Title 4. P1C4. HSC PREP Papers 26_With Solve.pdf ✅

wbDUwbqvb ejwe` ̈v  HSC Prep Papers 1 wbDUwbqvb ejwe` ̈v Newtonian Mechanics PZz_© Aa ̈vq Topicwise CQ Trend Analysis UwcK 2016 2017 2018 2019 2021 2022 2023 †gvU wbDU‡bi e‡ji m~‡Îi g‡a ̈ m¤úK© Ñ 1 1 Ñ Ñ 2 2 6 wbDU‡bi e‡ji Z...Zxq m~‡Îi e ̈envi I fi‡e‡Mi wbZ ̈Zv Ñ Ñ Ñ 1 Ñ Ñ Ñ 1 msNl©- (w ̄’wZ ̄’vcK msNl© I Aw ̄’wZ ̄’vcK msNl©) Ñ Ñ Ñ 1 Ñ 1 1 3 RoZvi åvgK I †K.wYK fi‡eM Ñ 1 1 3 7 3 2 17 †K.wYK ivwkgvjv (miY, †eM, Z¡iY, fi‡eM, MwZkw3) 2 5 Ñ 3 4 2 Ñ 16 UK© 1 Ñ Ñ 1 Ñ Ñ Ñ 2 †K›`agyLx I †K›`awegyLx ej Ñ 1 Ñ Ñ 1 3 2 7 iv ̄Ívi e ̈vswKs 2 2 1 2 6 3 4 20 * we.`a.: 2020 mv‡j GBPGmwm cixÿv AbywôZ nqwb| weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| 5 m cÖk ̄Í GKwU iv ̄Ívi GKwU wbw`©ó ̄’v‡bi euv‡Ki eμZvi e ̈vmva© 80 m| iv ̄Ívi Dfq cv‡ki D”PZvi cv_©K ̈ 0.4 m| euvK AwZμ‡gi c~‡e© GKwU Mvwo 54 kmh–1 †e‡M PjwQj| [Xv. †ev. 23] (K) cÖZ ̈qbx ej Kv‡K e‡j? (L) GKRb bZ©Kx bvPvi mgq nvZ msKzwPZ Ki‡j N~Y©‡b Kx myweav cvq? e ̈vL ̈v K‡iv| (M) euv‡Ki ̄’v‡b iv ̄Ívi e ̈vswKs †KvY wbY©q K‡iv| mgvavb: 0.4 m 5 m  Avgiv Rvwb, e ̈vswKs †KvY,  = sin–1     0.4 5   = 4.588 (Ans.) (N) DÏxcK Abyhvqx MvwowU D3 †e‡M wbivc‡` euvK wb‡Z cvi‡e wK? MvwYwZKfv‡e we‡kølY K‡iv| mgvavb: ÔMÕ n‡Z cvB, e ̈vswKs †KvY,  = 4.588 Avgiv Rvwb, tan = v 2 m rg Mvwoi eZ©gvb †eM, v0 = 54 3.6 ms–1 = 15 ms–1  vm = tan(4.588) × 80 × 9.8  vm = 7.931 ms–1 < 15 ms–1  v0 > vm nIqvq MvwowU wbivc‡` euvK wb‡Z cvi‡e bv| (Ans.) 2| 400 kg f‡ii GKwU Mvwo 60 kmh–1 mg‡e‡M 10 †Kv‡Y bZ Zj eivei Dc‡i D‡V| [Nl©Y ̧Yv1⁄4  Gi gvb 0.3 Ges g Gi gvb 9.8 ms–2 ] [Xv. †ev. 23] (K) WvBfvi‡RÝ Kv‡K e‡j? (L) GKwU w ̄úas-†K LwÐZ Ki‡j Gi aaæeK cwieZ©b n‡e wK? e ̈vL ̈v K‡iv| (M) Mvwoi Dci wμqvkxj weiæ× e‡ji gvb wbY©q K‡iv| mgvavb: F f mgsin v Mvwoi Dci wμqvkxj †gvU weiæ× ej = mgsin + f = mgsin + mgcos [R = mgcos] = 400 × 9.8 × sin(10) + 0.3 × 400 × 9.8 × cos(10) = 1838.83 N (Ans.) (N) Mvwoi Bwćbi ÿgZv KZ n‡j MvwowU mg‡e‡M bv P‡j eis Z¡iY cÖvß n‡e? MvwYwZK we‡kølY `vI| mgvavb: F f mgsin v [mg‡e‡M Pjvq, F = mgsin + mgcos] eZ©gv‡b Mvwoi BwÄb KZ...©K cÖhy3 ÿgZv, P = Fv = (mgsin + mgcos)v  P = 1838.83 × 60 3.6  P = 30647.246 W  P > 30647.246 W n‡j MvwowU mg‡e‡M bv P‡j eis Z¡iY cÖvß n‡e| 3| GKwU e„ËvKvi PvKwZ Gi Z‡ji mv‡_ ga ̈we›`yMvgx j¤^ Aÿ AB mv‡c‡ÿ wPÎ-1 Abyhvqx Nyi‡Q| Avevi wPÎ-2 Abyhvqx PvKwZwUi mv‡_ j¤^fv‡e ̄’vwcZ XY A‡ÿi mv‡c‡ÿ Nyi‡Q| [iv. †ev. 23] m = 2 kg r = 0.5 m h = 0.25 m B G A Y G X h (K) RoZvi åvgK Kv‡K e‡j? (L) e„ËvKvi c‡_ euvK cvi nIqvi mgq GKRb mvB‡Kj Av‡ivnx †n‡j hvq †Kb?


4  HSC Physics 1st Paper Chapter-4 11| [w`. †ev. 23] A P Q 1m B P Q 0.5 kg f‡ii AB `ÐwU PQ Aÿ mv‡c‡ÿ cÖwZ wgwb‡U 60 evi †Nviv‡bv n‡jv| cieZ©x‡Z `ÐwU‡K Gi •`‡N© ̈i mgvb e ̈v‡mi GKwU cvZjv PvKwZ‡Z cwiYZ K‡i PQ Aÿ mv‡c‡ÿ wgwb‡U 70 evi †Nviv‡bv n‡jv| (K) GK wbDUb ej Kx? (L) e ̈vswKs †Kv‡Yi gvb evov‡j iv ̄Ívi euv‡K Mvwo Pvjv‡bvi MwZmxgv ev‡oÑe ̈vL ̈v K‡iv| (M) PvKwZwUi †KŠwYK fi‡eM wbY©q K‡iv| mgvavb: †K.wYK fi‡eM, L = I  L = 1 2 mr2 × 2N t  L = 1 2 × 0.5 × 0.52 × 2 × 70 60  L = 0.458 kgm2 s –1 (Ans.) (N) †Kvb †ÿ‡Î N~Y©‡bi Rb ̈ Kg kw3i cÖ‡qvRb n‡eÑ MvwYwZKfv‡e e ̈vL ̈v K‡iv| mgvavb: `‡Ði †ÿ‡Î, MwZkw3, E1 = 1 2 I1 2 1  E1 = 1 2 ×     1 3 mL2 ×     2N t 2  E1 = 1 2 ×     1 3 × 0.5 × 12 ×     2 × 60 60 2  E1 = 3.289 J PvKwZi †ÿ‡Î, MwZkw3, E2 = 1 2 I2 ×  2 2  E2 = 1 2 ×     1 2 mr2 ×     2 × 70 60 2  E2 = 1.68 J  PvKwZwU Nyiv‡Z Kg kw3 cÖ‡qvM Ki‡Z n‡e| (Ans.) 12| 8 m cÖ‡ ̄’i iv ̄Ív w`‡q GKwU Mvwo h_vμ‡g 100 m I 80 m e ̈vmv‡a©i `yBwU euvK AwZμg Ki‡jv| iv ̄Ívi wfZ‡ii I evB‡ii cÖv‡šÍi D”PZvi euvK AwZμg Ki‡jv| iv ̄Ívi wfZ‡ii I evB‡ii cÖv‡šÍi D”PZvi cv_©K ̈ 0.4 m| [g. †ev. 23] (K) PμMwZi e ̈vmva© Kv‡K e‡j? (L) U‡K©i w`K Kxfv‡e wbY©q Ki‡e? (M) DÏxc‡K DwjøwLZ iv ̄Ívi e ̈vswKs †KvY wbY©q K‡iv| mgvavb: 0.4 m 8 m  C A B e ̈vswKs †KvY,  = sin–1     BC AC   = sin–1     0.4 8   = 2.87 (Ans.) (N) MvwowU Dfq euvK wK mgvb †e‡M AwZμg Ki‡Z cvi‡eÑ MvwYwZKfv‡e e ̈vL ̈v K‡iv| mgvavb: 100 m e ̈vmv‡a©i iv ̄Ívi †ÿ‡Î, tan = v 2 1 r1g  v1 = tan × r1 × g  v1 = tan(2.87) × 100 × 9.8  v1 = 7 ms–1 Avevi, 80 m e ̈vmv‡a©i ev‡Ki †ÿ‡Î, tan = v 2 2 r2g  v2 = tan × r2 × g = tan(2.87) × 80 × 9.8  v2 = 6.27 ms–1  Mvwoi †eM  6.27 ms–1 n‡j Dfq evK mgvb †e‡M AwZμg Ki‡Z cvi‡e| 13| 1000 kg f‡ii GKwU evm 75000 J MwZkw3 wb‡q Pjvi mgq 100 m e ̈vmva© wewkó GKwU evu‡Ki m¤§yLxb n‡jv| iv ̄Ívi cÖ ̄’ 10 m Ges iv ̄Ívi cÖvšÍ؇qi ga ̈eZ©x D”PZvi e ̈eavb 0.2m. [Xv. †ev. 22] (K) e‡ji NvZ Kv‡K e‡j? (L) big gvwU‡Z jvd w`‡j AvNvZ cvIqvi m¤¢vebv KgÑ e ̈vL ̈v Ki| (M) evmwUi fi‡eM wbY©q Ki| DËi: 12247.45 kgms–1 (N) MwZ‡eM bv Kwg‡q evmwU wbivc‡` evuKwU AwZμg Ki‡Z cvi‡e wK? MvwYwZKfv‡e we‡kølY Ki| DËi: cvi‡e bv 14| 72 kmh–1 †e‡M Pjgvb 1800 kg f‡ii GKwU eo Mvwo mvg‡b `vuwo‡q _vKv 1000 kg f‡ii GKwU †QvU Mvwo‡K wcQb w`K †_‡K m‡Rv‡i av°v w`‡jv| av°vi ci Mvwo `yÕwU GKwÎZ n‡q 60 m wM‡q †_‡g †M‡jv| iv ̄Ívi cv‡k `vuwo‡q _vKv weÁv‡bi QvÎ gvnx, `yN©UbvwU ch©‡eÿY K‡i ejj GwU GKwU Aw ̄’wZ ̄’vcK msNl©| [P. †ev. 22] (K) UK© Kx? (L) iv ̄Ívi evu‡K mvB‡Kj Av‡ivnx‡K †fZ‡ii w`‡K †n‡j evuK wb‡Z nq †Kb? e ̈vL ̈v Ki| (M) DÏxc‡Ki Mvwo `yÕwU _vgv‡Z †h evuav`vbKvix ej wμqvkxj wQj Zvi gvb wbY©q Ki| mgvavb: •iwLK fi‡e‡Mi wbZ ̈Zv Abymv‡i, m1u1 + m2u2 = (m1 + m2)v  1800 × 72 3.6 + 1000 × 0 = (1800 + 1000) × v  v = 12.857 ms–1  F = (m1 + m2) × a  F = (m1 + m2) × v 2 2s  F = (1800 + 1000) × 12.8572 2 × 60  F = 3857.05 N (Ans.)