In a metre-bridge when a resistance in the left gap is 2Ω and...

In first circuit:

For balanced condition, we get $\dfrac{2}{40} = \dfrac{X}{60} ⇒ X = 3Ω$

In second case:

Net resistance of the right side due to shunting becomes, ${X}^{'} = \dfrac{2 \times 3}{2 + 3} = 1 . 2Ω$

Again for balanced condition, 

$\dfrac{2}{ℓ} = \dfrac{1 . 2}{100 - ℓ}$

$⇒ 200 - 2ℓ = 1 . 2ℓ$

$⇒ ℓ = \dfrac{200}{3 . 2} = 62 . 5cm$

Clearly, balance length changes by $22 . 5cm$.