Title 6. P1C6.-Gravitation and Gravity-Engg._With Solve.pdf ✅

2  Physics 1st Paper Chapter-6 Qw›`Z ̄ú›`‡b ̄úw›`Z n‡Z jvM‡jv| c„w_ex‡K GKwU mylg †MvjK g‡b K‡i Ges evav`vbKvix mKj ej D‡cÿv K‡i c„w_exi †K›`a †_‡K 5 × 105 m `~i‡Z¡ e ̄‘wUi Z¡iY I †`vj‡bi ch©vqKvj wbY©q Ki| [†`Iqv Av‡Q, c„w_exi e ̈vmva© R = 6.4 × 106 m Ges g = 9.8 ms–2 ] [BUET 16-17] mgvavb: c„ô †_‡K h MfxiZvq A_©vr †K›`a †_‡K x `~i‡Z¡ cv_‡ii Dci Z¡iY a = – 4 3 Gx    ∵  g = 4 3 GR  a g = – x R  a = – g x R = – 9.8 × 5 × 105 6.4 × 106 ms–2 = – 0.765 ms–2 a = – 4 3 Gx  –  2 x = – 4 3 Gx   = 4G 3  2 T = 4G 3  T = 3 G = 3 5.5 × 103 × 6.673 × 10–11 = 5067.5 s (Ans.) 8| GKwU 20 kg f‡ii K...wÎg DcMÖn ARvbv f‡ii GKwU MÖ‡ni Pviw`‡K 8.0 × 106 m e ̈vmv‡a©i e„ËvKvi c‡_ AvewZ©Z n‡j Zvi ch©vqKvj 2.4 h nq| MÖnc„‡ôi AwfKl©R Z¡i‡Yi gvb 8.0 ms–2 n‡j MÖnwUi e ̈vmva© KZ? [BUET 14-15] mgvavb: r =     T 2GM 4 2 1 3  r =     (2.4 × 3600) 2 × gR2 4 2 1 3  (8 × 106 ) 3 = (2.4 × 3600) 2 × 8 × R2 4 2  R = 5.82 × 106 m (Ans.) 9| GKwU wi‡gvU †mwÝs m ̈v‡UjvBU c„w_exi Pvwiw`‡K f~-c„ô n‡Z 250 km Dc‡i e„ËvKvi c‡_ Nyi‡Q| GB c‡_ m ̈v‡UjvBUwU MwZ‡eM Ges N~Y©b Kvj wbY©q Ki| [Re = 6400 km, g = 9.8 ms–2 ] [BUET 10-11] mgvavb: v = GM R + h = gR2 R + h = 6.4 × 106 9.8 6.4 × 106 + 250 × 103 ms–1 = 7769.3 ms–1 (Ans.) T = 2(R + h) v = 2(6400 × 103 + 250 × 103 ) 7769.3 s = 5377.98 s (Ans.) 10| f~-c„ô n‡Z Lvov Dc‡ii w`‡K GKwU i‡KU‡K 5 kms–1 `aæwZ‡Z Dr‡ÿcY Kiv nj| i‡KUwU wVK wdievi g~û‡Z© f~-c„ô †_‡K KZ D”PZvq †cuŠQv‡e Zv †ei Ki| [c„w_exi fi = 6.0 × 1024 kg, c„w_exi e ̈vmva© = 6.4 × 106 m, G = 6.67 × 10–11 Nm2 kg–2 ] [BUET 05-06] mgvavb: kw3i msiÿYkxjZv Abyhvqx, Einitial = Efinal  – GMm R + 1 2 mv 2 = – GMm R + h  – G × 6 × 1024 6.4 × 106 + 1 2 × (5 × 103 ) 2 = – G × 6 × 1024 6.4 × 106 + h  h = 1.598 × 106 m (Ans.) 11| g1⁄2j MÖ‡ni fi c„w_exi f‡ii 0.11 ̧Y Ges Gi e ̈vmva© c„w_exi e ̈vmv‡a©i 0.532 ̧Y| g1⁄2j MÖ‡ni f~-c„ô †_‡K GKwU gnvk~b ̈hvb‡K b~ ̈bZg KZ †e‡M Dr‡ÿcY Ki‡j gnvk~b ̈hvbwU g1⁄2j MÖ‡ni ga ̈vKl©Y e‡ji evB‡i P‡j †h‡Z cvi‡e? [c„w_exi fi 5.975 × 1024 kg, c„w_exi e ̈vmva© = 6.371 × 106 m, G = 6.673 × 10–11 Nm2 kg–2 ] [BUET 03-04] mgvavb: g1⁄2j MÖ‡n gyw3‡eM, ve = 2GMm Rm = 2 × G × 0.11 × 5.975 × 1024 6.371 × 106 × 0.532 ms–1 = 5087.23 ms–1 (Ans.) 12| f~-c„‡ôi PZzw`©‡K wbiÿe„Ë eivei e„ËvKvi c‡_ AveZ©bkxj GwU f~-w ̄’i †hvMv‡hvM DcMÖ‡ni e„ËvKvi c‡_i e ̈vmva© KZ? DcMÖnwU f~-c„ô n‡Z KZ D”PZvq Nyi‡Q? [BUET 01-02] mgvavb: h =     T 2GM 4 2 1 3 – R =     (24 × 3600) 2 × 6.673 × 10–11 × 6 × 1024 4 2 1 3 – 6.4 × 106 = 35.9 × 106 m (Ans.) 13| c„w_ex †_‡K 2600 wK‡jvwgUvi D”PZvq GKwU K...wÎg DcMÖn c„w_ex‡K †K›`a K‡i e„ËvKvi c‡_ cÖ`wÿY Ki‡Q| Gi †eM N›Uvq KZ n‡e? [c„w_exi e ̈vmva© = 6400 wK‡jvwgUvi, fi = 6 × 1024 †KwR Ges G = 6.67 × 10–11 wbDUb-wg2 /†KwR2 ] [KUET 19-20] mgvavb: K...wÎg DcMÖ‡ni †eM, v = GM R + h = 6.67 × 10–11 × 6 × 1024 6400 × 103 + 2600 × 103 = 6668.33 ms–1 = 24005.98 kmh–1 (Ans.)
gnvKl© I AwfKl©  Mastery Practice Sheet 3 14| c„w_ex c„‡ô GKRb †jv‡Ki IRb 80 kg-wt| c„w_exi fi P‡›`ai f‡ii 81 ̧Y n‡j P›`a †jvKwUi IRb KZ n‡e? (c„w_ex Ges P‡›`ai e ̈vmv‡a©i AbycvZ 4:1) [KUET 04-05] mgvavb: Wm We = gm ge = G Mmm Rm 2 G Mem Re 2 = Mm Re 2 81 Mm Rm 2 = 1 81 × 42 = 16 81  Wm = 16 81 × 80 kg-wt = 15.8 kg-wt (Ans.) 15| g‡b Ki c„w_exi †K›`a w`‡q Gi e ̈v‡mi GK cÖvšÍ †_‡K Aci cÖvšÍ ch©šÍ GKwU Uv‡bj Lbb Kiv n‡jv| †`LvI †h, GB Uv‡b‡j GKwU cv_i †dj‡j Gi MwZ mij †`vjb MwZ n‡e| cv_iwUi mij †`vjbMwZi ch©vqKvj wbY©q Ki| (c„w_exi NbZ¡ = 5.5 × 103 kgm–3 ) [RUET 19-20] mgvavb: c„w_exi c„ô †_‡K h MfxiZvq A_©vr †K›`a †_‡K x `~i‡Z¡ cv_‡ii Dci ej F = ma = – 4 3 Gx.m [∵ ej I mi‡Yi w`K wecixZ]  a  – x  hv mij †`vjb MwZi kZ©|  cv_iwUi MwZ mij †`vjb MwZ n‡e| a = – 4 3 Gx  –  2 x = 4 3 Gx   = 4G 3  2 T = 4G 3  T = 3 G = 3 5.5 × 103 × 6.673 × 10–4 = 5068.64 s (Ans.) 16| †`LvI †h, †Kvb e ̄‘‡K b~ ̈bZg 11.2 kms–1 †e‡M gnvk~‡b ̈i w`‡K Qz‡o gvi‡j e ̄‘wU c„w_exi AwfKl©R e‡ji AvKl©Y KvwU‡q DV‡Z cv‡i| c„w_exi e ̈vmva© = 6.4 × 106 m [RUET 17-18] mgvavb: c„w_exi AwfKl©R e‡ji AvKl©Y KvwU‡q gnvk~‡b ̈ e ̄‘wUi †gvU kw3 k~b ̈ n‡e| Einitial = Etotal  – GMm R + 1 2 mv2 = 0  v = 2GM R  v = 2 × 6.673 × 10–11 × 6 × 1024 6.4 × 106 ms–1 = 11200 ms–1 = 11.2 kms–1 (Ans.) 17| GKwU K...wÎg DcMÖn c„w_ex c„ô †_‡K 700 km D”PZvq e„ËvKvi c‡_ Nyi‡Q| Gi Avbyf~wgK †eM wbY©q Ki| [c„w_exi e ̈vmva© = 6300 km Ges c„w_ex c„‡ô g = 9.8 ms–2 ] [RUET 17-18] mgvavb: GM R + h = gR2 R + h = R g R + h = 6.3 × 106 × 9.8 6.3 × 106 + 7 × 105 = 7454.26 ms–1 (Ans.) 18| c„w_exi e ̈vmva© R = 6.4 × 106 m Ges AwfKl©R Z¡iY 9.8 ms–2 n‡j c„w_exi c„ô n‡Z †Kvb e ̄‘i gyw3‡eM KZ n‡e? [RUET 13-14, 09-10; CUET 13-14; KUET 05-06] mgvavb: gyw3‡eM, ve = 2gR = 2 × 9.8 × 6.4 × 106 = 1.12 × 104 ms–1 (Ans.) 19| c„w_exi fi P‡›`ai f‡ii 81 ̧Y Ges Zv‡`i †K‡›`ai ga ̈eZ©x `~iZ¡ 38.6 × 104 km| P›`a I c„w_exi ms‡hvMKvix †iLvi †Kv_vq †Kvb e ̄‘i Dci Df‡qi Uvb mgvb n‡e? [RUET 08-09] mgvavb: F1 = F2  GMm x 2 = GMm 81 × (38.6 × 107 – x) 2 [†hLv‡b x = c„w_ex †_‡K `~iZ¡]  x = 9(38.6 × 107 – x)  x = 34.74 × 107 m (Ans.) 20| c„w_ex I m~h© mgvb e‡j G‡K Aci‡K AvKl©Y K‡i| ZeyI c„w_ex m~‡h©i Pvwiw`‡K †Nv‡i †Kb? [RUET 06-07] mgvavb: c„w_ex I m~h© mgvb e‡j G‡K Aci‡K AvKl©Y K‡i| wKš‘ c„w_exi Zzjbvq m~‡h©i fi A‡bK †ewk nIqvq c„w_ex m~‡h©i Pviw`‡K †Nv‡i| c„w_ex I m~‡h©i ga ̈eZ©x AvKl©Y ej G‡ÿ‡Î c„w_ex‡K †K›`agyLx e‡ji †RvMvb †`q| (Ans.) 21| c„w_exi fi I G Gi gvb h_vμ‡g 6 × 1024 kg I 6.7 × 10–11 Nm2 kg–2 Ges e ̈vmva© 6.4 × 106 m n‡j, f~-w ̄’i DcMÖ‡ni D”PZv I †eM KZ? [RUET 03-04] mgvavb: h =     GMT2 4 2 1 3 – R =     6.7 × 10–11 × 6 × 1024 × 864002 4 2 1 3 – 6.4 × 106 = 3.596 × 107 m v = GM (R + h)
4  Physics 1st Paper Chapter-6 = 6.7 × 10–11 × 6 × 1024 6.4 × 106 × 6 + 35.96 × 106 = 3080 ms–1 (Ans.) 22| c„w_exc„‡ô GKwU †jv‡Ki IRb 90 kg n‡j g1⁄2j c„‡ô Zvi IRb KZ n‡e? g1⁄2j Gi fi c„w_ex f‡ii 1 9 Ask Ges g1⁄2‡ji e ̈vmva© c„w_exi e ̈vmv‡a©i A‡a©K| [CUET 03-04] mgvavb: Wm We = gm ge = G Mmm Rm 2 G Mem Re 2 = Me Re 2 9 Mm Rm 2 = 1 9 × 22 = 4 9  Wm = 4 9 × 90 × 9.8 N = 392 N (Ans.) 23| Avgv‡`i c„w_exi e ̈vm 12800 km| GKwU DcMÖn e„ËvKvi K‡ÿ 7.8 kms–1 MwZ †e‡M Ny‡i| e„ËvKvi K‡ÿ AwfKl©R Z¡iY 9.0 ms–2 n‡jÑ [CUET 04-05] (K) e„ËvKvi K‡ÿi D”PZv (L) GKevi c~Y© N~Y©‡bi mgqKvj wbY©q Ki| mgvavb: (K) mg = mv2 (R + h)  9 = (7800) 2 6.4 × 106 + h  h = 360 km (Ans.) (L) T = 2(R + h) v = 2(6.4 × 106 + 360 × 103 ) 7800 = 5445.42 s (Ans.) 24| GKwU e ̄‘‡K f~-c„ô n‡Z Dc‡i DVv‡bv nj| m‡e©v”P Ae ̄’v‡b AwfKl©R Z¡i‡Yi gvb f~-c„‡ôi AwfKl©R Z¡i‡Yi gv‡bi 64 fv‡Mi 1 fvM| e ̄‘wU‡K f~-c„ô †_‡K KZ D”PZvq DVv‡bv n‡qwQj? (c„w_exi e ̈vmva© R = 6.4 × 106 m) [BUTex 22-23] mgvavb: gh =     R R + h 2 g  1 64 = R R + h = R + h R = 8  h = 7R = 4.48 × 107 m (Ans.) 25| c„w_exi N~Y©b †eM eZ©gvb N~Y©b †e‡Mi KZ ̧Y n‡j wbiÿxq A‡j †Kv‡bv e ̄‘ fvinxb n‡e? [BUTex 20-21] mgvavb: eZ©gvb N~Y©b †eM, 1 = 2 24 × 3600 rad/s = 7.27 × 10–5 rad/s wbiÿxq A‡j †Kv‡bv e ̄‘ finxb n‡j, 0 = g – 2 2 R  2 = g R = 9.8 6.4 × 106 = 1.23 × 10–3 rad/s  2 1 = 1.23 × 10–3 7.27 × 10–5 = 17.02 (Ans.) 26| gnvKl© aaæeK ‘G’ Gi gvÎv wjL| [BUTex 10-11] mgvavb: G = F.r2 m1m2 = [MLT–2 ] [L2 ] [M2 ] = [L3M –1T –2 ] (Ans.) 27| c„w_ex †_‡K 1600 km D”PZvq K...wÎg DcMÖn c„w_ex‡K †K›`a K‡i e„ËvKvi c‡_ cÖ`wÿY Ki‡Q| DcMÖnwUi cÖwZN›Uvq †eM KZ? [ME = 6 × 1024 kg, RE = 6.4 × 106 m, G = 6.67 × 10–11 Nm2 kg–2 ] [BUTex 07-08] mgvavb: Avgiv Rvwb, v = GME RE + h = 6.67 × 10–11 × 6 × 1024 6.4 × 106 + 1600 × 103 = 7072.84 ms–1 = 25462.2 kmh–1 (Ans.) 28| ZvcgvÎv evo‡j gnvKl©xq aaæe‡Ki gvb wK n‡e? [BUTex 06-07] mgvavb: †Kvb cwieZ©b n‡e bv| (Ans.) weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv eûwbe©vPbx cÖkœmg~n 1. m ̈v‡UjvB‡Ui MwZkw3i mgxKiY †KvbwU? [BUET Preli 22-23] Ek = GMm R + h Ek = GMm 2(R + h) Ek = –GMm (R + h) Ek = –GMm 2(R + h) DËi: Ek = GMm 2(R + h) 2. g1⁄2jMÖ‡ni fi I e ̈vmva© h_vμ‡g c„w_exi fi I e ̈vmv‡a©i 16 ̧Y Ges 49 ̧Y| g1⁄2jMÖ‡ni gyw3‡eM KZ? [BUET Preli 22-23] 5.4 kms–1 6.4 kms–1 7.4 kms–1 8.4 kms–1