See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

  4 1 1 1 KE PE mkr 2 2 2          4 2 1 1 KE mkr mv 4 2   4 2 4 mkr .2 1 v kr 4.m 2    … (1) Since, 2 2 2 2 2 2 nh n h mvr v 2 4 m r      … (2) From (1) and (2) 2 2 4 2 2 2 1 n h kr 2 4 m r   6 2 3 r n r n     AITS-PT-I-PCM(Sol.)-JEE(Main)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 10