Physics - Current Electricity Question with Solution | TestHub

For a meter bridge, the balance condition is given by $\frac{X}{R}=\frac{l}{100-l}$, where $X$ is the unknown resistance and $R$ is the standard resistance.

Given $R=100\Omega$ and $l=16$ cm, we can find $X$:

$\frac{X}{100}=\frac{16}{100-16}=\frac{16}{84}$

$X=\frac{1600}{84}=\frac{400}{21}\approx 19.04\Omega$

To obtain the null point near the middle of the wire, the balancing length should be $l^{\prime }\approx 50$ cm. Using the balance condition again with the new standard resistance $R^{\prime }$:

$\frac{X}{R^{\prime }}=\frac{50}{100-50}=1$

$R^{\prime }=X\approx 19.04\Omega$

Thus, the value of the standard resistance should be chosen as $19\Omega$.

Answer: $19\Omega$