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(C) In case $I$, the resistances are $R_1 = 2 \ \Omega$ and $R_2 = 3 \ \Omega$. The null point is at $l$. Using the meter bridge formula: $\frac{R_1}{

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$6$

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(C) In case $I$, the resistances are $R_1 = 2 \ \Omega$ and $R_2 = 3 \ \Omega$. The null point is at $l$. Using the meter bridge formula: $\frac{R_1}{R_2} = \frac{l}{100-l} \implies \frac{2}{3} = \frac{l}{100-l}$.Solving this, $200 - 2l = 3l \implies 5l = 200 \implies l = 40 \ cm$.In case $II$, the resistance in the right gap becomes $R_2' = \frac{3x}{3+x}$ because $x$ is connected in parallel to $3 \ \Omega$. The null point shifts by $10 \ cm$ to the right, so the new position is $l' = 40 + 10 = 50 \ cm$.Using the formula: $\frac{R_1}{R_2'} = \frac{l'}{100-l'} \implies \frac{2}{\frac{3x}{3+x}} = \frac{50}{100-50} = \frac{50}{50} = 1$.Therefore, $\frac{2(3+x)}{3x} = 1 \implies 6 + 2x = 3x \implies x = 6 \ \Omega$.

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