In a metre-bridge when a resistance in the left gap is $$2 \Omega$$ and unknown resistance in the right gap, the balance length is found to be 40 cm. On shunting the unknown resistance with $$2 \Omega$$, the balance length changes by:

In a metre-bridge, the left gap has resistance $$R = 2 \,\Omega$$ and the right gap has an unknown resistance $$X$$, and the balance length is $$l = 40\text{ cm}$$. Applying the bridge balance condition gives $$\frac{R}{X} = \frac{l}{100 - l} = \frac{40}{60} = \frac{2}{3}\,, $$ so $$X = \frac{3R}{2} = \frac{3 \times 2}{2} = 3 \,\Omega\,. $$

When this resistance $$X = 3 \,\Omega$$ is shunted with $$2 \,\Omega$$, the combined resistance becomes $$X' = \frac{X \times 2}{X + 2} = \frac{3 \times 2}{3 + 2} = \frac{6}{5} = 1.2 \,\Omega\,. $$ For the new balance condition one has $$\frac{R}{X'} = \frac{l'}{100 - l'} = \frac{2}{1.2} = \frac{5}{3}\,, $$ leading to $$3l' = 5(100 - l') = 500 - 5l'\,, $$ so $$8l' = 500\,, $$ and hence $$l' = 62.5\text{ cm}\,. $$

The change in balance length is $$\Delta l = l' - l = 62.5 - 40 = 22.5\text{ cm}\,, $$ which is the required result.