The resistance in the two arms of a meter bridge are $5\,\Omega $ and $R\,\Omega $, respectively. When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6\, l_1$. The resistance $‘R’$ is ................. $\Omega$
Medium
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$\frac{5}{\mathrm{R}}=\frac{\ell_{1}}{100-\ell_{1}}$ and $\frac{5}{\mathrm{R} / 2}=\frac{1.6 \ell_{1}}{100-1.6 \ell_{1}}$
$\Rightarrow \mathrm{R}=15\, \Omega$
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