Question

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

• A. $10.2ohm$
• B. $10.6ohm$
• C. $10.8ohm$
• D. $11.1ohm$

Answer 1

Answer: (B)

$10.6ohm$

Applying the condition of balanced Wheatstone bridge, we get
$\frac{X}{10\Omega }=\frac{(52+1)cm}{(48+2)cm}=\frac{53}{50}$,
$X=10\Omega \times \frac{53}{50}=10.6\Omega$.