A) 720 J
B) 2.41 J
C) 24.12 J
D) \[4.34\times {{10}^{4}}\text{J}\]
Correct Answer: D
Solution :
Current in silver voltmeter \[{{i}_{1}}=\frac{{{m}_{1}}}{{{z}_{1}}{{t}_{1}}}\] \[=\frac{1}{(11.2\times {{10}^{-4}})}\times (30\times 60)\] \[=0.5A\] \[{{i}_{2}}=\frac{{{m}_{2}}}{{{z}_{2}}{{t}_{2}}}\] \[{{i}_{2}}=\frac{1.8}{(6.6\times {{10}^{-4}})\times (30\times 60)}\] \[=1.82A\] So, the total current gives by battery \[i={{i}_{1}}+{{i}_{2}}=2.01A\] Energy supplied by battery \[=EiL\] \[\text{W}=\left( \text{12} \right)\times \left( \text{2}.0\text{1} \right)\times \left( \text{3}0\times \text{6}0 \right)\] \[=4.34\times {{10}^{4}}J\]
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