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

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)