Question

The resistance in two arms of meterbridge are $R=5\Omega$ and $S=1\Omega$ respectively. If error in measurement of null point is $1\text{cm}$, find the error in measurement of resistance.

• A. $\frac{2}{25}\Omega$
• B. $\frac{4}{25}\Omega$
• C. $\frac{9}{25}\Omega$
• D. $\frac{8}{25}\Omega$

Answer 1

The correct option is C $\frac{9}{25}\Omega$

For balanced meterbridge, we know that

$R=\frac{S{l}_1}{100-l_1}$

$5=\frac{1\times l_1}{100-l_1}\Rightarrow 500=6l_1$

$\therefore l_1=\frac{250}{3}$

Now, for error in measurement of resistance

$\frac{\Delta R}{R}=\left[\frac{\Delta l_1\times 100}{l_1(100-l_1)}\right]$

putting the value,

$\Delta R=\frac{5\times 1\times 100}{\frac{250}{3}(100-\frac{250}{3})}=\frac{500\times 3\times 3}{250\times 50}$

$\therefore \Delta R=\frac{9}{25}\Omega$

Hence, option $\left(c\right)$ is correct.