Tiêu đề 8. P1C8. Phy. For FRB-2024_With Solve_Sha_30.03.24 (1).pdf ✅

ch©ve„wËK MwZ  Final Revision Batch 1 Aóg Aa ̈vq ch©ve„wËK MwZ Periodic Motion Topicwise CQ Trend Analysis UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU SHM wewkó KYvi mgxKiY kw3, †gvU hvwš¿K kw3 Ges kw3i msiÿYkxjZv 1 3 Ñ Ñ 1 4 3 12 mij †`vjK 3 5 Ñ 4 8 9 6 35 e ̄‘i †eM Ñ 3 Ñ 1 Ñ 3 Ñ 7 cÖZ ̈qbx ej Ñ 2 Ñ Ñ Ñ 1 Ñ 3 cvnv‡oi D”PZv Ñ 1 Ñ 4 5 4 2 16 w ̄úas, Dj¤^ w ̄úas, Abyf~wgK w ̄úas Ges mgwš^Z w ̄úas Gi †`vjbKvj, w ̄úas Gi kw3 Ñ Ñ Ñ 1 5 2 1 9 *we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| GKwU †m‡KÛ †`vjK †Kv‡bv GKwU cvnv‡oi cv`‡`‡k mwVK mgq †`q| wKš‘ †`vjKwU‡K cvnv‡oi P‚ovq wb‡q †M‡j cÖwZ NÈvq 30 †m‡KÛ mgq ax‡i P‡j| c„w_exi e ̈vmva© 6400 km Ges AwfKl©R Z¡iY 9.8 ms–2 | [Xv. †ev. 23] (K) GKK †f±i Kx? (L) gnvKl©xq wef‡ei gvb FYvZ¥K nq †Kb? e ̈vL ̈v K‡iv| (M) cvnv‡oi P‚ovq †`vjKwUi †`vjbKvj wbY©q K‡iv| mgvavb: †`vjbKvj (t)  1 Aa©‡`vjb msL ̈v T  1 n  T2 T1 = n1 n2  T2 = T1  n1 n2 = 2  3600 3570 = 2.0168 sec A_©vr †`vjKwUi †`vjbKvj 2.0168 sec Here, n = t T 2 n1 = 3600 2 2 wU n2 = t T 2 = 3600 – 30 2 2 = 3570 wU (N) DÏxc‡Ki Z‡_ ̈i wfwˇZ cvnvowUi D”PZv wbY©q Kiv hv‡e wKbv, Zv MvwYwZK we‡køl‡Yi gva ̈‡g e ̈v ̈L ̈v K‡iv| mgvavb: Th Te = R + h R  1 + h R = 2.0168 2  h = 53.76 km (Ans.) 2| c„w_ex c„‡ô GKwU mij †`vj‡Ki MwZi mgxKiY x = 10 sin     t +  4 | mij †`vjKwU webv evavq `yj‡Q|[iv. †ev. 23] (K) KvwjK chh©vqμg Kx? (L) kxZKv‡j †`vjK Nwo `aæZ P‡j †Kb? e ̈vL ̈v Ki| (M) 2s ci †`vjKwUi Z¡iY †ei Ki| mgvavb: x = 10 sin    t +  4  v = dx dt = 10 cos    t +  4  a = dv dt= 10 2 sin    t +  4 2 sec ci Z¡iY, a = – 10 2 sin    2 +  4 = – 69.789 ms–2 (Ans.) (N) †`vjKwU‡K f~-c„ô n‡Z 100 m Mfx‡i wb‡q †M‡j †`vjb msL ̈vi cwieZ©b n‡e wK? hvPvB Ki| mgvavb: Tn Te = R + h R = ne nh n = t T  6400  103 + 100 6400  103 = ne nh  ne nh = 1.0000156  ne – nh nh  100 = 1.0000156 – 1 1  100 = 0.00156% evo‡e †`vjbmsL ̈v 0.00156% evo‡e| (Ans.) 3| m f‡ii GKwU e ̄‘‡K c„w_exi †K›`aMvgx GKwU myo‡1⁄2i ga ̈ w`‡q †Q‡o †`qv nj| c„w_exi c„‡ô g = 9.8 ms–1 Ges c„w_exi e ̈vmva© 6.4  106 m. [OP = 5  105 m] [Kz. †ev. 23] m O P R (K) ̄’vwbK ch©ve„wË Kx? (L) mij Qw›`Z MwZm¤úbœ KYvi Z¡iY mi‡Yi wecixZgyLxÑ e ̈vL ̈v K‡iv|
2  HSC Physics 1st Paper Chapter-8 (M) P we›`y‡Z AwfKl©R Z¡i‡Yi gvb wbY©q K‡iv| mgvavb: g = g    1 – h R = 9.8       1 –     6.4  106 – 5  105 6.4  106  g = 0.766 ms –1 (Ans.) (N) DÏxc‡K myo1⁄2 c‡_ †Q‡o †`Iqv e ̄‘wUi †`vjbKvj wbY©q Kiv m¤¢e wK-bv? MvwYwZKfv‡e we‡kølY K‡iv| mgvavb: awi, f‚-c„ô n‡Z h MfxiZvi Z¡iY, gh = 4 3  G(R – h) GB (R – h) n‡jv c„w_exi †K›`a †_‡K miY, G‡K x Øviv cÖwZ ̄’vcb K‡i cvB, gh = 4 3  Gx  M f‡ii e ̄‘i Dci wμqvkxj ej, F = Mgh = – 4 3  Gmx = – kx     4 3 Gm = aaæeK = k a‡i  F = – kx  gh = – k m x = –  2 x  myo1⁄2c‡_ MwZ GKwU SHM  T = 2  = 2 m k = 2 m 4 3 Gm = 2 3 4G = 3 G = 5068.64 sec (Ans.) 4| c`v_©weÁvb j ̈v‡e e ̈eüZ GKwU †m‡KÛ †`vj‡Ki MÖx®§Kv‡j •`N© ̈ cvIqvq †`vjbKvj 2.1 s nq| mwVK mgq cvIqvi Rb ̈ GKRb QvÎ Gi •`N© ̈ 2% Kwg‡q †`q| (g = 9.81 ms–2 ) [Kz. †ev. 23] (K) m¤ú›`b MwZ Kv‡K e‡j? (L) ̄ú›`iZ KYvi †g.wjK we ̄Ívi 4 Gi g‡a ̈ ivL‡Z nq †Kb, e ̈vL ̈v K‡iv| (M) ̄^vfvweK Ae ̄’vq †`vjKwUi Kvh©Kix •`N© ̈ wbY©q K‡iv| mgvavb: T = 2 L g  L = gT2 4 2 = 9.8  (2) 2 4 2  L = 0.993 m (Ans.) (N) DÏxc‡Ki Av‡jv‡K QvÎwU mdj n‡e wK bv MvwYwZKfv‡e hvPvB K‡iv| mgvavb: Avw` •`N© ̈ 1.0958 m bZzb •`N© ̈ = 1.0958 – 1.0958  2 100 = 1.0738 m  T = 2 l g T = 2  1.0738 9.81 = 2.078 s  T  2s  QvÎwU mdj n‡e bv| (Ans.) 5| GKwU †cÛzjvg Nwo f‚-c„‡ô cÖwZ †m‡K‡Û 1wU Aa©‡`vjb †`q| wKš‘ cvnv‡oi Dci wb‡q †M‡j •`wbK 100 s mgq nvivq| c„w_exi e ̈vmva© 6400 km Ges c„w_ex c„‡ô AwfKl©R Z¡iY 9.8 ms–2 | [h. †ev. 23] (K) ̄’vwbK ch©vqe„wË Kv‡K e‡j? (L) GKwU f‚w ̄’i DcMÖ‡n w ̄úas fi e ̈e ̄’vq †`vjbKvj †Kgb n‡e? e ̈vL ̈v K‡iv| (M) DÏxc‡Ki Z‡_ ̈i Av‡jv‡K cvnv‡oi D”PZv †ei K‡iv| mgvavb: T T = R + h R  86400 86400 – 100 = 1 + h R  1 + h R = 864 863  h = 7.41 km A_©vr cvnv‡oi D”PZv 7.41 km (Ans.) (N) cvnv‡oi Dc‡i †`vjKwU‡K †m‡KÛ †`vj‡K ciYZ Ki‡Z Kx e ̈e ̄’v Ki‡Z n‡e? MvwYwZKfv‡e we‡kølY K‡iv| mgvavb: T = 86400  2 86400 – 100 = 2.00231 sec T  L; L = mij †`vj‡Ki Kvh©Kix •`N© ̈  T T = L L  L L =     2.00231 2 2  L L =1.0023  L – L L = 1.0023 – 1 1  L L = 0.00231  •`N© ̈ evov‡Z n‡e L L  100% = 0.23% (Ans.) 6| GKwU †m‡KÛ †`vj‡Ki wmwjÛvi AvK...wZi ee cvwbc~Y© Ae ̄’vq Av‡Q| e‡ii •`N© ̈ 8 cm| [e. †ev. 23] (K) ̄ú›`b MwZ Kv‡K e‡j? (L) mKj ch©vq e„Ë MwZ mij †`vjb MwZ bq †Kb? e ̈vL ̈v Ki|
ch©ve„wËK MwZ  Final Revision Batch 3 (M) †`vjKwUi Kvh©Kix •`N© ̈ wbY©q Ki| mgvavb: T = 2 sec (second pendulum)  T = 2 L g  L = T 2 4 2  g  L = 2 2 4 2  9.8 GLv‡b, L = Kvh©Kix •`N© ̈  L = 0.9929 m A_©vr Kvh©Kix •`N© ̈ 0.9929 m (Ans.) (N) eewU A‡a©K Lvwj Ki‡j ZLb †`vjKwU `aæZ bv ax‡i Pj‡e? MvwYwZKfv‡e hvPvB K‡i gZvgZ `vI| mgvavb: e‡ei •`N© ̈ 8 cm  e‡ei e ̈vmva© 4 cm = r ee Lvwj Kivi c~‡e©i †`vjbKvj T = 2s eewU A‡a©K Lvwj Kivi ci fi‡K›`a r 2 cwigvY wb‡P †b‡g hvq| A_©vr Kvh©Kix •`N© ̈ e„w× cv‡e,  L = l + r + r 2 = l + 3r 2 T = 2 L g T  L  T2 T1 = L2 L1 r 2 r  T2 = l + 3r 2 l + r .T1 2r = 8  r = 4 = 0.9929 + 0.04  3 2 0.9929 + 0.4  2 = 2.019 sec  T2 > T1  †`vjbKvj †e‡o †M‡Q gv‡b †`vjKwU ax‡i Pj‡e| (Ans.) 7| GKwU w ̄úas Gi Dci 10 N ej cÖ‡qvM Kivq GwU 4 cm cÖmvwiZ nq| ̄úaxswU‡K cÖ_‡g 6 cm Ges cieZ©x‡Z Av‡iv 6 cm cÖmvwiZ Kiv n‡jv| [w`. †ev. 23] (K) Kg©`ÿZv Kx? (L) †K›`agyLx ej Øviv K...Z KvR k~b ̈ nqÑe ̈vL ̈v K‡iv| (M) w ̄úaswUi w ̄úas aaæeK wbY©q K‡iv| mgvavb: Spring constant, k = mg x = 10 4  10–2 = 250 Nm–1 A_©vr w ̄úas aaæeK 250 Nm–1 (Ans.) (N) cÖ_g I wØZxq †ÿ‡Î mgvb cÖmvi‡Yi Rb ̈ Kv‡Ri cwigvY mgvb n‡e wK? MvwYwZKfv‡e e ̈vL ̈v K‡iv| mgvavb: cÖ_g †ÿ‡Î, W1 = 1 2 k (x ) 2 2 – x 2 1 x1 = 0.04 m x2 = (0.04 + 0.06) = 0.1 m = 1 2  250  {(0.1) } 2 – (0.04) 2  W1 = 1.05 J wØZxq †ÿ‡Î, W2 = 1 2 k (x ) 2 2 – x 2 1 x 1 = 0.1 m x 2 = (0.1 + 0.06) = 0.16 m = 1 2  250  {(0.16) } 2 – (0.1) 2  W2 = 1.95 J  W1  W2 A_©vr Dfq‡ÿ‡Î Kv‡Ri cwigvY mgvb n‡e bv| (Ans.) 8| ivwb I ewb `yB fvB‡qi fi h_vμ‡g 32 kg Ges 40 kg| Zviv `yRb GKwU †`vjbvq GKB †K.wYK mi‡Y †`vj Lvw”Qj| †`vjbvwUi Mo Kvh©Kix •`N© ̈ 3 m Ges Gi mvg ̈ve ̄’vb A| [g = 9.8 ms–1 ] [w`. †ev. 23] O 1m A B 3m (K) †K.wYK K¤úvsK Kv‡K e‡j? (L) gnvKv‡k GKRb b‡fvPvixi wbKU GKwU †m‡KÛ †`vj‡Ki K¤úvs‡Ki cwieZ©b nqÑe ̈vL ̈v K‡iv| (M) iwbi Dci myZvi Uvb wbY©q K‡iv| mgvavb: A B 3m  1m mg mg cos   tan = 1 3   = tan–1     1 3  iwbi Dci myZvi Uvb, T = mg cos = 32  9.8  cos     tan–1     1 3 = 297.50 N A_©vr myZvi Uvb 297.50 N (Ans.)
4  HSC Physics 1st Paper Chapter-8 (N) DÏxc‡Ki B Ae ̄’v‡b iwb I ewb mgvb Kvh©Ki ej Abyfe Ki‡jv wKbvÑMvwYwZKfv‡e e ̈vL ̈v K‡iv| mgvavb: A  1m mg mg cos  T mg sin iwbi Dci Kvh©Ki ej, F1 = m1g sin  = tan–1 1 3  F1 = 99.169 N ewbi Dci Kvh©Kix ej F2 = m2gsin = 123.96 N  F1  F2 A_©vr B Ae ̄’v‡b Df‡q mgvb Kvh©Ki ej Abyfe Ki‡e bv| (Ans.) 9| GKwU mij †`vj‡Ki Kvh©Ki •`N© ̈ 1 m Ges 6 cm| †`vjKwU‡K cÖ_‡g P ̄’v‡b wb‡q hvIqv n‡j †`vjbKvj 1.5 sec Ges Gici Q ̄’v‡b wb‡q †M‡j †`vjbKvj 2 sec cvIqv †Mj| [g. †ev. 23] (K) `kv Kv‡K e‡j? (L) Ôw ̄úas ej cÖZ ̈qbx ejÕÑ e ̈vL ̈v K‡iv| (M) †`vj‡Ki mvg ̈ve ̄’v‡b †eM wbY©q K‡iv| mgvavb: T = 2 L g = 2 1 9.8  T = 2.007 sec v = A = 2 T × A = 2 2.007 × 0.06  v = 0.188 ms–1 A_©vr mvg ̈ve ̄’v‡b †eM 0.188 ms–1 (Ans.) (N) †Kvb ̄’v‡bi D”PZv †ewk wQj? MvwYwZKfv‡e e ̈vL ̈v K‡iv| mgvavb: T 1 g Avevi, g 1 r 2 [r = †K›`a †_‡K D”PZv]  T r A_©vr c„w_exi †K›`a †_‡K P I Q ̄’v‡bi `~iZ¡ h_vμ‡g r1 I r2 n‡j, T1 T2 = r1 r2  1.5 2 = r1 r2  r1 = 0.75 r2  r2 > r1 A_©vr Q ̄’v‡bi D”PZv †ewk wQj| 10| `ywU w ̄úas Gi w ̄úas aaæeK K1 = 1000 Nm–1 Ges K2 = 2000 Nm–1 | m = 4.5 kg I m f‡ii `ywU e ̄‘ wPÎ Abymv‡i hy3 †_‡K gm„Y †g‡S‡Z `yj‡Z mÿg| [Xv. †ev. 22] k1 m k2 wPÎ-1 k2 m wPÎ-2 k1 m k2 wPÎ-3 (K) f~w ̄’i DcMÖn Kv‡K e‡j? (L) mij †`vjMwZ ch©ve„ËMwZ; wKš‘ ch©ve„ËMwZ mij †`vjMwZ bqÑ e ̈vL ̈v Ki| (M) DÏxc‡Ki 2bs wP‡Î w ̄úas Gi †K.wYK K¤úv1⁄4 KZ n‡e? mgvavb: †K.wYK K¤úv1⁄4,  = k m = 2000 4.5  = 21.08 rad/s A_©vr 2bs wP‡Î w ̄úas-Gi †K.wYK K¤úv1⁄4 21.08 rad/s (Ans.) (N) m = 1kg n‡j DÏxc‡Ki 1bs I 3bs wP‡Îi w ̄úas ̧‡jvi ch©vqKvj mgvb n‡eÑ e3e ̈wU hvPvB Ki| mgvavb: 1bs wP‡Î Rb ̈: Equivalent spring constant Keq = K1 + K2 (mgvšÍivj mgev‡q hy3)  keq = 3000 Nm–1  T = 2 m Keq = 2 4.5 3000 T = 0.243 s 3bs wP‡Îi Rb ̈: Equivalent spring constant keq = k1k2 k1 + k2 (†kÖwY mgev‡q hy3) = 1000  2000 3000 = 666.7 Nm–1  T = 2 m keq = 0.243 s = T  T = T A_©vr 1bs I 3bs wP‡Îi w ̄úas- ̧‡jvi ch©vqKvj mgvb n‡e| (Ans.)