Fig. shows rough sketch of meter bridge. $(G)$ deflects zero at length $\ell \, cm$. Now $R_1$ and $R_2$ are interchanged then balancing length increases by $25\, cm$. Find $R_1/R_2$
A$\frac{3}{5}$
B$\frac{2}{5}$
C$\frac{2}{3}$
D$\frac{5}{2}$
Medium
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A$\frac{3}{5}$
a $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\ell}{100-\ell}$ and $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\ell+25}{75-\ell}$
So $\frac{\ell}{100-\ell}=\frac{\ell+25}{75-\ell}$ or $\ell=\frac{75}{2} \mathrm{\,cm}$
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