See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 2 2 r b a e   …(1)   2 2 2 b a 1 e   …(2) From equation (1) & (2) r = a Now from conservation of energy 2 GMm 1 GMm mv 2a 2 a    GM v a  B D C A ae b r 23.   2 2 3 ucos v ucos 5    2 2 2 2 3 u cos v 5 5  …(1) 2 2 2 2 2 1 u sin v u sin 2g 3 2g          2 2 2 2 v u sin 3   …(2) H v ucos ucos u AITS-FT-VIII-PCM (Sol.)-JEE(Main)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 6 24.     0 2 10cos60 2 2 v   0 5 v m / s 2  …(1)   5 4 4 16 v 2    1 v m / s 2  …(2) Now from conservation of energy   2 2 1 5 1 1 4 4 16 2 2 2 2                  4 10 1 1 cos      3 cos 4  1 3 cos 4        25. F0cos t  kA cos t = mA2 cos t  A = 0 2 F k m   26. Heat produced  r2 but change in temperature  r 27. Gravitation potential due to disc is V = 2 2 2 2Gm ( R ) R      So, i Gmm U   (R<<) f 2Gmm U R  (  0) i f 2 1 U U U Gmm R             2 2 1 1 2 1 mv mv Gmm 2 2 R           2 1 v Gm R          So, relative velocity R 2 1 v 2v 4Gm R           28. log P = m log V + C …(i) Where C is constant and m is slope