See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Given a b c 2b         a b cos 8     4 cos a   a 4   and  = 0  b c 0      Either b c    or b c     2a b c 2a        or 2a b c 2a 2b 3a            For More Material Join: @JEEAdvanced_2024 AITS-FT-VIII (Paper-2)-PCM(Sol.)-JEE(Advanced)/2024 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 12 SECTION – B