In the fig. shown for given values of $R_1$ and $R_2$ the balance point for jockey is at $40\, cm$ from $A$. When $R_2$ is shunted by a resistance of $10\,\Omega $, balance shifts to $50\, cm. R_1$ and $R_2$ are $(AB = 1\,m)$
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Let resistance per unit length of wire is $\rho$ then from given information
$\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\rho \times 40}{\rho \times 60}$
$\frac{\frac{\mathrm{R}}{10 \mathrm{R}_{2}}}{\frac{10 \mathrm{R}_{2}}{10+\mathrm{R}_{2}}}=\frac{50 \times \rho}{50 \times \rho}$
Solving above equation
$\mathrm{R}_{1}=\frac{10}{3}\, \Omega \quad \mathrm{R}_{2}=5\, \Omega$
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