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(B) In a meter bridge, the balancing condition is given by $\frac{P}{Q} = \frac{\ell}{100 - \ell}$.Initially, $\frac{5}{R} = \frac{\ell_1}{100 - \ell_

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$15$

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(B) In a meter bridge, the balancing condition is given by $\frac{P}{Q} = \frac{\ell}{100 - \ell}$.Initially, $\frac{5}{R} = \frac{\ell_1}{100 - \ell_1}$ --- $(1)$When $R$ is shunted with an equal resistance $R$, the equivalent resistance becomes $R' = \frac{R 	imes R}{R + R} = \frac{R}{2}$.The new balance point is at $1.6 \ell_1$, so:$\frac{5}{R/2} = \frac{1.6 \ell_1}{100 - 1.6 \ell_1} \implies \frac{10}{R} = \frac{1.6 \ell_1}{100 - 1.6 \ell_1} \implies \frac{5}{R} = \frac{0.8 \ell_1}{100 - 1.6 \ell_1}$ --- $(2)$Equating $(1)$ and $(2)$:$\frac{\ell_1}{100 - \ell_1} = \frac{0.8 \ell_1}{100 - 1.6 \ell_1}$$100 - 1.6 \ell_1 = 0.8(100 - \ell_1)$$100 - 1.6 \ell_1 = 80 - 0.8 \ell_1$$20 = 0.8 \ell_1 \implies \ell_1 = \frac{20}{0.8} = 25 \, cm$.Substituting $\ell_1 = 25$ in $(1)$:$\frac{5}{R} = \frac{25}{100 - 25} = \frac{25}{75} = \frac{1}{3}$$R = 15 \, \Omega$.

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