Q. E.C.E. of $Cu$ and $Ag$ are $7 \times 10^{- 6}$ and $1.2 \times 10^{- 6}.$ A certain current deposits $14\ gm$ of $Cu.$ Amount of $Ag$ deposited is
A) $1.2\ gm$
B) $1.6\ gm$
C) $2.4\ gm$
D) $1.8\ gm$
Explanation
(c) $\frac{m_{1}}{m_{2}} = \frac{Z_{1}}{Z_{2}} \Rightarrow m_{2} = \frac{m_{1}Z_{2}}{Z_{1}} = \frac{14 \times 1.2 \times 10^{- 6}}{7 \times 10^{- 6}} = 2.4g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.