JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2024
- Hard
- MCQ
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is \(60 \mathrm{~cm}\), the length of the closed pipe will be _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(15 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
\(\frac{\lambda}{4}=\mathrm{L}_1 \quad 2\left(\frac{\lambda}{2}\right)=\lambda\)
\(\mathrm{v}=\mathrm{f} \lambda \quad \mathrm{f}_2=\frac{2 \mathrm{v}}{2 \mathrm{~L}_2}\)
\(\mathrm{v}=\mathrm{f}_1\left(4 \mathrm{~L}_1\right) \quad \mathrm{f}_2=\frac{\mathrm{v}}{\mathrm{L}_2}\)
\({v}=\mathrm{f}_1\left(4 \mathrm{~L}_1\right)\)
\({f}_2=\frac{\mathrm{v}}{\mathrm{L}_2}\)
\({f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1}\)
\({f}_1=\mathrm{f}_2\)
\(\frac{\mathrm{V}}{4 \mathrm{~L}_1}=\frac{\mathrm{V}}{\mathrm{L}_2}\)
\(\Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1\)
\(60=4 \times \mathrm{L}_1\)
\({~L}_1=15 \mathrm{~cm} \)
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