1. Use the metre-bridge balance condition

For a metre-bridge at balance,

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

where:

• $R$ = resistance in left gap
• $X$ = resistance in right gap
• $l$ = balance length from left end

Here,

$R=2\,\Omega, \qquad l=40\,\text{cm}$

So,

$\frac{2}{X}=\frac{40}{60}=\frac{2}{3}$

Hence,

$X=3\,\Omega$

2. Find the new resistance when $X$ is shunted by $2\,\Omega$

Shunting means parallel combination:

$X' = \frac{3\times 2}{3+2} = \frac{6}{5}=1.2\,\Omega$

3. Find the new balance length

Now the balance condition becomes

$\frac{2}{1.2}=\frac{l'}{100-l'}$

$\frac{5}{3}=\frac{l'}{100-l'}$

Cross-multiplying,

$3l' = 5(100-l')$

$3l' = 500-5l'$

$8l' = 500$

$l' = 62.5\,\text{cm}$

4. Find the change in balance length

Initial balance length = $40\,\text{cm}$

New balance length = $62.5\,\text{cm}$

So change is

$62.5-40=22.5\,\text{cm}$

5. Evaluate options

• A: $62.5$ → this is the new balance length, not the change
• B: $22.5\,\text{cm}$ → correct
• C: $20\,\text{cm}$ → incorrect
• D: $65\,\text{cm}$ → incorrect

Therefore, the correct option is B.