200k 05. GKwU Kv‡b©v BwÄb 500K ZvcgvÎvi Zvc Drm †_‡K 1250J Zvc MÖnb K‡i Ges Zvc MÖvn‡K 700J Zvc eR©b K‡i| Zvc MÖvn‡Ki ZvcgvÎv KZ ? [RU. 2011-12, CUET: 2010-11] A. 280 K B. 300 K C. 290 K D. 310 K Q1 Q2 = T1 T2 ev, 1250 700 = 500 T2 ev, T2 = 280K 06. GKwU Kv‡b©v Bwćbi Zvc Dr‡mi ZvcgvÎv 2270 C Ges ZvcMÖvn‡Ki ZvcgvÎv 270C n‡j Bwćbi `¶Zv KZ? [RU. 2010-11, JUST. 15-16] A. 20% B. 88% C. 44% D. 40% 100% 1 1 2 T T T 100% 500 227 27 100% 40% 500 200 07. 90˚C I 180˚C ZvcgvÎvi g‡a¨ Kvh©iZ GKwU cÖZ¨vMvgx Zvc Bwćbi `¶Zv KZ? [RU. 2007-08] A. 5.4% B. 12% C. 19.8% D. 25% Bwćbi `¶Zv, = 1 – T2 T1 100% = 1 – 90 + 273 180 + 273 100% = 19.8% STEP 04 ANALYSIS OF CU QUESTION 01. GKwU cÖZ¨veZ©x BwÄb 167C Ges 57C ZvcgvÎvq KvR Ki‡j Zvi m‡e©v”P `ÿZv KZ n‡e? [CU-A, Set-3. 20-21] A. 25% B. 75% C. 5% D. 50% S A info = 1 T2 T1 100% = 1 (57 + 273) (167 + 273) 100% = 25% 02. GKwU h‡š¿i Kg©`¶Zv mgvb KZ? [CU. 2015-16] A. wb‡ek KvR Drcvw`Z KvR 100% B. Drcvw`Z KvR wb‡ek KvR 100% C. wb‡ek KvR wb‡ek KvR + Drcvw`Z KvR 100% D. Drcvw`Z KvR wb‡ek KvR + Drcvw`Z KvR 100% E. †Kv‡bvwUB bq Drcvw`Z KvR (out put) wb‡ek KvR (in put) 03. GKwU Zvc Bwćbi Kg©`¶Zv 80%| MÖvn‡Ki ZvcgvÎv 1270C n‡j Dr†mi ZvcgvÎv KZ? [CU. 2013-14] A. 2000K B.2100K C.2200K D.2300K E. †KvbwUB bq Dr†mi ZvcgvÎv T1 = T2 1 – = 273 + 127 1 – 0.80 = 400 0.20 = 2000K 04. GKwU Kv‡b©vBwćbi `¶Zv 60% hw` Zvc Dr†mi ZvcgvÎv 400k nq, MÖvn‡Ki ZvcgvÎv KZ? [CU. 08-09; CU: 2012-13] A. 110 K B. 120 K C. 130 K D. 150 K E. 160 K MÖvn‡Ki ZvcgvÎv T2=(1-)T1=(1-0.60)400=0.40400 =160K 05. GKwU Zvc BwÄb 175C I 75C – Gi g‡a¨ wµqv K‡i| Gi Kg©`¶Zv KZ? [CU. 2006-07] A. 30% B. 25% C. 22.3% D. 28.8% Bwćbi `¶Zv, 1– T2 T1 100% = 1– 75 + 273 175 + 273 100% = 22.3% STEP 05 ANALYSIS OF GST QUESTION 01. GKwU K‡b©v Bwćbi Zvc Drm I Zvc MÖvn‡Ki ZvcgvÎv h_vµ‡g 327C I 127C| BwÄbwU Zvc Drm †_‡K 4500 J Zvc MÖnY K‡i wKQz Zvc Kv‡R iƒcvšÍwiZ K‡i Ges Aewkó Zvc MÖvc‡K eR©b K‡i| ewR©Z Zvc‡i cwigvY KZ Ryj (J)| [GST-A. 20-21] A. 1500 B. 2000 C. 2500 D. 3000 S D info Q2 = T2 T1 × Q1 = 400 600 × 4500 = 3000J PART A Analysis of General University Question JnU 01. GKwU Kv‡b©vi BwÄb (Carnot’s engine) 327o I 27o ZvcgvÎvq KvR K‡i| Gi Kg©`¶Zv (efficiency) KZ? [JnU. 2014-15, DU.Tec. 20-21] A. 50% B. 0% C. 100% D. 92% = 327 – 27 600 100% = 300 600 100% = 50% 02. GKwU Kv‡Yv© Bwćbi Kg©`¶Zv 48% ZvcMÖvn‡Ki ZvcgvÎv 100C n‡j Dr‡mi ZvcgvÎv KZ? [JnU. 2009-10] A. 271.230C B. 272.00C C. 277.50C D. 273.60C 0.52 283 1 0.48 283 1- T T 2 1 = 544.23k = 271.230C 03. GKwU BwÄb Zvc Drm †_‡K 300C ZvcgvÎvq ZvcMÖnY K‡i Ges wbgœ AvKv‡i 100C ZvcgvÎvq Zvc eR©b K‡i| BwÄbwUi `¶Zv KZ? [JnU. 2005-06] A. 6.6% B. 16% C. 32% D. 66% = 30 – 10 303 100% = 6.6% KU 01. GKwU Kv‡b©v Bwćbi `¶Zv 60%| hw` Zvc Dr‡mi ZvcgvÎv 500K nq, Z‡e Zvc MÖvn‡Ki ZvcgvÎv KZ? [KU. 2010-11] A. 833.33K B. 333.33K C. 180K D. 300K S Blank info MÖvn‡Ki ZvcgvÎv T2 = (1 – ).T1 ev, T2 = (1 – 0.60) 500 = 0.4 × 500 = 200 K 02. GKwU Bwćbi Kg©`¶Zv 40% Gi MÖvn‡Ki ZvcgvÎv 7 0C Dr‡mi ZvcgvÎv KZ? [KU. 2009-10] A.466.7K B. 467.66K C. 566.7K D. 465.7K T1 = T2 1 – = 280 1 – 0.4 = 280 0.6 = 466.67K 03. GKwU BwÄb 3400J Zvc MÖnY K‡i Ges 2400J Zvc eR©b K‡i| BwÄbwUi `¶Zv KZ ? [KU. 2006-07, IU. 14-15] A. 32.41% B. 27.41% C. 28.00% D. 29.41% = 3400 – 2400 3400 100% = 29.41% 04. GKwU BwÄb 7.35105 J Zvc Drm †_‡K 600 †Kjwfb ZvcgvÎvq 1.47 x 106 J kw³ eR©b K‡i| wbgœ ZvcgvÎvi Avav‡ii ZvcgvÎv KZ Ges Bwćbi `¶Zv wbY©q Ki| [KU. 2003-04] A. 50% B. 60% C. 40% D. 30% T1 T2 = Q1 Q2 ev, T2 = Q2 Q1 T1 ev, T2 = 7.35 105 1.47 106 600 = 300K `¶Zv, = T1 – T2 T1 100% = 600 – 300 600 100% = 50%
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 423 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES IU 01. GKwU cÖZ¨vMvgx BwÄb 27C ZvcgvÎvq 900J Zvc MÖnY K‡i Ges wm‡¼ 540J Zvc eR©b K‡i| Bwćbi `¶Zv- [IU-D, 19-20] A. 40% B. 50% C. 60% D. 70% Bwćbi `¶Zv, = Q1 – Q2 Q1 100% = 900 – 540 900 100% = 40% PART B Analysis of Science & Technology Question SUST 01. GKwU Kv‡b©v BwÄb hLb 27˚C ZvcgvÎvq ZvcMÖvn‡K _v‡K ZLb Gi Kg©`¶Zv 50%| G‡K 60% `¶ Ki‡Z n‡j Gi Dr‡mi ZvcgvÎv KZ wWMÖx evov‡Z n‡e? [SUST: 2016–17. KU. 11-12] A. 50 B. 150 C. 250 D. 350 E. 600 cÖ_g †¶‡Î Dr‡mi ZvcgvÎv, T1 = T2 1 – = 300 1 – 0.5 = 600K w`¡Zxq †¶‡Î Dr‡mi ZvcgvÎv, T1 = T2 1 – = 300 1 – 0.6 = 750K Dr‡mi ZvcgvÎv cwieZ©b, T 750 600 150K 02. 35% Kg©`¶Zvwewkó GKwU Zvcxq BwÄb cÖwZ c~Y©Pµ †k‡l Drm †_‡K M„nxZ Zv‡ci kZKiv KZfvM eR©b K‡i? [SUST: 2015-16] A. 35 B. 45 C. 50 D. 65 E. 55 eR©b K‡i = 100 – 35 = 65 03. GKwU Kv‡b©v BwÄb hLb 37oC ZvcgvÎvi ZvcMÖvn‡K _v‡K ZLb Gi Kg©`¶Zv 30%| G‡K 40% `¶ Ki‡Z n‡j Dr†mi ZvcgvÎv KZ K cwieZ©b Ki‡Z n‡e ? [SUST: 2014–15] A. 50.25 B. 53.75 C. 64.85 D. 73.81 E. 87.13 ZvcgvÎvi cwieZ©b = cÖ`Ë ZvcgvÎv `¶Zvi cv_©K¨ (1 n1) (1 n1) 1 0.301 0.40 310 0.1 0.7 0.6 310 0.1 = 73.81 04. BwÄb A KvR Ki‡Q 500K I 450K ZvcgvÎvq Ges BwÄb B KvR Ki‡Q 450K I 400K ZvcgvÎvq| BwÄb B Gi `¶Zv BwÄb A †_‡K KZUzKz †ewk? [SUST: 2012-13] A. 0% B. 1.0% C. 1.5% D. 1.75% E. 2.0% B A = 500 450 = 1.110 = 1% 05. GKwU Kv‡b©v Bwćbi Zvc MÖvn‡Ki ZvcgvÎv 300C. `¶Zv 30% Bwćbi `¶Zv 35% Ki‡Z n‡j ZvcMÖvn‡Ki ZvcgvÎv KZ Kgv‡Z n‡e| [SUST: 10-11] A. 22C B. 44C C. 33.29C D. 70C T = T2 (1 –1) (1 – 2) = 0.05 303 (1 – 0.30) (1 – 0.35) – 0.05 303 0.7 0.65 = 33.29k = 33.29C 06. 10000C Ges 5000C ZvcgvÎvq Kvh©iZ GKwU Bwćbi m¤¢ve¨ wK cwigvY hvwš¿K `¶Zv n‡q‡Q| [SUST: 2009-10] A. 40% B. 60% C. 80% D. 100% = T1 – T2 T1 100% = 1000 – 500 1273 100% = 500 100% 1273 = 39.27% = 40% 07. GKwU BwÄb KZ©„K M„nxZ I ewR©Z Zv‡ci cwigvY h_vµ‡g 3500J Ges 4000j. BwÄb KZ©„K Kv‡Ri cwigvY- [SUST: 2007-08] A. 500 J B. 1000 J C. 100 J D. 2000 J W = Q1Q2 = 35004000 = 500 J JUST 01. GKwU Kv‡Y©v BwÄb 230˚C I 29˚C ZvcgvÎvi g‡a¨ KvR Ki‡j Gi Kg©`ÿZv KZ? [JUST-C, 19-20] A. 40% B. 50% C. 60% D. 70% = T1 – T2 T1 × 100% = 503 – 302 503 × 100% = 39.96 40% 02. hw` †Kvb Kv‡b©v BwÄb 120˚C I 30˚C ZvcgvÎvi g‡a¨ KvR K‡i Z‡e Zvi `ÿZv kZKiv KZ n‡e? [JUST-A, Set-Ka 19-20, Xv. †ev. 2017] A. 20.9 B. 22.9 C. 24.9 D. 26.9 = T1–T2 T1 × 100% = 393–303 393 × 100% = 22.9% MBSTU 01. GKwU Zvc BwÄb 147C ZvcgvÎvq Zvc Drm †_‡K 1260 Ryj Zvc MÖnY K‡i Ges 37C ZvcgvÎvq Zvc MÖvn‡K 930 Ryj Zvc eR©b K‡i| Bwćbi `ÿZv KZ? [MBSTU-A, Set-2 19-20] A. 25.8% B. 26.2% C. 25.8% D. 26.2% = Q1 – Q2 Q1 100% = 1260 – 930 1260 100% = 26.2% 02. 25% `¶Zvi GKwU Kv‡b©vi BwÄb 270C ZvcgvÎvq Zvc eR©b K‡i| GwU KZ ZvcgvÎvq Zvc †kvlY Ki‡e? [MBSTU. 2015-16] A. 127°C B. 227°C C. 327°C D. 427°C T1 = T1 1 – 27 + 273 1 – 0.25 = 300 0.75 = 400k = 127C STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. 1270C I 4270C ZvcgvÎvi g‡a¨ Kvh©iZ GKwU Bwćbi m¤¢ve¨ m‡e©v”P `¶Zv KZ n‡e ? [BUET. 2010-11] A. 48% B. 42% C. 29.74% D. 70.25% = 427 – 127 273 + 427 100% = 42.85% 02. 200C Ges 1000C ZvcgvÎv ؇qi g‡a¨ Kvh©iZ GKwU Kv‡Y©v Bwćbi `¶Zv n‡e- [BUET. 2009-10] A. 0.8 B. 4 C. – 0.8 D. – 4 S Blank info = 100 – 20 373 100% = 21.44% KUET 01. GKwU Av`k© Kv‡bv© Bwćbi Kg©`¶Zv 40% Ges Bnv 500k ZvcgvÎvq Zvc MÖnY K‡i| hw` Kg©`¶Zv 50% nq, Zvn‡j GKB ewn©Mvgx ZvcgvÎvi Rb¨ AšÍ:Mvgx ZvcgvÎv KZ n‡e| [KUET: 2009-10] A. 500 k B. 600 k C. 700 k D. 800 k E. 900 k S Blank info T = T (1 – 1) (1 – 2) ev, T = 0.1 500 0.6 0.5 = 166.666K 02. GKwU Kv‡b©vBwÄb 400k ZvcgvÎvi Zvc Drm †_‡K 200 cal Zvc MÖnY K‡i Ges Zvc MÖvn‡K 150 cal Zvc eR©b K‡i| Zvc MÖvn‡Ki ZvcgvÎv KZ? [KUET: 2008-09] A. 400k B. 200 k C. 150 k D. 450 k S Blank info T2 T1 = Q2 Q1 ev, T2 400 = 150 200 ev, T2 = 300 k CUET 01. GKwU Kv‡b©v BwÄb 500K ZvcgvÎvi Zvc Drm †_‡K 300cal Zvc MÖnY K‡i Ges Zvc MÖvn‡K 225cal Zvc eR©b K‡i| Zvc MÖvn‡Ki ZvcgvÎv KZ? [CUET: 2014-15] A. 666.67K B. 135K C. 300K D. 375K T2 T1 = Q2 Q1 ev, T2 = Q2.T1 Q1 = 225 500 300 = 375K
424 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 07 ANALYSIS OF HSC BOARD QUESTION DÏxc‡Ki Av‡jv‡K 01 I 02 bs cÖ‡kœi DËi `vI: GKwU Kv‡b©vBwÄb hLb 72C ZvcgvÎvi ZvcMÖvn‡K _v‡K ZLb Gi Kg©`ÿZv 40%| 01. DÏxcK Abymv‡i Bwćbi Dr‡mi ZvcgvÎv KZ? [Xv. †ev. 2021] A. 138K B. 207K C. 575K D. 863K S C info = 1 T2 T1 0.4 = 1 72 + 273 T1 0.6 = 345 T1 T1 = 575K 02. MÖvn‡Ki ZvcgvÎv w¯’i †i‡L BwÄbwU‡K 60% `ÿ Ki‡Z n‡j- [Xv. †ev. 2021] i. Dr‡mi cwiewZ©Z ZvcgvÎv n‡e 862.5K ii. Dr‡mi ZvcgvÎv e„w× cv‡e 287.5K iii. Dr‡mi ZvcgvÎv n«vm 287.5K cv‡e wb‡Pi †KvbwU mwVK? A. i B. iii C. i I ii D. i I iii Ans C wb‡Pi DÏxc‡Ki Av‡jv‡K 03 I 04 bs cÖ‡kœi DËi `vI: [Xv. †ev. 2015] GKwU Zvc BwÄb 327C ZvcgvÎvq 500 J Zvc MÖnY K‡i Ges 27C ZvcgvÎvq Zvc eR©b K‡i| wKQz mgq ci Zvc MÖvn‡Ki ZvcgvÎv 177C G DbœxZ nq| 03. BwÄb KZ…©K m¤úvw`Z Kv‡Ri cwigvY KZ? A. 1500 J B. 1000 J C. 500 J D. 250 J S D info Q2 Q1 = T2 T1 ev, Q2 = T2 T1 Q1 = 300 K 500 J 600 K = 250J Bwćbi K…ZKvR, W = Q1 – Q2 = (500 – 250) J = 250 J 04. `yB Ae¯’vq Bwćbi Kg© `ÿZvi AbycvZ KZ? A. 3 : 4 B. 1 : 1 C. 2 : 3 D. 2 : 1 S D info cÖ_g Ae¯’vq, = T1 – T2 T1 100% = (600 – 300) 600 100% = 50% hLb, T2 = 27C Gi cwieZ©b K‡i T2 = 177C-G DbœxZ nq, ZLb Kg©`ÿZv, = T1 – T2 T1 100% = 600 – 450 600 = 25% wb‡Pi DÏxc‡Ki Av‡jv‡K 05 I 06 bs cÖ‡kœi DËi `vI: [iv. †ev. 2021] GKwU Kv‡b©v BwÄb 327C ZvcgvÎvq 800 J Zvc MÖnY K‡i Ges 127C ZvcgvÎvi ZvcMÖvn‡K Zvc eR©b K‡i| cieZ©x‡Z Zvc MÖvn‡Ki ZvcgvÎv 227C G DbœxZ Kiv nq| 05. BwÄb KZ…©K m¤úvw`Z KvR n‡eÑ A. 250 J B. 267 J C. 500 J D. 800 J S B info Q1 Q2 = T1 T2 800 Q2 = 600 400 Q2 = 533.33 w = Q1 Q2 = 800 533.33 = 266.967 = 267 J 06. cieZ©x Ae¯’vq `ÿZv c~‡e©iÑ A. A‡a©K B. mgvb C. wظY D. wZb¸Y S A info 1 = 1 T2 T1 100% = 1 400 600 100% = 33.33% 2 = 1 500 600 100% = 20% 2 1 = 0.16 0.33 2 = 2 1 myZivs cÖvq wظY n‡e| wb‡Pi Z‡_¨i wfwˇZ 07 I 08 bs cÖ‡kœi DËi `vI : [w`. †ev. 2021] GKwU Kv‡Y©v BwÄb 327C I 27C ZvcgvÎv cwim‡i KvR K‡i| BwÄbwU Drm n‡Z Q cwigvY Zvc MÖnY K‡i wms‡K 3000J Zvc eR©b K‡i| 07. Q Gi gvb KZ? A. 1000J B. 1500J C. 2000J D. 6000J S D info Q1 Q2 = T1 T2 Q1 = T1 Q2 T2 = 600 3000 300 = 600 08. Bwćbi `ÿZv KZ? A. 100% B. 75% C. 50% D. 25% S C info = 1 Q2 Q1 100% = 1 3000 6000 100% = 50% DÏxc‡Ki Av‡jv‡K wb‡Pi 09 I 10 bs cÖ‡kœi DËi `vI| [h. †ev. 2015] GKwU Zvc Bwćbi Kvh©Ki e¯‘ 600K ZvcgvÎvq Drm n‡Z 1200 J Zvc MÖnY K‡i kxZj Avav‡i 300 J Zvc eR©b K‡i| 09. kxZj Avav‡ii ZvcgvÎv KZ? A. 150 K B. 300 K C. 600 K D. 2400 K S A info T2 T1 = Q2 Q1 T2 600 = 300 1200 T2 = 150K 10. BwÄbwUi `ÿZv KZ? A. 44% B. 75% C. 60% D. 50% S B info = Q1 – Q2 Q1 100% = 1200 – 300 1200 100% = 75% 11. †Kv‡bv Zvc BwÄb n‡Z A‡a©K Zvc eR©b n‡j Bwćbi `ÿZv KZ n‡e? [Kz. †ev. 2021] A. 25 % B. 50 % C. 75 % D. 80 % Ans B 12. †Kv‡bv Kv‡b©v Bwćbi `ÿZv 75% Ges ZvcMÖvn‡Ki ZvcgvÎv 67C| Zvc Dr‡mi ZvcgvÎv KZ n‡e? [Kz. †ev. 2015] A. 85C B. 840C C. 1087C D. 1360C S C info = 1 – T2 T1 T2 T1 = 1 – T1 = T2 1 – = 340 1 – 75% = 340 0.25 = 1360K = 1087C Concept 6 GbUªwci cwieZ©b msµvšÍ MvwYwZK cÖ‡qvM FORMULA 01. w¯’i ZvcgvÎvi †¶‡Î GbUªwc, dS = dQ T ; dS = mLf T ; dS = mLv T 02. ZvcgvÎv T1 †_‡K T2 †Z cwiewZ©Z n‡j, GbUªwc, 1 T 2 T dS ms n CONCEPTUAL MATH MEx 01 1000C ZvcgvÎvi 2kg cvwb‡K 1000C ZvcgvÎvi ev‡®ú cwiYZ Ki‡Z GbUªwci cwieZ©b KZ? 373 2 2.26 106 T mL ds = 1.21 104 J kg–1 MEx 01 0 0C ZvcgvÎvi 10gm cvwb‡K 100C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z cÖ‡qvRbxq GbUªwci e„w× nq KZ? 273 283 0.01 4200 ln 1 2 ln T T ds ms = 1.5109J K–1 v
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 425 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES NOW START PRACTICE 01. 0C ZvcgvÎvi 100 MÖvg cvwb‡K 100C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z G›Uªwci e„w× KZ nq wbY©q Ki| 02. 0C G 0.350kg eid M‡j GKB ZvcgvÎvq cvwb nq| G cÖwµqvq GbUªwci cwieZ©b wbY©q Ki| [L = 336000JK –1 ] 03. 0°C ZvcgvÎvi 273kg eid‡K 0°C ZvcgvÎvi cvwb‡Z cwiYZ Kiv n‡q‡Q, GbUªwci cwieZ©b wbY©q Ki| NOW PRACTICE SOLVE : 01. dS = msln T2 T1 = 0.1 4200 ln 373 273 = 131.08JK–1 02. dS = mL T = 0.35 3.36 105 273 = 43077JK –1 03. dS= T mL f = 1 336000Jk 273 273 336000 REAL TEST ANALYSIS OF PREVIOUS YEAR QUESTIONS STEP 01 ANALYSIS OF DU QUESTION 01. 100C ZvcgcvÎvi 373 kg cvwb‡K 100C ZvcgvÎvi ev‡®ú cwiYZ Kiv n‡j GbUªwci cwieZ©b n‡e [cvwbi ev®úxfe‡bi myßZvc = 2.26106 J/kg] [DU. 2017-18] A. 2.26106 J/K B. 842.98 106 J/K C. 165.04 106 J/K D. 847.01106 J/K GbUªwci cwieZ©b, dS = dQ T = mLv T 373 373 2.26 106 = 2.26106 J/K 02. 0C ZvcgvÎvi 273 kg eid‡K 0C ZvcgvÎvi cvwb‡Z iƒcvšÍi Kiv n‡j GbUªwci cwieZ©b KZ n‡e? eid Mj‡bi Av‡cw¶K myßZvc n‡jv 3.36105 J/kg [DU. 2016-17] A. 917.28105 J/K B. 3.36105 J/K C. 273105 J/K D. 0 J/K GbUªwci cwieZ©b, dS = dQ T = mLf T 273 273 3.36 105 = 3.36105 J/K STEP 02 ANALYSIS OF JU QUESTION 01. 100oC ZvcgvÎvi 4 kg cvwb‡K 100oC ZvcgvÎvi ev‡¯ú cwiYZ Ki‡j GbUªwci cwieZ©b KZ n‡e? [JU-A, Set-C. 20-21. iv. †ev. 2021] A. 2.42×104 JK-1 B. 3.42X104 JK-1 C. 4.42x104 JK-1 D.†KvbwUB bq S A info S = mlv T = 4 2.26 106 373 = 2.42 × 104 JK-1 02. GKwU Kv‡b©vP‡µ †gvU G›Uªwci cwieZ©b- [JU-A, Set-D. 19-20] A. TV B. T1 – T2 T1 C. 1 – Q1 Q2 D. †KvbwUB bq Ans D STEP 03 ANALYSIS OF RU QUESTION 01. 1000C ZvcgvÎvq 2kg cvwb‡K 1000C ZvcgvÎvq ev‡®ú cwibZ Ki‡Z GbUªwci cwieZ©b KZ? [RU. 2012-13] A. 4.52104 JK-1 B. 1.21104 JK-1 C. 165.6104 JK-1 D. 1.13104 JK-1 dS = mLv T = 2 226 104 373 = 1.21 104 JK–1 02. 1000C ZvcgvÎvq wKQz cwigvb cvwb‡K GKB ZvcgvÎvq ev‡®ú cwibZ GbUªwci cwieZ©b 1.21104 JK-1 . cvwbi cwigvb KZ? [RU. 2009-10] A. 2kg B. 2.5 kg C. 1.5 kg D. 2.4 kg dQ = mLv T ev, m = dQ T Lv = 1.21 104 373 226 104 = 2kg STEP 04 ANALYSIS OF CU QUESTION 01. 3kg cvwb‡K 100C ZvcgvÎvi ev‡®ú cwiYZ Ki‡Z GbUªwci cwieZ©b KZ n‡e? (cvwbi ev®úxfe‡bi Av‡cw¶K myßZvc 2.26106 J/kg) [CU. 11-12] A. 1.21109 J/k B. 1.21106 J/k C. 1.81106 J/k D. 1.81104 J/k E. 1.81105 J/k GbUªwci cwieZ©b = j T mL 4 6 1.81 10 373 3 2.26 10 /k STEP 05 ANALYSIS OF GST QUESTION PART A Analysis of Science & Technology Question SUST 01. 60C ZvcgvÎvi 10 kg cvwb‡K 100C ZvcgvÎvi ev‡®ú cwiYZ Ki‡j G›Uªwci cwieZ©b KZ J.K–1 ? (cvwbi ev®úxfe‡bi myß Zvc 2.26×106 J.kg–1 ) [SUST-B, 19-20] A. 8.05 × 104 B. 6.04 × 104 C. 7.26 × 104 D. 0.48 × 104 E. 6.54 × 104 S E info G›Uªwci cwieZ©b, dS = S1 + S2 = msln T2 T1 + mlv T = 10 4200 ln 373 333 + 10 2.26 106 373 = 10(476.43 + 6,058.98) = 10 6,534.98 = 6.54 104 JUST 01. 27C ZvcgvÎvi 20 gm cvwb‡K 50C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z cÖ‡qvRbxq GbUªwci e„w× nq KZ? [JUST: 2015-16] A. 6.21 J/k B. 2.61 J/k C. 0.62 J/k D. 3.21 J/k E. 5.21 J/k GbUªwci cwieZ©b, ds = msln T2 T1 = 20 10–3 420ln 323 300 = 6.20 J/k MBSTU 01. 10C ZvcgvÎvi 5 kg cvwb‡K 100C ZvcgvÎvq DbœxZ Ki‡Z GbUªwci cwieZ©b KZ n‡e? [MBSTU-C, Set-2 19-20] A. 5900 JK1 B. 6000 JK1 C. 5800 JK1 D. 5798 JK1 dS = msln T2 T1 = 5 4200 ln 373 283 = 5798 Jk1 PUST 01. 0C Gi 0.5 kg f‡ii eid‡K 0C Gi cvwb‡Z iæcvšÍi Kiv n‡j GbUªwci cwieZ©b KZ? [PUST-A, 19-20] A. 615.3 J/K B. 612.5 J/K C. 615.2 J/K D. 610.2 J/K G›Uªwci cwieZ©b, dS = mLf T = 0.5 336000 273 = 615.38 J/K
426 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 06 ANALYSIS OF ENGINEERING & BUTex QUESTION BUET 01. hLb 10g cvwb 0 0C †_‡K 400C ZvcgvÎvq DËß Kiv nq ZLb GbUªwci cwieZ©b n‡e? [BUET. 2010-11] A. 5.43 calk-1 B. 2.83 cal k-1 C. 1.37 calk-1 D.10.58 calk-1 ds = ms ln T2 T1 = 0.014200 ln 313 273 = 5.74 Jk1 = 5.74 4.2 = 1.367 =1.37 cal k1 CKRuet. Combind 01. –10C ZvcgvÎvq 10 kg eid‡K 0C ZvcgvÎvi cvwb‡Z cwiYZ Ki‡Z G›Uªwci cwieZ©b KZ n‡e? [eid Mj‡bi Av‡cwÿK myßZvc 80,000 cal/kg, ei‡di Av‡cwÿK Zvc 0.5 Jkg–1K –1 Ges cvwbi Av‡cwÿK Zvc 4200 Jkg–1K –1 |] [CKRUET. 2020-21] A. 3.12 kcalK–1 B. 2.93 kcalK–1 C. 312 JK–1 D. 0.187 kcalK-1 E. 3.12 calK–1 S B info S = S2 + S2 = mSiceln 273 273–10 + mlf 273 = 10 × 0.5 × ln 273 263 + 10×80000×42 273 = 12307.88JK–1 = 12.31 kJK–1 = 2.93 kcalK–1 [1 cal = 4.2J] RUET 01. 0 0C ZvcgvÎvi 3kg eid‡K 0 0C ZvcgvÎvq cvwb‡Z cvwb‡Z Ki‡Z GbUªwci cwieZ©b KZ ? [RUET. 2009-10] A. 3592 J/K B. 3582 J/K C. 377J/K D. 369.23 J/K S Blank info dQ = mL T = 3 336000 273 = 3692.3J/K STEP 07 ANALYSIS OF HSC BOARD QUESTION 01. 20gm cvwb‡K 0C †_‡K 80C ZvcgvÎvq DrZß Kiv n‡jv| G›Uªwci cwieZ©b KZ n‡e? [P. †ev. 2021, g. †ev. 2021] A. 21.59 JK–1 B. 24.02 JK–1 C. 40.20 JK–1 D. 46.20 JK–1 S A info S = msln T2 T1 = 0.02 4200 ln 353 273 = 21.59 Jk1 02. 0.01 kg cvwb‡K 0C †_‡K 10C G DËß Kiv n‡jv| G›Uªwc cwieZ©b n‡jv- [w`. †ev. 2016; wm. †ev. 2016; P. †ev. 2015] A. 3.5 JK1 B. 4.5 JK1 C. 2.5 JK1 D. 1.5 JK1 S D info S = 283 273 ms dT T = 0.01 4200 {ln(283) – ln(273)} = 1.5JK–1 SAQ Short Answer Questions WRITTEN PART BAQ Broad Answer Questions 01. wc÷bhy³ GKwU wmwjÛv‡i wKQz M¨vm Ave× Av‡Q| M¨v‡mi Pvc 500 c¨vm‡K‡j w¯’i †i‡L 750 Ryj Zvc kw³ Lye ax‡i ax‡i mieivn Kivq 1250 Ryj KvR m¤úvw`Z n‡j M¨v‡mi AvqZb I AšÍt¯’ kw³i cwieZ©b wbY©q Ki| Solve Avgiv Rvwb, dQ = dU + dW ev, dU = (750 – 1250)J dU = 500J (Ans.) Avevi, dW = PdV ev, dV = (1250/500)m3 dV = 2.5m3 †`Iqv Av‡Q, P = 500 Pa dQ = 750J dW = 1250J dU = ? dV = ? 02. Kve©b WvB A·vBW M¨v‡mi Rb¨ w¯’i AvqZ‡b I w¯’i Pv‡c †gvjvi Av‡cwÿK Zvc wbY©q Ki| †`Iqv Av‡Q, = 1.33 Ges R = 8.314J mol–1K –1 Solve Avgiv Rvwb, CP – CV = R Ges CP CV = CV = R – 1 = 8.314 1.33 – 1 = 25.18 J mol–1 K –1 †`Iqv Av‡Q, = 1.33 R = 8.314 J mol–1 K –1 CP = ? CV = ? CP = CV + R = 25.18 + 8.314 = 33.49 J mol–1 K –1 03. GKwU wmwjÛv‡ii g‡a¨ 1 †gvj M¨vm Av‡Q| M¨vmwU‡K iƒ×Zvcxq cÖwµqvq Ggbfv‡e msKzwPZ Kiv n‡jv hv‡Z ZvcgvÎv 27 †mjwmqvm †_‡K 97 †mjwmqv‡m DbœxZ nq| GB cÖwµqvq m¤úvw`Z KvR Ges Drcbœ Zv‡ci cwigvY wbY©q Ki| ( = 1.5) Solve Avgiv Rvwb, W = R – 1 (T1 – T2) = 8.314 1.5–1 (300 – 370) = – 11.62 10 2 J GLv‡b, T1 = 27C = 300K T2 = 97C = 370K W = ? H = ? Drcvw`Z Zvc, H = W J = 11.62102 4.2 = 276.7 Cal 04. 327 †mjwmqvm cÖv_wgK ZvcgvÎvi 1 †gvj evqy Øviv GKwU Kv‡b©v BwÄb KvR m¤úv`b Ki‡Q| Kv‡b©v P‡µi cÖwZwU av‡c ms‡KvPb-cÖmviY AbycvZ 1:6 n‡j Ñ (i) Kv‡b©v P‡µi me©wb¤œ ZvcgvÎv wbY©q Ki| (i) Bwćbi `ÿZv wbY©q Ki| (i) cÖwZwU P‡µ wbU Kv‡Ri cwigvY wbY©q Ki| ( = 1.4) Solve (i) Avgiv Rvwb, T1V1 – 1 = T2V2 – 1 ev, T2 = T1 V1 V2 – 1 ev, T2 = 600 1 6 1.4 – 1 T2 = 20C = 293 K V1 V2 = 1 6 = 1.4 T1 = 327C = 600 K T2 = ? (ii) Avgiv Rvwb, = T1 – T2 T1 100% = 1 – 293 600 100% = 51.2% (iii) Avgiv Rvwb, W = W1 – W3 = RT1ln(V2/V1) – RT2ln(V3/V4) = RT1ln(V2/V1) – RT2ln(V2/V1) = Rln(V2/V1) (T1 – T2) = 8.314 ln6 (600 – 293) = 4572.2J 05. GKwU Kv‡b©v Bwćbi Zvc MÖvnK hLb 27 ‡mjwmqvm ZvcgvÎvq _v‡K, ZLb Gi Kg©`ÿZv 40%| BwÄbwUi cÖviw¤¢K Kg©`ÿZvi 10% e„w× Ki‡Z n‡j Gi Dr‡mi ZvcgvÎv KZ evo‡Z n‡e? Solve cÖ_g I wØZxq †ÿ‡Îi Dr‡mi ZvcgvÎv h_vµ‡g T1 I T1| cÖ_g †ÿ‡Î = T1 – T2 T1 ev, T1 = T2 1 – = 300 40 = 5 3 300 K T1 = 535.7K dT = 537.7 – 500 K = 37.7K Dr‡mi ZvcgvÎv 37.7K evov‡Z n‡e| GLv‡b, cÖ_g †ÿ‡Î T2 = 27C = 300K = 40% T1 = ? wØZxq †ÿ‡Î, = 40% + 10% of 40% = 44% T1 = ? dT = ?
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 427 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 06. †Kv‡bv Kv‡b©v Bwćbi Zvcvavi؇qi ZvcgvÎv h_vµ‡g 320 †Kjwfb I 260 †Kjwfb| Drm †_‡K 500 Ryj Zvc MÖnY Ki‡j BwÄb KZ…©K m¤úvw`Z Kv‡Ri cwigvY KZ? Solve Avgiv Rvwb, Kv‡b©v Bwćbi †ÿ‡Î, Q1 T1 = Q2 T2 GLv‡b, T1 = 320K T2 = 260 K Q1 = 500 J; Q2 = ? Q2 = T2 T1 Q1 = 260 500 320 J = 406.25J Avevi, W = Q1 – Q2 = (500 – 406.25)J = 93.75J 07. GKwU cÖZ¨veZ©x BwÄb Zv‡ci 1/6th As‡k Kv‡R iƒcvšÍi K‡i| hLb Dr‡mi ZvcgvÎv wVK †i‡L MÖvn‡Ki ZvcgvÎv 60C Kgv‡bv nq, ZLb Bwćbi `ÿZv wظY nq| MÖvnK Gi ZvcgvÎv wbY©q Ki| Solve awi, MÖvn‡Ki ZvcgvÎv T1 I Dr‡mi ZvcgvÎv T2| = 1 – T1 T2 1 6 = 1 – T1 T2 Avevi, 2 6 = 1 – T1 – 62 T2 1 3 = 1 – T1 T2 + 62 T2 1 3 = 1 6 + 62 T2 T2 = 372 K = 99C & T1 = 310K = 37C MÖvn‡Ki ZvcgvÎv 37C 08. 100C ZvcgvÎvq 0.02kg Rjxq ev®ú Nbxf~Z n‡q – 10C ZvcgvÎvq cwiYZ Ki‡Z KZ Zvc eR©b Ki‡Z n‡e? ev‡®úi Nbxfe‡bi Av‡cwÿK myßZvc = 2268000Jkg–1 , ei‡di Av‡cwÿK Zvc = 2100Jkg–1 Ges ei‡di Mj‡bi Av‡cwÿK myßZvc = 336000Jkg–1 | Solve 100C Rjxqev®ú Q1 – 100C cvwb Q2 – 0C cvwb Q3 – 10C eid Q4 – –10C eid ; Q1 = mlv = 0.02 2268000 = 45360 J; Q2 = mwSw = 0.02 4200 (100–0) = 8400J Q3 = mlf = 0.02 336000 = 6720 J; Q4 = miSi = 0.02 2100 10 = 420 J Zvc eR©b Ki‡Z n‡e = Q1 + Q2 + Q3 + Q4 = 60900J 09. GKwU Kv‡b©v P‡µi `ÿZv 1 6 | ZvcMÖvn‡Ki ZvcgvÎv 70C Kgv‡bvi d‡j `ÿZv nq 1 3 | Kv‡b©vi PµwU GLb †Kvb cÖviw¤¢K Ges PzovšÍ ZvcgvÎv؇qi Kvh©iZ? Solve 1 6 = 1 – T2 T1 ................ (i) 1 3 = – T2 – 70 T1 = 1 – T2 T1 + 70 T1 = 1 6 + 70 T1 [from (i)] 70 T1 = 1 3 – 1 6 = 1 6 T1 = 420K T2 = T1 1 – 1 6 = 350K eZ©gv‡b cÖviw¤¢K ZvcgvÎv 420K I P~ovšÍ ZvcgvÎv (350 – 70)K = 280K 10. T1 Avw` ZvcgvÎvi GKwU Av`k© M¨v‡mi Av`k© Avw` AvqZb 2m3 | iƒ×Zvcxq cÖwµqvq cÖmvi‡Yi d‡j Gi AvqZb 4m3 nq Zvici m‡gv cÖwµqvq cÖmvwiZ Kivq AvqZb 10m3 nq, cieZ©x av‡c iƒ×Zvcxq cÖwµqvq ms‡KvP‡bi d‡j Gi ZvcgvÎv cybivq T1 nq| Gi P~ovšÍ AvqZb KZ? Solve T1V1 – 1 = T2V2 – 1 T1 T2 = V2 V1 –1 = 2–1 GLb, T2V3 – 1 = T1V4 – 1 T2 T1 = 1 2 y–1 = V4 V3 –1 V4 = v3 2 = 10 2 m 3 = 5m3 (Ans.) 11. GKwU Kv‡b©v BwÄb hLb 27C ZvcgvÎvq Zvc MÖvn‡K _v‡K Gi Kg© `ÿZv 50%| G‡K 60% `ÿ Ki‡Z n‡j Gi Dr‡mi ZvcgvÎv KZ evov‡Z n‡e? Solve awi, Dr‡mi ZvcgvÎv T MÖvn‡Ki ZvcgvÎv T1 = 27 + 273 = 300K = 1 – T1 T 0.5 = 1 – 300 T T = 600K = 60% n‡j, 0.6 = 1 – T1 T = 1 – 300 T T = 750K ZvcgvÎv evov‡Z n‡e (750 – 600)K = 150K 12. GKwU Kv‡b©v BwÄb AšÍM©vgx Zv‡c 1 4 Ask Kv‡R iƒcvšÍi K‡i| Gi Zvc MÖvn‡Ki ZvcgvÎv Av‡iv 70C n«vm Ki‡j Zvi `ÿZv wظY nq| Drm ZvcgvÎv I Zvc MÖvn‡Ki ZvcgvÎv †ei Ki| Solve 1g †ÿ‡Î, = 1 4 = T1–T2 T1 ev, 1 4 = 1 – T2 T1 ev, T2 T1 = 3 4 2q †ÿ‡Î, T1 – (T2 – 70) T1 = 1 4 1 2 ev, 1 – T2 T1 + 70 T1 = 1 2 ev, 70 T1 = 1 2 + 3 4 – 1 = 1 4 ev, T1 = 280K T1 = 280 K T2 = 3 4 280 = 210K 13. GKwU Kv‡b©v Bwćbi Zvc Drm I Zvc MÖvn‡Ki ZvcgvÎv h_vµ‡g 500K I 375K| hw` BwÄbwU cÖwZ P‡µ 252 104 J Zvc †kvlY K‡i Z‡e, (i) Bwćbi `ÿZv, (ii) cÖwZP‡µ Kv‡Ri cwigvY I (iii) cÖwZ P‡µ ewR©Z Zv‡ci cwigvY wbY©q Ki| Solve T1 = 500K, T2 = 375K, Q1 = 252 104 J, Q2 = ?, = ?, W = ? (i) = 1 – T2 T1 = 1 – 375 500 = 0.25 = 25% (ii) W = Q1 – Q2 = (252 – 189) 104 = 6.3 105 J (iii) Q2 Q1 = T2 T1 Q2 = 375 500 252 104 = 1.89 106 J 14. GKwU wd«‡Ri gU‡ii ÿgZv 200W| hw` VvÛv cÖ‡Kv‡ôi ZvcgvÎv 270K Ges evwn‡ii ZvcgvÎv 300K nq Zvn‡j 100min G VvÛv cÖ‡Kvô †_‡K m‡ev©”P Kx cwigvY Zvc cvIqv hv‡e? Solve GLv‡b, T1 = 300K, T2 = 270K = T1 – T2 T1 = 300 – 270 300 = 0.1; Pin = Qin t Qin = Pin t ......... (i) from (i), Qin = Pin t = 200 100 60 = 1.2 106 J = Qin – Qout Qin Qout= Qin–Qin=1.2 106 – 0.1 1.2 106 = 1.08 106 J 15. 0C ZvcgvÎvq 1g ei‡d cÖwZ †m‡K‡Û 10J Zvc cÖ`vb Kiv n‡j KZÿY ci m¤ú~Y© eid ev®úxf~Z n‡e? [JnU-A, Set-2. 19-20] Solve GLv‡b Zvc, H = mlice + ms + mlstem = 1 10–3 (3.36 105 + 4.2 103 102 + 2.26 106 ) H = 10–3 105 (3.36 + 4.2 + 22.6) = 102 30.16 J eid ev®úxf~Z nIqvi mgq, t = 102 30.16 10 = 10 30.16 sec = 5.02 min 16. 1 †gv‡ji †Kvb M¨vm‡K 27C m‡gvò cÖwµqvq cÖmvwiZ n‡Z †`Iqv n‡jv hZÿY bv ch©šÍ Gi AvqZb wظb nq| Zvici iæØZvcxq cÖwµqvi G‡K Avevi Av‡Mi AvqZ‡b wdwi‡q Avbv nj| †gvU K…Z Kv‡Ri cwigvY wbY©q Ki? [BUTex. 2020-21] Solve m‡gvò cÖwµqvq K…ZKvR, W1 = nRT ln V2 V1 W1 = 1 8.4 300 ln2 W1 = 1746.73J n = 1 R = 8.4Jmol–1 T = 300K V2 = 2V1 GLb, 300K T1 2V1 iæ×Zvc V1 m~Î: T1V1 –1 = T2V2 –1 T1 = V2 V1 –1 T2 1 = (2)1.4–1 300 = 395.86K iæ×Zvcxq cÖwµqvq K…ZKvR, W2 = nR 1– [T1 – T2] = 18.4 1–1.4 [395.86 – 300] W2 = – 2013.06J [eySvq KvR wm‡÷g wb‡R K‡i‡Q] †gvU Kv‡Ri cwigvY, W = W1 + W2= –266.169J
428 An Exclusive Parallel Text Book of Physics ASPECT PHYSICS ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES 17. 40 wWMÖx †mjwmqvm ZvcgvÎvq Ges 76 †mwg evqy Pv‡c wKQz cwigvY evqy‡K iæ×Zvcxq cÖwµqvq AvqZb wظY Kiv n‡j ZvcgvÎv KZ n‡e? Avevi hw` cwiewZ©Z ZvcgvÎv †_‡K 20 wWMÖx †mjwmqvm e„w× Kiv nq, Z‡e Gi Pvc KZ n‡e? [KUET; 19-20] Solve GLv‡b, T1 = 40 + 273 = 313K, V1 = V, V2 = 2V, T2 = ? GLb, T1V1 –1 = T2V2 –1 T2 = T1 V1 V2 –1 = 313 V 2V 1.4–1 = 237.21 K = – 35.79C Gici ZvcgvÎv 20C e„w× Ki‡j ZvcgvÎv T2 = (–35.79 + 20)C = –15.79C = 257.21K GLb, P1 = 76cm (Hg) = 1atm, T1 = 313K, T2 = 257.21 K T1P1 1– = T2P2 1– P2 P1 1– = T1 T2 P2 = P1 T1 T2 1– = 76 313 257.21 1.4 1–1.4 = 38.23 cm (Hg) = 0.503 atm 18. T1 Avw` ZvcgvÎvi GKwU Av`k© M¨v‡mi Av`k© Avw` AvqZb 2m3 | iæ×Zvcxq cÖwµqvq cÖmvi‡Yi d‡j Gi AvqZb 4m3 nq Zvici m‡gvò cÖwµqvq cÖmvwiZ Kivq AvqZb 10m3 nq, cieZx© av‡c iæ×Zvcxq cÖwµqvq ms‡KvP‡bi d‡j Gi ZvcgvÎv cybivq T1 nq| Gi P~ovšÍ AvqZb KZ? [BUET:2018-19] Solve T1V1 –1 = T2V2 –1 T1 T2 = V0 V1 – 1 = 2 – 1 GLb, T2V3 –1 = T1V4 –1 T2 T1 = 2 1 – 1 = V4 V3 – 1 V4= Vs 2 = 10 2 = 5m3 19. 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 62oC Kgv‡bv nq, ZLb Bwćbi `ÿZv wظY nq| MÖvnK Gi ZvcgvÎv wbY©q Ki| [BUET; 19-20] Solve awi, Dr‡mi ZvcgvÎv T1 I MÖvn‡Ki ZvcgvÎv T2| = 1 – T2 T1 1 6 = 1 – T2 T1 Avevi, 2 6 = 1 – T2 – 62 T1 1 3 = 1 – T2 T1 + 62 T1 1 3 = 1 6 + 62 T1 62 T1 = 1 3 – 1 6 = 1 6 T1 = 372 K = 99oC T2 = 1 6 – 1 T1 = 310K = 37oC MÖvn‡Ki ZvcgvÎv 37oC ASPECT SPECIAL: T1 = 6 62 = 372 K Ges T2 = 372 – 62 = 310 K 20. ¯^vfvweK ZvcgvÎv I Pvc Kv‡K e‡j? 0˚C ZvcgvÎvi 2kg f‡ii cvwbi mv‡_ mgvb f‡ii 90˚C ZvcgvÎvi cvwb wgwkÖZ Kiv n‡jv| G›Uªwci cwieZ©b wbY©q Ki| [NSTU-A, 19-20] Solve 0˚C ev 273K ZvcgvÎv‡K ¯^vfvweK ZvcvgÎv I 760 mm HgP ev 1 atm Pvc‡K ¯^vfvweK Pvc ejv nq| (NTP = Natural Temperature and Pressure) GLv‡b, cÖ_g cvwbi ZvcgvÎv, T1 = 0˚C = 273 K; fi, m1 = 2 Kg wØZxq cvwbi ZvcgvÎv, T2 = 90˚C = (90 + 273)K = 363 K fi, m2 = 2 Kg Dfq cvwbi Av‡cwÿK Zvc, s = 4200 J Kg–1 k –1 awi, wgkÖ‡Yi me©‡kl ZvcgvÎv = T †Kjwfb kZ©g‡Z, m1S (T–T1) = m2S(T2 – T) T – T1 = T2 –T 2T = 363 + 273 T = 318 K T = (318 – 273)K = 45K A_©vr M„nxZ/ewR©Z Zvc, Q = m1ST = (2420045)J = 3,78,000J G›Uªwci cwieZ©b, S = – dQ T2 + dQ T1 = 378000 1 273 – 1 363 = 343.293J/K 21. 100C ZvcgvÎvq 5kg cvwb‡K 1000C ZvcgvÎvq DbœxZ Ki‡Z GbUªwci cwieZ©b wbb©q Ki? [PUST. 2011-12] Solve GbUªwci cwie©Zb, dS = ms ln T2 T1 = 54200ln 373 283 = 5798.76 JK -1 22. 0˚C ZvcgvÎvi 1kg eid‡K 100˚C ZvcgvÎvi ev‡®ú cwiYZ Kiv n‡jv| G›Uªwci cwieZ©b wbY©q Ki| [BUTEX; 19-20] Solve s = mlf T1 + msln T2 T1 + mlv T2 = 1 336000 273 + 1 4200 ln373 273 + 12268000 373 = 8622.046 JK–1 23. dv‡ibnvBU †¯‹‡ji †Kvb ZvcgvÎv †mw›U‡MÖW †¯‹‡ji cv‡Vi wZb¸Y? [Zdv¾j m¨vi] Solve Avgiv Rvwb, C 5 = F 32 9 aiv hvK, dv‡ibnvB‡Ui x ZvcgvÎv †mw›U‡MÖW †¯‥‡ji cv‡Vi wZb¸Y, †mw›U‡MÖW †¯‥‡j ZvcgvÎv x 3 x 3 5 = x 32 9 ev, x 15 = x 32 9 ev, 9x = 15x 480 ev, 6x = 480 x = 80 24. KZ D”PZv n‡Z GKLÛ eid co‡j m¤ú~Y©iƒ‡c M‡j hv‡e? [Zdv¾j m¨vi] Solve h m D”PZvq m fiwewkó GKwU eidL‡Ûi wefe kw³Z = mgh J Avevi eidLÐwU m¤ú~Y©fv‡e M‡j †h‡Z cÖ‡qvRbxq Zvc = m Lf J. mgh = mLf ev, h = 3.36 105 Jkg1 9.8 Nkg1 = 34285.7 m 34285.7 m GLv‡b, g = 9.8 Nkg1 Lf = eid Mj‡bi myßZvc = 3.36 Jkg1 25. GKwU ÎæwUc~Y© _v‡g©vwgUv‡i wb¤œ w¯’i we›`y 4 Ges EaŸ w¯’i we›`y 98| H _v‡g©vwgUv‡i 51 cvV w`‡j dv‡ibnvBU I †Kjwfb †¯‥‡j cvV KZ n‡e? [Zcb m¨vi] Solve Avgiv Rvwb, mKj †¯‥‡ji †ÿ‡Î, ZvcgvÎv wb¤œw¯’i we›`y EaŸ©w¯’i we›`y wb¤œw¯’i we›`yAbycvZ mgvb dv‡ibnvBU †¯‥‡ji †ÿ‡Î, F 31 212 32 = S M B M ev, F 32 180 = 51 4 98 4 F = 122 F GLv‡b, ÎæwUc~Y© _v‡g©vwgUv‡i, ZvcgvÎv, S = 51 wb¤œw¯’i we›`y, M = 4 DaŸ©w¯’i we›`y, B = 98 dv‡ibnvBU †¯‥‡j cÖK…Z ZvcgvÎv, F = ? †Kjwfb †¯‥‡j cÖK…Z ZvcgvÎv, K = ? Avevi, †Kjwfb †¯‥‡ji †ÿ‡Î, K 273 373 273 = S M B M ev, K 273 100 = 51 4 98 4 K = 323 K 122 F ; 323 K 26. 100˚C ZvcgvÎvq 0.02 kg Rjxq ev®ú Nbxf~Z n‡q –10˚C ZvcgvÎvq cwiYZ Ki‡Z KZ Zvc eR©b Ki‡Z n‡e? ev‡®úi Nbxfe‡bi Av‡cwÿK myßZvc = 2268000 Jkg–1 , ei‡di Av‡cwÿK Zvc = 2100 Jkg –1 Ges ei‡di Mj‡bi Av‡cwÿK myßZvc = 336000 Jkg–1 | [RUET: 19-20] Solve 100˚C Rjxq ev®ú Q1 100˚C cvwb Q2 0˚C cvwb Q3 0˚C eid Q4 – 10˚C eid Q1 = mlv = 0.02 2268000 = 45360 J Q2 = mwSw = 0.02 4200 (100 – 0) = 8400 J Q3 = mlf = 0.02 336000 = 6720 J Q4 = miSi = 0.02 2100 (0 + 10) = 420 J Zvc eR©b Ki‡Z n‡e = Q1 + Q2 + Q3 + Q4 = 60900 J
ASPECT PHYSICS wØZxq cÎ ZvcMwZwe`¨v 429 ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES ASPECT SERIES STEP 3 mKj cvV¨eB‡qi NCTB QUESTIONS ANALYSIS cÖkœ e¨vLvmn mgvavb cÖkœ 01. ‡Kvb wm‡÷g cwi‡ek †_‡K 800 J Zvckw³ †kvlY Kivq Gi AšÍt¯’ kw³ 500 J e„w× cvq| wm‡÷g KZ…K cwi‡e‡ki Dci K…Z Kv‡Ri cwigvY KZ? [Bm&nvK m¨vi] A. 200J B. 400J C. 1500J D. 300J dW = dQ – du = 800 – 500=300J 02. wb‡Pi †KvbwU ZvcMwZwe`¨vi wØZxq m~‡Îi MvwYwZK iƒcÑ [Bm&nvK m¨vi] A. T dS dQ B. T dQ dS C. dQ T dS D. Q1 W Ans B 03. 0 0C ZvcgvÎvi 1kg eid‡K 0 0C ZvcgvÎvi 1kg cvwb‡Z cwiYZ Ki‡Z GbUªwci Kx cwieZ©b n‡e? [Bm&nvK m¨vi] A.1.20103 JK–1 B.1.24103 JK–1 C.1.23103 JK–1 D.1.5103 JK–1 dS = mlf T = 3.36 105 273 = 1.23 103 JK1 04. wbgœ ùzUbv‡¼i †Kvb Zij cwicvk¦© n‡Z jxb Zvc ev myß Zvc MÖnb K‡i cwicvk¦©‡K kxZj K‡i Zv‡K Kx e‡j? [Bm&nvK m¨vi] A.wngvqb B.wngvqK C.Zvcxq BwÄb D.†iwd«Rv‡iUi Ans A 05. GbUªwci S.I GKK †KvbwU? [Bm&nvK m¨vi] A. kJ–1 B. JK–1 C. JK D. J2K Ans B 06. Av`k© M¨v‡mi iæ×Zvcxq m¤úªmvi‡Yi †ÿ‡Î †Kvb †jLwPÎwU mwVK? [Bm&nvK m¨vi, AvRMi m¨vi] A. ln T ln P O B. ln T ln P O C. ln T ln P O D. ln T ln P O S B info 1 TP = constant lnT + 1 lnP = lnK lnT = 1 lnP + lnK y = mx + c 07. GKwU GK cvigvYweK M¨v‡mi w¯’i AvqZ‡b †gvjvi Av‡cwÿK Zv‡ci †jLwPÎ- [Bm&nvK m¨vi] A. Cv T O B. Cv T O C. Cv T O D. Cv T O Ans B 08. `ywU e¯‘i Nl©‡bi d‡j Zvc Drcbœ nq, GwU †Kvb ai‡bi cÖwµqv? [Zcb m¨vi] A. cÖZ¨veZ©x cÖwµqv B. AcÖZ¨veZ©x cÖwµqv C. iƒ×Zvcxq cÖwµqv D. m‡gvò cÖwµqv `ywU e¯‘i g‡a¨ Nl©‡bi Rb¨ †h Zvc m„wó nq Zv AcÖZ¨veZ©x cÖwµqv KviY, Nl©‡bi weiæ‡× †h KvR nq ZvB Zv‡c cwiYZ nq Ges H Zvc‡K Kv‡R cwiYZ Kiv hvq bv| GKwU Kv‡b©v BwÄb 500K ZvcgvÎvi Zvc Drm †_‡K 1250J Zvc MÖnb K‡i Ges ZvcMÖvn‡K 700J Zvc eR©b K‡i| wb‡¤œi 09 bs 10 cÖ‡kœi DËi `vI: 09. Zvc MÖvn‡Ki ZvcgvÎv KZ? [Zdv¾j m¨vi] A. 240K B. 250K C. 280K D. 300K 2 2 1 1 T Q T Q 500 280k 1250 700 T Q Q T 1 1 2 2 10. BwÄbwUi `¶Zv KZ? [Zdv¾j m¨vi] A. 22% B. 44% C. 40% D. 50% = 1 – T2 T1 100% = 1 – 280 500 100% = 44% 11. Zv‡ci MZxq gZev` †K cÖeZ©b K‡ib? [Mwb m¨vi] A. K¨vjwiK B. KvD›U ivg‡dvW© C. wbDUb D. AvBb÷vBb KvD›U ivg‡dvW©, n¨vg‡d †Wfx, †Rgm †cÖmKU Ryj cÖgvY K‡ib †h, KvR Z_v hvwš¿K kw³ †_‡K Zvc Drcbœ nq Ges Zvc MwZiB GKwU iƒc| 12. Wv³vwi _v‡g©vwgUv‡i GKRb †jv‡Ki †`‡ni ZvcgvÎvi cvV 98.4°F †mjwmqvm †¯‥‡j GB ZvcgvÎv KZ? [Mwb m¨vi] A. 36.88°C B. 40°C C. 39°C D. 39°C 36.88 C 9 98.4 32 C 5 9 F 32 5 C 13. GbUªwc- [Mwb m¨vi] i. Zvc cÖev‡ni w`K wb‡`©k K‡i ii. AcÖZ¨veZ©x cÖwµqvq Bnv AcwiewZ©Z _v‡K iii. wek„•Ljvi wnmve K‡i wb‡Pi †KvbwU mwVK? A. i I ii B. ii I iii C. ii D. i Iiii Ans D `vM bv KvUv GKwU cvi` _v‡g©vwgUvi‡K ch©vqµ‡g MwjZ eid, ev®ú I mgy‡`ªi cvwb‡Z Wzwe‡q hLb cvi` ¯Í¤¢ w¯’i n‡jv ZLb †¯‥j w`‡q †g‡c wb¤œwjwLZ wPÎ Abyhvqx cvV cvIqv †Mj| 12cm 4cm 2cm ev®ú mgy`ª cvwb eid Dc‡ii Z_¨ †_‡K 14 I 15 cÖ‡kœi DËi `vI : 14. mgy‡`ªi cvwbi ZvcgvÎv KZ? [Mwb m¨vi] A. 2°C B. 20°C C. 40°C D. 80°C 100 20 C 12 2 4 2 15. wPÎ Abyhvqx ev‡®úi ZvcgvÎv- [Mwb m¨vi] A. 98°C B. 273K C. 373K D. 100K 273.16k 273.16k 12 12 T 16. iƒ×Zvcxq cÖwµqvi Rb¨ PVγ= aªæeK GwU‡K cÖwZôv Ki‡Z wb‡Pi †KvbwU e¨eüZ n‡q‡Q? [kvgmyi m¨vi] A. Cp=Cv B. Cp=RCv C. PdV=RdT= VdP D. dQ=CvdT=CvdT+PdV=0 dQ = dU + dW, dW = pdV Ges Cv = dT dU dQ = CvdT + PdV 17. ‣Îawe›`yi cwi‡cÖw¶‡Z _v‡g©vwgwZi g~j mgxKiY †KvbwU? [nvwjg m¨vi] A. K X X T 273.16 tr B. K X X T 273 tr C. 2 E a b D. C X X T 273.16 0 tr Ans A