Waves

JEE Main level, single correct, medium

102 of 142

Question 102 of 142

The end correction of a resonance column is 1cm1cm. If the shortest length resonating with the tuning fork is 10cm10cm, the next resonating length should be

Show answer and solution

Answer: (A) 32cm32cm

Given: e=1cme=1cm. For first resonance, λ4=l1+e=11cm\dfrac{\lambda }{4}={l}_{1}+e=11cm. For second resonance, 3λ4=l2+e⇒l2=3×11−1=32cm\dfrac{3\lambda }{4}={l}_{2}+e\Rightarrow {l}_{2}=3\times 11-1=32cm.

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.