Tiêu đề 1. P2C1. HSC PREP Papers 26_With Solve.pdf ✅

ZvcMwZwe` ̈v  HSC Prep Papers 1 cÖ_g Aa ̈vq ZvcMwZwe` ̈v Thermodynamics Topicwise CQ Trend Analysis UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU ZvcgvÎv cwigv‡ci bxwZ (_v‡g©vwgUvi) Ñ Ñ Ñ Ñ 1 Ñ Ñ 1 ZvcMwZwe` ̈vi 1g m~Î Ñ 2 Ñ 1 3 2 3 11 iƒ×Zvcxq cÖwμqvi m~Î 1 Ñ Ñ 2 3 6 1 13 m‡gvò I iæ×Zvcxq cÖwμqvq K...ZKvR Ñ Ñ Ñ 1 2 6 6 15 Zvc, Af ̈šÍixY kw3 I KvR Ñ 2 Ñ 4 Ñ 1 1 8 Kv‡b©v Pμ I Kv‡b©v BwÄb ev cÖZ ̈vMvgx BwÄb 2 3 2 1 9 2 5 24 Zvcxq BwÄb: †iwd«Rv‡iU Ñ Ñ Ñ Ñ 1 Ñ Ñ 1 Bwćbi `ÿZv 3 5 1 2 9 2 1 23 GbUawc I wek„•Ljv Ñ 2 1 4 3 4 1 15 * we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| wb‡Pi DÏxcKwU jÿ ̈ Ki: [Xv. †ev. 23; w`. †ev. 23] P(N/m2 ) 40 20 R Q S T P V(m3 ) 4 8 wP‡Î M ̈v‡mi Pvc I ZvcgvÎvi cwieZ©b †`Lv‡bv n‡q‡Q| GLv‡b Q †_‡K R G †h‡Z ZvcMZxq e ̈e ̄’vq 80 J Zvckw3 mieivn Kiv n‡q‡Q| (K) Af ̈šÍixY kw3 Kv‡K e‡j? (L) iæ×Zvcxq cÖmvi‡Y wm‡÷g kxZj nqÑ e ̈vL ̈v Ki| (M) DÏxcK Abymv‡i R Ae ̄’v‡b Avm‡Z ZvcMZxq e ̈e ̄’vwU‡Z AšÍt ̄’ kw3i cwieZ©b KZ? DËi: K...ZKvR, dW = PdV  dW = 0 [∵ dV = 0] mieivnK...Z Zvckw3, dQ = dU + dW  dU = dQ – 0 = 80 J A_©vr R Ae ̄’v‡b Avm‡Z AšÍt ̄’ kw3i cwieZ©b 80 J| (Ans.) (N) DÏxcK Abymv‡i, PQRP P‡μi cÖwZwU av‡c Kv‡Ri Zzjbv Ki| mgvavb: dWPQ = PPQdVPQ = 20  (8 – 4)  dWPQ = 80 J dWQR = 0 [∵dV = 0] dWRP = PQRS Gi †ÿÎdj = 1 2  PQ  (PS + RT) = 1 2  4  (20 + 40)  dWRP = 120 J  dWRP > dWPQ > dWQR (Ans.) 2| 56 g bvB‡Uav‡Rb M ̈vm‡K GKwU Bwćbi mvnv‡h ̈ cÖ_‡g m‡gvò cÖwμqvq I c‡i iæ×Zvcxq cÖwμqvq AvqZb wZb ̧Y Kiv n‡jv| BwÄbwU 127C Ges 27C ZvcgvÎvq Kvh©Ki Av‡Q| (bvB‡Uav‡R‡bi AvYweK fi 28 g) [Xv. †ev. 23; w`. †ev. 23] (K) ZvcMwZwe` ̈vi k~b ̈Zg m~ÎwU wee„Z Ki| (L) ms‡e`bkxj •e`y ̈wZK h‡š¿ mv‡›Ui e ̈envi Riæwi †Kb?Ñ e ̈vL ̈v Ki| (M) BwÄbwUi Kg©`ÿZv wbY©q Ki| mgvavb: Kg©`ÿZv,  =     1 – T2 T1  100% =     1 – 27 + 273 127 + 273  100%   = 25% A_©vr BwÄbwUi Kg©`ÿZv 25%| (Ans.) (N) DÏxc‡Ki †Kvb cÖwμqvq K...ZKvR †ewk n‡e?Ñ MvwYwZK we‡køl‡Yi gva ̈‡g gZvgZ `vI| mgvavb: m‡gvò cÖwμqvq, W1 = nRT1ln     V2 V1 = 2  8.314  400  ln     3V V  W1 = 7307.09 J (Ans.) n = 56 28 = 2 mol iæ×Zvcxq cÖwμqvq, W2 = nR  – 1 (T1 – T2) = 2  8.314 1.4 – 1 (400 – 300)  W2 = 4157 J  W1 > W2 A_©vr m‡gvò cÖwμqvq K...ZKvR †ewk n‡e| (Ans.)
2  HSC Physics 2nd Paper Chapter-1 3| GKwU ZvcMZxq e ̈e ̄’vq 14 g bvB‡Uav‡Rb M ̈vm 30C ZvcgvÎvq I 1 evqygÐjxq Pv‡c iwÿZ Av‡Q| w ̄’iPv‡c G‡Z mieivn Kiv n‡j ZvcgvÎv 35C nq| cieZ©x‡Z Dc‡iv3 cÖwμqvwU m‡gvò cÖwμqvq GKwU Avw` Ae ̄’v‡b n‡Z GKB AvqZ‡bi cwieZ©b K‡i K...Z Kv‡Ri cwigvY Kiv n‡jv (R = 8.3 Jmol–1K –1 , CV = 20.8 Jmol–1K –1 ) [iv. †ev. 23] (K) Zv‡ci hvwš¿K mgZv Kx? (L) Mvwoi Uvqvi we‡ùvi‡Yi mgq Kx ai‡bi ZvcMZxq cÖwμqv msNwUZ nq? e ̈vL ̈v K‡iv| (M) DÏxc‡K w ̄’i Pv‡ci †ÿ‡Î Af ̈šÍixY kw3i cwieZ©b wbY©q K‡iv| mgvavb: PdV = nRdT  PdV = 14 28  8.3  (35 – 30)  PdV = 20.75 Nm ZvcMwZwe` ̈vi cÖ_g m~Îvbyhvqx, dQ = dU + dW  dU = dQ – dW = nCPdT – PdV = 14 28  (20.8 + 8.3)  (35 – 30) – 20.75  dU = 52 J A_©vr w ̄’i Pv‡c Af ̈šÍixY kw3i cwieZ©b 25 J| (Ans.) (N) w ̄’i Pvc cÖwμqv Ges m‡gvò cÖwμqvq DÏxc‡K wb‡Y©q K...Z Kv‡Ri gvb mgvb n‡e wK? MvwYwZK we‡kølY Ki| mgvavb: w ̄’i Pvc cÖwμqvq, W1 = PdV = 20.75 J m‡gvò cÖwμqvq, W2 = nRT1 ln     V2 V1 w ̄’iPv‡c, V2 V1 = T2 T1 = 35 + 273 30 + 273  V2 V1 = 1.017  W2 = 0.5  8.3  (30 + 273)  ln(1.017)  W2 = 20.581 J  W1 > W2 A_©vr Dfq‡ÿ‡Î K...Z Kv‡Ri gvb mgvb n‡e bv| 4| P A D C B T1 = 827 K T2 = 827 K V Q1 = 500 J [Kz. †ev. 23] wP‡Î GKwU K‡Y©v Bwćbi P-V †jLwPÎ †`Lv‡bv n‡jv| (K) Zvc BwÄb Kv‡K e‡j? (L) M ̈v‡mi †gvjvi Av‡cwÿK Zvc `yB cÖKvi †Kb? (M) BwÄb KZ...©K K...Z Kv‡Ri cwigvY wbY©q K‡iv mgvavb: Kv‡b©v Bwćbi †ÿ‡Î, Q2 Q1 = T2 T1  Q1 – Q2 Q1 = T1 – T2 T1  W 500 = 827 – 387 827 [∵ W = Q1 – Q2]  W = 266.022 J A_©vr, BwÄb KZ...©K K...ZKv‡Ri cwigvY 266.022 J (Ans.) (N) DÏxc‡Ki Kv‡b©v Bwćbi ZvcMÖvn‡Ki ZvcgvÎv wØ ̧Y Ki‡j `ÿZv A‡a©K n‡e wKbvÑ MvwYwZK we‡kølY K‡iv| mgvavb: Kg©`ÿZv,  =     1 – T2 T1  100% =     1 – 387 827  100%   = 53.204% ZvcMÖvn‡Ki ZvcgvÎv wØ ̧Y Ki‡j,  =     1 – 2T2 T1  100% =       1 –     2  387 827  100%   = 6.409% GLb,   = 6.409 53.204    = 0.12  0.5 A_©vr ZvcMÖvn‡Ki ZvcgvÎv wØ ̧Y Ki‡j `ÿZv A‡a©K n‡e bv| (Ans.) 5| wØ-cvigvYweK M ̈vm m¤^wjZ GKwU Kv‡b©v BwÄb 500 K ZvcgvÎvi Drm n‡Z Zvc MÖnY K‡i| cÖwZ cÖmvi‡Y Gi AvqZb wZb ̧Y nq| [h. †ev. 23] (K) Zwor w؇giæ Kv‡K e‡j? (L) iæ×Zvcxq ms‡KvP‡bi mgq wm‡÷‡gi Af ̈šÍixY kw3 e„w× cvqÑ e ̈vL ̈v Ki| (M) DÏxc‡Ki BwÄbwUi cÖv_wgK `ÿZv wbY©q Ki| mgvavb: iæ×Zvcxq cÖmvi‡Yi †ÿ‡Î, T1V –1 1 = T2V –1 2  T2 =     V1 V2  – 1  T1  T2 =     1 3 1.4–1  500  T2 = 322.197 K  =     1 – T2 T1  100% =     1 – 322.197 500  100%   = 35.561% A_©vr BwÄbwUi cÖv_wgK `ÿZv 35.561%| (Ans.) (N) Bwćbi `ÿZv 60% Ki‡Z n‡j Kx e ̈e ̄’v wb‡Z n‡e? MvwYwZK we‡kølY Ki| mgvavb: Kg©`ÿZv 60% Kiv hv‡e `ywU Dcv‡q| ZvcMÖvn‡Ki ZvcgvÎv cwieZ©b K‡i,  =     1 – T 2 T1  100%  60% =     1 – T 2 500  100%  T 2 500 = 0.4  T 2 = 200 K  ZvcgvÎv Kgv‡Z n‡e = (322.197 – 200) K = 122.197 K
ZvcMwZwe` ̈v  HSC Prep Papers 3 ZvcDr‡mi ZvcgvÎv cwieZ©b K‡i,  =     1 – T2 T 1  100%  60% =     1 – 322.197 T 1  100%  322.197 T 1 = 0.4  T 1 = 805.493 K  ZvcgvÎv evov‡Z n‡e = (805.493 – 500) K = 305.493 K  BwÄbwUi `ÿZv 60% Ki‡Z n‡j Zvc Dr‡mi ZvcgvÎv 305.493 K e„w× A_ev Zvc MÖvn‡Ki ZvcgvÎv 122.197 K n«vm Ki‡Z n‡e| (Ans.) 6| K¬wmqvm wc÷b wmwjÛv‡i GK †gvj nvB‡Wav‡Rb M ̈vm wb‡q P-V Gi †jLwPÎ wb‡¤œ cÖ`wk©Z PμwUi Abyiƒc GKwU Pμ †c‡jb| K¬wmqv‡mi g‡Z GwU GKwU cÖZ ̈veZ©x Pμ| [P. †ev. 23] 4 2.5 (0, 0) 2.5 4.5 X Q2 D Q1 (P4, V4, T2) C(P3, V3, T2) B(P2, V2, T1) A(P1, V1, T1) Y P  10 5 Nm –2 T1 = 600 K T2 = 400 K V  10–3 m 3 (K) ZvcMwZwe` ̈vi 2q m~Î wee„Z Ki| (L) iƒ×Zvcxq cÖwμqvq M ̈vm‡K msKzwPZ Ki‡j ZvcgvÎv e„w× cvqÑ e ̈vL ̈v K‡iv| (M) DÏxcK Abymv‡i nvB‡Wav‡Rb M ̈vm‡K B n‡Z C †Z Avb‡Z K...ZKv‡Ri cwigvY wbY©q K‡iv| mgvavb: WBC = nR  – 1 (T1 – T2) = 1  8.314 (1.4 – 1) (600 – 400)  WBC = 4157 J A_©vr nvB‡Wav‡Rb M ̈vm‡K B n‡Z C-†Z Avb‡Z K...ZKv‡Ri cwigvY 4157 J| (Ans.) (N) K¬wmqv‡mi `vwewU †h.w3K wK bv e ̈vL ̈v K‡iv| mgvavb: Q1 = WAB [∵ m‡gvò cÖwμqvq, dQ = dW] = nRT1ln     V2 V1 = 1  8.314  600  ln     4.5 2.5  Q1 = 2932.115 J Q2 = – WCD [∵ m‡gvò ms‡KvPb]  Q2 = – nRT2ln     V4 V3 ...... (i) BC Ask n‡Z,     V3 V2 –1 = T2 T1      V3 4.5  10–3 (1.4–1) = 400 600  V3 = 5.292  10–3 m 3 Avevi, DA As‡k,     V4 V1 –1 = T2 T1      V4 2.5  10–3 (1.4–1) = 400 600  V4 = 2.94  10–3 m 3 (i) n‡Z cvB, Q2 = – 1  8.314  400 ln     2.94  10–3 5.292  10–3 = 1954.743 J  Q2 Q1 = 0.667 Avevi, T2 T1 = 0.667  Q2 Q1 = T2 T1 A_©vr K¬vwmqv‡mi `vwewU †hŠw3K| 7| [e. †ev. 23] Pvc (P) AvqZb (V) A B D C T1 = 500C T2 = 200C Q2 = 600 J Dc‡ii P-V wPÎwU GKwU cÖZ ̈veZ©x Zvc Bwćbi| (K) e× wm‡÷g Kx? (L) m‡gvò cÖwμqvq Af ̈šÍixY kw3i cwieZ©b k~b ̈ †Kb? e ̈vL ̈v Ki| (M) DÏxc‡Ki Zvc Bwćb m‡gvò cÖmvi‡Y G›Uawci cwieZ©b wbY©q Ki| mgvavb: G›Uawci cwieZ©b, dS = Q1 T1 = 600 473  ds = 1.268 J/K cÖZ ̈veZ©x BwÄb nIqvq,  Q1 T1 = Q2 T2 = 600 (200 + 273)  Q1 T1 = 600 473 A_©vr m‡gvò cÖmvi‡Y G›Uawci cwieZ©b 1.268 J/K. (Ans.) (N) Dr‡mi ZvcgvÎv w ̄’i †i‡L BwÄbwU `ÿZv 1.5 ̧Y Kiv m¤¢e wKbv? MvwYwZKfv‡e we‡kølY Ki| mgvavb: BwÄbwUi `ÿZv,  =     1 – T2 T1  100% =       1 –     200 + 273 500 + 273  100%   = 38.81% `ÿZv 1.5 ̧Y Ki‡j,  = 1.5  38.81% = 58.215%
4  HSC Physics 2nd Paper Chapter-1   =     1 – T 2 T1  100%  58.215% =       1 – T 2 500 + 273  100%  T 2 773 = 0.418  T 2 = 323 K = 50C A_©vr Dr‡mi ZvcgvÎv w ̄’i †i‡L BwÄbwUi `ÿZv 1.5 ̧Y Kiv m¤¢e| (Ans.) 8| Y O 1 2 3 4 6 (0,0) 1 2 3 4 5 X 5 T2 = 300 K T1 = 400 K Pvc P(10 5 Nm –2 ) AvqZb D C A B wP‡Î 1 mole cwigvY †Kv‡bv M ̈v‡mi †ÿ‡Î `ywU m‡gvò †jL †`Lv‡bv n‡q‡Q| M ̈vmwUi w ̄’i AvqZb †gvjvi Av‡cwÿK Zvc 25.18 J mol–1K –1 | [wm. †ev. 23] (K) GbUawc Kv‡K e‡j? (L) Zvc Drm I ZvcMÖvn‡Ki ZvcgvÎvi g‡a ̈ cv_©K ̈ K‡g †M‡j Bwćbi `ÿZvI K‡g hvqÑ e ̈vL ̈v Ki| (M) CD As‡k K...ZKv‡Ri cwigvY wbY©q K‡iv| mgvavb: m‡gvò cÖmvi‡Y, K...ZKvR W = nRT1 ln     VD VC = 1  8.314  400 ln     4 2  W = 2305.13 J  CD As‡k K...ZKv‡Ri cwigvY 2305.13 J| (Ans.) (N) A n‡Z C †Z wb‡Z Zvckw3i cwieZ©b, B n‡Z D †Z wb‡Z Zvckw3i cwieZ©‡bi mgvb n‡e wK bv? mgvavb: CP = CV + R = 25.18 + 8.314  CP = 33.494 J mole–1 K –1 A n‡Z C †Z, Zvckw3i cwieZ©b, dQAC = dU + dWAC = nCP(T1 – T2) + PAC  (VC – VA) = 1  33.494  (400 – 300) + 4  105  1  dQAC = 403349.4 J B n‡Z D †Z, Zvckw3i cwieZ©b, dQBD = dU + dWBD = nCP(T1 – T2) + PBD  (VD – VB) = 1  33.494  (400 – 300) + 2  105  2 dQBD = 403349.4 J  dQAC = dQBD A_©vr A n‡Z C †Z wb‡Z Zvckw3i cwieZ©b, B n‡Z D †Z wb‡Z Zvckw3i cwieZ©‡bi mgvb n‡e| (Ans.) 9| DÏxcK wP‡Îi Dfq K‡b©vP‡μ Kvh©wbe©vnK e ̄‘ wn‡m‡e 1 †gvj wØcvigvYweK M ̈vm e ̈eüZ n‡q‡Q| Pμ `ywUi cÖwZ P‡μ ms‡KvPb I cÖmvi‡Yi AbycvZ h_vμ‡g 1 : 3 Ges 1 : 4| (R = 8.31 Jmol–1K –1 ) [g. †ev. 23] (P) X A(P1, V1) B(P2, V2) T1 = 60C C(P3, V3) D(P4, V4) T2 = 211K  V X A(P1  , V1  ) B(P2  , V2  ) T1  = 70C C(P3  , V3  ) D(P4  , V4  ) T2 = 211K  V (P) Y 1g Kv‡b©vPμ Y 2q Kv‡b©vPμ  (K) †iwd«Rv‡iU‡ii Kg©m¤úv`b mnM Kv‡K e‡j? (L) mgAvqZb cÖwμqvq wm‡÷‡g cÖ`Ë Zvc m¤ú~Y©UvB Af ̈šÍixY kw3 e„w×i Kv‡R e ̈eüZ nq| e ̈vL ̈v K‡iv| (M) DÏxc‡Ki 1g Kv‡b©vP‡μi Kvh©wbe©vnK e ̄‘‡K B †_‡K C †Z wb‡Z †gvU K...ZKvR wbY©q K‡iv| mgvavb: iæ×Zvcxq cÖwμqvq, W = nR  – 1 (T1 – T2) =     1  8.31 1.4 – 1  {(60 + 273) – 211}  W = 2534.55 J A_©vr 1g Kv‡b©vP‡μ B †_‡K C-†Z wb‡Z †gvU K...ZKvR 2534.55 J| (Ans.) (N) DÏxc‡Ki Abymv‡i, †Kvb Kv‡b©v PμwU †ewk Kvh©Ki, MvwYwZK we‡kølY K‡i gZvgZ `vI| mgvavb: Kg©`ÿZv,  =     1 – T2 T1  100%  1 =       1 –     211 60 + 273  100%  1 = 36.637% 2 =       1 –     211 70 + 273  100%  2 = 38.484% ∵ 2 > 1 A_©vr wØZxq Kv‡b©v PμwU †ewk Kvh©Ki| (Ans.) 10| 0C ZvcZgvÎvi 0.07 kg eid‡K GKwU wbw`©ó D”PZv †_‡K †d‡j †`qv n‡jv| G‡Z wefe kw3i 55% Zv‡c iƒcvšÍwiZ n‡jv Ges GB Zvc mg ̄Í eid‡K Mwj‡q w`‡jv| wKQz mgq ci eidMjv cvwbi ZvcgvÎv 5C G DbœxZ n‡jv| †`qv Av‡Q, eid Mj‡bi Av‡cwÿK myßZvc 3.36  105 J kg–1 Ges cvwbi Av‡cwÿK Zvc 4200 Jkg–1 k –1 . [Xv. †ev. 22] (K) ZvcMwZwe` ̈vi wØZxq m~ÎwU wee„Z Ki| (L) mgAvqZb cÖwμqvq KvR k~b ̈ †Kb? e ̈vL ̈v `vI| (M) eid LÐwU KZ D”PZv †_‡K †djv n‡qwQj? mgvavb: cÖkœg‡Z, 55%  wefekw3 = cÖ‡qvRbxq Zvckw3  0.55  mgh = mlf  h = lf 0.55  g = 3.36  105 0.55  9.8  h = 62337.662 m A_©vr eid LÐwU †djv n‡qwQj 62337.662 m D”PZv †_‡K| (Ans.)