JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2025
- Medium
- MCQ
In the resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(360 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Fundamental frequency in close/organ pipe
\(\begin{aligned} & (\mathrm{f})=\frac{\mathrm{v}}{4 \ell} \\ & \mathrm{f}_1=\frac{\mathrm{v}}{4 \ell_1} \& \mathrm{f}_2=\frac{\mathrm{v}}{4 \ell_2}\end{aligned}\)
Beat \(=\left(\mathrm{f}_1-\mathrm{f}_2\right)=\frac{\mathrm{v}}{4}\left(\frac{1}{\ell_1}-\frac{1}{\ell_2}\right)\)
\(\begin{aligned} & 15=\frac{\mathrm{v}}{4}\left(\frac{1}{1}-\frac{1}{1.2}\right) \\ & \mathrm{v}=\left(\frac{15 \times 4 \times 1.2}{0.2}\right)=60 \times 6=360 \mathrm{~m} / \mathrm{s}\end{aligned}\)
\(\begin{aligned} & (\mathrm{f})=\frac{\mathrm{v}}{4 \ell} \\ & \mathrm{f}_1=\frac{\mathrm{v}}{4 \ell_1} \& \mathrm{f}_2=\frac{\mathrm{v}}{4 \ell_2}\end{aligned}\)
Beat \(=\left(\mathrm{f}_1-\mathrm{f}_2\right)=\frac{\mathrm{v}}{4}\left(\frac{1}{\ell_1}-\frac{1}{\ell_2}\right)\)
\(\begin{aligned} & 15=\frac{\mathrm{v}}{4}\left(\frac{1}{1}-\frac{1}{1.2}\right) \\ & \mathrm{v}=\left(\frac{15 \times 4 \times 1.2}{0.2}\right)=60 \times 6=360 \mathrm{~m} / \mathrm{s}\end{aligned}\)
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