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(A) Let the two resistances be $r_1$ and $r_2$. In a metre bridge,the balance condition is $\frac{r_1}{r_2} = \frac{l}{100-l}$.Given $l = 20 \ cm$,so

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$9$

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(A) Let the two resistances be $r_1$ and $r_2$. In a metre bridge,the balance condition is $\frac{r_1}{r_2} = \frac{l}{100-l}$.Given $l = 20 \ cm$,so $\frac{r_1}{r_2} = \frac{20}{80} = \frac{1}{4}$,which implies $r_2 = 4r_1$. Thus,$r_1$ is the smaller resistance.When $15 \ \Omega$ is connected in series with $r_1$,the new balance length is $40 \ cm$.The new condition is $\frac{r_1 + 15}{r_2} = \frac{40}{60} = \frac{2}{3}$.Substituting $r_2 = 4r_1$ into the equation: $\frac{r_1 + 15}{4r_1} = \frac{2}{3}$.Cross-multiplying gives $3(r_1 + 15) = 2(4r_1) \Rightarrow 3r_1 + 45 = 8r_1$.$5r_1 = 45 \Rightarrow r_1 = 9 \ \Omega$.

The unknown resistances are connected in two gaps of a metre bridge. The null point is at $20 \ cm$ from the zero end. $A$ resistance of $15 \ \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \ cm$. The smaller resistance is (in $\Omega$)

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