Tiêu đề 7. P1C7 Structural Properties of Matter_With Solve.pdf ✅

c`v‡_©i MvVwbK ag©  Mastery Practice Sheet 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 cm2 /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  d 5.3 = 6.5 × 106 3 × 1010

c`v‡_©i MvVwbK ag©  Mastery Practice Sheet 3  Y1 Y2 = 2 2 × 1 3 = 4 3  Y1 : Y2 = 4 : 3 (Ans.) 15| B ̄úv‡Zi Bqs-Gi ̧Yv1⁄4 2.1 × 1011 Nm–2 I cvwbi AvqZ‡bi w ̄’wZ ̄’vcK ̧Yv1⁄4 2.1 × 109 Nm–2 | B ̄úvZ I cvwbi NbZ¡ h_vμ‡g 7.8 × 103 kgm–3 I 103 kgm–3 | B ̄úvZ I cvwb‡Z k‡ãi †e‡Mi Zzjbv Ki| [KUET 04-05] mgvavb: Avgiv Rvwb, B ̄úv‡Z k‡ãi †eM, vs = Y s = 2.1 × 1011 7.8 × 103  vs = 5188.745 ms–1 cvwb‡Z k‡ãi †eM, vw = B w = 2.1 × 109 103  vw = 1449.137 ms–1 (Ans.) 16| mgvb •`‡N© ̈i `ywU wfbœ c`v‡_©i Zv‡ii •`N© ̈ eivei mgvb ej cÖ‡qvM Kiv nj| d‡j wØZxq ZviwU cÖ_gwUi 2.5 ̧Y cÖmvwiZ nj| Zvi `ywUi Bqs Gi ̧Yv1⁄4 h_vμ‡g 1.8 × 1011 Nm–2 I 1.6 × 1011 Nm–2 | G‡`i e ̈v‡mi AbycvZ wbY©q Ki| [RUET 11-12, 03-04] mgvavb: Y = FL r 2 l  r 2 = FL Yl  r 2  1 Yl      r1 r2 2 = Y2 Y1 × l2 l1  d1 d2 = 1.6 1.8 × 2.5 l1 l1 = 2 5 3  d1 : d2 = 2 5 : 3 (Ans.) 17| 2 mm2 cÖ ̄’‡”Q‡`i GKwU Zv‡ii mv‡_ 15 kg fi Szjv‡bv Av‡Q| fi Szjv‡bv Ae ̄’vq ZviwUi •`N© ̈ 4 m| Zv‡`i Dcv`v‡bi Bqs-Gi ̧Yv1⁄4 1.3 × 1010 Nm–2 | fiwU mwi‡q wb‡j ZviwUi •`N© ̈ Kx cwigvY msKzwPZ n‡e? [RUET 08-09] mgvavb: Y = FL Al  1.3 × 1010 = 15 × 9.8 × (4 –x) 2 × 10–6 × x awi, •`N© ̈ e„w×, l = x  Avw` •`N© ̈, L = 4 – x  x = 22.488 mm msKzwPZ n‡e (Ans.) 18| †Kvb Zv‡ii •`N© ̈ 3 m Ges fi 20 gm| 50 N Uv‡b Gi •`N© ̈ 1 mm ev‡o| Bqvs Gi ̧Yv1⁄4 wbY©q Ki| Zv‡ii NbZ¡ 7.5 × 103 kgm–3 | [RUET 07-08, 05-06] mgvavb: Y = FL Al  = m v = m AL  Y = FL m L × l = FL 2 ml  Y = 50 × 7.5 × 103 × 32 20 × 10–3 × 0.001 = 1.6875 × 1011 Nm–2 (Ans.) 19| `yBwU Zv‡ii cÖ‡Z ̈KwUi •`N© ̈ 3 m Ges G‡`i Bqs Gi ̧Yv1⁄4 h_vμ‡g 1.6 × 1011 Nm–2 Ges 1.8 × 1011 Nm–2 | Zvi `yBwUi •`N© ̈ eivei mgvb ej cÖ‡qvM Kiv n‡j †`Lv hvq wØZxqwU cÖ_gwUi wØ ̧Y cÖmvwiZ n‡q‡Q| Zuvi `yBwUi e ̈vmv‡a©i AbycvZ wbY©q Ki| [RUET 06-07] mgvavb: Y = FL r 2 l  r 2  1 Yl      r1 r2 2 = Y2 Y1 × l2 l1 = 1.8 1.6 × 2l1 l1  r1 r2 = 1.8 × 2 1.6 = 3 2  d1 d2 = r1 r2 = 3 : 2 (Ans.) 20| `yBwU wbw`©ó cÖvšÍ we›`yi ga ̈eZ©x 30 m j¤^v GKwU A ̈vjywgwbqv‡gi Zv‡ii g‡a ̈ MÖx®§Kv‡j Uvb ej 200 kN| MÖx®§ I kxZ Kv‡ji g‡a ̈ cvwicvwk¦©K ZvcgvÎvi e ̈eavb 25C| hw` Zv‡ii e ̈vmva© 1.5 cm, Zv‡ii Dcv`v‡bi Zvcxq •`N© ̈ cÖmvivsK 24 × 10–6 /C Ges Bqs Gi ̧Yv1⁄4 7 × 107 Ncm–2 nq Z‡e kxZKv‡j Zv‡ii g‡a ̈ m„ó AwZwi3 e‡ji cwigvY Ges GB AwZwi3 e‡ji aib wbY©q Ki| ZLb Zv‡ii g‡a ̈ me©mvKz‡j ̈ e‡ji cwigvY KZ? [RUET 04-05] mgvavb: Avgiv Rvwb, F = YA  kxZKv‡j m„ó AwZwi3 ej, F = 7 × 107 10–4 ×  × 0.0152 × 24 × 10–6 × 25  F = 296.88 kN  me©mvKz‡j ̈ kxZKv‡j e‡ji cwigvY = MÖx®§Kv‡j e‡ji cwigvY + kxZKv‡j e„w× cÖvß e‡ji cwigvY = 200 + 296.88 kN = 496.88 kN (Ans.) 21| cvi‡`i AvqZb ̧Yv1⁄4 2.2 × 1010 Nm–2 | 1 m3 cvi‡`i AvqZb 2 × 10–6 m 3 n«vm Ki‡Z (i) Kx cwigvY KvR Ki‡Z n‡e? (ii) cvi‡`i Kx cwigvY w ̄’wZkw3 mwÂZ n‡e? [RUET 06-07] mgvavb: W = 1 2 × B × (V) 2 V B =  V V  W = 1 2 × 2.2 × 1010 × (2 × 10–6 ) 2 1  W = 0.044 J (Ans.) mwÂZ w ̄’wZkw3, V = W = 0.044 J 22| 1 mm2 cÖ ̄’‡”Q` wewkó GKwU B ̄úv‡Zi Zv‡ii •`N© ̈ 5% e„w× Ki‡Z KZ ej cÖ‡qvM Ki‡Z n‡e? (Y = 2 × 1011 Nm–2 for steel) [CUET 09-10, 03-04] mgvavb: Y = FL Al  2 × 1011 = F 10–6 × 5%  F = 104 N (Ans.) l = 5% × L  l L = 5%