In a meter bridge, null point is 20 cm, when the known resistance R is shunted by $10\Omega$ resistance, null point is found to be shifted by 10 cm. Find the unknown resistance X.

$\frac{R}{X}=\frac{l}{100-l}$

$X=\left(\frac{100-l}{l}\right)R$

$X=\left(\frac{100-20}{20}\right)R=4R$

When known resistance R is shunted, its net resistance will decrease. Therefore, resistance parallel to this (i.e., P ) should also decrease or its new null point length should also decrease.

$\frac{R^{\text{'}}}{X}=\frac{l^{\text{'}}}{100-l^{\text{'}}}$

$=\frac{20-10}{100-(20-10)}=\frac{1}{9}$

$X=9R^{\text{'}}$

$4R=9R^{\text{'}}=9\left[\frac{10R}{10+R}\right]$

Solving this equation, we get

$R=\frac{50}{4}\Omega$

$X=4R=4\left(\frac{50}{4}\right)$

$X=50\Omega$