Two resistors 2Ω and 3Ω are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with 3Ω resistor, the null point is shifted by 22.5 cm toward Y. The resistance of unknown resistor is ____ Ω .

Let the initial balancing length from end $$X$$ be $$l_1$$.

$$ \frac{R_1}{l_1} = \frac{R_2}{100 - l_1} $$

$$ \frac{2}{l_1} = \frac{3}{100 - l_1} $$

$$ 2(100 - l_1) = 3l_1 $$

$$ 200 - 2l_1 = 3l_1 $$

$$ 5l_1 = 200 \implies l_1 = 40 \text{ cm} $$

An unknown resistor $$R$$ is connected in parallel with the $$3 \, \Omega$$ resistor. The new equivalent resistance in the right gap ($$R_2'$$) is:

$$ R_2' = \frac{3R}{3 + R} $$

The null point shifts by $$22.5 \text{ cm}$$ towards $$Y$$. This means the new balancing length $$l_2$$ increases:

$$ l_2 = l_1 + 22.5 = 40 + 22.5 = 62.5 \text{ cm} $$

Applying the meter bridge principle for the new state:

$$ \frac{2}{l_2} = \frac{R_2'}{100 - l_2} $$

$$ \frac{2}{62.5} = \frac{R_2'}{100 - 62.5} $$

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

$$ R_2' = 2 \times \left( \frac{37.5}{62.5} \right) $$

$$ R_2' = 2 \times \left( \frac{3}{5} \right) = \frac{6}{5} \, \Omega $$

$$ \frac{3R}{3 + R} = \frac{6}{5} $$

$$ 15R = 6(3 + R) $$

$$ 9R = 18 $$

$$ R = 2 \, \Omega $$