Tiêu đề 7. P1C7 Structural Properties of Matter_(With Solve).pdf ✅

c`v‡_©i MvVwbK ag©  Engineering Practice Sheet............................................................................................................... 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. 1 mm e ̈vmva©wewkó GKwU Zv‡i 20 kg f‡ii e ̄‘ Szwj‡q w`‡j ZvcgvÎv KZ e„w× cv‡e? [e ̄‘i NbZ¡ 9000 kgm–3 , Av‡cwÿK Zvc 0.42 kJkg–1K –1 , Bqs Gi ̧Yv1⁄4 2  1011 Nm–2 ] [BUET 23-24 cÖ‡kœi Abyiƒc] mgvavb: mS = 1 2 Fl  mS = 1 2 F  FL AY    ∵ Y =  FL Al  AL  S = 1 2 F 2L AY   = F 2 2A2YS = (20  9.8) 2 2  (  0.0012 ) 2  2  1011  9000  420 = 0.00257C (Ans.) 2. 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f g × m i m i – m f  1 5 × 10–11 = (3 × 103 × f × 9.8) × f f – 1000  f = 1001.47 kgm–3 (Ans.) 3. 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: 2T sin(15) = W  T = 45 × 9.8 2 × sin15  T = 851.9465 N W 15 T 15 T GLb, Y = TL r 2 l  l L = T r 2Y  l L = T r 2Y  l L = 851.9465  × 0.0052 × 2 × 1011 = 5.423 × 10–5 (Ans.) 4. `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.) 5. 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 Poisson's ratio 0.25] [BUET 17-18] mgvavb: cqm‡bi AbycvZ,  = d d L L  d =  × L L × d = 0.25 × 15% × 2  d = 0.075 mm  e ̈vm n«vm cv‡e 0.075 mm (Ans.) Avevi, Y = FL Al  F = YA × l L = 2 × 1011 ×  × 0.0012 × 15%  F = 94.247 kN (Ans.)