A) 60 W
B) 47 W
C) 57 W
D) 35 W
Correct Answer: C
Solution :
Potential gradient \[x=\frac{e}{(R+{{R}_{h}}+r)}.\frac{R}{L}\] \[\Rightarrow \,\,\,\,\,\,\frac{{{10}^{-3}}}{{{10}^{-2}}}=\frac{2}{(3+{{R}_{h}}+0)}\times \frac{3}{1}\Rightarrow {{R}_{h}}=57\Omega \].
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