A) 3.54g
B) 3.0g
C) 1.36g
D) 2.16g
Correct Answer: D
Solution :
\[+3.12\times {{10}^{15}}\times 1.6\times {{10}^{-19}}\] As 276g of \[=1\times {{10}^{-3}}=1mA\] will give = 216 g of Ag So. 2.76 g of \[=q\times {{n}^{2/3}}\] will give \[R=1k\Omega \]
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