Answer

Correct option: B.
$0.02 V / cm$
(b) : Here $P Q=1 m , E_2=1.02 V$
When $S$ is open, $P J=P Q-J Q=100-49=51 cm$
Also, $E_2=k P J$
Potential gradient in the wire, $k=\frac{E_2}{P J}$
$
k=\frac{1.02 V }{51 cm }=0.02 V / cm
$

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