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(D) Let $\rho$ be the resistance per unit length of the wire $AB$. The total length is $L = 200 \, cm$.For the first case, the null point is at $l_1 =

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$10 \, \Omega, 15 \, \Omega$

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(D) Let $\rho$ be the resistance per unit length of the wire $AB$. The total length is $L = 200 \, cm$.For the first case, the null point is at $l_1 = 80 \, cm$ from $A$. The length $AJ = 80 \, cm$ and $JB = 200 - 80 = 120 \, cm$.Using the Wheatstone bridge principle: $\frac{R_1}{R_2} = \frac{R_{AJ}}{R_{JB}} = \frac{\rho \cdot 80}{\rho \cdot 120} = \frac{80}{120} = \frac{2}{3}$.So, $3R_1 = 2R_2$ --- $(1)$In the second case, $R_2$ is shunted with $30 \, \Omega$. The new resistance $R_2' = \frac{R_2 \cdot 30}{R_2 + 30}$.The null point is at $100 \, cm$ from $B$, so $JB = 100 \, cm$ and $AJ = 200 - 100 = 100 \, cm$.Using the principle again: $\frac{R_1}{R_2'} = \frac{100}{100} = 1$.So, $R_1 = R_2' = \frac{30R_2}{R_2 + 30}$ --- $(2)$From $(1)$, $R_2 = 1.5R_1$. Substitute this into $(2)$:$R_1 = \frac{30(1.5R_1)}{1.5R_1 + 30} \implies 1.5R_1 + 30 = 45 \implies 1.5R_1 = 15 \implies R_1 = 10 \, \Omega$.Then $R_2 = 1.5(10) = 15 \, \Omega$.Thus, $R_1 = 10 \, \Omega$ and $R_2 = 15 \, \Omega$.

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