In a metre bridge experiment the balance point is obtained if the gaps are closed by $$2$$ $$\Omega$$ and $$3$$ $$\Omega$$. A shunt of $$X$$ $$\Omega$$ is added to $$3$$ $$\Omega$$ resistor to shift the balancing point by $$22.5$$ cm. The value of $$X$$ is ______.

We have a metre bridge with $$R_1 = 2 \Omega$$ and $$R_2 = 3 \Omega$$. At balance:

$$\frac{l}{100 - l} = \frac{R_1}{R_2} = \frac{2}{3}$$

Solving, $$3l = 200 - 2l$$, which gives $$5l = 200$$, so $$l = 40$$ cm.

Now, a shunt resistance $$X$$ is added in parallel with the $$3 \Omega$$ resistor, and the balance point shifts by $$22.5$$ cm. Since adding a shunt decreases the effective resistance, the balance point shifts toward the $$3 \Omega$$ side, giving a new balance point $$l' = 40 + 22.5 = 62.5$$ cm.

The effective resistance with the shunt is $$R_2' = \frac{3X}{3 + X}$$. At the new balance point:

$$\frac{62.5}{37.5} = \frac{2}{R_2'}$$

So $$R_2' = \frac{2 \times 37.5}{62.5} = \frac{75}{62.5} = 1.2 \Omega$$.

Now we solve for $$X$$:

$$\frac{3X}{3 + X} = 1.2$$

$$3X = 3.6 + 1.2X$$

$$1.8X = 3.6$$

$$X = 2 \Omega$$

So, the answer is $$2$$.