Tiêu đề 8. P1C8_পর্যাবৃত্ত গতি_Swapan Merged Ok-Swapan 25.4.24 (Nashita) Ok.pdf ✅

ch©ve„Ë MwZ  Varsity Question Bank 1 ch©ve„Ë MwZ Periodic Motion Aóg Aa ̈vq weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| 55 gm f‡ii e ̄‘ w ̄úas G mshy3 Ki‡j 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.) 2| 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] 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.) 3| 4000 kV wefe cv_©‡K ̈i ga ̈ w`‡q GKwU B‡jKUab Z¡wiZ n‡j MwZkxj Ae ̄’vq KYvi fi I †eM wbY©q Ki| [BUET 21-22] mgvavb: Ek = (m – m0) c2  4000  103  1.6  10–19 = (m – 9.11  10–31)  (3  108 ) 2  m = 8.022  10–30 kg (Ans) m0 = m 1 – v 2 c 2  9.11  10–31 = 8.022  10–30  1 – v 2 c 2  v = 2.98  108 ms –1 (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 (ii) Potertial Energy + kinetic Energy = Total Energy  1 2 kx2 + 1 2 mv 2 = 1 2 kA2  1 2  0.3  9.8 0.1  (5  10–2 ) 2 + 1 2  0.3v 2 = 1 2  0.3  9.8 0.1 (810–2 ) 2  v = 0.618 ms–1 (Ans.) Alternative way:  = 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 = 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 = 2  105  rads (Ans.) (b) vmax = A  1  103 = 2  105   A  A = 1.59  10–3 m (Ans.) 6| GKwU jvDW w ̄úKv‡ii k1⁄4z (cone) 262 Hz K¤úv‡1⁄4 mij Qw›`Z ̄ú›`‡bi ̄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]


4  Physics 1st Paper Chapter-8 x = 10 cos    5  2 +  4  x = 7.07 m (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 (M) a = dv dt = – 50  5 cos    5t +  4 = –250 2 cos    10 +  4 = –1744.71 ms–2 (Ans.) 20| 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.) 21| 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¤úvs1⁄4, 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  y = 20 sin(t + )  5 = 20 sin(  0 + )  sin = 0.25   = 0.2527 rad  Avw` `kv = 0.2527 rad 10 s c‡i `kv = (t + ) = (0.2094  10 + 0.2527) rad = 2.3467 rad (Ans.) 22| 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 = 3.60555 s (Ans.) 23| †Kvb †m‡KÛ †`vj‡Ki •`N© ̈ 200% evov‡j-Gi †`vjbKvj KZ n‡e? [RUET 06-07; BUTex 05-06] mgvavb: T2 T1 = L2 L1  T2 = 300 100  2 = 2 3 s (Ans.) 24| GKwU e›`y‡Ki w ̄úas‡K 4 cm msKzwPZ K‡i 10 gm f‡ii GKwU ̧wj †Quvov n‡jv| ̧wji †eM wbY©q K‡iv hLb w ̄úaswU c~e©ve ̄’vq wd‡i Av‡m| w ̄úas aaæeK 200 Nm–1 | [RUET 05-06] mgvavb: Kinetic Energy of Bullet = Total Energy of Spring  1 2 mv 2 = 1 2 kx2  v = kx2 m = 200  0.042 10  10–3 = 5.65685 ms–1 (Ans.) 25| 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 Kx n‡e? [RUET 05-06; BUTex 01-02] mgvavb: T  1 g  R M Tm Te = ge gm = Rm Re  Me Mm = 1 4  81 = 9 4  Tmoon =     9 4  2 s = 4.5 s (Ans)  Puv‡` 2wU Aa‡`vjb †`q 4.5 s G  3600wU Ó Ó 4.5  3600 2 s G = 8100 s G = 2.25 hour (Ans) (a) mg‡e‡M Pj‡j, a = 0  †`vjbKvj Gi †Kv‡bv cwieZ©b n‡e bv| (b) K...wÎg DcMÖ‡ni Af ̈šÍ‡i wb‡q †M‡j T = 2 L g – g =  n‡e (c) c„w_exi †K‡›`a g = 0, ZvB, T = 2 L 0 = 