In a meter bridge experiment,S is a standard | vedclass

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(D) For a meter bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100 - l}$.Given $l = 25 \; cm$,we have $\frac{R}{S} = \frac{25}{100

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$40$

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(D) For a meter bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100 - l}$.Given $l = 25 \; cm$,we have $\frac{R}{S} = \frac{25}{100 - 25} = \frac{25}{75} = \frac{1}{3}$.Thus,$S = 3R$.The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \frac{4 \rho l}{\pi d^2}$.When the length is halved $(l^{\prime} = l/2)$ and the diameter is halved $(d^{\prime} = d/2)$,the new resistance $R^{\prime}$ becomes:$R^{\prime} = \frac{4 \rho (l/2)}{\pi (d/2)^2} = \frac{4 \rho l / 2}{\pi d^2 / 4} = 2 \left( \frac{4 \rho l}{\pi d^2} \right) = 2R$.Now,for the new balancing length $l^{\prime}$,the condition is $\frac{R^{\prime}}{S} = \frac{l^{\prime}}{100 - l^{\prime}}$.Substituting $R^{\prime} = 2R$ and $S = 3R$:$\frac{2R}{3R} = \frac{l^{\prime}}{100 - l^{\prime}}$$\frac{2}{3} = \frac{l^{\prime}}{100 - l^{\prime}}$$200 - 2l^{\prime} = 3l^{\prime}$$5l^{\prime} = 200$$l^{\prime} = 40 \; cm$.

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