If the length of the potentiometer wire is increased by keeping constant potential difference across the wire, then

Only $4 \%$ of the total current in the circuit passes through a galvanometer. If the resistance of the galvanometer is G , then the shunt resistance connected to the galvanometer is

The potential difference $\left(V_A-V_B\right)$ between the points A and B in the given figure is

A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by $5 \Omega$. When a resistance of $15 \Omega$ is used for shunting, null point moves to 300 cm . The internal resistance of the cell is