Q1 / 90 · KVPY 2013 (SB/SX) · Chemistry
The density of eq. wt of a metal are $10.5 \mathrm{gcm}^{-3}$ and 100 , respectively. The time required for a current of 3 amp to deposit a $0.005 \mathrm{~mm}$ thick layer of the same metal on an area of $80 \mathrm{~cm}^{2}$ is closest to
- A$120 \mathrm{~s}$
- Correct answerB$135 \mathrm{~s}$
- C$67.5 \mathrm{~s}$
- D$270 \mathrm{~s}$
Solution
$\mathrm{w}=\frac{\mathrm{E} \times \mathrm{i} \times \mathrm{t}}{96500}$
$\mathrm{w}=\frac{100 \times 3 \times \mathrm{t}}{96500}$
$\mathrm{~d}=\frac{\mathrm{m}}{\mathrm{V}}$
$\mathrm{w}=10.5 \times 80 \times 5 \times 10^{-4}=42 \times 10^{-2}$
$42 \times 10^{-2}=\frac{100 \times 3 \times \mathrm{t}}{96500}$
$\mathrm{t}=135 \mathrm{sec}$
Hence, the correct answer is option: 'B'.