MHT CET · Physics · Waves and Sound
A sonometer wire stretched by weight ' \(w\) ' is in unison with a tuning fork. The corresponding resonating length is ' \(\mathrm{L}_1\) ' if the weight is increased by ' \(3 w\) ' the corresponding resonating length of the sonometer in unison tuning fork becomes ' \(\mathrm{L}_2\).' . The ratio \(\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)\) is
- A \(4: 1\)
- B \(2: 2\)
- C \(1: 2\)
- D \(1: 4\)
Answer & Solution
Correct Answer
(C) \(1: 2\)
Step-by-step Solution
Detailed explanation
Velocity of the sound on the string under tension \(\mathrm{T}\) and with mass per unit length \(\mu\) is given by
\(\begin{aligned} & \mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}} \\ & \therefore \mathrm{v}_1=\sqrt{\frac{\mathrm{w}}{\mu}} \mathrm{v}_2=\sqrt{\frac{(3 \mathrm{w}+\mathrm{w})}{\mu}}\end{aligned}\)
We know \(\lambda \propto L \& f=\frac{v}{\lambda}\)
Therefore, \(\mathrm{f}_1 \propto \frac{\sqrt{\mathrm{w}}}{\mathrm{L}_1}\) and \(\mathrm{f}_2 \propto \frac{\sqrt{4 \mathrm{w}}}{\mathrm{L}_2}\)
\(\because \mathrm{f}_1=\mathrm{f}_2 \Rightarrow \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{1}{\sqrt{4}}=\frac{1}{2}\)
\(\begin{aligned} & \mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}} \\ & \therefore \mathrm{v}_1=\sqrt{\frac{\mathrm{w}}{\mu}} \mathrm{v}_2=\sqrt{\frac{(3 \mathrm{w}+\mathrm{w})}{\mu}}\end{aligned}\)
We know \(\lambda \propto L \& f=\frac{v}{\lambda}\)
Therefore, \(\mathrm{f}_1 \propto \frac{\sqrt{\mathrm{w}}}{\mathrm{L}_1}\) and \(\mathrm{f}_2 \propto \frac{\sqrt{4 \mathrm{w}}}{\mathrm{L}_2}\)
\(\because \mathrm{f}_1=\mathrm{f}_2 \Rightarrow \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{1}{\sqrt{4}}=\frac{1}{2}\)
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