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________.
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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Resistance of a wire is \(\rho L/A\propto L/d^2\). Then use the metre bridge condition \(R/S = l/(100 - l)\).
Updated On: Nov 5, 2026
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Correct Answer: 40
Solution and Explanation
Metre bridge balance: \(\frac{R}{S} = \frac{l}{100 - l} = \frac{25}{75} = \frac13\), so \(S = 3R\).
New wire: length halved, diameter halved so area becomes one quarter: \(R' = \rho\frac{L/2}{A/4} = 2R\).
New balance: \(\frac{R'}{S} = \frac{2R}{3R} = \frac23 = \frac{l'}{100 - l'}\), so \(3l' = 200 - 2l'\) and \(l' = 40\) cm.
Answer: 40
New wire: length halved, diameter halved so area becomes one quarter: \(R' = \rho\frac{L/2}{A/4} = 2R\).
New balance: \(\frac{R'}{S} = \frac{2R}{3R} = \frac23 = \frac{l'}{100 - l'}\), so \(3l' = 200 - 2l'\) and \(l' = 40\) cm.
Answer: 40
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