See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 1 KE mv 2    2 1 d KE d mv 2             d KE mv dv   … (1) Since, h dv dx 4 m      h dv 4 m dx    … (2) From (1) and (2)     h d KE mv. 4 m. dx   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 13   8 34 12 1 3 10