A) 2.4 m
B) \[2.4\,\,\mu m\]
C) \[2.4\,\,\mu m\]
D) None of these
Correct Answer: C
Solution :
\[\operatorname{Mass} deposited = Density \times Volume of the metal\] \[m=p\times A\times X\] ... (i) Hence from Faraday?s first law \[\operatorname{m} = Zit\] ?.(ii) So from equations (i) and (ii) \[Zit\,\,=\,\,\rho \times Ax\Rightarrow \,\,x=\frac{Zit}{\rho A}\] \[=\,\,\frac{3.04\times {{10}^{-4}}\times {{10}^{-3}}\times 1\times 3600}{9000\times 0.05}\] \[=2.4\times 1{{0}^{-\,6}}m\,\,=\,\,2.4\,\,\mu m\]You need to login to perform this action.
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