Question:
In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance \(P = 4\,\Omega\) and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is :
In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance \(P = 4\,\Omega\) and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is :
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At balance \(P/Q = l/(100-l)\); find Q from the first balance, then use it with \(P+R\) for the second balance.
Updated On: Nov 11, 2026
- \(\frac{33}{5}\,\Omega\)
- \(6\,\Omega\)
- \(7\,\Omega\)
- \(\frac{20}{3}\,\Omega\)
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The Correct Option is D
Solution and Explanation
In the figure P is in the left gap (towards A) and Q in the right gap (towards B). Balance condition: \(\frac{P}{Q} = \frac{l}{100-l}\).
Initially: \(\frac{4}{Q} = \frac{60}{40} = \frac32\), so \(Q = \frac{8}{3}\,\Omega\).
After connecting R in series with P: \(\frac{4+R}{Q} = \frac{80}{20} = 4\), so \(4+R = 4\times\frac{8}{3} = \frac{32}{3}\).
\(R = \frac{32}{3} - 4 = \frac{20}{3}\,\Omega\).
Answer: (4) \(\frac{20}{3}\,\Omega\)
Initially: \(\frac{4}{Q} = \frac{60}{40} = \frac32\), so \(Q = \frac{8}{3}\,\Omega\).
After connecting R in series with P: \(\frac{4+R}{Q} = \frac{80}{20} = 4\), so \(4+R = 4\times\frac{8}{3} = \frac{32}{3}\).
\(R = \frac{32}{3} - 4 = \frac{20}{3}\,\Omega\).
Answer: (4) \(\frac{20}{3}\,\Omega\)
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