Tiêu đề 7. P1C7 Structural Properties of Matter Merge Ok_Swapan 2.5.24 (Mahee) - Ok.pdf ✅

c`v‡_©i MvVwbK ag©  Practice Content 1 c`v‡_©i MvVwbK ag© Structural Properties of Matter mßg Aa ̈vq weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| weï× cvwbi msbg ̈Zv 5 × 10–12 cm2 /dyne Ges NbZ¡ 1 gm/cc n‡j 3 km MfxiZvq NbZ¡ KZ? [BUET 22-23] mgvavb: B = P × V V = P × Vi Vi – Vf K = 5 × 10–12 cm 2 /dyne = 5 × 10–11 m 2 /N  B = hfg × m i m i – m f  1 5 × 10–11 = (3 × 103 × f × 9.8) × f f – 1000  f = 1001.47 kgm–3 (Ans.) 2| mgvb •`N© ̈ I r = 0.5 cm e ̈vmv‡a©i `ywU B ̄úvZ Zv‡ii mvnv‡h ̈ 45 kg f‡ii GKwU UavwdK jvBU Szjv‡bv Av‡Q| hw` Zvi `ywU Abyf~wg‡Ki mv‡_ 15 †KvY •Zwi K‡i, Zvn‡j UavwdK jvB‡bi IR‡bi Rb ̈ Zvi `ywUi •`N© ̈ weK...wZi cwigvY KZ n‡e? †`Iqv Av‡Q, B ̄úv‡Zi Bqs-Gi ̧Yv1⁄4 2 × 1011 Nm–2 | [BUET 19-20] mgvavb: 2Tsin(15) = W  T = 45 × 9.8 2 × sin15  T = 851.9465 N GLb, Y = TL r 2 l  l L = T r 2Y W 15 T 15 T  l L = T r 2Y  l L = 851.9465  × 0.0052 × 2 × 1011 = 5.423 × 10–5 (Ans.) 3| `ywU Zv‡ii •`N© ̈ mgvb wKš‘ e ̈vm h_vμ‡g 3 mm Ges 6 mm| Zvi `yBwU‡K mgvb e‡j Uvb‡j cÖ_gwUi •`N© ̈ e„w× wØZxqwUi •`N© ̈ e„w×i wZb ̧Y nq| Zvi `ywUi g‡a ̈ †KvbwU †ewk w ̄’wZ ̄’vcK? MvwYwZK we‡køl‡Yi gva ̈‡g †Zvgvi gZvgZ e ̈3 Ki| [BUET 18-19] mgvavb: Y = FL r 2 l  Y  1 r 2 l  1 d 2 l  Y1 Y2 =     d2 d1 2 ×     l2 l1 =     6 3 2 ×     l2 3l2  Y1 Y2 = 4 3 > 1  Y1 > Y2 myZivs cÖ_g ZviwU AwaK w ̄’wZ ̄’vcK| (Ans.) 4| 2 mm e ̈v‡mi GKwU B ̄úv‡Zi Zv‡ii •`N© ̈ 15% e„w× Ki‡Z KZ kN ej cÖ‡qvM Ki‡Z n‡e? Gi d‡j Zv‡ii e ̈v‡mi KZUv cwieZ©b n‡e? [B ̄úv‡Zi Young’s Modulus 2 × 1011 Nm–2 Ges Poisso’'s ratio is 0.25] [BUET 17-18] mgvavb: cqm‡bi AbycvZ, 6 = d d L L  d = 6 × L L × d = 0.25 × 15% × 2  d = 0.075 mm  e ̈vm n«vm cv‡e 0.075 mm Avevi, Y = FL Al  F = YA × l L = 2 × 1011 ×  × 0.0012 × 15%  F = 94.247 kN (Ans.) 5| GKwU †`qvj n‡Z 4.8 cm e ̈v‡mi GKwU A ̈vjywgwbqv‡gi `Û Abyf~wgKfv‡e 5.3 cm cÖ‡ÿwcZ Av‡Q| `ÛwUi †kl cÖv‡šÍ 1200 kg f‡ii GKwU e ̄‘ †Svjv‡bv nj| A ̈vjywgwbqv‡gi e ̈eZ©b ̧Yv1⁄4 3 × 1010 Nm–2 | `ÛwUi fi‡K D‡cÿv K‡i (a) `ÛwUi Dci e ̈eZ©b cxob, Ges (b) `ÛwUi cÖv‡šÍi Djø¤^ wePz ̈wZ wbY©q Ki| [BUET 16-17] mgvavb: 5.3 cm 5.3 cm d  F e ̈eZ©b cxob = F A = 1200 × 9.8  × 0.0242 = 6.5 × 106 Nm–2 Avevi,  = F A tan  tan = 6.5 × 106 3 × 1010


4  Physics 1st Paper Chapter-7 23| 2 wU wewfbœ Dcv`v‡bi •Zwi Zv‡ii cÖ‡Z ̈KwUi •`N© ̈ 10 wg. Ges Bnv‡`i e ̈vm h_vμ‡g 2 wg. wg. Ges 4 wg. wg.| cÖ_g c`v‡_©i Bqs-Gi w ̄’wZ ̄’vcK ̧Yv1⁄4 wØZxq c`v‡_©i Bqs-Gi w ̄’wZ ̄’vcK ̧Yv1⁄4-Gi †P‡q Pvi ̧Y †ewk| •`N© ̈ eivei Df‡qi DciB 100 wK‡jvMÖvg IR‡bi ej cÖ‡qvM Ki‡j, cÖ_g I wØZxq Zv‡ii cÖmv‡ii Zzjbv evwni Ki| [CUET 05-06] mgvavb: Y = FL r 2 l  l  1 Yr 2  l1 l2 =    Y2  Y1 ×     r2 r1 2 = 1  l1 : l2 = 1 : 1 (Ans.) 24| `ywU wbw`©ó cÖvšÍwe›`yi ga ̈eZ©x 50 wgUvi j¤^v GKwU A ̈vjywgwbqvg Zv‡ii g‡a ̈ kxZKv‡j Uvbv ej 100 kN| kxZ I MÖx®§ Kv‡ji g‡a ̈ cvwicvwk¦©K ZvcgvÎv e ̈eavb 20C| hw` Zv‡ii e ̈vmva© 1 †m. wg., Dcv`v‡bi Zvcxq •`N© ̈ cÖmvivsK 20 × 10–6 /C Ges Bqs Gi ̧Yv1⁄4 1.1 × 107 Ncm–2 nq Z‡e MÖx®§Kv‡j Zv‡ii g‡a ̈ m„ó e‡ji cwigvY wbY©q Ki| [CUET 04-05] mgvavb: Avgiv Rvwb, MÖx®§Kv‡j Uvb ej n«vm cv‡e| F = YA  F = 1.1 × 107 10–4 ×  × 0.012 × 20 × 10–6 × 20  F = 13.823 kN  MÖx®§Kv‡j Zv‡i †gvU ej = kxZKv‡j Uvb e‡ji cwigvY – n«vmcÖvß Uvb ej = 100 – 13.823 = 86.177 kN (Ans.) 25| 20 cm2 cÖ ̄’‡”Q‡`i †ÿÎdjwewkó GKwU B ̄úvZ `ЇK 100C ZvcgvÎvq DËß K‡i 10C ZvcgvÎvq kxZj Kiv n‡jv| kxZjxKi‡Yi mgq `ÐwU‡Z GKwU ms‡KvPb e‡ji DTM¢e nq| B ̄úv‡Zi cÖmviY ̧YvsK 12  10–6 K –1 Ges Bqs ̧YvsK 2  1011 Nm–2 n‡j ms‡KvPb e‡ji gvb †ei K‡iv| [BUTex 23-24] mgvavb:L2 = L1(1 + T)  L2 – L = L1T  L L1 = T Avevi, Y = FL1 AL  Y = F Al L1 = F AT  F = YAT = 2  1011  2  10–4  12  10–6  90 = 4320 N (Ans.) 26| 3 m `xN© I 3 mm e ̈vmv‡a©i GKwU Zvi‡K Uvb‡j Zv‡ii •`N© ̈ e„w× nq 0.2 mm| ZviwUi e ̈vm KZUzKz Kg‡e? [cqm‡bi AbycvZ  = 0.2] [BUTex 20-21] mgvavb: cqm‡bi AbycvZ,  = d d L L  d = 0.2 × 0.2 3 × 103 × (3 × 2) × 10–3  d = 8 × 10–8 m  e ̈vm 8 × 10–8 n«vm cv‡e| (Ans.) 27| 1 m j¤^v Ges 1 mm e ̈v‡mi GKwU Zvi‡K GKwU û‡K †e‡a Aci cÖv‡šÍ ej cÖ‡qvM Kivq GwU 0.025 cm cwigvY j¤^v n‡jv| ZviwUi Bqs Ges ̧Yv1⁄4 2 × 1011 Nm–2 n‡j Kx cwigvY KvR m¤úbœ n‡qwQj? [BUETex 18-19] mgvavb: K...ZKvR, W = YAl 2 2L  W = 2 × 1011 × (0.0005) 2 × (0.025 × 10–2 ) 2 2 × 1  W = 4.91 × 10–3 J (Ans.) 28| 5 m •`N© ̈ Ges 1 mm e ̈vm wewkó Zv‡i 25 kg f‡ii d‡j •`N© ̈ 0.1 mm cÖmvwiZ n‡j ZviwUi mwÂZ kw3i cwigvY wbY©q Ki| [BUET 14-15] mgvavb: mwÂZ kw3i = 1 2 Fx = 1 2 × 25 × 9.8 × 0.1 × 10–3 = 0.01225 J (Ans.) 29| weK...wZi (Strain) gvÎv wjL| [BUTex 10-11] mgvavb: weK...wZi †Kv‡bv gvÎv †bB †Kbbv GwU `ywU GKB cÖKvi ivwki AbycvZ| (Ans.) 30| GKwU c`v‡_©i Dci cÖhy3 AvqZb cxob 3 × 108 Nm–2 Ges AvqZb weK...wZ 1.5 × 10–3 n‡j H c`v‡_©i Dcv`v‡bi AvqZb ̧Yv1⁄4 wbY©q Ki| [BUTex 06-07] mgvavb: AvqZb ̧Yv1⁄4 = AvqZb cxob AvqZb weK...wZ = 3 × 108 1.5 × 10–3 = 2 × 1011 Nm–2 (Ans.) weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv eûwbe©vPbx cÖkœmg~n 1. †Kv‡bv `‡Ûi Dcv`v‡bi Bqs Gi ̧Yv1⁄4 800 GPa| `‡Ûi •`N© ̈ 20% e„wׇZ cÖhy3 cxob = ? [BUET Preli. 22-23] 1.6  1010 Nm–2 1.6  1011 Nm–2 1.6  1012 Nm–2 1.6  1013 Nm–2 DËi: 1.6  1011 Nm–2 e ̈vL ̈v: Y = cxob weK...wZ  cxob = 800  109  20% = 1.6  1011 Nm–2