Question:
A sonometer wire of length \(l_1\) is in resonance with a frequency 250 Hz.
If the length of wire is increased to \(l_2\), then 2 beats per second are heard.
The ratio of lengths \( \dfrac{l_1}{l_2} \) of wire will be
A sonometer wire of length \(l_1\) is in resonance with a frequency 250 Hz.
If the length of wire is increased to \(l_2\), then 2 beats per second are heard.
The ratio of lengths \( \dfrac{l_1}{l_2} \) of wire will be
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In sonometer problems, frequency varies inversely with length of the wire.
Updated On: Feb 11, 2026
- \(1:2\)
- \(124:125\)
- \(2:1\)
- \(1:250\)
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The Correct Option is B
Solution and Explanation
Step 1: Relation between frequency and length.
For a stretched string under constant tension,
\[ f \propto \frac{1}{l} \]
Step 2: Frequencies involved.
Initial frequency \(f_1 = 250\,\text{Hz}\). After increasing length, frequency becomes
\[ f_2 = 250 \pm 2 \]
Since length is increased, frequency decreases:
\[ f_2 = 248\,\text{Hz} \]
Step 3: Taking ratio of lengths.
\[ \frac{l_1}{l_2} = \frac{f_2}{f_1} = \frac{248}{250} \]
Step 4: Simplifying.
\[ \frac{l_1}{l_2} = \frac{124}{125} \]
Step 5: Conclusion.
The required ratio is \(124:125\).
For a stretched string under constant tension,
\[ f \propto \frac{1}{l} \]
Step 2: Frequencies involved.
Initial frequency \(f_1 = 250\,\text{Hz}\). After increasing length, frequency becomes
\[ f_2 = 250 \pm 2 \]
Since length is increased, frequency decreases:
\[ f_2 = 248\,\text{Hz} \]
Step 3: Taking ratio of lengths.
\[ \frac{l_1}{l_2} = \frac{f_2}{f_1} = \frac{248}{250} \]
Step 4: Simplifying.
\[ \frac{l_1}{l_2} = \frac{124}{125} \]
Step 5: Conclusion.
The required ratio is \(124:125\).
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