A meter bridge is set up as shown, to determine an unknown resistance ‘X’ using a standard 10 ohm resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of ‘X’ is

10.6 ohm

At Null point

$\frac{X}{l_1}=\frac{10}{l_2}$

Here $l_1$ = 52 + End correction = 52 + 1 = 53 cm

         $l_2$ = 48 + End correction = 48 + 2 = 50 cm

∴ $\frac{X}{53}=\frac{10}{50}$            ∴ $X=\frac{53}{5}=10.6 \Omega$