In the process of electroplating, $m g$ of silver is deposited when $4$ ampere of current flows for $2$ minutes. The amount (in g) of silver deposited by $6$ ampere of current flowing for $40$ seconds will be:

4 m

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\frac{m}{2}

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\frac{4 m}{3}

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3 m

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Solution

The correct option is B $\frac{m}{2}$
$\frac{W_{1}}{W_{2}} = \frac{Q_{1}}{Q_{2}}$
$\frac{W_{1}}{W_{2}} = \frac{I_{1} \times t_{1}}{I_{2} \times t_{2}}$
$\frac{m}{W_{2}} = \frac{4 \times 2 \times 60}{6 \times 40}$
$W_{2} = \frac{m}{2} g$ .

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