Q. A current of 1.5 A flows through a copper voltameter. The thickness of copper deposited on the electrode surface of size $50\ cm \times 10\ cm$ is 20 min will be (density of copper $= 9000\ kg - m^{- 3}\ and\ ECE\ of\ copper = 0.00033gC^{- 1})$
A) $3.3 \times 10^{- 6}$ m
B) 6.6 $\times 10^{- 6}$ m
C) 1.3 $\times 10^{- 5}$ m
D) 2.6 $\times 10^{- 5}$ m
Explanation
(b) Thickness$\ d = \frac{V}{A} = \frac{m}{\rho A} = \frac{zlt}{\rho A}$
$= \frac{\left( 3.3 \times 10^{7} \right) \times (1.5) \times (20 \times 60)}{9000 \times (50 \times 10 \times 2) \times 10^{- 4}} = 6.6 \times 10^{- 6}$ m.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.