(N/A) meter bridge is based on the principle of the Wheatstone bridge.
$1$. Connect the unknown resistor $R$ in one gap and a known resistance $S$ from a resistance box in the other gap of the meter bridge.
$2$. $A$ jockey is slid along the wire $AC$ until the galvanometer shows zero deflection at a point $D$. This is called the null point.
$3$. Let the length of the wire $AD$ be $l$. Then the length of the wire $DC$ is $(100 - l) \text{ cm}$.
$4$. Let $\rho$ be the resistance per unit length of the wire. The resistance of segment $AD$ is $P = \rho l$ and the resistance of segment $DC$ is $Q = \rho(100 - l)$.
$5$. According to the Wheatstone bridge principle,at the balanced condition:
$\frac{R}{S} = \frac{P}{Q} = \frac{\rho l}{\rho(100 - l)}$
$6$. Simplifying this,we get:
$\frac{R}{S} = \frac{l}{100 - l}$
$7$. Therefore,the unknown resistance $R$ is given by:
$R = S \left( \frac{l}{100 - l} \right)$
$8$. By measuring the length $l$ and knowing $S$,the value of $R$ can be calculated. To minimize errors,the null point should be obtained near the center of the wire (between $40 \text{ cm}$ and $60 \text{ cm}$).