In a meter bridge circuit, resistance in the left hand gap is $2\Omega$ and an unknown resistance X is in the right hand gap as shown in Figure. The null point is found to be $40$cm from the left end of the wire. What resistance should be connected to X so that the new null point is $50$cm from the left end of the wire?

Case 1 :

Null point is found at $l_1 = 40$ cm

Using balanced Wheatstone bridge condition, $\dfrac{2}{X} = \dfrac{l_1}{100 - l_1}$

$\therefore$ $\dfrac{2}{X} = \dfrac{40}{100 - 40}$

$\implies$ $X = 3\Omega$

Case 2 :

New null point is found at $l_2 = 50$ cm

Using balanced Wheatstone bridge condition, $\dfrac{2}{X'} = \dfrac{l_2}{100 - l_2}$

$\therefore$ $\dfrac{2}{X'} = \dfrac{50}{100 - 50}$

$\implies$ $X' = 2\Omega$

Since the unknown resistance gets decreased, thus we have to connect a resistance $R$ in parallel to $X$ so that $X'$ comes out to be $2\Omega$.

Thus resistance of parallel combination $X' = \dfrac{XR}{X+R}$

$\therefore$ $2= \dfrac{3R}{3+R}$

$\implies$ $R =6\Omega$

Hence $6\Omega$ must be connected in parallel to unknown resistance $X$.