Question

The density and equivalent weight 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 $80 \,cm^2$ is closest to

• A. 120 s
• B. 135 s
• C. 67.5 s
• D. 270 s

Answer 1

Answer: (B)

Given,
density of metal = $10.5\,g cm^{-3}$
Equivalent weight of metal = $100$
Current, $I = 3A$
Area of cross section = $80\,cm^2$, length
$= 0.005\,mm = 5 \times 10^{-4}\,cm$
According to Faraday’s second law,
$W=\frac{eq.wt.\times I\times t}{n\times96500} $
$W=\frac{100\times3\times t}{1\times96500} ....\left(i\right)$
Also, $W$ = density x volume
= density x area x length
$=10.5\times80\times5\times10^{-4}=42\times10^{-2} ...\left(ii\right)$
Substituting the value of $W$ in Eq. (i),
$\frac{100\times3\times t}{1\times96500}=42\times10^{-2}$
$t=\frac{42\times10^{-2}\times96500}{100\times3}=135\,s$