Tiêu đề 3.-Phy-1st-Paper-Gotibidda-Solve Sheet.pdf ✅

MwZwe` ̈v  HSC Preparation 1 MwZwe` ̈v Dynamics Z...Zxq Aa ̈vq ACS Physics Department Gi g‡bvbxZ cÖkœmg~n 1| GKwU †ejyb n‡Z GKwU †QvU †evZj c‡o hvIqvi 20 s ci GwU f~wg‡Z AvNvZ K‡i| †ejybwUi D”PZv wbY©q Ki hLbÑ (K) †ejybwU evZv‡m w ̄’i wQj| (L) †ejybwU 50 ms–1 `aæwZ‡Z DaŸ©Mvgx wQj| mgvavb: (K) †ejy‡bi Avw` †eM, u = 0 m/s [ evZv‡m w ̄’i] Avgiv Rvwb, h = ut + 1 2 gt2  h = 0  20 + 1 2  9.8  202  h = 1960 m  h = 1.96 km (Ans.) (L) †ejybwU 50ms–1 †e‡M DaŸ©Mvgx n‡j, u = 50 ms–1 h = – ut + 1 2 gt2 + ut  h = 1 2  9.8  202  h = 1960 m  h = 1.96 km (Ans.) 2| `yBwU †QvU †MvjK wfbœ wfbœ D”PZv n‡Z †Q‡o †`Iqvi ci GKB mg‡q f~wg‡Z G‡m AvNvZ K‡i| hw` cÖ_g †MvjKwU †Q‡o †`Iqvi 1.5 s ci wØZxq †MvjKwU †Qu‡o †`qv nq Ges cÖ_g †MvjKwU Quvovi 5 s ci †MvjKØq f~wg‡Z AvNvZ K‡i Z‡eÑ †MvjKØq †h †h D”PZv n‡Z †Q‡o †`qv n‡qwQj †mB D”PZv؇qi cv_©K ̈ wbY©q Ki| mgvavb: 1g †Mvj‡Ki †ÿ‡Î, h1 = u1t1 + 1 2 gt1 2 = 0  5 + 1 2  9.8  5 2  h1 = 122.5m 2q †Mvj‡Ki †ÿ‡Î, h2 = u2t2 + 1 2 gt2 2 = 0  3.5 + 1 2  9.8  (3.5)2  h2 = 60.025 m myZivs †MvjK؇qi D”PZvi cv_©K ̈ = (h1 – h2) = (122.5 – 60.025) m = 62.475 m (Ans.) 3| f~wg n‡Z h MfxiZvi GKwU Kzqvi GKwU cv_i‡K †Q‡o †`qvi t sec ci cv_iwU covi kã †kvbv †M‡jv| hw` k‡ãi `aæwZ v nq Z‡e t wbY©q Ki| mgvavb: awi, cv_iwU co‡Z cÖ‡qvRbxq mgq = t1 I kã Avm‡Z cÖ‡qvRbxq mgq = t2 cv_‡ii †ÿ‡Î, h = 1 2 gt1 2 [†h‡nZz cv_‡ii Avw`‡eM k~b ̈]  t1 = 2h g Avevi, k‡ãi †ÿ‡Î, h = vt2 = v(t – t1)  t = t1 + h v  t = 2h g + h v (Ans.) 4| GKwU e ̄‘‡K h D”PZv n‡Z †Q‡o w`‡j GwU f~wg‡Z 3 ms–1 `aæwZ‡Z cwZZ nq| GKB D”PZv n‡Z GKB f‡ii Av‡iKwU e ̄‘‡K 4 ms–1 `aæwZ‡Z wb‡Pi w`‡K wb‡ÿc Ki‡j wØZxq e ̄‘wU KZ †e‡M f~wg‡Z AvNvZ Ki‡e? mgvavb: cÖ_g e ̄‘i †ÿ‡Î, v 2 1 = u2 1 + 2gh  3 2 = 0 2 + 2  9.8  h  h = 45 98 m 2q e ̄‘i †ÿ‡Î, v 2 1 = 42 + 2gh  v 2 = 4 2 +     2  9.8  45 98 = 5 m/s (Ans.) 5| GKB mg‡q `ywU e ̄‘ Lvov Dc‡ii w`‡K wbwÿß nj| Zv‡`i GKwU AciwU A‡cÿv 44.2 m Dc‡i D‡V Ges 2 s c‡i gvwU‡Z c‡o| e ̄‘ `ywUi wbwÿß †eM wbY©q Ki| mgvavb: 1g e ̄‘i †ÿ‡Î, h = u1 2 2g ........ (1) T = 2u1 g ........ (2) 2q e ̄‘i †ÿ‡Î, h + 44.2 = u2 2 2g ... (3) T + 2 = 2u2 g ..... (4) mgxKiY (4) n‡Z mgxKiY (2) we‡qvM K‡i cvB, 2 = 2u2 g – 2u1 g  1 = 1 g (u2 – u1)  u2 – u1 = g ......... (5) Avevi, mgxKiY (3) n‡Z mgxKiY (1) we‡qvM K‡i cvB, 44.2 = u 2 2 2g – 2u2 1 2g  44.2 = u 2 2 – u 2 1 2g
2  Physics 1st Paper Chapter-3  (u2 + u1) (u2 – u1) 2g = 44.2  (u2 + u1)  g 2g = 44.2  u2 + u1 = 88.4 m ............... (6) Avevi, mgxKiY (5) I (6) †hvM K‡i cvB, 2u2 = g + 88.4  u2 = 9.8 + 88.4 2  u2 = 49.1 ms–1 (Ans.) u2 Gi gvb mgxKiY (6) ewm‡q cvB, 49.1 + u1 = 88.4  u1 = 88.4 – 49.1  u1 = 39.3 ms–1 (Ans.) 6| 44.1m Mfxi GKwU Kz‡c GKwU cv_i wbwÿß n‡jv| Kz‡ci g‡a ̈ k‡ãi †eM 340ms–1 n‡j cv_i wb‡ÿ‡ci gyûZ© †_‡K GwU cvwb‡Z cZ‡bi kã ïb‡Z AwZμvšÍ mgq †ei Ki| mgvavb: cv_i Kzqv‡Z co‡Z t1 mgq jvM‡j, s = ut1 + gt2 1  44.1 = 1 2  9.81  t 2 1 [ cv_‡ii Avw`‡eM k~b ̈]  t 2 1 = 9  t1 = 3 s GB `~iZ¡ AwZμg Ki‡Z k‡ãi t2 mgq jvM‡j, s = vt2  t2 = s v = 44.1 340 = 0.13 s wb‡ÿ‡ci gyn~Z© n‡Z kã †kvbv ch©šÍ †gvU AwZμvšÍ mgq = (t1 + t2) = (3 + 0.13) = 3.13 s (Ans.) 7| GKwU wgbv‡ii kxl© n‡Z 30 ms–1 †e‡M GKwU cv_i‡K Luvov Dc‡ii w`‡K wb‡ÿc Kiv n‡jv| 4 †m‡KÛ c‡i wØZxq GKwU cv_i‡K wb‡P †Qu‡o †`qv n‡jv Ges cv_i `ywU GKB m‡1⁄2 gvwU ̄úk© Kij| wgbv‡ii D”PZv I wØZxq cv_‡ii cZbKvj KZ? (g = 10ms–2 ) mgvavb: mgvavb: g‡bKwi, 2q e ̄‘wU wb‡ÿ‡ci t mgq ci h D”PZvi wgbv‡ii cv`‡`‡k e ̄‘Øq wgwjZ n‡e| 1g e ̄‘i †ÿ‡Î, cZbKvj = t + 4  h = – 30 (t + 4) + 1 2 g (t + 4)2 ..... (1) h u 2q e ̄‘i †ÿ‡Î, h = 1 2 gt2 ................ (2) [∵ Avw`‡eM 0 m/s] (1) I (2) bs n‡Z cvB, – 30 (t + 4) + 1 2 g (t + 4)2 = 1 2 gt2  1 2 g (t2 + 8t + 16) – 1 2 gt2 – 30t – 120 = 0  1 2 10 (t2 + 8t + 16) –     1 2  10t2 – 30t – 120 = 0  5t2 + 40t + 80 – 5t2 – 30t – 120 = 0  10t = 40  t = 40 10  t = 4 s (Ans.) wgbv‡ii D”PZv = 1 2 gt2 = 1 2  10  4 2 = 80 m (Ans.) 8| GKwU e›`y‡Ki ̧wj 100 km/hr †e‡M Dc‡ii w`‡K †Qvovq 15 cm cyiæ Qv` †f` Kivi ci A‡a©K †eM nvivq| ̧wjwU cybivq Dci n‡Z wb‡P G Qv`wU‡K AvNvZ Kivi ci Avi KZ`~‡i Qv`wU †f` Ki‡e? mgvavb: awi, ̧wjwU cÖ_‡g Qv‡`i wb‡Pi c„ô w`‡q x †e‡M cÖ‡ek K‡i Dc‡ii c„ô w`‡q x 2 †e‡M †ei nq| Gici ̧wjwU wKQz`~i Dc‡i D‡V Avevi Qv‡`i Dc‡ii c„‡ô x 2 †e‡MB wd‡i Av‡m| GLb x 2 †e‡M ̧wjwU Avevi Qv‡`i Dc‡ii c„ô w`‡q cÖ‡ek K‡i y `~iZ¡ AwZμg K‡i †_‡g hvq| u1 = x v1 = x 2 GLv‡b, g›`b a| Avgiv Rvwb, v1 2 = u1 2 – 2as1      x 2 2 = x 2 – 2a  0.15  x 2 –     x 2 2 = 2a  0.15 ............ (i)  2q evi hLb ̧wjwU Qv‡`i Dc‡ii c„ô w`‡q cÖ‡ek K‡i ZLb, v2 2 = u2 2 – 2as2 [ g›`b]  0 2 =     x 2 2 – 2a  y      x 2 2 = 2a  y ............ (ii) (ii) †K (i) bs Øviv fvM K‡i cvB, y 0.15 =     x 2 2 x 2 –     x 2 2  y = x 2 4 x 2     1 – 1 4  0.15 = 0.05 m = 5 cm (Ans.) 9| wP‡Î cÖ_g †dvUv gvwU‡Z ̄úk© Ki‡j PZz_© †dvUv co‡Z ïiæ K‡i| Zvn‡j cÖ_g †dvUv gvwU‡Z co‡j evwK †dvUv ̧‡jvi Ae ̄’vb †Kv_vq? [BUET 19-20]
MwZwe` ̈v  HSC Preparation 3 4 3 2 1 2.45 m mgvavb: (K) cÖ_g †duvUvi Rb ̈, h  1 2 gt2  2.45  1 2 gt2  t  0.707 s cici `yBwU †duvUv covi ga ̈eZ©x mgq  t 3  wØZxq †duvUvi Rb ̈ mgq  t – t 3  2t 3  f‚wg n‡Z wØZxq †duvUvi D”PZv, h2  h – 1 2 g     2t 3 2 = (2.45 – 1.089) m  1.361 m (Ans.) Z...Zxq †duvUvi Rb ̈ mgq  2t 3 – t 3  t 3  f‚wg n‡Z Z...Zxq †duvUvi D”PZv, h2  h – 1 2 g     t 3 2  (2.45 – 0.272) m  2.178 m (Ans.) 10| GKRb QvÎx c ̈vivmyU covi ci Nl©Ynxb fv‡e 50 m cwZZ nq| c ̈vivmyU †Lvjvi ci †_‡K †m 2.0 ms–2 g›`‡b wb‡Pi w`‡K cwZZ nq| f~wg‡Z †cuŠQvi gyû‡Z© Zvi `aæwZ 3.0 m/s| QvÎx KZÿY evqy‡Z wQj? [BUET 06-07] mgvavb: 50 m (g Z¡i‡Y) t1 mgq 2 m/s2 g›`‡b t2 mgq QvÎx v0 v1 v2 f‚wg mgvavb: c ̈vivmy ̈U †Lvjvi Av‡M, v 2 1 = (0)2 + 2  9.81  50  v1 = 31.32 ms–1 v1 = 0 + 9.81  t1  t1 = 3.192 s c ̈vivmyU †Lvjvi ci, v2 = v1 – at2  3 = 31.32 – 2  t2  t2 = 14.16 s †gvU mgq, t = t1 + t2 = 14.16 + 3.192 = 17.35 s (Ans.) 11| GKwU cv_i GKwU wbw`©ó D”PZv †_‡K 5 †m‡K‡Û f~wg‡Z cwZZ nq| cv_iwU‡K 3 †m‡KÛ ci _vwg‡q w`‡q Avevi co‡Z †`qv n‡jv| evvwK `~iZ¡ AwZμg K‡i cv_iwUi f~wg‡Z †cuŠQv‡Z KZ mgq jvM‡e? [RUET 15-16] mgvavb: h = 1 2 gt2 = 1 2  9.8  5 2 = 122.5m 3 s G AwZμvšÍ `~iZ¡ = 1 2  g  3 2 = 44.1m Aewkó `~iZ¡ = (122.5 – 44) m = 78.4 m  G `~iZ¡ AwZμ‡g cÖ‡qvRbxq mgq t n‡j, 1 2  g  t 2 = 78.4 [_vwg‡q †`Iqvq Avw`‡eM k~b ̈]  t 2 = 2  78.4 9.8 = 16  t = 4 s (Ans.) 12| GKwU i‡KU Lvovfv‡e Dc‡ii w`‡K 20 m/s2 Z¡i‡Y Pj‡Z ïiæ Ki‡jv| 10 sec c‡i i‡KUwUi Bwćbi myBP nVvr eÜ Kiv n‡j i‡KUwU m‡e©v”P KZ D”PZvq †cuŠQv‡e Zv wbY©q Ki| AwfKl©R Z¡iY-Gi gvb 10 m/sec2 [BUET 07-08] mgvavb: 10 s ci i‡K‡Ui †eM, v = at = 20  10 = 200 ms–1 Ges D”PZv, h = 1 2  20  (10)2 = 1000 m myZivs, 1000 m D”PZv IVvi ci myBP eÜ Ki‡j g g›`‡b i‡KUwU Dc‡i DV‡e|  m‡e©v”P D”PZv, H = 2002 2  10 = 2000 m A_©vr, i‡KUwU f~wg n‡Z me©vwaK (2000 + 1000) m ev, 3000 m D”PZvq DV‡e| (Ans.) 13| GKwU e ̄‘ †Kv‡bv UvIqv‡ii Dci w ̄’ive ̄’v n‡Z wb‡P cwZZ nIqvi mgq †kl GK †m‡K‡Û †gvU D”PZvi A‡a©K AwZμg K‡i| cZ‡bi mgq I UvIqv‡ii D”PZv wbY©q Ki| [g = 9.8 ms–2 ] [BUET 07-08] mgvavb: awi, UvIqv‡ii D”PZv h Ges †gvU cZbKvj t|  t Zg †m‡K‡Û, h 2 = u + 1 2  g  (2t – 1)  h = g(2t – 1) [∵ u = 0 m/s] Avgiv Rvwb, h = 1 2 gt2  1 2 gt2 = g(2t – 1)  t 2 2 = 2t – 1  t 2 – 4t + 2 = 0  t1 = 3.4142 s t2 = 0.585 s < 1 (MÖnY‡hvM ̈ bq)  h = 1 2  9.8  (3.4142)2 [∵ t > 1 n‡ZB n‡e] = 57.118m (Ans.)
4  Physics 1st Paper Chapter-3 14| x = 3t3 + 4t2 + 3t Øviv GKwU e ̄‘i miY wgUv‡i wb‡`©wkZ nq| 4 sec c‡i e ̄‘wUi Z¡iY wbY©q Ki| [KUET 07-08] mgvavb: x = 3t3 + 4t2 + 3t  v = dx dt = 9t2 + 8t + 3  a = dv dt = 18t + 8  4 sec ci Z¡iY = 18  4 + 8 = 80 ms–2 (Ans.) 15| GKwU e ̄‘KYvi †eM‡K vs = 0.10 ms–1 + (0.02 ms–1 )t 2 Øviv cÖKvk Kiv hvq| 2 sec I 5 sec Gi e ̄‘KYvi Mo Z¡iY KZ? [KUET 09-10] mgvavb: vs = 0.10ms–1 + (0.02ms–1 )t2 as = dvs dt = d dt [0.1 + 0.02 t2 ] as = 0.02  2t = 0.04t Mo Z¡iY, – a = a2 + a5 2 = (0.04  2) + (0.04  5) 2 = 0.08 + 0.2 2 = 0.14 ms–2 (Ans.) Alternative:  Mo Z¡iY, – a = v5 – v2 5 – 2 = 0.14 ms–2 (Ans.) GLv‡b, vt = 0.1 + 0.02t2 v2 = 0.18 ms–1 v5 = 0.6 ms–1 16| GKwU e ̄‘i Z¡iY ‘a’ m/sec2 mgq ‘t’ sec Gi mv‡_ a = 3t – 1 mgxKiY Abyhvqx cwiewZ©Z nq| t = 2 sec mg‡q e ̄‘wUi MwZ KZ n‡e? [RUET 11-12, BUET 07-08] mgvavb: a = dv dt  dv = adt   dv =  adt  v = 3  2 0 t dt –  2 0 dt = 3    t 2 2 2 0 – [t] 2 0 = 3 2  4 – 2 = 4 ms–1 (Ans.) 17| wgUv‡i cÖKvwkZ GKwU e ̄‘i Ae ̄’vb x(t) = 16t – 3t3 †hLv‡b mgq t †m‡K‡Û cÖKvwkZ| e ̄‘wU ÿwY‡Ki Rb ̈ w ̄’ive ̄’vq _v‡K hLb t Gi gvbÑ [BUET 12-13] mgvavb: x(t) = 16t – 3t3  v = dx dt = 16 – 9t2 w ̄’i _vK‡j, v = 0  16 – 9t2 = 0  t = 16 9 = 1.3333 s  1.30 s (Ans.) 18| †Kv‡bv e ̄‘i Ae ̄’v‡bi mgxKiY x = 3t2 – t 3 n‡j e ̄‘wU abvZ¥K x Aÿ eivei m‡e©v”P KZ `~i hv‡e? mgvavb: v = dx dt = 6t – 3t2 v Gi gvb k~b ̈ nIqvi AvM ch©šÍ e ̄‘wU x A‡ÿi abvZ¥K w`‡K hv‡e|  6t – 3t2 = 0  t = 2  m‡e©v”P miY, x = 3  2 2 – 2 3 = 4m (Ans.) 19| cÖ`Ë †jLwP‡Îi a – t MÖvd †_‡K v – t I x – t MÖvd AuvK| 0 1 0 5 sec Z¡iY (ms–2 ) 10 mgvavb: MwZkxj e ̄‘wUiwUi Z¡iY Amg hvi mgxKiYÑ a(t) = 2t; (0  t  5) Ges a(t) = 10; (5 < t < 10) GLb, (v – t) †jL Gi Rb ̈Ñ (0  t  5) sec mgq Gi g‡a ̈ e ̄‘wUi †eM, v =  5 0 adt =  5 0 2tdt =     2t2 2 5 0 = 25 ms–1 Abyiƒc fv‡e, (5 < t < 10) sec Gi g‡a ̈ t = 10 sec G e ̄‘wUi †eM, v = u + at = 25 + (10  5) = 75 ms–1  v – t †jLwPÎÑ 0 10 t(s) v (m/s) 5 25 75 x – t †jL Gi Rb ̈: (0 < t < 5) sec G e ̄‘wU KZ...©K AwZμvšÍ c_, x(5) =  5 0 vdt –  5 0 2tdt =     t 3 3 5 0 = 41.67 m Abyiƒcfv‡e, (5 < t < 10) sec G e ̄‘wUi AwZμvšÍ `~iZ¡ G mg‡qi g‡a ̈ v – t †jL G Ave× †ÿÎd‡ji mgvb| ZvB (5 < t < 10) sec G e ̄‘wUi AwZμvšÍ `~iZ¡, x = 1 2  (25 + 75)  5 = 250 m.  x – t †jLwPÎÑ 0 10 t(s) x (m) 5 41.67 291.67