Question

In a meter bridge the point $D$ is a neutral point as shown in figure.

• A. The meter bridge can have no other neutral point for this set of resistances
• B. When the jockey contacts a point on meter wire left of D, current flows to B from the wire
• C. When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.
• D. When R is increased, the neutral point shifts of left

Answer 1

Answer: (A), (B), (C)

When there is a neutral point at $D$ in meter-bridge, then
$\frac{R}{S}=\frac{l_1}{(100-l_1)}$
For the given values of $R$ and $S$, there will be only one value of $l_1$ for which we shall get the neutral point on bridge wire.
In this case $V_A-V_B=V_A-V_D$ or $V_B=V_D$.
Therefore the galvanometer shows no deflection when jockey contacts a point at $D$. There is no current in galvanometer arm.
When jockey contacts a point $D_1$ on meter-bridge wire left of $D$, the resistance of arm $A{D}_1$ becomes smaller than previous value.
Due to it, $V_A-V_B>V_A-V_{D_1}$ or $V_{D_1}>V_B$.
Therefore current flows to $B$ from the wire through galvanometer.
When jockey contacts a point $D_2$ on meter bridge wire right of $D$, the resistance of arm $A{D}_2$ becomes more than first value.
Due to it, $V_A-V_{D_2}>V_A-V_B$ or $V_B>V_{D_2}$ Therefore the current will flow from $B$ to the wire through galvanometer.
When $R$ is increased, the neutral point will shift to the right instead of left on bridge wire.