A meter bridge is set-up as shown, to determine an unknown resistance X using a standard 10Ω 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:

10.6Ω

Applying the condition of balanced Wheatstone bridge, we get

$\frac{X}{10 \Omega}=\frac{(52+1) ~cm}{(100-52+2) ~cm}=\frac{53}{50}$

X = 10Ω × $\frac{53}{50}$ = 10.6 Ω