Current Electricity Charging Discharging Of Capacitors Question 82
Question: 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
[Orissa PMT 2004]
Options:
A) 1.2 gm
B) 1.6 gm
C) 2.4 gm
D) 1.8 gm
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{m _{1}}{m _{2}}=\frac{Z _{1}}{Z _{2}} $
Therefore $ m _{2}=\frac{m _{1}Z _{2}}{Z _{1}}=\frac{14\times 1.2\times {{10}^{-6}}}{7\times {{10}^{-6}}} $ = 2.4 g
BETA
