In a metre bridge experiment the balance point is obtained if the...

Case $1$: No shunt is added

The initial balancing length $l$ of the bridge is,

$\dfrac{2}{3} = \dfrac{l}{100 - l}$

$⇒ 200 - 2l = 3l ⇒ l = 40cm$

Case $2$: When shunt is added

The new balancing length is, ${l}_{1} = 22 . 5 + 40 = 62 . 5cm$

So, $\dfrac{62 . 5}{37 . 5} = \dfrac{2}{\left(\dfrac{3X}{3 + X}\right)}$

$⇒ X = 2$.