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(C) In a meter-bridge,the balance condition is given by $\frac{P}{Q} = \frac{\ell}{100-\ell}$.For the first case: $\frac{18}{\ell_{1}} = \frac{R}{100-

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$18$

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(C) In a meter-bridge,the balance condition is given by $\frac{P}{Q} = \frac{\ell}{100-\ell}$.For the first case: $\frac{18}{\ell_{1}} = \frac{R}{100-\ell_{1}}$ ... $(1)$For the second case: $\frac{18}{1.5 \ell_{1}} = \frac{R/3}{100-1.5 \ell_{1}}$ ... $(2)$From equation $(1)$,we have $\frac{R}{18} = \frac{100-\ell_{1}}{\ell_{1}} = \frac{100}{\ell_{1}} - 1$.From equation $(2)$,$\frac{R/3}{18} = \frac{100-1.5 \ell_{1}}{1.5 \ell_{1}} = \frac{100}{1.5 \ell_{1}} - 1$.Dividing equation $(1)$ by equation $(2)$: $\frac{18/\ell_{1}}{18/(1.5 \ell_{1})} = \frac{R/(100-\ell_{1})}{(R/3)/(100-1.5 \ell_{1})}$$1.5 = 3 	imes \frac{100-1.5 \ell_{1}}{100-\ell_{1}}$$0.5 = \frac{100-1.5 \ell_{1}}{100-\ell_{1}}$$50 - 0.5 \ell_{1} = 100 - 1.5 \ell_{1}$$1.0 \ell_{1} = 50 \implies \ell_{1} = 50 	ext{ cm}$.Substituting $\ell_{1} = 50$ in equation $(1)$: $\frac{18}{50} = \frac{R}{100-50} \implies \frac{18}{50} = \frac{R}{50} \implies R = 18 \Omega$.

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