In a meter bridge,the point D is the neutral | vedclass

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(B) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{l_1}{100 - l_1}$.At the null point $D$,the potential at $B$ is equal to t

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When the jockey contacts a point on the meter wire to the left of $D$,current flows to $B$ from the wire.

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(B) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{l_1}{100 - l_1}$.At the null point $D$,the potential at $B$ is equal to the potential at $D$,so no current flows through the galvanometer.If the jockey is moved to the left of $D$,the resistance of the left segment decreases,making the potential at $D$ higher than at $B$. Thus,current flows from $D$ to $B$ (i.e.,to $B$ from the wire).If the jockey is moved to the right of $D$,the potential at $D$ becomes lower than at $B$,so current flows from $B$ to $D$ (i.e.,from $B$ to the wire).Therefore,option $B$ is correct.

In a meter bridge,the point $D$ is the neutral point (null point) as shown in the figure.

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