Tiêu đề 8. P1C8_পর্যাবৃত্ত গতি_Ridoy_(With Solve).pdf ✅

ch©ve„Ë MwZ  Engineering Practice Sheet.......................................................................................................................... 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. mij Qw›`Z ̄ú›`‡b ̄úw›`Z †Kv‡bv KYvi we ̄Ívi 10 cm| mvg ̈ve ̄’v †_‡K 6 cm `~‡i _vKv Ae ̄’vq evZv‡mi Kvi‡Y KYvwUi †eM wØ ̧Y n‡j bZzb we ̄Ívi KZ? [BUET 23-24 cÖ‡kœi Abyiƒc] mgvavb: v2 v1 = A 2 2 – x 2 A 2 1 – x 2  2 2 = A 2 2 – 6 2 102 – 6 2  A2 = 17.088 cm (Ans.) 2. 55 gm f‡ii GKwU e ̄‘ w ̄úas G mshy3 Ki‡j w ̄úaswU 8.1 cm msKzwPZ nq| k = 1.8 N/m n‡j (i) mvg ̈ve ̄’v †_‡K 5 cm `~i‡Z¡ Gi †eM KZ? (ii) e ̄‘i m‡e©v”P †eM KZ? [BUET 22-23] mgvavb: (i) v =  A 2 – x 2 = k m × A 2 – x 2  v = 1.8 55 × 10–3 × 0.0812 – 0.052  v = 0.3645 ms–1 (Ans.) (ii) vmax = A = k m A = 1.8 55 × 10–3 × 0.081 = 0.4633 ms–1 (Ans.) 3. 0.1 kg f‡ii GKwU KYv 0.1 m we ̄Ív‡ii mv‡_ mij †`vjb MwZ m¤úbœ Ki‡Q| hLb KYvwU mvg ̈ve ̄’vq _v‡K, ZLb Zvi MwZkw3 8  10–3 J| KYvwUi K¤ú‡bi Avw` `kv 45 n‡j, Gi MwZi mgxKiY †ei Ki| [BUET 21-22, 19-20] mgvavb: Ekmax = 8  10–3  1 2 m 2A 2 = 8  10–3  1 2  0.1   2  0.12 = 8  10–3   = 4 rads–1  MwZi mgxKiY, x = 0.1 sin     4t +  4 m (Ans.) 4. GKwU w ̄úas‡qi AMÖfv‡M 0.3 kg fiwewkó GKwU e ̄‘ Szjv‡bv n‡j w ̄úaswU 0.10 m j¤^v nq| w ̄úawU‡K GB mvg ̈ve ̄’v †_‡K AviI 8  10–2 m j¤^v K‡i †Q‡o †`qv n‡jv| (i) e ̄‘i †gvU kw3 KZ? (ii) mvg ̈ve ̄’v †_‡K 5  10–2 m `~‡i Ae ̄’vbKv‡j e ̄‘wUi †eM KZ? [BUET 20-21] mgvavb: (i) †gvU kw3 = 1 2 kA2 = 1 2  F x  A 2 = 1 2  0.3  9.8 0.10  (8  10–2 ) 2 = 0.094 J (Ans.) (ii)  = g e = 9.8 0.1   = 7 2 rads–1 v =  A 2 – x 2 = 7 2 (8  10–2 ) 2 – (5  10–2 ) 2  v = 0.618 ms–1 (Ans.) 5. 1.00  10–20 kg f‡ii GKwU KYvi mij Qw›`Z MwZi †`vjbKvj 1.00  10–5 s Ges Zvi m‡e©v”P MwZ‡eM 1.00  103 ms–1 | KYvwUi (a) †KŠwYK K¤úv1⁄4 Ges (b) m‡e©v”P miY wbY©q K‡iv| [BUET 18-19] mgvavb: (a) T = 1  10–5 s  = 2 T = 2 10–5   = 6.28  105 rads–1 (Ans.) (b) vmax = A  1  103 = 2  105   A  A = 1.59  10–3 m (Ans.)
2 .........................................................................................................................................  Physics 1st Paper Chapter-8 6. GKwU jvDW w ̄úKv‡ii k1⁄4z (cone) 262 Hz K¤úv‡1⁄4 mij Qw›`Z ̄ú›`‡b ̄úw›`Z nq| k1⁄4zi †K‡›`ai we ̄Ívi A = 1.5  10–4 m Ges t = 0 mg‡q miY x = A nq| k1⁄4zi †K‡›`ai MwZ eY©bvKvix mgxKiYwU wbY©q K‡iv| k1⁄4zi †eM I Z¡iY‡K mg‡qi dvskb wn‡m‡e cÖKvk K‡iv| [BUET 14-15] mgvavb: f = 262 Hz,  = 2f MwZi ïiæ we ̄Ívi Ae ̄’vb †_‡K nvIqvq, x = A cos (t)  x = 1.5  10–4 cos (2  262t) v = dx dt = – 1.5  10–4  2  262 sin (2  262t) v = – 0.0786 sin (524t) ms–1 (Ans.)  a = dv dt  a = – 0.0786  524 cos (524t)  a = –41.1864 2 cos (524t) ms–2 (Ans.) 7. hLb 1 kg Av`k© fi GKwU Pjgvb cøvUd‡g©i Dci ivLv nq ZLb Zvi ̄ú›`‡bi nvi 125 vibmin–1 | ARvbv KZ f‡ii Rb ̈ ̄ú›`‡bi nvi 243 vibmin–1 n‡e? Pjgvb cøvUd‡g©i fi AMÖvn ̈ Ki| [BUET 10-11] mgvavb: T = 2 m k = 1 f  f  1 m  f1 f2 = m2 m1  125 243 = m2 1  m2 = 0.2646 kg (Ans.) 8. GKwU w ̄’i wjd‡Ui g‡a ̈ ivLv GKwU mij †`vj‡Ki †`vjbKvj T| hw` wjdUwU Dc‡ii w`‡K g 4 Z¡iY wb‡q D‡V, Zvn‡j †`vjKwUi †`vjbKvj KZ n‡e? [BUTex 08-09; BUET 03-04] mgvavb: T = 2 L g  T  1 g  T2 T = g g2 = g g + g 4  T2 = 2 5 T (Ans.) [wjdU hLb Dc‡i D‡V ZLb wμqviZ Z¡iY (g + a)] 9. †m‡KÛ †`vj‡Ki •`N© ̈ 1% e„w× Ki‡j, D3 †`vjK w`‡b KZ mgq nviv‡e? [BUET 07-08] mgvavb: T  L  T T = 86400 86400 – x = 101 100  x = 428.78677 s (Ans.) 10. GKwU †m‡KÛ †cÛzjvgwewkó Nwo cÖwZw`b Avav wgwbU (30 sec) jvf K‡i| †cÛzjvgwU mwVK mgq w`‡Z n‡j Dnvi mij †`vj‡Ki •`‡N© ̈i Kx cwigvY n«vm-e„w× NUv‡Z n‡e? (g = 980 cm–2 ) [BUET 07-08] mgvavb: bZzb •`N© ̈ L2 n‡j, L2 L1 =     T2 T1 2 =       2 2  86400 86400 + 30 2 =     86400 + 30 86400 2  L2 – L1 L1  100% = 0.0694%  •`N© ̈ e„w× Ki‡Z n‡e 0.0694%| (Ans.) 11. mij Qw›`Z ̄ú›`bm¤úbœ GKwU e ̄‘i †eM 3 ms–1 hLb miY 4 m Ges †eM 4 ms–1 hLb miY 3 m| (a) †`vj‡bi we ̄Ívi I ch©vqKvj wbY©q Ki| (b) e ̄‘wUi fi 50 kg n‡j †`vj‡bi †gvU kw3 wbY©q Ki| [BUET 05-06] mgvavb: (a) v1 =  A 2 – x 2 1  3 =  A 2 – 16 ....... (i) v2 =  A 2 – x 2 2  4 =  A 2 – 9 .......... (ii)  3 4 = A 2 – 16 A 2 – 9  9 16 = A 2 – 16 A 2 – 9  A 2 = 25  A = 5 m (Ans)  3 =  25 – 16   = 1 rads–1  T = 2  = 2 s (Ans.) (b) Etotal = 1 2 m 2A 2 = 1 2  50  1  5 2  Etotal = 625 J (Ans.) 12. GKwU IRb gvcvi w ̄úas wbw3i Dci `vov‡bvi ci Zzwg jÿ ̈ Ki‡j †h mvg ̈v ̄’vq Avmvi c~‡e© wbw3i KuvUvwU mvg ̈ve ̄’vi `ycv‡k K‡qKevi †`vj Lvq| †`vjbKvj 0.8 †m‡KÛ n‡j Ges †Zvgvi fi 64 kg n‡j wbw3i w ̄úas aaæeK KZ?[BUET 01-02] mgvavb: T = 2 m k  0.8 = 2 64 k  k = 3947.84 Nm–1 (Ans.)
ch©ve„Ë MwZ  Engineering Practice Sheet......................................................................................................................... 3 13. GKwU e ̄‘i mij Qw›`Z MwZ x = 6.0 cos     3t +  3 m mgxKiY Øviv wee„Z Kiv hvq| t = 2 sec mg‡q (i) miY, (ii) †eM Ges (iii) Z¡iY †ei K‡iv| [BUET 00-01] mgvavb: (i) x(t = 2s) = 6 cos     3  2 +  3 = 3 m (Ans.) (ii) v(t = 2s) = dx dt = – 6  3 sin     6 +  3 = –48.98 ms–1 (Ans.) (iii) a(t = 2s) = dv dt = – 6  3  3 cos     3  2 +  3 = –266.479 ms–2 (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 1. AwfKl©R Z¡iY g = 9.8 ms–2 Gi Rb ̈ GKwU †m‡KÛ †`vj‡Ki •`N© ̈ wbY©q K‡iv| AwfKl©R Z¡iY 10% Kwg‡q D3 †`vj‡Ki †`vjbKvj wVK ivL‡Z n‡j •`‡N© ̈i Kx cwieZ©b n‡e? [KUET 19-20] mgvavb: T = 2 L g  L = gT2 4 2 = 9.8  2 2 4 2 = 0.9929 m (Ans.)  L  g L2 L1 = g2 g1 = 90 100 = 0.9  L1 – L2 = (1 – 0.9) × L1 = 0.1 × 0.9929 = 0.09929 m n«vm cv‡e| (Ans.) 2. GKwU †m‡KÛ †`vjK Nwo cvnv‡oi cv`‡`‡k wVK mgq †`q wKš‘ cvnv‡oi P‚ovq DVv‡j 2 NÈvq 8 †m‡KÛ mg‡qi cv_©K ̈ †`Lvq| c„w_exi e ̈vm 12800 km n‡jÑ [KUET 17-18] (i) cvnv‡oi D”PZv wbY©q K‡iv| (ii) cvnv‡oi P‚ovq mwVKfv‡e KvR Ki‡Z n‡j †`vj‡Ki •`N© ̈ KZ % cwieZ©b Ki‡Z n‡e? mgvavb: (i) 2 NÈvq 8 s ax‡i P‡j  24 NÈvq (8  12) s = 96 s ax‡i P‡j T T = 86400 86400 – x Avevi, T  1 g  R  T T = 86400 86400 – x = R + h R  86400 86400 – 96 = 6.4  106 + h 6.4  106  h = 7119.02 m (Ans.) (ii) T = 2 L g  T  L L2 L1 =     T2 T1 2 =       2 2 × 86400 86400 – 96 2 =     86400 – 96 86400 2  L1 – L2 L1 × 100% = (1 – 0.997779) × 100% = 0.222% n«vm Ki‡Z n‡e| (Ans.) 3. GKwU wjd‡Ui Qv` †_‡K GKwU mij †`vjK Szjv‡bv Av‡Q| wjdU Pjvi mgq GB †`vj‡Ki †`vjbKvj wjd‡Ui w ̄’i Ae ̄’vi Zzjbvq hw` A‡a©K nq, Z‡e wjd‡Ui Z¡i‡Yi w`K I gvb wbY©q K‡iv| [KUET 05-06] mgvavb: T  1 g T2 T1 = g1 g2 = 1 2  g1 g2 = 1 2  g2 > g1  wjdU Dc‡ii w`‡K MwZkxj, Z¡iY a n‡j, g g + a = 1 4  a = 29.4 ms–2 (Ans.) 4. GKwU cvnv‡oi cv`‡`‡k GKwU †m‡KÛ †`vjK mwVK mgq †`q| GwU‡K cvnv‡oi m‡e©v”P k„‡1⁄2 wb‡q †M‡j cÖwZw`b 2 wgwbU ax‡i P‡j| cvnv‡oi D”PZv wbY©q K‡iv| (c„w_exi e ̈vmva© 6400 km) [KUET 04-05] mgvavb: T T = 86400 86400 – 2  60 = g g = R + h R  86400 86400 – 120 = 6.4  106 + h 6.4  106  h = 8901.251739 m (Ans.)
4 .........................................................................................................................................  Physics 1st Paper Chapter-8 weMZ mv‡j RUET-G Avmv cÖkœvejx 1. c„w_exc„‡ô GKwU mij †`vj‡Ki †`vjbKvj 2 sec| G‡K P›`ac„‡ô wb‡j Gi †`vjbKvj nq 4.5 sec| c„w_exi fi I P‡›`ai f‡ii AbycvZ 81 n‡j c„w_exi e ̈vmva© I P‡›`ai e ̈vmv‡a©i AbycvZ wbY©q K‡iv| [RUET 11-12, 03-04] mgvavb: T = 2 L g ; g = GM R 2  g  M R 2  T  1 g  R M  Tearth Tmoon = Rearth Rmoon  Mmoon Mearth  Rearth Rmoon = 2 4.5  81 = 4 (Ans.) 2. hw` †Kvb ̄’v‡b GKwU †m‡KÛ †`vj‡Ki •`N© ̈ 1 m nq, Z‡e †h †`vjK †mB ̄’v‡b wgwb‡U 20 evi †`vj †`q, Zvi •`N© ̈ †ei K‡iv| [RUET 10-11] mgvavb: 20 evi †`vj †`q 60 s G  1 evi †`vj †`q 60 20 s G = 3 s G  T2 = 3 s, T1 = 2 s ∵ T  L  T2 T1 = L2 L1  L2 =     T2 T1 2  L1  L2 = 9 4  L1  L2 = 2.25 m (Ans.) 3. GKwU e ̄‘i mij Qw›`Z MwZ x = 10.0 cos    5t +  4 m mgxKi‡Yi mvnv‡h ̈ cÖKvk Kiv hvq| t = 2 s mg‡q D3 e ̄‘i (K) miY (L) †eM I (M) Z¡iY wbY©q K‡iv| [RUET 09-10] mgvavb: x = 10.0 cos     5t +  4 m (K) t = 2 s G, x = 10 cos     5  2 +  4  x = 7.07 m (Ans.) (L) v = dx dt = – 10  5 sin     5t +  4 = – 50 sin     5t +  4 t = 2s G, v = – 50 sin     10 +  4  v = –111.07 ms–1 (Ans.) (M) a = dv dt = – 50  5 cos     5t +  4 = –250 2 cos     10 +  4  a = –1744.71 ms–2 (Ans.) 4. GKwU †m‡KÛ †`vj‡Ki •`N© ̈ ivRkvnx‡Z 95 cm Ges PÆMÖv‡g 100 cm| †Kvb e ̄‘i IRb ivRkvnx‡Z 95 gm-wt n‡j, PÆMÖv‡g Dnvi IRb KZ? [RUET 09-10, 06-07; CUET 04-05] mgvavb: T = 2 L g  L  g  W WC WR = LC LR  WC = 100 95  95 = 100 gm-wt (Ans.) 5. mij Qw›`Z MwZm¤úbœ GKwU KYvi MwZi mgxKiY, y = 20 sin (t + ) cm| GB MwZi ch©vqKvj 30 s Ges Avw` miY 5 cm n‡j KYvwUi †KŠwYK K¤úv1⁄4, Avw` `kv I 10 s c‡ii `kv wbY©q K‡iv| [RUET 07-08] mgvavb: †KŠwYK K¤úv1⁄4,  = 2 T = 2 30 = 0.2094 rads–1 (Ans.)  y = 20 sin (t + )  5 = 20 sin (  0 + )  sin = 0.25   = 0.2527 rad  Avw` `kv = 0.2527 rad (Ans.) 10 s c‡i `kv = (t + ) = (0.2094  10 + 0.2527) rad = 2.3467 rad (Ans.) 6. GKwU †m‡KÛ †`vj‡Ki •`N© ̈ 225% evov‡bv n‡j Gi †`vjbKvj KZ n‡e wbY©q K‡iv| [RUET 06-07; BUTex 05-06] mgvavb: T  L  T2 T1 = L2 L1  T2 = 325 100  2  T2 = 3.60555 s (Ans.) 7. GKwU †m‡KÛ †`vjK f‚c„‡ô mwVK mgq †`q| P‡›`a wb‡q †M‡j Gi †`vjbKvj KZ n‡e? c„w_exi fi P‡›`ai f‡ii 81 ̧Y Ges c„w_exi e ̈vmva© P‡›`ai e ̈vmv‡a©i 4 ̧Y| c„w_ex‡Z KZ mgq AwZμvšÍ n‡j P‡›`ai †`vjKwU 1 NÈv †`Lv‡e? †`vjKwU‡K (a) mg‡e‡M PjšÍ wjd‡U wb‡j (b) N~Y©vqgvb K...wÎg DcMÖ‡ni Af ̈šÍ‡i wb‡j Ges (c) c„w_exi †K‡›`a wb‡j Kx n‡e? [RUET 05-06; BUTex 01-02]