Title 8. P2C8. আধুনিক পদার্থবিজ্ঞানের সুচনা (With Solve).pdf ✅

2 ........................................................................................................................................  Physics 2nd Paper Chapter-8 7. GKwU w ̄’i e ̄‘ we‡ùvi‡Yi gva ̈‡g `ywU 1 kg wbðj fiwewkó L‡Ð wef3 nj Ges ci ̄úi 0.6c (GLv‡b c = Av‡jvi †eM) †e‡M `~‡i m‡i †Mj| g~j e ̄‘wUi wbðj fi wbY©q K‡iv| [BUET 18-19] mgvavb: awi, LÐ `ywUi fi mx kw3i wbZ ̈Zv Abyhvqx, ETz = ETx + ETx [ET = †gvU kw3] Ep z + 0 = mxc 2 + mxc 2  mz0 c 2 = 2  mx0 c 2 1 – v 2 c 2  mz0 = 2  1 1 – 0.62 kg = 2.5 kg (Ans.) 8. 3000 A  Zi1⁄2‣`‡N© ̈i GKwU AwZ †e ̧wb iwk¥ 2.28 eV Kvh©v‡cÿK wewkó GKwU e ̄‘i c„‡ô AvcwZZ n‡q GKwU B‡jKUab wbM©Z Kij| wbM©Z B‡jKUa‡bi †eM KZ n‡e? [h = 6.62  10–34 Js, B‡jKUa‡bi fi = 9.1  10–31 kg] [BUET 18-19] mgvavb: E = W0 + Ek  hc  = W0 + 1 2 mv 2 max  6.62  10–34  3  108 3  10–7 = (1.6  10–19  2.28) + 1 2  9.1  10–31  v 2 max  vmax = 8.082  105 ms–1 (Ans.) 9. GKwU B‡jKUab (wbðj fi 9.1  10–31 kg) Av‡jvi `aæwZi 90% `aæwZ‡Z Pj‡Q| AvBb÷vB‡bi Av‡cwÿK ZË¡ Abymv‡i B‡jKUabwUi MwZkw3 wbY©q K‡iv| [BUET 17-18] mgvavb: MwZkw3, Ek = (m – m0) c2 = m0c 2       1 1 – v 2 c 2 – 1  Ek = 9.1  10–31  9  1016     1 1 – 0.92 – 1  Ek = 1.06  10–13 J (Ans.) 10. 0.4 nm Zi1⁄2‣`‡N© ̈i GKwU †dvUb w ̄’ive ̄’vq _vKv GKwU B‡jKUa‡bi mv‡_ msN‡l©i ci †dvUbwU c~‡e©i MwZc‡_i mv‡c‡ÿ 150 †Kv‡Y wewÿß nq| wewÿß †dvU‡bi †eM I Zi1⁄2‣`N© ̈ wbY©q K‡iv| [BUET 17-18; IUT 17-18] mgvavb: wewÿß †dvU‡bi †eM = 3  108 ms–1 (Ans.) Zi1⁄2‣`N© ̈,  =  + h m0c (1 – cos)   = 0.4  10–9 + 6.626  10–34 9.11  10–31  (1 – cos150)   = 4.0452  10–10 m (Ans.) 11. g‡b Ki, GKRb wcZv Zvi Kb ̈v †_‡K 20 eQ‡ii eo| wZwb c„w_ex †_‡K AwZ `aæZMwZi gnvKvkhv‡b `~ieZ©x †Kv‡bv MÖ‡n †h‡Z PvB‡jb| D3 MÖ‡n Zvi †h‡Z 2 eQi Ges Avm‡Z 2 eQi jvMj (Zvi wb‡Ri KvVv‡gv‡Z gvcv)| c„w_ex‡Z G‡m wZwb †`L‡jb Zvi Kb ̈v †_‡K wZwb 20 eQi †QvU n‡q †M‡Qb| G hvÎvq gnvKvkhvbwUi †eM KZ n‡e? [BUET 16-17] mgvavb: g‡b Kwi, hvÎvi c~‡e© Kb ̈vi eqm x wcZvi eqm x + 20 Avevi, hvÎvi c‡i, wcZvi eqm = (x + 20) + t0 = x + 24 [t0 = 4 years] Kb ̈vi eqm = x + t  (x + t) – (x + 24) = 20  t = 44 years  t = t0 1 –     v c 2  44 = 4 1 –     v c 2  v = 0.9958c (Ans.) 12. cvi‡`i ev®ú 140 nm Zi1⁄2‣`‡N© ̈i GKwU †dvUb ï‡l wb‡q cieZ©x‡Z `ywU †dvUb wbtmiY K‡i| GKwU †dvUb Gi Zi1⁄2‣`N© ̈ 180.5 nm n‡j Aci †dvUbwUi Zi1⁄2‣`N© ̈ KZ? [1 nm = 10–9 m] [BUET 16-17] mgvavb: E = E1 + E2  hc  = hc 1 + hc 2  1 140 = 1 180.5 + 1 2  2 = 623.95 nm (Ans.) 13. wba©vwiZ Zi‡1⁄2i GKwU wewKiY †Kvb avZec„‡ôi Dci AvcwZZ n‡j wbe„wË wef‡ei gvb 4.8 V nq| D3 avZec„‡ô wØ ̧Y Zi1⁄2‣`‡N© ̈i GKwU wewKiY AvcwZZ n‡j wbe„wË wef‡ei gvb 1.6 V cvIqv hvq| avZe c„ôwUi m~Pb Zi1⁄2‣`N© ̈ cÖ_‡g AvcwZZ Zi1⁄2‣`‡N© ̈i mv‡c‡ÿ KZ n‡e? [BUET 16-17] mgvavb: hc  = hc 0 + eV1  hc     1  – 1 0 = eV1 .........(i) hc     1 2 – 1 0 = eV2...... (ii) (i)  (ii) K‡i, 1  – 1 0 1 2 – 1 0 = V1 V2 = 4.8 1.6 = 3  1  – 1 0 = 3 2  1  – 3 0  2 0 = 1 2  1   0 = 4 (Ans.)
AvaywbK c`v_©weÁv‡bi m~Pbv  Engineering Question Bank & Practice Book .................................................................... 3 14. †Kvb GKwU 1.8 eV Kvh©v‡cÿK wewkó avZz‡Z 400 nm Zi1⁄2‣`N© ̈ wewkó Av‡jv AvcwZZ n‡jÑ (K) wbM©Z nIqv B‡jKUab ̧‡jvi wbe„wË wefe KZ n‡e? (L) wbM©Z B‡jKUab ̧‡jvi m‡e©v”P MwZ‡eM KZ? [BUET 14-15] mgvavb: (K) E = W0 + Ek  hc  = W0 + eV0  6.626  10–34  3  108 400  10–9 =1.8  1.6  10–19 + 1.6  10–19  V0  V0 = 1.306 V (Ans.) (L) Avevi, eV0 = 1 2 mv 2 max  vmax = 2eV0 m = 2  1.6  10–19  1.306 9.11  10–31  vmax = 6.773  105 ms–1 (Ans.) 15. GKRb gnvk~b ̈Pvix 25 eQi eq‡m 1.8  108 ms–1 †e‡M MwZkxj GKwU gnvk~b ̈hv‡b P‡o gnvKvk åg‡Y †M‡jb| c„w_exi wn‡m‡e wZwb 30 eQi gnvKv‡k KvwU‡q G‡j Zvi eqm KZ n‡e? [RUET 12-13, 05-06, 04-05; BUET 08-09; KUET 03-04; BUTex 03-04] mgvavb: t = t0 1 –     v c 2  t0 = 30 1 –     1.8 3 2 = 24 years  eZ©gvb eqm = 25 + 24 = 49 years (Ans.) 16. †Kv‡bv GK ai‡bi RxevYy cÖwZ 20 w`‡b Zvi msL ̈v e„w× K‡i wØ ̧Y nq| GB ai‡bi `yBwU RxevYy‡K GKwU b‡fvhv‡b K‡i gnvKv‡k cvVv‡bv nj Ges 1000 w`b c‡i c„w_ex‡Z wdwi‡q Avbv nj| hw` b‡fvhvbwUi MwZ †m‡K‡Û Av‡jvi MwZi 0.995 ̧Y nq, Z‡e b‡fvhvbwU c„w_ex‡Z wd‡i Avmvi ci G‡Z KZ ̧‡jv RxevYy cvIqv hv‡e? [BUET 11-12] mgvavb: t = t0 1 –     v c 2  t0 = 1000 1 – 0.9952 = 99.875 days RxevYyi e„w×i Rb ̈, N = N0e t = N0e ln2 T1/2 t  N = 2  e ln2 20  99.875 = 63.72  64 wU  c„w_ex‡Z wd‡i Avmvi ci 64 wU RxevYy cvIqv hv‡e| (Ans.) 17. 1.67  10–27 kg w ̄’wZ f‡ii GKwU †cÖvU‡bi fi wØ ̧Y n‡Z n‡j KZ `aæwZi cÖ‡qvRb n‡e? †cÖvUbwUi D3 `aæwZ AR©b Ki‡Z †h kw3i cÖ‡qvRb n‡e Zv †ei K‡iv| [c = 3  108 ms–1 ] [IUT 11-12; BUET 03-04] mgvavb: m = m0 1 – v 2 c 2  2m0 = m0 1 – v 2 c 2  v = 3 2 c (Ans.) cÖ‡qvRbxq MwZkw3, Ek = (m – m0) c2  Ek = (2m0 – m0) c2 = 1.67  10–27  9  1016  Ek = 1.503  10–10 J (Ans.) 18. wfbœ MÖ‡ni GKwU b‡fvhvb 0.6c MwZ‡Z (gv‡Vi †L‡jvqvo‡`i cwigvc Abyhvqx) ey‡qU dzUej gv‡Vi •`N© ̈ eivei AwZμg K‡i| dzUej gvVwU 110 wgUvi j¤^v Ges 50 wgUvi cÖk ̄Í| b‡fvhv‡bi wfbœ MÖnevmxi cwigvc Abyhvqx dzUej gvVwUi •`N© ̈ I cÖ ̄’ KZ n‡e? [BUET 09-10] mgvavb: cÖ ̄’ AcwiewZ©Z _vK‡e|  bZzb cÖ ̄’ = 50 m (Ans.) Avevi, L = L0 1 – v 2 c 2 =110 1 – 0.62  L = 88 m  bZzb •`N© ̈ = 88 m (Ans.) 19.  = 4000 A  Zi1⁄2‣`‡N© ̈i †dvUb †Kvb avZe c„‡ô AvcwZZ n‡q m‡e©v”P 0.4 eV MwZkw3i d‡UvB‡jKUab wbM©Z K‡i| H avZzi Kvh© A‡cÿK wbY©q K‡iv| [BUET 07-08] mgvavb: E = W0 + Ek  W0 = hc  – eV  W0 = 6.626  10–34  3  108 4  10–7 – 0.4  1.6  10–19  W0 = 4.3295  10–19 J = 2.706 eV (Ans.) 20. GKwU wgUvi † ̄..j‡K Zvi •`N© ̈ eivei gnvk~‡b ̈ 2.6  108 ms–1 †e‡M wb‡ÿc Kiv nj| c„w_ex †_‡K 1 m † ̄..jwUi •`N© ̈ KZ g‡b n‡e wbY©q K‡iv| [BUET 06-07] mgvavb: L = L0 1 – v 2 c 2 = 1 1 –     2.6 3 2  L = 0.498 m (Ans.)
4 ........................................................................................................................................  Physics 2nd Paper Chapter-8 21. w ̄’i Ae ̄’vq GKwU KYvi fi 10–30 kg| hw` KYvwU GKwU †eM wb‡q MwZkxj nq, Z‡e fi nq 1.25  10–30 kg| KYvwUi †eM KZ? [BUET 04-05] mgvavb: m = m0 1 –     v c 2  1.25 = 1 1 –     v c 2  v c = 0.6  v = 0.6c (Ans.) 22. †Kvb avZzi c„‡ô Av‡jvK iwk¥ AvcwZZ nIqvq wbtm„Z B‡jKUab m¤ú~Y©iƒ‡c _vgv‡Z 3 V weiwZ wefe Gi cÖ‡qvRb nq| D3 avZzi Av‡jvK Zwor wμqv 6  1014 Hz K¤úv‡1⁄4i Av‡jvK iwk¥ Øviv m~wPZ nq| AvcwZZ Av‡jvKiwk¥i K¤úv1⁄4 I avZzi Kvh© A‡cÿK wbY©q K‡iv| [h = 6.63  10–34 Js, e = 1.6  10–19 C, 1 eV = 1.6  10–19J] [BUET 03-04] mgvavb: E = W0 + eV  hf = hf0 + eV  f = f0 + eV h = 6  1014 + 1.6  10–19  3 6.63  10–34  f = 1.324  1015 Hz (Ans.) 23. GKwU B‡jKUab 107 ms–1 MwZ‡eM wb‡q Pj‡Q| Gi MwZkw3 B‡jKUab-†fv‡ë wbY©q Ki| [B‡jKUa‡bi fi, m = 9.1  10–31 kg, B‡jKUa‡bi PvR©, e = 1.6  10–19 coul] [BUET 03-04] mgvavb: MwZkw3, Ek = (m – m0) c2 = m0c 2       1 1 – v 2 c 2 – 1  Ek = 9.1  10–31  9  1016         1 1 –     107 3 108 2 – 1  Ek = 4.55  10–17 J = 4.55  10–17 1.6  10–19 eV = 284.612 eV (Ans.) 24. `aæZ MwZm¤úbœ GKwU e ̄‘i `aæwZ KZ n‡j e ̄‘wUi MwZkw3 Zvi †gvU kw3i 1 5 Ask n‡e? [BUET 02-03] mgvavb: Ek = 1 5 ET  (m – m0) c2 = 1 5 mc2  m0 = 4 5 m = 4 5 m0 1 –     v c 2  v c = 0.6  v = 0.6c (Ans.) 25. 8.3  107 ms–1 MwZ‡Z MwZkxj GKwU †cÖvU‡bi MwZkw3 KZ? mbvZb MwZkw3i mv‡_ G gv‡bi Zzjbv K‡iv| [w ̄’i Ae ̄’vq †cÖvU‡bi fi = 1.67265 10–27 kg, c = 3  108 ms–1 ] [BUET 02-03] mgvavb: Av‡cwÿK ZË¡xq MwZkw3, Ek1 = m0c 2       1 1 – v 2 c 2 – 1  Ek1 = 1.67265  10–27  9  1016         1 1 –     0.83 3 2 – 1 Ek1 = 6.115  10–12 J mbvZb MwZkw3, Ek2 = 1 2 mv2  Ek2 = 1 2  1.67265  10–27  (8.3  107 ) 2  Ek2 = 5.77  10–12 J  Ek1 Ek2 = 6.115  10–12 5.77 – = 1.06  Ek1  Ek2 = 1.06 : 1 (Ans.) 26. GKwU †cÖvUb 2.4  108 ms–1 MwZ‡Z Pj‡j Zvi MwZkw3 KZ n‡e? mbvZb MwZkw3i mv‡_ G gv‡bi Zzjbv K‡iv| w ̄’i Ae ̄’vq †cÖvU‡bi fi = 1.7  10–27 kg; Av‡jvi †eM (k~b ̈ gva ̈‡g) = 3.0  108 ms–1 | [BUET 01-02] mgvavb: Av‡cwÿK ZË¡xq MwZkw3, Ek1 = m0c 2       1 1 – v 2 c 2 – 1  Ek1 = 1.7  10–27  9  1016         1 1 –     2.4 3 2 – 1  Ek1 = 1.02  10–10 J mbvZb MwZkw3, Ek2 = 1 2 mv2  Ek2 = 1 2  1.7  10–27  (2.4  108 ) 2  Ek2 = 4.896  10–11 J  Ek1 Ek2 = 1.02  10–10 4.896  10–11 = 2.083  Ek1 : Ek2 = 2.083 : 1 (Ans.)