Title 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
2  Physics 1st Paper Chapter-7  d 5.3 = 6.5 × 106 3 × 1010  d = 1.148 × 10–3 cm  Dj¤^ wePz ̈wZ, d = 1.148 × 10–3 cm (Ans.) 6| cvwbi DcwiZ‡j cvwbi NbZ¡ 1.03 × 103 kgm–3 n‡j 800 atm Pvc MfxiZvq cvwbi NbZ¡ KZ n‡e? [†`Iqv Av‡Q, cvwbi msbg ̈Zv = 45.8 × 10–11 Pa–1 Ges 1 atm = 1.013 × 105 Pa] [BUET 16-17] mgvavb: B =  . V V =  . Vi Vi – Vf  B =  . m i m i – m f  1 K =  . f f – i  1 45.8 × 10–11 = 800 × 1.013 × 105 × f f – 1.03 × 103  f = 1.069 × 103 kgm–3 (Ans.) 7| 2 × 108 Nm–2 Pv‡c mxmvi NbZ¡ KZ n‡e? [mxmvi ̄^vfvweK NbZ¡  = 11.4 gcm–3 Ges mxmvi AvqZb ̧Yv1⁄4 0.80 × 1010 Nm–2 ] [BUET 08-09; KUET 03-04] mgvavb: B =  . f f – i  0.8 × 1010 = 2 × 108 × f f – 11.4  f = 11.692 gcm–3 (Ans.) 8| GKB •`‡N© ̈i `ywU B ̄úv‡Zi Zv‡ii e ̈vm h_vμ‡g 1.0 mm I 2.0 mm| Zvi `ywU‡K h_vμ‡g 40 N I 80 N ej Øviv Uvbv nj, G‡`i cÖmvi‡Yi AbycvZ wbY©q Ki| [BUET 08-09] mgvavb: Y = FL r 2 l  l = FL Yr 2  l  F r 2  l1 l2 = F1 F2 ×     r2 r1 2 = 40 80 ×     2 1 2 = 1 2 × 4 = 2  l1 : l2 = 2 : 1 (Ans.) 9| GKB c`v‡_©i `yBwU Zvi A Ges B Gi •`‡N© ̈i AbycvZ 1 : 2 Ges e ̈v‡mi AbycvZ 2 : 1 hw` GKB ej Øviv Zvi `y‡Uv‡K Uvbv nq, ZLb A Ges B Gi •`N© ̈ e„w×i AbycvZ wK n‡e? [BUET 06-07] mgvavb: l = FL Yr 2  l  L r 2  l1 l2 = L1 L2 ×     r2 r1 2 = 1 2 ×     1 2 2 = 1 8  l1 : l2 = 1 : 8 (Ans.) 10| 1 m j¤^v Ges 1 mm e ̈v‡mi GKwU Zv‡ii GK cÖv‡šÍ 80 N Gi GKwU IRb Szjv‡Z n‡e| hw` G‡ÿ‡Î 1 mm Gi †ewk •`N© ̈-e„w× MÖnY‡hvM ̈ bv nq, Zvn‡j Zvgvi Zvi e ̈envi Kiv hv‡e wK? [Zvgvi Bqs ̧Yv1⁄4 = 13 × 1010 Nm–2 ] [BUET 04-05] mgvavb: Y = FL r 2 l  l = FL r 2Y = 80 × 1  × (0.5 × 10–3 ) 2 × 13 × 1010  l = 7.835 × 10–4 m = 0.7835 mm < 1 mm  •`N© ̈ e„w× 1 mm n‡Z Kg nIqvq Zvgvi Zvi e ̈envi Kiv hv‡e| (Ans.) 11| 130 cm `xN© Ges 1.1 mm e ̈v‡mi GKwU B ̄úv‡Zi Zvi‡K 830C ZvcgvÎvq DËß K‡i Zv‡ii `yB cÖvšÍ `ywU `„p eÜbxi mv‡_ Giƒc AuvUwK‡q †`qv n‡jv †hb ZviwU `ycv‡k Uvbv Ae ̄’vq _v‡K| ZviwU VvÛv n‡q 20C G †b‡g Avm‡j Zv‡i wK cwigvY Uvb m„wó n‡e? [B ̄úv‡Zi Bqvs ̧Yv1⁄4, Y = 200 × 109 Nm–2 , cÖmviv1⁄4,  = 11 × 10–6 /C] [BUET 03-04] mgvavb: Zv‡i m„ó Uvb, F = YA  F = 200 × 109 ×  ×     1.1 × 10–3 2 2 × 11 × 10–6 × (830 – 20)  F = 1693.5 N (Ans.) 12| GKwU w ̄’wZ ̄’vcK Zvi‡K †U‡b j¤^v Kiv n‡q‡Q| Zv‡ii g~j •`N© ̈ L, •`‡N© ̈i e„w× x; Zv‡ii cÖ ̄’‡”Q` A Ges e ̄‘i Bqs‡qi aaæeK Y n‡j, KZUv w ̄’wZ ̄’vcK kw3 G‡Z Rgv n‡q‡Q? [BUET 01-02] mgvavb: Y = FL Al  F = YAx L  K...ZKvR ev mwÂZ wefekw3, W =  x2 x1 Fdx  W =  x2 x1 YA L xdx  W = 1 2 . YAx2 L (Ans.) 13| B ̄úvZ I ivev‡ii g‡a ̈ †KvbwU †ekx w ̄’wZ ̄’vcK? †Zvgvi Dˇii e ̈vL ̈v `vI| [KUET 04-05] mgvavb: GKB •`N© ̈ I cÖ ̄’‡”Q` wewkó ivevi I B ̄úv‡Zi Zvi mgvb •`N© ̈ e„w× Ki‡Z B ̄úv‡Z †ewk ej cÖ‡qvM Ki‡Z nq| ZvB B ̄úvZ ivevi A‡cÿv †ewk w ̄’wZ ̄’vcK| (Ans.) 14| `ywU mgvb •`‡N© ̈i Zv‡ii e ̈vmv‡a©i AbycvZ 1 : 2| G‡`i Dci GKwU mgvb ej cÖ‡qvM Kiv nj| hw` Zvi `ywUi •`‡N© ̈ e„w×i AbycvZ 3 : 1 nq Z‡e Zvi `ywUi Dcv`v‡bi Bqs-Gi ̧Yv‡1⁄4i AbycvZ wbY©q Ki| [KUET 04-05] mgvavb: Y = FL r 2 l  Y  1 r 2 l
c`v‡_©i MvVwbK ag©  Practice Content 3 Y1 Y2 =     r2 r1 2 × l2 l1  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%
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