Tiêu đề পদার্থের গাঠনিক ধর্ম (Structural Properties of Matter) (With Solve).pdf ✅

c`v‡_©i MvVwbK ag©  Varsity Practice Content 1 mßg Aa ̈vq c`v‡_©i MvVwbK ag© Structural Properties of Matter ACS Physics Department Gi g‡bvbxZ wjwLZ cÖkœmg~n 1| GKwU mylg Zv‡ii •`N© ̈ L Ges IRb W| ZviwUi GKcÖvšÍ Qv‡` Szwj‡q Aci cÖv‡šÍ W1 IR‡bi GKwU e ̄‘ Szwj‡q †`Iqv n‡jv| ZviwUi cÖ ̄’‡”Q‡`i †ÿÎdj s n‡j ZviwUi wb¤œcÖvšÍ n‡Z 3L 4 D”PZvi we›`ywU‡Z cxob wbY©q Ki| [Medium] mgvavb: A we›`y‡Z ej = Szjv‡bv IRb + Zv‡ii IR‡bi 3 4 Ask = W + 3 4 W1 = 4W + 3W1 4 3L 4 A L  A we›`y‡Z cxob = ej cÖ ̄’‡”Q‡`i †ÿÎdj = 4W + 3W1 4S (Ans.) 2| r e ̈vmv‡a©i GKwU Zv‡ii 2 cÖvšÍ A I B we›`y‡Z AvUKv‡bv| †hLv‡b Zv‡ii Avw` •`N© ̈ = AB = 2l| ZviwUi ga ̈we›`y‡Z GKwU fi Szwj‡q w`‡j Uvb ej wbY©q Ki| [Zv‡ii Dcv`v‡bi Bqs ̧Yv1⁄4 Y Ges d < l|] [Medium] T T A B C d mgvavb: AC = BC = l 2 + d2 Szjv‡bvi ci •`N© ̈ e„w× = AC + BC – AB = 2 l 2 + d2 – 2l Zv‡i Uvb ej T n‡j, cxob = T r 2 Ges weK...wZ = 2 l 2 + d2 – 2l 2l = l 2 + d2 – l l  Y = T r 2 × l l 2 + d2 – l  T = Yr 2 ( l 2 + d2 – l) l = Yr 2 l     l 1 + d 2 l 2 – l = Yr 2     1 + d 2 l 2 – 1 (Ans.) 3| wP‡Îi wjdUwU 2 ms–2 Z¡i‡Y EaŸ©Mvgx| wjd‡Ui wmwjs-G GKwU Zvi Szjv‡bv hvi cÖ ̄’‡”Q‡`i †ÿÎdj 2 cm2 | ZviwUi gy3cÖv‡šÍ 10 kg f‡ii GKwU e ̄‘ Szjv‡bv n‡jv| ZviwUi Bqs ̧Yv1⁄4 2 × 1011 Nm–2 n‡j Aby‣`N© ̈ weK...wZ wbY©q Ki| [g = 10 ms–2 ] [Medium] a0 = 2 m/s2 10 kg mgvavb: G‡ÿ‡Î Zv‡i Uvb ej F = m(g + a) = 10 × (10 + 2) = 120 N cxob = F A = 120 2×10–4 = 60 × 104 Nm–2 Y = cxob weK...wZ  2 × 1011 = 60 × 104 weK...wZ  weK...wZ = 30 × 10–7 = 3 × 10–6 (Ans.) 4| 1 kg I 2 kg f‡ii 2wU e ̄‘ GKwU avZe Zvi w`‡q wP‡Îi b ̈vq hy3| 2 kg f‡ii e ̄‘‡K 10 N e‡j Uvbv n‡jv| ZviwUi Amn cxob 2 × 109 Nm–2 | Zv‡ii me©wb¤œ †ÿÎdj KZ n‡j ZviwU wQo‡e bv? [Medium] F = 10 N 1 kg 2 kg gm„Y Zj mgvavb: 1 kg f‡ii e ̄‘i Rb ̈, T = ma  T = a – (i) 2 kg f‡ii e ̄‘i Rb ̈, F – T = ma  10 – T = 2a 10 – a = 2a [(i) n‡Z]  a = 10 3 ms–2 T = 10 3 N F = 10 N 1 kg 2 kg a T T  cxob = T A  2 × 109 = 10 3A  A = 5 3 × 10–9 m 2 (Ans.)
2  Physics 1st Paper Chapter-7 5| GKwU finxb w ̄úas G m kg fi Szjv‡j mvg ̈ve ̄’vq w ̄úaswU x cm cÖmvwiZ nq| w ̄úaswU‡K †K‡U `yB UzKiv Kiv n‡jv| 60 cm •`‡N© ̈i eo UzKivwU‡Z m kg fi Szjv‡j 7.5 cm cÖmvwiZ nq| †QvU UzKivwU‡Z m kg fi Szjv‡j 5 cm cÖmvwiZ nq| x Gi gvb KZ? [Hard] mgvavb: awi, g~j w ̄úas Gi •`N© ̈ L Ges w ̄úas aaæeK k eo UzKivi w ̄úas aaæeK k1 †QvU UzKivi w ̄úas aaæeK k2  mg = kx = k1(7.5) = k2(5) – (i) Avevi, k1.60 = k2(L – 60) = kL  k1 = kL 60 .... (ii) k2 = kL (L – 60) .... (iii) GLb, (i) G k1 I k2 Gi gvb emvB k1 × 7.5 = k2 × 5  kL 60 × 7.5 = kL L–60×5  L – 60 60 = 5 7.5 = 2 3  L 60 = 2 3 + 1 = 5 3  L = 100 cm k1 = kL 60 = k × 100 60  k1 = 5 3 k GLb, k1 × 7.5 = kx  5 3 k × 7.5 = kx  x = 12.5 cm (Ans.) 6| GKwU Zv‡ii Ici ej cÖ‡qvM Ki‡j cÖ ̄’‡”Q‡`i †ÿÎdj 2% n«vm cvq| ZviwUi Dcv`v‡bi cqm‡bi AbycvZ 0.4 n‡j kZKiv •`N© ̈ e„w× KZ? [Easy] mgvavb: A = d 2 4  A A = 2 d d  2% = 2 × d d  d d = 1%  = d/d L/L  0.4 = 1% L/L  L L = 2.5%  •`N© ̈ kZKiv 2.5% e„w× cv‡e| (Ans.) 7| cvwbi AvqZb ̧Yv1⁄4 2 × 109 Nm–2 | cvwbi NbZ¡ 0.1% evov‡Z wK cwigvY Pvc cÖ‡qvRb n‡e? [Medium] mgvavb: df – di di = 0.1 100  df di = 1 1000 + 1  di df = 1000 1001 .... (i) v V = AvqZb weK...wZ = Pvc AvqZb ̧Yv1⁄4 = P 2 × 109 di = M V ; df = M V – v  di df = V – v V = 1 – v V  1000 1001 = 1 – P 2 × 109  P 2 × 109 = 1 1001  P = 2 × 106 Nm–2 (Ans.) 8| 5 mm2 cÖ ̄’‡”Q‡`i GKwU Zv‡ii mv‡_ 15 kg fi Szjv‡bv Av‡Q| fi Szjv‡bv Zv‡ii •`N© ̈ 4 m| Zv‡ii Dcv`v‡bi Bqs ̧bv1⁄4 0.98 × 1010 Nm–2 | fiwU mwi‡q wb‡j Zv‡ii •`N© ̈ wK cwigvY msKzwPZ n‡e? [Medium] mgvavb: Y = FL A(4 – L)  Y = mgL A(4 – L) Avw` •`N© ̈ = L msKzwPZ •`N© ̈ = l  L + l = 4  l = 4 – L  0.98 × 1010 = 15 × 9.8 × L 5 × 10–6 (4 – L)  4 – L L = 3 × 10–3  4 L = 1.003  L = 4 1.003  l = 4 – 4 1.003 = 0.012 1.003 = 0.012 (cÖvq) = 12 mm (Ans.) 9| wewfbœ Dcv`v‡bi mgvb •`N© ̈ wewkó `yBwU Zv‡ii GKwUi e ̈vm 1 mm I AciwUi e ̈vm 4 mm| Dfq‡K mgvb ej w`‡q Uvb‡j cÖ_gwUi •`N© ̈ e„w× wØZxqwUi •`N© ̈ e„w×i 4 MyY nq| †Kvb ZviwU †ewk w ̄’wZ ̄’vcK? [Easy] mgvavb: l1 = 4l2 F1 = F2  Y1A1l1 L = Y2A2l2 L  Y1d 2 1 l1 = Y2d 2 2 l2  Y1 Y2 = l2 l1 ×     d2 d1 2 = 1 4 × 42 = 4 1gwUi Bqs ̧bv1⁄4 2qwUi Pvi ̧Y|  1g ZviwU †ewk w ̄’wZ ̄’vcK| (Ans.)
c`v‡_©i MvVwbK ag©  Varsity Practice Content 3 10| GKB Dcv`vb w`‡q •Zwi `yBwU mylg Zv‡ii •`N© ̈ 5 cm I 7 cm| mgvb cwigvb •`N© ̈ e„w×i Rb ̈ Zvi `ywUi Ici h_vμ‡g 0.03 J I 0.07 J KvR Kivi cÖ‡qvRb nq| Zvi `ywUi cÖ ̄’‡”Q‡`i AbycvZ KZ? [Easy] mgvavb: W1 = Y1A1l 2 1 2L1 W2 = Y2A2l2 2 2L1  W1 W2 = Y1A1l1 2 × L2 L1 × Y2A2l2 2  0.03 0.07 = A1 × 7 × 10–2 A2 × 5 × 10–2 [∵ Y1 = Y2 I l1 = l2]  A1 A2 = 15 49  A1 : A2 = 15 : 49 (Ans.) 11| GKwU ̧jwZi iev‡ii e ̈v‡Ûi cÖ ̄’‡”Q‡`i †ÿÎdj 16 mm2 Ges cÖv_wgK •`N© ̈ 8 cm| e ̈vÛwU‡K 2 cm †U‡b 400 g f‡ii GKwU wXj wb‡ÿc Kiv n‡jv| hw` iev‡ii Bqs ̧bv1⁄4 5 × 108 Nm–2 nq Z‡e wX‡ji †eM KZ? mgvavb: ievi m¤úamvi‡Y K...ZKvR = wX‡ji MwZkw3  1 2 × YAl 2 L = 1 2 mv2  5 × 108 × 16 × 10–6 × (0.02) 2 0.08 = 0.4 × v2  v 2 = 100  v = 10 m/s (Ans.) 12| ivev‡ii AvqZb weK...wZ ̧Yv1⁄4 10 × 108 Nm–2 | ivev‡ii GKwU †MvjK‡K GKwU n«‡`i Zj‡`‡k KZ Mfx‡i wb‡j AvqZb 0.4% n«vm cv‡e? [g = 10 m/s2 ] [Medium] mgvavb: v V = 0.4% = 4 × 10–3 K = PV v = hg × V v  h = K g . v V = 10 × 108 103 × 10 × 4 × 10–3 = 400 m (Ans.) 13| GKwU Zv‡ii Ici cÖhy3 ej 28 N †_‡K 49 N Ki‡j •`N© ̈ e„w× 0.4 mm †_‡K †e‡o 0.7 mm nq| Zv‡ii GB •`N© ̈ e„wׇZ K...ZKvR KZ? [Medium] mgvavb: Avw` Ae ̄’vq, U1 = 1 2 F1l1 †kl Ae ̄’vq, U2 = 1 2 F2l2 W = U2 – U1 = 1 2 F1l1 – 1 2 F2l2 =     1 2 × 49 × 0.7 × 10–3 –     1 2 × 28 × 0.4 × 10–3 J = 1 2 (0.0343 – 0.0112) J = 1 2 × 0.0231 J = 0.01155 J (Ans.) 14| GKwU 5 m `xN© Ges 2 × 103 kgm–3 Nb‡Z¡i GKwU fvix ivev‡ii iwki GKcÖvšÍ wmwjs †_‡K Szwj‡q w`‡j iwki wbR ̄^ IR‡bi Kvi‡Y KZUzKz •`N© ̈ e„w× NU‡e? ivev‡ii Bqs ̧Yv1⁄4 5 × 108 Nm–2 | [Medium] mgvavb: Y = FL Al  Y = mgL Al  Y = VgL Al  Y = AgL2 Al  Y = gL2 l  5 × 108 = 2 × 103 × 9.8 × 52 l  l = 9.8 × 10–4 m (Ans.) 15|    F F   l Ô GKwU Nb‡Ki cÖwZ evûi •`N© ̈ 0.5 m Ges Dci I wb‡Pi Z‡j F = 2 × 105 N gv‡bi `yBwU mgvšÍivj wKš‘ wecixZgyLx ̄úk©Kxq ej cÖ‡qvM Kivq Dc‡ii Z‡ji 0.5 cm miY N‡U| hw` 1 m evû wewkó NbK‡K GKB cwigvY ejØq cÖ‡qvM Kiv nq Z‡e mi‡Yi cwigvY wbY©q Ki| [Medium] mgvavb: e ̈eZ©b weK...wZ,  = x L = 0.5 × 10–2 0.5 = 10–2 e ̈eZ©b ̧bv1⁄4,  = F A = 2 × 105 0.52 × 10–2 = 8 ×107 Nm–2 Nb‡Ki evûi •`N© ̈ cwieZ©b K‡i ej AcwiewZ©Z ivL‡j e ̈eZ©b ̧Yv1⁄4 AcwiewZ©Z _vK‡e,  = F A = 2 × 105 1 2 × 8 × 107   = x L  x = L = 2.5 × 10–3 = 2.5 mm (Ans.) 16| 3 m `xN© I 5 mm2 cÖ ̄’‡”Q‡`i †ÿÎdj wewkó GKwU mylg Zvi‡K Nl©Ynxb cywji Ici w`‡q †bIqv n‡jv| ZviwUi 2 cÖv‡šÍ 5 kg Gi 2wU fi mshy3 Kiv n‡jv| ZviwU cywji Ici w`‡q MwZkxj n‡j Gi cÖmviY KZ n‡e? ZviwUi Dcv`v‡bi Bqs ̧Yv1⁄4 2 × 1011 Nm–2 | [Medium] mgvavb: Y = FL Al  l = FL AY = 2mg L AY = 2 × 5 × 9.8 × 3 5 × 10–6 × 2 × 1011 = 2.94 × 10–4 m (Ans.) 5 kg 5 kg
4  Physics 1st Paper Chapter-7 17| 0.5 m •`‡N© ̈i GKwU Zv‡ii GKcÖv‡šÍ 2kg f‡ii GKwU e ̄‘ AvUwK‡q e ̄‘wU‡K Abyf~wgK mgvšÍivj Z‡j Nyiv‡bv n‡”Q| hw` Zv‡ii cÖ ̄’‡”Q‡`i †ÿ‡Îdj 1 × 10–6 m 2 Ges Amn cxob 4.9 × 107 Nm–2 nq Z‡e ZviwU bv wQ‡o e ̄‘wU m‡e©v”P KZ †K.wYK `aæwZ‡Z Nyiv‡bv hv‡e? [Medium] mgvavb: P = F A  P = m 2 r A   2 = PA mr   2 = 4.9 × 107 × 10–6 2 × 0.5 = 49  max = 7 rads–1 (Ans.) 18| 5 mm2 cÖ ̄’‡”Q‡`i †ÿÎdj wewkó GKwU Zvi 25 N ej cÖ‡qv‡M †f‡1⁄2 hvq| Zv‡ii NbZ¡ 250 kg/m3 , Dcv`v‡bi Bqs ̧Yv1⁄4 1.5 × 1011 Nm–2 Ges Av‡cwÿK Zvc 300 Jkg–1K –1 | Zvi †f‡1⁄2 hvIqvi mgq ZvcgvÎv e„w× KZ n‡e? [Hard] mgvavb: W = H  1 2 YA l 2 L = mS  ALS = 1 2 AY l 2 L  AS = 1 2 AY l 2 L 2  S = 1 2 Y ×     F AY 2   = F 2 2A2 SY = 252 2(5 × 10–6 ) 2 × 250 × 300 × 1.5 × 1011 = 1 900 = 1.11  10–3 K (Ans.) 19| mxmvi NbZ¡ 11.4 gcm–3 Ges AvqZb ̧Yv1⁄4 0.8 × 1010 Nm–2 n‡j 6 × 109 Nm–2 Pv‡c mxmvi NbZ¡ KZ? [Medium] mgvavb: B = PV v  v V = P B ......(i) mxgvi UyKivi fi AcwiewZ©Z _vK‡e| m1 = m2  V11 = V22  1V = 2(V – v)  1 = 2     V – v V  1 = 2     1 – v V  1 = 2     1 – P B  11.4 = 2     1 – 6 × 109 8 × 109  11.4 = 2 ×     1 – 3 4  2 = 11.4 × 4 = 45.6 gcm–3 (Ans.) 20| hw` mvaviY wkjvi w ̄’wZ ̄’vcK mxgv 1.5 × 108 Nm–2 Ges Mo NbZ¡ 50 × 103 kg/m3 nq Z‡e c„w_ex c„‡ô †Kv‡bv ce©‡Zi m‡e©v”P D”PZv KZ n‡Z cv‡i? (g = 10 m/s2 ) [Medium] mgvavb: P = hg  h = p g = 1.5 × 108 50 × 103 × 10 = 300 m (Ans.) 21| 10 m `xN© I 2 × 10–6 m 2 cÖ ̄’‡”Q‡`i †ÿÎdj wewkó GKwU Abyf~wgK B ̄úvZ Zv‡ii `yB cÖvšÍ‡K ci ̄úi wecixZ w`K †_‡K 20 kg-wt Gi mgvb e‡j Uvb‡j Dfq cÖv‡šÍi w`‡K •`N© ̈ 5 × 10–4 m e„w× cvq| B ̄úv‡Zi Bqs ̧Yv1⁄4 wbY©q Ki| [Medium] mgvavb: Y = FL Al = 20 × 9.8 × 10 2 × 10–6 × 2 × 5 × 10–4 Nm–2 = 9.8 × 1011 Nm–2 (Ans.) 22| GKB •`N© ̈ I cÖ ̄’‡”Q‡`i †ÿÎdj wewkó 2wU `‡Ûi Dcv`v‡bi Bqs ̧Yv1⁄4 Y1 I Y2| `Û؇qi GKwU‡K AciwUi GKcÖv‡šÍ hy3 Ki‡j Zzj ̈ Bqs ̧Yv•K KZ? [Medium] mgvavb: awi, Zvi؇qi •`N© ̈ L, cÖ ̄’‡”Q‡`i †ÿÎdj A, F cwigvY ej cÖ‡qvM •`N© ̈e„w× l1 I l2  Y1 = FL Al1  l1 = FL AY1 I l2 = FL AY2 GLb, †kÖwY‡Z hy3 _vKvq, †gvU •`N© ̈e„w× l n‡j, l = l1 + l2  F.2L AY = FL AY1 + FL AY2  2 Y = 1 Y1 + 1 Y2  2 Y = Y1 + Y2 Y1Y2  Y 2 = Y1Y2 Y1 + Y2  Y = 2Y1Y2 Y1 + Y2  Zzj ̈ Bqs ̧Yv1⁄4 2Y1Y2 Y1 + Y2 (Ans.)
c`v‡_©i MvVwbK ag©  Varsity Practice Content 5 23| 2 mm2 cÖ ̄’‡”Q‡`i †ÿÎdj wewkó GKwU Zv‡ii •`N© ̈ 1 mm evov‡Z 10.2 kg fi Szjv‡Z nq| Zv‡ii Avw` •`N© ̈ KZ? Y = 2 × 1011 Nm–2 [Easy] mgvavb: Y = FL Al  2 × 1011 = 10.2 × 9.8 × L 2 × 10–6 × 1 × 10–3  10  10  L 2  10–9  L = 4 m (Ans.) 24| 2.5 m `xN© Zv‡ii GKcÖvšÍ `„pfv‡e AvUwK‡q Aci cÖv‡šÍ fi Pvcv‡j 2 mm •`N© ̈ cÖmvwiZ nq| Zv‡ii e ̈vm 1 mm Ges Dcv`v‡bi cqm‡bi AbycvZ 0.24 n‡j cÖmvwiZ Ae ̄’vq ZviwUi e ̈vmva© ms‡KvPb wbY©q Ki| [Easy] mgvavb:  = dL Dl  d = Dl L = 0.24 × 1 × 10–3 × 2 × 10–3 2.5 = 1.92 × 10–7 m  e ̈vmva© ms‡KvPb = d 2 = 9.6 × 10–8 m (Ans.) 25| `yBwU wfbœ c`v‡_©i •Zwi mgvb •`N© ̈ (1 m) I mgvb cÖ ̄’‡”Q‡`i †ÿÎdj (1 mm2 ) wewkó `yBwU Zvi wmwi‡R mshy3 Av‡Q| wK cwigvY ej cÖ‡qv‡M G‡`i mw¤§wjZ •`N© ̈ 0.9 mm e„w× cv‡e wbY©q Ki| [Y1 = 2 × 1011 Nm–2 , Y2 = 7 × 1011 Nm–2 ] [Medium] mgvavb: wmwi‡R msyh3 _vK‡j Dfq Zv‡i mgvb ej cÖhy3 n‡e| Y = FL Al  Yl = FL A = const  Y1l1 = Y2l2  l1 l2 = Y2 Y1 = 7 × 1011 2 × 1011 = 7 2  l1 l2 = 7 2  2l1 – 7l2 = 0 .... (i) Avevi, l1 + l2 = 0.9 .... (ii)  l1 = 0.7 mm l2 = 0.2 mm  F = Y1Al1 L = 2 × 1011 × 10–6 × 0.7 × 10–3 1 = 140 N (Ans.) ACS Physics Department Gi g‡bvbxZ eûwbe©vPwb cÖkœmg~n 1. GKwU Zv‡i me©wb¤œ 20 kg f‡ii e ̄‘ Szjv‡j ZviwU wQ‡o hvq| H Zv‡ii wØ ̧Y e ̈vmva© wewkó Aci GKwU Zv‡i (GKB Dcv`v‡bi) m‡e©v”P KZ fi Szjv‡bv hv‡e? [Medium] 20 kg 5 kg 80 kg 160 kg DËi: 80 kg e ̈vL ̈v: GKB Dcv`v‡bi nIqvq Amn fi mgvb|  F1 A1 = F2 A2  F2 = F1 A2 A1 = F1 × 4  m2 = 4m1 = 4 × 20 kg = 80 kg 2. GKwU Zv‡ii •`N© ̈ L Ges cÖ ̄’‡”Q‡`i †ÿÎdj A| ZviwUi Dcv`v‡bi Bqs ̧Yv1⁄4 Y n‡j ZviwUi •`N© ̈ x cwigvY m¤úamvi‡Y K...ZKvRÑ [Easy] YAx2 L YAx 2L YAx 2L2 YAx2 2L DËi: YAx2 2L 3. eavm Ges ÷x‡ji Bqs gWyjvm h_vμ‡g 1 × 1010 Nm–2 I 2 × 1010 Nm–2 | mgvb •`‡N© ̈i GKwU eav‡mi Zvi I GKwU ÷x‡ji Zvi‡K mgcwigvb ej Øviv 1mm cÖmvwiZ Kiv n‡jv| eavm I ÷x‡ji Zvi؇qi e ̈vmva© h_vμ‡g RB I Rs n‡jÑ [Medium] Rs = 2 RB Rs = RB 2 Rs = 4RB Rs = RB 4 DËi: Rs = RB 2 e ̈vL ̈v: FS = FB  YsAs l L = YBABl L  YsRs 2 = YBRB 2  2 × 1010 Rs 2 = 1 × 1010 RB 2  2 Rs = RB 4. hw` GKwU M ̈v‡mi Pvc 1.01 × 105 Pa †_‡K 1.165 × 105 Pa Kivi d‡j AvqZb 10% n«vm cvq Z‡e H M ̈v‡mi AvqZb ̧Yv1⁄4Ñ [Medium] 15.5 × 105 Pa 1.4 × 105 Pa 1.55 × 105 Pa 0.0155 × 105 Pa DËi: 1.55 × 105 Pa e ̈vL ̈v: B = – Vd dV = 100 10 × (1.165 × 105 – 1.01 × 105 ) = 1.55 × 105 Pa