A) 0.6 W
B) 1W
C) 1.25W
D) 1.67W
Correct Answer: D
Solution :
As we know that work done \[\frac{2\sqrt{2}{{\mu }_{0}}i}{\pi a}\] When potential difference increases from 5V to 10 V then \[\frac{{{\mu }_{0}}i}{\sqrt{2}\pi a}\] ...(i) When potential difference increases form 10 to 15V, then\[{{R}_{1}}=1\Omega \] ...(ii) On solving Eqs. (i) and (ii) we get W' = 1.67 WYou need to login to perform this action.
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