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(D) Let the unknown resistance be $R$ and the initial balance point be at $l 	ext{ cm}$ from the left end.Using the metre bridge principle,$\frac{R_1

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$30$

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(D) Let the unknown resistance be $R$ and the initial balance point be at $l 	ext{ cm}$ from the left end.Using the metre bridge principle,$\frac{R_1}{R_2} = \frac{l}{100-l}$.Case $1$: $R_1 = 20 \Omega$ and $R_2 = R$. So,$\frac{20}{R} = \frac{l}{100-l} \quad --- (1)$Case $2$: $R_1 = R$ and $R_2 = 20 \Omega$. Since $R > 20 \Omega$,the balance point shifts towards the right,so $l' = l + 20 	ext{ cm}$.Thus,$\frac{R}{20} = \frac{l+20}{100-(l+20)} = \frac{l+20}{80-l} \quad --- (2)$From $(1)$,$\frac{R}{20} = \frac{100-l}{l}$. Substituting this into $(2)$:$\frac{100-l}{l} = \frac{l+20}{80-l}$$(100-l)(80-l) = l(l+20)$$8000 - 100l - 80l + l^2 = l^2 + 20l$$8000 = 200l \Rightarrow l = 40 	ext{ cm}$.Substituting $l = 40$ into $(1)$:$\frac{20}{R} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$$R = 30 \Omega$.

$A$ resistance of $20 \Omega$ is connected in the left gap of a metre bridge and an unknown resistance greater than $20 \Omega$ is connected in the right gap. When these resistances are interchanged,the balance point shifts by $20 \text{ cm}$. The unknown resistance is (in $\Omega$)

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