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(A) For a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{\ell}{100 - \ell}$.When $R_1$ and $R_2$ are interchanged,the new b

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$\frac{3}{5}$

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(A) For a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{\ell}{100 - \ell}$.When $R_1$ and $R_2$ are interchanged,the new balancing length becomes $\ell + 25 \, cm$. The new condition is $\frac{R_2}{R_1} = \frac{\ell + 25}{100 - (\ell + 25)} = \frac{\ell + 25}{75 - \ell}$.Taking the reciprocal of the first equation,we get $\frac{R_2}{R_1} = \frac{100 - \ell}{\ell}$.Equating the two expressions for $\frac{R_2}{R_1}$:$\frac{100 - \ell}{\ell} = \frac{\ell + 25}{75 - \ell}$$(100 - \ell)(75 - \ell) = \ell(\ell + 25)$$7500 - 100\ell - 75\ell + \ell^2 = \ell^2 + 25\ell$$7500 = 200\ell$$\ell = \frac{7500}{200} = 37.5 \, cm$.Now,substitute $\ell = 37.5$ into the first equation:$\frac{R_1}{R_2} = \frac{37.5}{100 - 37.5} = \frac{37.5}{62.5} = \frac{375}{625} = \frac{3}{5}$.

The figure shows a rough sketch of a meter bridge. The galvanometer $(G)$ shows zero deflection at a length $\ell \, cm$. Now,$R_1$ and $R_2$ are interchanged,then the balancing length increases by $25 \, cm$. Find the ratio $R_1/R_2$.

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