Question:
When two wires are connected in the two gaps of a meter bridge, the balancing point is obtained at a distance of \(48.4\) cm from the left end of the bridge wire. If the wire in the left gap is stretched so that its resistance increases by \(3.2\%\) and the wire in the right gap is stretched so that its resistance increases by \(10\%\), then the new balancing length from the left end of the bridge wire is nearly
When two wires are connected in the two gaps of a meter bridge, the balancing point is obtained at a distance of \(48.4\) cm from the left end of the bridge wire. If the wire in the left gap is stretched so that its resistance increases by \(3.2\%\) and the wire in the right gap is stretched so that its resistance increases by \(10\%\), then the new balancing length from the left end of the bridge wire is nearly
Show Hint
In a meter bridge,
\[
\frac{R}{S}=\frac{l}{100-l}.
\]
If resistances are modified, first calculate the new ratio
\[
\frac{R'}{S'},
\]
and then use the balance condition again to find the new balancing length.
Updated On: Jul 29, 2026
- \(33.3\) cm
- \(66.6\) cm
- \(44.4\) cm
- \(55.5\) cm
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The Correct Option is C
Solution and Explanation
Concept:
For a meter bridge at balance,
\[
\frac{R}{S}
=
\frac{l}{100-l},
\]
where \(R\) and \(S\) are the resistances in the left and right gaps respectively, and \(l\) is the balancing length from the left end.
Step 1: Find the initial ratio of resistances. Given, \[ l=48.4\ \text{cm}. \] Hence, \[ \frac{R}{S} = \frac{48.4}{100-48.4}. \] \[ = \frac{48.4}{51.6}. \] \[ = \frac{121}{129}. \]
Step 2: Calculate the new resistance ratio. The left resistance increases by \(3.2\%\), \[ R'=1.032R. \] The right resistance increases by \(10\%\), \[ S'=1.10S. \] Therefore, \[ \frac{R'}{S'} = \frac{1.032}{1.10} \cdot \frac{R}{S}. \] \[ = \frac{1.032}{1.10} \cdot \frac{121}{129}. \] \[ = 0.872. \]
Step 3: Determine the new balancing length. Let the new balancing length be \(l'\). Then \[ \frac{l'}{100-l'} = 0.872. \] \[ l' = 0.872(100-l'). \] \[ l' = 87.2-0.872l'. \] \[ 1.872l' = 87.2. \] \[ l' = 46.58\ \text{cm}. \] Using the nearest value among the options, \[ l' \approx 44.4\ \text{cm}. \] Therefore, \[ \boxed{l' \approx 44.4\ \text{cm}} \] \[ \boxed{\text{Answer = (C)}} \]
Step 1: Find the initial ratio of resistances. Given, \[ l=48.4\ \text{cm}. \] Hence, \[ \frac{R}{S} = \frac{48.4}{100-48.4}. \] \[ = \frac{48.4}{51.6}. \] \[ = \frac{121}{129}. \]
Step 2: Calculate the new resistance ratio. The left resistance increases by \(3.2\%\), \[ R'=1.032R. \] The right resistance increases by \(10\%\), \[ S'=1.10S. \] Therefore, \[ \frac{R'}{S'} = \frac{1.032}{1.10} \cdot \frac{R}{S}. \] \[ = \frac{1.032}{1.10} \cdot \frac{121}{129}. \] \[ = 0.872. \]
Step 3: Determine the new balancing length. Let the new balancing length be \(l'\). Then \[ \frac{l'}{100-l'} = 0.872. \] \[ l' = 0.872(100-l'). \] \[ l' = 87.2-0.872l'. \] \[ 1.872l' = 87.2. \] \[ l' = 46.58\ \text{cm}. \] Using the nearest value among the options, \[ l' \approx 44.4\ \text{cm}. \] Therefore, \[ \boxed{l' \approx 44.4\ \text{cm}} \] \[ \boxed{\text{Answer = (C)}} \]
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