A) 2.4m
B) 2.4 \[\mu \]m
C) 2.4 mm
D) None of these
Correct Answer: B
Solution :
Mass deposited = density \[\times \] volume of the metal \[m=\rho \times A\times x\] ?...(i) Hence, from Faradays first law m = Zit ... (ii) So, from Eqs. (i) and (ii) \[Zit=\rho Ax\] \[\Rightarrow \] \[x=\frac{Zit}{\rho A}\] \[=\frac{3.04\times {{10}^{-4}}\times {{10}^{-3}}\times 1\times 3600}{9000\times 0.05}\] \[=2.4\times {{10}^{-6}}m=2.4\mu m\]You need to login to perform this action.
You will be redirected in
3 sec