Tiêu đề 6. মহাকর্ষ ও অভিকর্ষ (Gravity And Gravitation) With Solve_Sha 7.10.23 PDF.pdf ✅

gnvKl© I AwfKl©  Varsity Practice Content 1 lô Aa ̈vq gnvKl© I AwfKl© Gravitation and Gravity ACS Physics Department Gi g‡bvbxZ wjwLZ cÖkœmg~n 1| f~-c„ô n‡Z Aí D”PZvq Ges f~-c„‡ôi mgvšÍiv‡j GKwU b‡fvhvb wK `aæwZ‡Z Pj‡j GKRb hvÎx IRbnxbZv Abyfe Ki‡eb? [g = 10 ms–2 , R = 6400 km] [Medium] mgvavb: mg mv2 r cÖkœg‡Z, mg = mv2 r  v = gr = 10 × 6.4 × 106  v = 64 × 106 = 8 × 103 ms–1 = 8000 ms–1 (Ans.) 2| cÖgvY Ki †h, c„w_exi †giæ AÂj Ges welyexq A‡ji AwfKl©R Z¡i‡Yi cv_©K ̈  2R Gi mgvb| †hLv‡b,  = c„w_exi †KŠwYK †eM Ges R = c„w_exi e ̈vmva©| [Easy] mgvavb: †giæ A‡j, g1 = g –  2Rcos2 (90) = g welyexq A‡j, g2 = g –  2Rcos2 (0) = g –  2R  g1 – g2 =  2R (Proved) 3| m~h© n‡Z e„n ̄úwZ MÖ‡ni `~iZ¡ c~‡e©i 1 4 ̧Y n‡j, AveZ©bKvj KZ ̧Y n‡e? [Easy] mgvavb: T 2  R 3      T2 T1 2 =     R2 R1 3  T2 T1 =     R2 R1 3 2  T2 T1 =     1 4 3 2 =     1 2 3 = 1 8 (Ans.) 4| wP‡Îi A, B, C we›`y‡Z AwfKl©R Z¡i‡Yi gvb wbY©q K‡iv| [Easy] mgvavb: A  f~-c„ô  g B  g =       R R + R 4 2 g = 16 25 g C  g =       1 – R 4 R g = 3 4 g (Ans.) A R 4 B R 4 C O 5| `ywU MÖn A I B Gi e ̈vmva© GKB Z‡e B MÖ‡ni NbZ¡ A MÖ‡ni Pvi ̧Y| Z‡e B MÖ‡ni gyw3‡eM A MÖ‡ni KZ ̧Y n‡e? [Medium] mgvavb: ve = 2gR = 2 × 4 3 GR × R = R 8 3 G  ve    vB vA = B A  vB = 4 vA = 2vA (Ans.) 6| GKwU wb‡iU †Mvj‡Ki fi I e ̈vmva© h_vμ‡g 10 kg I 5 m| †MvjKwUi †K›`a n‡Z 10 m `~i‡Z¡ †Kv‡bv we›`y‡Z cÖvej ̈ KZ? [Easy] mgvavb: 10kg 10m P E = GM r 2 = 6.673 × 10–11 × 10 100 = 0.673 × 10–11 Nkg–1 (Ans.) 7| `ywU e ̄‘i fi h_vμ‡g 5 kg I 10 kg| e ̄‘ `ywU 2m `~i‡Z¡ Ae ̄’vb Ki‡j Zv‡`i ga ̈eZ©x ga ̈vKl©Y e‡ji gvb wbY©q K‡iv| [G = 6 × 10–11 Nm2 kg–2 ] [Easy] mgvavb: F = Gm1m2 r 2 = 6 × 10–11 × 5 × 10 4 = 75 × 10–11 N (Ans.) 8| GKwU mgevû wÎfz‡Ri wZbwU kx‡l© m f‡ii wZbwU e ̄‘ ̄’vcb Kiv n‡jv| wÎfz‡Ri evûi •`N© ̈ a n‡j, wÎfz‡Ri fi‡K‡›`a gnvKl©xq wef‡ei gvb wbY©q K‡iv| [Medium] mgvavb: cos30 = a 2 r  r = a 3  V = V1 + V2 + V3 = – 3 × Gm a 3 = –3 3Gm a (Ans.) m a O a m a m 30 a 2 9| 2m evû wewkó GKwU e‡M©i kxl© ̧‡jv‡Z h_vμ‡g 1kg, 2kg, 3kg I 4kg f‡ii e ̄‘‡K ̄’vcb Kiv n‡j, e‡M©i †K‡›`a gnvKl©xq wef‡ei gvb xG (G gnvKl©xq aaæeK) n‡j x = ? [Medium] mgvavb: AB = 2 2 m  AO = 2 2 2 = 2 m  V = – G r [m1 + m2 + m3 + m4] = – G 2 [1 + 2 + 3 + 4]  V = – 10G 2 = – 5 2 G 2m O B A  x = –5 2 (Ans.)
2  Physics 1st Paper Chapter-6 10| gyw3‡e‡Mi mgxKiY cÖwZcv`b K‡iv| [Easy] mgvavb: 1 2 mv2 = GMm     1 ri – 1 rf  1 2 v 2 = GM     1 R – 1  = GM R  v = 2GM R (Ans.) 11| †`LvI †h, †Kv‡bv DcMÖ‡ni †gvU kw3 Zvi wefekw3i A‡a©K| [Easy] mgvavb: wefekw3, U = – GMm r †gvU kw3 = wefekw3 + MwZkw3 = – GMm r + 1 2 m GM r = – GMm 2r = 1 2 U (Ans.) 12| f~-c„ô n‡Z 1600 km DuPz‡Z AwfKl©R Z¡i‡Yi gvb wbY©q K‡iv| [g = 10 ms–2 , R = 6400 km] [Medium] mgvavb: g =     R R + h 2 g =       R R + R 4 2 × 10 | 1600km = 1 4 R  g = 16 25 × 10 = 16 × 2 5 = 32 5 = 32 × 2 5 × 2 = 64 10 = 6.4 ms–2 (Ans.) 13| f~-c„‡ôi Lye KvQ †_‡K GKwU K...wÎg DcMÖn c„w_ex‡K cÖ`wÿY Ki‡j †`LvI †h, Zvi cÖ`wÿY `aæwZ = gR | [Easy] mgvavb: v = GM R + h GLb, h << R n‡j, v = GM R = gR2 R = gR (Ans.) 14| P›`ac„‡ô AwfKl©R Z¡i‡Yi gvb f~-c„‡ô AwfKl©R Z¡i‡Yi gv‡bi 1 6 ̧Y| hw` c„w_exi fi P‡›`ai f‡ii 81 ̧Y nq Z‡e c„w_exi e ̈vmva© P‡›`ai e ̈vmv‡a©i KZ ̧Y? [Medium] mgvavb: g  M R 2  gm ge = Mm Me ×     Re Rm 2 = Mm 81Mm ×     Re Rm 2  1 6 = 1 81 ×     Re Rm 2  Re Rm = 27 2  Re = 3 3 2 Rm (Ans.) 15| c„w_exi †K›`a n‡Z R 4 `~i‡Z¡ AwfKl©R Z¡i‡Yi gvb wbY©q K‡iv| [g = 10 ms–1 ] [Easy] mgvavb: g =     1 – h R g = R – h R g = d R g = R 4 R = 1 4 × 10  g = 2.5 ms –2 (Ans.) R–h=d h 16| c„w_exc„‡ô gyw3‡eM 11.2 km/s| †Kv‡bv MÖ‡ni e ̈vmva© hw` c„w_exi e ̈vmv‡a©i wØ ̧Y nq Ges fi c„w_exi f‡ii AvU ̧Y nq Z‡e †mLv‡b gyw3‡eM? [Easy] mgvavb: 11.2 kms–1 = 2GM R  vnew = 2 × G × 8M 2R = 2 2GM R  vnew = 2 × 11.2 km/s = 22.4 km/s (Ans.) 17| hw` f~-c„ô n‡Z h D”PZvq †Kv‡bv e ̄‘i IRb f~c„‡ôi IR‡bi 1% nq Z‡e h †K R (c„w_exi e ̈vmva©) Gi gva ̈‡g cÖKvk K‡iv| [Medium] mgvavb: W  g  g g =     R R + h 2  W W =     R R + h 2  1 100 = R R + h  10 = R + h R  h R = 9  h = 9R (Ans.) 18| c„w_exi N~Y©b wμqvi Kvi‡Y AwfKl©R Z¡iY, g Gi ZviZ‡g ̈i mgxKiY cÖwZcv`b K‡iv| [Medium] mgvavb: r = Rcos ..... (i) jwä ej, mg = mg – m 2 rcos  g = g –  2Rcos2  [(i) n‡Z] r   mg m 2 rcos m 2 r R  19| wP‡Î S n‡jv m~‡h©i Ae ̄’vb| Gi PZzw`©‡K Dce„ËvKvi Kÿc‡_ GKwU MÖ‡ni A n‡Z B we›`y‡Z Avm‡Z 20 w`b mgq jvM‡j, C n‡Z D we›`y‡Z Avm‡Z KZ mgq jvM‡e? [ASB Gi †ÿÎdj = 4 × CSD Gi †ÿÎdj|] [Medium] mgvavb: †Kcjv‡ii †ÿÎdj m~Î n‡Z, †ÿÎdj ASB †ÿÎdj CSD = t1 t2 = 20 t2  4 = 20 t2  t2 = 5 w`b A B C S D
gnvKl© I AwfKl©  Varsity Practice Content 3 20| GKwU K...wÎg DcMÖn c„w_ex c„ô n‡Z 3600 km D”PZvq c„w_ex‡K AveZ©b Ki‡Q| Gi Avbyf~wgK †eM wbY©q K‡iv| [g = 10ms–2 , R = 6400km] [Medium] mgvavb: v = GM R + h = gR2 R + h     g = GM R 2  v = R g R + h = 6.4 × 106 10 6.4 × 106 + 3.6 × 106 = 6.4 × 106 10 10 × 106  v = 6.4 × 106 × 10–3 = 6.4 km/s (Ans.) 21| f~-c„ô n‡Z KZ Mfx‡i AwfKl©R Z¡i‡Yi gvb f~-c„‡ôi AwfKl©R Z¡i‡Yi 1 n ̧Y n‡e? [Easy] mgvavb: g g = 1 – h R  1 n = 1 – h R  h R = 1 – 1 n = n –1 n  h = R(n – 1) n (Ans.) 22| GKwU e ̄‘‡K f~-c„ô †_‡K 1600 km D”PZvq D‡Ëvjb Kiv n‡j e ̄‘wUi Dci c„w_exi AvKl©Y ej c~‡e©i KZ ̧Y n‡e? [Easy] mgvavb: F  1 R 2  F2 F1 =     R1 R2 2 =       R R + R 4 2 [1600 = 6400 4 = R 4 ]  F2 = 16 25 F1 (Ans.) 23| GKwU MÖ‡ni fi I e ̈vmva© DfqB c„w_exi fi I e ̈vmv‡a©i wØ ̧Y| c„w_exc„‡ô g Gi gvb 9.8 ms–2 n‡j H MÖ‡ni c„‡ô g Gi gvb KZ n‡e? [Easy] mgvavb: g  M R 2  gp ge = Mp Me ×     Re Rp 2 = 2 ×     1 2 2 = 1 2  gp = 1 2 × ge = 1 2 × 9.8 = 4.9 ms–2 (Ans.) 24| c„w_ex‡Z GKwU e ̄‘i IRb 270 kg-wt| g1⁄2j MÖ‡ni fi c„w_exi f‡ii 1 9 ̧Y Ges e ̈vmva© 1 2 ̧Y n‡j g1⁄2jMÖ‡n e ̄‘wUi IRb KZ? [Easy] mgvavb: gm ge = Wm We = Mm Me ×     Re Rm 2  Wm = 1 9 × (2)2 × We  Wm = 4 9 × 270 = 120 kg-wt (Ans.) 25| f~-c„ô †_‡K KZ Mfx‡i †M‡j †mLvbKvi AwfKl©R Z¡i‡Yi gvb, f~-c„‡ôi gv‡bi 75% n‡e? [R = 6400 km] [Easy] mgvavb: g g = 1 – h R  75 100 = 1 – h R  h R = 1 – 3 4  h = 1 4 × 6400 = 1600 km (Ans.) 26| m1 I m2 f‡ii `ywU e ̄‘ R `~i‡Z¡ Aew ̄’Z| Z...Zxq m f‡ii GKwU e ̄‘‡K m2 fi n‡Z KZ `~i‡Z¡ ̄’vcb Ki‡j e ̄‘wUi Dci wbU ej k~b ̈ n‡e? [Easy] mgvavb: R – x x m1 F1 m F2 m2 R wbU ej k~b ̈ n‡j, F1 = F2  Gmm1 (R – x) 2 = Gmm2 x 2  R – x x = m1 m2  R x = m1 m2 + 1  x = R m1 m2 + 1 (Ans.) 27| M fi Ges a e ̈vmva© wewkó wZbwU mylg wb‡iU †MvjK‡K Ggbfv‡e ivLv n‡jv †hb †MvjK wZbwU ci ̄úi‡K ̄úk© K‡i| †h‡Kv‡bv GKwU †Mvj‡Ki Dci Aci `ywU †MvjK Øviv cahy3 gnvKl©xq ej wbY©q K‡iv| [Hard] mgvavb: †h‡Kv‡bv `ywU †Mvj‡Ki ga ̈eZ©x AvKl©Y ej, F = GM2 (2a) 2 = GM2 4a2 Fnet = 2F cos    2 a F 60 a F M [`ywU mgvb †f±‡ii jwä, R = 2P cos     2 ]  Fnet = 2 × GM2 4a2 × cos30 = GM2 2a2 × 3 2 = 3 4 × GM2 a 2 N (Ans.)
4  Physics 1st Paper Chapter-6 28| c„w_exi c„ô n‡Z †Kv‡bv e ̄‘i gyw3‡eM ve| hw` f~c„ô n‡Z †Kv‡bv e ̄‘‡K 2ve †e‡M wb‡ÿc Kiv nq Z‡e wK †e‡M e ̄‘wU Amx‡g c„w_exi gnvKlx©q †ÿ·K AwZμg Ki‡e? [Easy] mgvavb: Ei = Ef  1 2 mv 2 1 – GmM R = 1 2 mv 2 2  1 2 m.4v2 e – GmM R = 1 2 mv 2 2 [v1 = 2ve]  1 2 v 2 2 = 2 × 2GM R – GM R = 3GM R  v2 = 3 × 2GM R = 3ve (Ans.) 29| GKwU e ̄‘‡K c„w_exi gnvKl©xq †ÿÎ AwZμg Kivi cÖ‡qvRbxq me©wb¤œ MwZkw3i A‡a©K MwZkw3 w`‡q Dc‡i wb‡ÿc Kiv n‡jv| e ̄‘wU f~c„ô n‡Z m‡e©v”P KZ D”PZvq DV‡e? [Medium] mgvavb: 1 2 Kmin = GMm     1 Ri – 1 Rf  1 2     1 2 mve 2 = GMm     1 R – 1 R + h  1 4 × 2GM R = GM     1 R – 1 R + h  1 2 R = 1 R – 1 R + h  1 R + h = 1 2R  h = R (Ans.) 30| m I M f‡ii `ywU e ̄‘‡K d `~i‡Z¡ ivLv n‡jv| e ̄‘ `ywUi ga ̈eZ©x †h we›`y‡Z gnvKl©xq cÖvej ̈ k~b ̈ nq, †m we›`y‡Z gnvKl©xq wefe wbY©q K‡iv| [Hard] mgvavb: x1 m x2 M d Avgiv Rvwb, m1 I m2 f‡ii ga ̈eZ©x `~iZ¡ r n‡j, m1 e ̄‘ n‡Z x = r m2 m1 + 1 `~i‡Z¡ cÖvej ̈ k~b ̈ n‡e|  x1 = d M m + 1 , x2 = d m M + 1  V = V1 + V2 = – Gm x1 – Gm x2  V = – Gm d    M  m + 1 – GM d    m  M + 1  V = – G d [ Mm + m + Mm + M]  V = – G d [M + m + 2 Mm ] = – G d ( M + m ) 2 (Ans.) 31| cÖwZwU m f‡ii AmsL ̈ KYv GKwU mij‡iLv eivei GKwU wbw`©ó we›`y O †_‡K r, 2r, 4r, 8r ..... BZ ̈vw` `~i‡Z¡ emv‡bv n‡jv| O we›`y‡Z gnvKl©xq wefe? [Hard] mgvavb: V = V1 + V2 + V3 ........  V = – Gm r – Gm 2r – Gm 4r – Gm 8r ......  V = – Gm r     1 + 1 2 + 1 4 + 1 8 .......  V = – Gm r × 1 1 – 1 2 = – 2Gm r [Amxg avivi mgwó, S = a 1 – r †hLv‡b a = cÖ_g c`, r = mvaviY AbycvZ]  V = – 2Gm r (Ans.) 32| †Kv‡bv GKwU ̄’v‡b gnvKl©xq †ÿÎ, E  = (10i )  + 10j  N/kg| evwn ̈K †Kv‡bv e ̈e ̄’v Øviv 2 kg f‡ii †Kv‡bv e ̄‘KYv‡K H †ÿ‡Î ax‡i ax‡i (0, 0) Ae ̄’vb †_‡K (5m, 4m) Ae ̄’v‡b wb‡Z K...ZKvR wbY©q K‡iv| [Medium] mgvavb: W = F  . r  | r  = rf  – ri  = 5i  + 4j   W = mE  .r   W = 2(10i )  + 10j  . (5i )  + 4j   W = (20i )  + 20j  . (5i )  + 4j   W = 100 + 80 = 180 J (Ans.) 33| f~-c„‡ô GKwU w ̄úas Zzjvq GKwU e ̄‘ Szjv‡j w ̄úaswUi 1 cm e„w× N‡U| f~-c„ô †_‡K 1600 km Dc‡i H GKB e ̈e ̄’vq w ̄úaswUi KZ e„w× NU‡e? [Medium] mgvavb: mg = Kx  g  x  g g =     R R + h 2 | 1600 = 6400 4 = R 4  x x =       R R + R 4 2 = 16 25  x = 16 25 × 1 cm = 16 × 4 25 × 4  x = 64 100 = 0.64 cm (Ans.) 34| GKwU K...wÎg DcMÖn e„ËvKvi Kÿc‡_ c„w_ex‡K cÖ`wÿY Ki‡Q| DcMÖnwUi cÖ`wÿY †eM c„w_ex †_‡K gyw3‡e‡Mi A‡a©K n‡j, f~-c„ô †_‡K DcMÖnwUi D”PZv wbY©q K‡iv| [Easy] mgvavb: v = GM R + h  1 2 × ve = GM R + h  1 4 × 2GM R = GM R + h  1 2R = 1 R + h  h = R (Ans.)
gnvKl© I AwfKl©  Varsity Practice Content 5 35| †Kv‡bv MÖ‡ni e ̈vmva© I Mo NbZ¡, c„w_exi e ̈vmva© I Mo Nb‡Z¡i A‡a©K n‡j, H MÖ‡ni c„‡ô AwfKl©R Z¡iY KZ? [f~- c„‡ô g = 980 cm/s2 ] [Easy] mgvavb: g = 4 3 GR  g  R  gp ge = p e × Rp Re = 1 2 × 1 2  gp = 1 4 × 980 = 1 2 × 490  gp = 245 cm/s2 (Ans.) ACS Physics Department Gi g‡bvbxZ eûwbe©vPwb cÖkœmg~n 1. gnvKl©xq wef‡ei †ÿ‡ÎÑ [Easy] (i) Gi m‡e©v”P gvb Amx‡g (ii) Gi m‡e©v”P gvb k~b ̈ (iii) GwU KL‡bv FYvZ¥K n‡Z cv‡i bv wb‡Pi †KvbwU mwVK? i I ii ii I iii i I iii i, ii I iii DËi: i I ii 2. f~-c„ô n‡Z h D”PZvq c„w_ex‡K cÖ`ÿxYiZ †Kv‡bv K...wÎg DcMÖ‡ni †eMÑ [Easy] v = GM2 R + h v = GM R + h v = 2GM R + h v = GM R DËi: v = GM R + h 3. g1⁄2j MÖ‡ni c„‡ô g = 4 ms–2 Ges e ̈vmva© 3000 km n‡j g1⁄2j c„‡ô gyw3‡eM KZ n‡e? [Easy] 2.5 km/s 3.7 km/s 4.8 km/s 6 km/s DËi: 4.8 km/s e ̈vL ̈v: ve = 2gR = 2 × 4 × 3 × 106 = 24 × 103  25 × 103 = 5 km/s (hv Option Gi me‡P‡q KvQvKvwQ) 4. f~w ̄’i DcMÖ‡ni AveZ©bKvjÑ [Easy] 1 N›Uv 365 w`b 12 NÈv 24 N›Uv DËi: 24 N›Uv 5. cvwK©s Kÿc_ n‡jvÑ [Easy] c„w_exi Kÿc_ f~w ̄’i DcMÖ‡ni Kÿc_ †cvjvi DcMÖ‡ni Kÿc_ †h c‡_ wegvb PjvPj K‡i DËi: f~w ̄’i DcMÖ‡ni Kÿc_ 6. DcMÖ‡ni AveZ©bKvj †KvbwUi Dci wbf©i K‡i bv? [Easy] DcMÖ‡ni Kÿc‡_i e ̈vmva© DcMÖ‡ni †eM DcMÖ‡ni fi c„w_exi e ̈vmva© DËi: DcMÖ‡ni fi 7. c„w_exi e ̈vm eivei myo‡1⁄2i g‡a ̈ †Kv‡bv e ̄‘i MwZÑ [Medium] (i) ch©vqe„Ë (ii) ̄ú›`b (iii) mij Qw›`Z ̄ú›`b wb‡Pi †KvbwU mwVK? ii I ii i I ii i I iii i, ii I iii DËi: i, ii I iii 8. c„w_ex‡Z gyw3‡e‡Mi gvb K‡Zv? [Easy] 11.2 km/s 11.2 ms–1 11.2 kms 1.12 km/s DËi: 11.2 km/s 9. `ywU e ̄‘i ga ̈Kvi `~iZ¡ A‡a©K Kiv n‡j Zv‡`i ga ̈eZ©x wμqviZ gnvKl©xq ej KZ ̧Y n‡e? [Easy] GKB _vK‡e 1 4 ̧Y 4 ̧Y 2 ̧Y DËi: 4 ̧Y e ̈vL ̈v: F  1 d 2 d  1 2 ̧Y F  1     1 2 2 = 4 ̧Y n‡e 10. KZ Aÿvs‡k g Gi gvb me©vwaK? [Easy] 90 60 30 0 DËi: 90 11. †Kv‡bv ̄’v‡bi gnvKl©xq cÖve‡j ̈i gvb †KvbwUi mgvb nq? [Easy] wefekw3 wefe AwfKl©R Z¡iY fi DËi: AwfKl©R Z¡iY 12. f~c„ô †_‡K c„w_exi e ̈vmv‡a©i wØ ̧Y D”PZvq †Kv‡bv e ̄‘i IRb f~c„‡ôi IR‡bi KZ ̧Y? [Easy] 4 ̧Y 1 2 ̧Y 2 ̧Y 1 9 ̧Y DËi: 1 9 ̧Y e ̈vL ̈v: g g = R 2 (R + h) 2  W W =     R R + 2R 2  W = 1 9 W