Question

One gram of copper is deposited in a copper voltameter when a current of $0.5$A flows for $30$ minutes. Then, the current required to deposit $2$g of silver in a silver voltmeter in the same time is (ece of copper $=3.3\times {10}^{-4}g{C}^{-1},$ ece of silver$=1.1\times {10}^{-4}g{C}^{-1}$)

• A. $4$A
• B. $6$A
• C. $2$A
• D. $5$A
• E. $3$A

Answer 1

The correct option is E $3$A

ECE of copper$=3.3\times {10}^{-4}gm{c}^{-1}$
ECE of silver$=1.1\times {10}^{-4}gm{c}^{-1}$
Time,$\left(t\right)=30$min
$l=0.5$A
Deposited mass$=2$gm
$m=Zlt$
$\Rightarrow l=\frac{m}{Zt}$
$\frac{l_2}{l_1}=\frac{m_2}{m_1}\times \frac{Z_1}{Z_2}\times \frac{t_2}{t_1}$
$t_1=t_2=30$min
$\Rightarrow \frac{l_2}{l_1}=\frac{2}{1}\times \frac{3.3\times {10}^{-4}}{1.1\times {10}^4}\times \frac{30}{30}$
$=\frac{2}{1}\times \frac{3.3}{1.1}=6$
$l_2=6\times l_1=6\times 0.5=3A$.