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(B) Initially, for a meter bridge, the balance condition is given by $\frac{P}{Q} = \frac{l}{100-l}$.Given $P = 2 \Omega$ and $Q = 3 \Omega$, we have

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$2$

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(B) Initially, for a meter bridge, the balance condition is given by $\frac{P}{Q} = \frac{l}{100-l}$.Given $P = 2 \Omega$ and $Q = 3 \Omega$, we have $\frac{2}{3} = \frac{l}{100-l}$.$2(100-l) = 3l \Rightarrow 200 - 2l = 3l \Rightarrow 5l = 200 \Rightarrow l = 40 	ext{ cm}$.When an unknown resistor $R$ is connected in parallel with $3 \Omega$, the new resistance $Q'$ becomes $\frac{3R}{3+R}$.The null point shifts by $22.5 	ext{ cm}$ toward $Y$, so the new length $l' = 40 + 22.5 = 62.5 	ext{ cm}$.The new balance condition is $\frac{2}{Q'} = \frac{62.5}{100-62.5} = \frac{62.5}{37.5} = \frac{5}{3}$.Substituting $Q'$, we get $\frac{2}{\frac{3R}{3+R}} = \frac{5}{3} \Rightarrow \frac{2(3+R)}{3R} = \frac{5}{3}$.$6(3+R) = 15R \Rightarrow 18 + 6R = 15R \Rightarrow 9R = 18 \Rightarrow R = 2 \Omega$.

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