A meter bridge is set-up as shown, to determine an unknown resistance $X$ using a standard $10 \mathrm{ohm}$…
A meter bridge is set-up as shown, to determine an unknown resistance $X$ using a standard $10 \mathrm{ohm}$ resistor. The galvanometer shows null point when tapping-key is at $52 \mathrm{~cm}$ mark. The end-corrections are $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively for the ends $A$ and $B$. The determined value of $X$ is
$10.2 \mathrm{ohm}$
$10.6 \mathrm{ohm}$
$10.8 \mathrm{ohm}$
$11.1 \mathrm{ohm}$
Solution
Applying the condition of balanced Wheatstone bridge, we get
$\begin{aligned} & \frac{X}{10 \Omega}=\frac{(52+1) \mathrm{cm}}{(48+2) \mathrm{cm}}=\frac{53}{50}, \\ & X=10 \Omega \times \frac{53}{50}=10.6 \Omega\end{aligned}$
