Waves

JEE Main level, numerical answer, medium

94 of 142

Question 94 of 142

The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20cm20cm. The length of the open organ pipe is _____ cmcm

Show answer and solution

Answer: 80

The fundamental frequency of the closed organ pipe is given by, f0=v4lc{f}_{0}=\dfrac{v}{4{l}_{c}}.
The frequency of the first overtone of a open organ pipe is given by, f1=2×v2lo{f}_{1}=2\times \dfrac{v}{2{l}_{o}}.
Since,
f0=f1⇒v4lc=vlo⇒lo=4×20=80cm{f}_{0}={f}_{1} \Rightarrow \dfrac{v}{4{l}_{c}}=\dfrac{v}{{l}_{o}} \Rightarrow {l}_{o}=4\times 20=80cm.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Standing waves on a string simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Waves (Organ Pipe (Vibration of Air Column)), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.