In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is $l = 25 c m$ . If R is replaced by a wire of half length and half diameter that of R of same material, then the balancing distance $l'$ (in cm) will now be_______

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answer is 40.

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Detailed Solution

$\begin{array}{l} \frac{R}{S} = \frac{L}{100 - L} = \frac{25}{75} \\ \Rightarrow R = \frac{1}{3} S \\ A l s o, R = \frac{ρ L}{A} = \frac{ρ L}{π r^{2}} \\ N e w v a l u e = R^{1} = \frac{ρ . \frac{L}{2}}{π {(\frac{r}{2})}^{2}} = 2 R \\ N o w, \frac{2 R}{S} = \frac{L^{1}}{100 - L^{1}} \\ \Rightarrow \frac{2. S / 3}{S} = \frac{L^{1}}{100 - L^{1}} \\ \Rightarrow \frac{2}{3} = \frac{L^{1}}{100 - L^{1}} \\ \Rightarrow \frac{200 - 2 L^{1}}{1} = \frac{3 L^{1}}{1} \\ \Rightarrow L^{1} = 40 c m \end{array}$