See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

54. 6 FIITJEE JEE(Advanced)-2018 ALL INDIA TEST SERIES AITS-PT-II (Paper-1)-PCM (Sol)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 2 Physics PART – I SECTION – A 1. F qv B      x y ˆ ˆ q(v i v j)B    2 0 y x 2 B F q x v   0 max v x 2 0 y 2 0 0 qB dv x dx m     3 0 max 0 2 qB x v 3 m   1/3 2 0 max 0 3m v x qB          (q, m) x y v0 vx vy v0 Alternate solution Impulse momentum theorem along x-axis max x 2 0 x 0 2 0 qB x v dt mv     max x 2 0 0 2 0 qB x dx mv     1/3 2 0 max 0 3m v x qB          2. Potential due to circular non conducting disc, VC = 0 r 2   (where r is radius of disc) So, P 0 v 2 2                  0 0 5 R 1 5 R 2 3 ( ) 2 12                          P   5 R 3  3. Work performed = 2R 4R 2 2 0 0 R 3R 1 1 E dV E dV 2 2      = 2 0 7 q 96 R  AITS-PT-II (Paper-1)-PCM (Sol)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 3 4. OAB ABCD   0 0 0 q q q 48 48 24       y  x z q A O B D C 5. Magnetic field due to the ring on its axis is 2 0 2 2 3/2 r i B 2 (r x )     (due to ring) 2 /2 0 3/2 2 2 /2 2 Rcos Rd (Rcos ) 2 / B 2 (Rcos ) (Rsin )                         Where  = 2 q 4 R  0q B 6 R     R  q O  Rd