Question

In a meter bridge, the null point is located at 20 cm from the left end of the wire when resistances \( R \) and \( S \) are connected in the left and right gaps respectively. If the resistance \( S \) is shunted with \( 60\,\Omega \) resistance, the null point shifted by 5 cm, then the values of \( R \) and \( S \) are respectively:

• A. \(24\,\Omega, 6\,\Omega\)
• B. \(6\,\Omega, 24\,\Omega\)
• C. \(5\,\Omega, 20\,\Omega\)
• D. \(20\,\Omega, 5\,\Omega\)

Hint:

In meter bridge problems, always apply the bridge balance condition \( \frac{R}{S} = \frac{l_1}{100 - l_1} \), and when resistances are modified by combinations, use equivalent resistance rules to proceed.

Answer 1

The Correct Option is C

Step 1: Use the meter bridge formula for initial condition: \[ \frac{R}{S} = \frac{l_1}{100 - l_1} = \frac{20}{80} = \frac{1}{4} \Rightarrow R = \frac{S}{4}
\text{(1)} \] Step 2: After shunting \( S \) with \( 60\,\Omega \), effective resistance \( S' \) is: \[ \frac{1}{S'} = \frac{1}{S} + \frac{1}{60} \Rightarrow S' = \frac{60S}{60 + S} \] New balance point at 25 cm: \[ \frac{R}{S'} = \frac{25}{75} = \frac{1}{3} \Rightarrow R = \frac{S'}{3}
\text{(2)} \] Step 3: Substitute equation (1) into (2): \[ \frac{S}{4} = \frac{1}{3} . \frac{60S}{60 + S} \Rightarrow \frac{1}{4} = \frac{60}{3(60 + S)} \Rightarrow 60 + S = 720 \Rightarrow S = 20\,\Omega \Rightarrow R = \frac{20}{4} = 5\,\Omega \]