A) \[115\text{ }sec\]
B) \[125\text{ }sec\]
C) \[135\text{ }sec\]
D) \[145\text{ }sec\]
Correct Answer: B
Solution :
Volume of \[Ag=800\times 5\times {{10}^{-3}}\text{ }c{{m}^{3}}\] Mass of \[Ag=V\times d=0.42\text{ }g\] \[A{{g}^{+}}+{{e}^{-}}\xrightarrow{{}}Ag\] \[\because \] \[108g\]\[Ag\] require \[96500\,C\] \[\therefore \]\[0.42\,g\]\[Ag\]require \[=\frac{96500}{108}\times 0.42=375.3C\] \[t=\frac{Q}{I}=\frac{375.3}{3}=125\sec \]You need to login to perform this action.
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