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

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

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

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

Or

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.

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

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

or $X=9R^{\text{'}}$

From Eqs. (i) and (ii), we have

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

Solving this equation, we get

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

Now, from Eq. (i), the unknown resistance

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

or $X=50\Omega$

Note $R^{\text{'}}$ is resultant of $R$ and $10\Omega$ in parallel.

$\therefore$ $\frac{1}{R^{\text{'}}}=\frac{1}{10}+\frac{1}{R}$

or $R^{\text{'}}=\frac{10R}{10+R}$