Question

If $m$ is the mass of silver deposited at the cathode by $2A$ current flowing for $25min$ through a silver voltameter, then the mass deposited by $1.5A$ current flowing for $600s$ is _________.

• A. $m$
• B. $3m$
• C. $0.3m$
• D. None

Answer 1

The correct option is C $0.3m$

$\frac{m^{\prime }}{m}=\frac{I^{\prime }{t}^{\prime }}{It}=\frac{1.5\times 600}{2\times 25\times 60}=\frac{900}{3000}=\frac{3}{10}=0.3$
$\Rightarrow m^{\prime }=0.3m$