A) \[5.49\times {{10}^{7}}C\] of electricity
B) \[1.83\times {{10}^{7}}C\] of electricity
C) \[5.49\times {{10}^{4}}C\] of electricity
D) \[5.49\times {{10}^{1}}C\] of electricity
Correct Answer: A
Solution :
27 gm of Al is obtained by passing a current of 3 × 96500 C. \[\because \]1 gm of Al is obtained by passing a current of \[3\times \frac{96500}{27}C.\] \ 5.12 × 103 gm of Al is obtained by passing a current of \[3\times \frac{96500}{27}\times 5.12\times 1000\] = 1.83 × 107 C × 3 = 5.49 × 107C.You need to login to perform this action.
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