See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

1.6 40        m We have to find d dt  when x = 30 m From AE 40 AEC, tan EC x    (1) Differentiating w.r.t. to t, 2 2 d 40 dx sec dt dt x     AITS-FT-IV-PCM(Sol.)-JEE (Main)/19 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 12 2 2 2 2 2 2 2 2 2 2 2 2 d 40 sec 2 dt x d 80 80 x x cos cos dt x x x 40 x 40 d 80 dt x 40                          when x = 30, 2 2 d 80 4 dt 125 30 40    radian/s.