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JEE MAIN 2020
09-01-2020 S2
Question
In a meter bridge experiment S is a standard resistance. R is a resistance wire. It is found that balancing length is l = 25 cm. 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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Solution
$\begin{aligned} & \frac{S}{R}=\frac{75}{25}=3 \\ & R^{\prime}=2 R \\ & \frac{S}{2 R}=\frac{3}{2}=\frac{100-I^{\prime}}{I^{\prime}} \\ & I^{\prime}=40 \mathrm{~cm}\end{aligned}$
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