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JEE MAINS 2024
31.01.24 S1
Question
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" " cm, the length of the closed pipe will be :
Select the correct option:
A
60 cm
B
45 cm
C
30 cm
D
15 cm
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{array}{ll}\frac{\lambda}{4}=\mathrm{L}_1 & 2\left(\frac{\lambda}{2}\right)=\lambda \\ \mathrm{v}=\mathrm{f} \lambda & \mathrm{f}_2=\frac{2 \mathrm{v}}{2 \mathrm{~L}_2} \\ \mathrm{v}=\mathrm{f}_1\left(4 \mathrm{~L}_1\right) & \mathrm{f}_2=\frac{\mathrm{v}}{\mathrm{L}_2}\end{array}$ $\begin{aligned} & \mathrm{f}_1=\frac{\mathrm{v}}{4 L_1} \\ & \mathrm{f}_1=\mathrm{f}_2 \frac{\mathrm{v}}{4 L_1}=\frac{\mathrm{v}}{L_2} \\ & \Rightarrow L_2=4 L_1 \\ & 60=4 \times L_1 \\ & L_1=15 \mathrm{~cm}\end{aligned}$
Question Tags
JEE Main
Physics
Medium
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