# The density of eq. wt of a metal are 10.5 g cm-3 and 100, respectively. The time required for a current of 3 amp to deposit a 0.005 mm thick layer of the same metal on an area of 80 cm2 is closest to:

$\\\text{The volume of deposited metal layer }=0.005 \times 0.1 \mathrm{cm} \times 80 \mathrm{cm}^{3}=0.04 \mathrm{cm}^{3} \\ the mass of the deposited metal =0.04 \times 10.5=0.42 g \\ now, W=\frac{E \times Q}{F} \\ \Rightarrow \quad 0.42=\frac{100 \times 3 \times t}{96485} \\ \Rightarrow \quad t=135 \mathrm{sec}$

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