See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

1 2 A Av v 4   v2 = 4v1 2 2 1 1 2 2 1 1 P gh Pv P Pv 2 2     2 2 2 1 2 2 v 1 (P P ) gh P v 2 16             2 1 2 2 15 (P P ) gh v 32      v2 = 7 m/s  dx Q  B  x  y(m) x(m) A B (1, 0.5) (0, 0) 3 m/s (0, 1.5) IAOR For More Material Join: @JEEAdvanced_2024 AITS-FT-VII-PCM(Sol.)-JEE(Main)/2024 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 6