In metre-bridge experiment the balance point is obtained if the gaps are closed by $2 \Omega$ and $3 \Omega$…

The metre bridge operates on the Wheatstone bridge principle, with the balance condition $\frac{R_1}{R_2} = \frac{l}{100 - l}$ where $l$ is the length from the $R_1$ side.

With $R_1 = 2\ \Omega$ and $R_2 = 3\ \Omega$, the initial balance point satisfies $\frac{2}{3} = \frac{l}{100 - l}$, yielding $l = 40$ cm.

When a shunt resistance $x$ is added in parallel to $3\ \Omega$, the equivalent resistance becomes $R_2' = \frac{3x}{x + 3}$.

The balance point shifts by 22.5 cm to $l' = 62.5$ cm. The new balance condition gives $\frac{2}{\frac{3x}{x + 3}} = \frac{62.5}{37.5}$, which simplifies to $\frac{2(x + 3)}{3x} = \frac{5}{3}$.

Solving $2(x + 3) = 5x$ yields $x = 2\ \Omega$.

Final answer: $\boxed{B}$