Navjyot Kalra
Last Activity: 11 Years ago
Sol. I = 15 A Surface area = 200 cm^2 , Thickness = 0.1 mm
Volume of Ag deposited = 200 × 0.01 = 2 cm^3 for one side
For both sides, Mass of Ag = 4 × 10.5 = 42 g
Z base Ag = E/F = 107.9/96500 m = ZIT
⇒ 42 = 107.9/96500 *15*T ⇒ T = 42 * 96500/107.9 * 15 = 2504.17 sec = 41.73 min = 42 min