JEE Main 2021 · Physics · Waves and Sound · medium

A closed organ pipe is excited by a tuning fork of frequency 500 Hz. Resonance is first obtained when the air column is 15 cm long, and the next…

Numerical answer

A closed organ pipe is excited by a tuning fork of frequency

500\ \mathrm{Hz}

. Resonance is first obtained when the air column is

15\ \mathrm{cm}

long, and the next resonance occurs at

49\ \mathrm{cm}

. The end correction of the pipe, in millimetres, is

Show answer and solution

Answer: 20

In a closed pipe successive resonances differ by

\lambda/2

\lambda/2=49-15=34\ \mathrm{cm}

, so

\lambda=68\ \mathrm{cm}

(and

v=500\times0.68=340\ \mathrm{m\,s^{-1}}

). The first resonance satisfies

l_1+e=\lambda/4=17\ \mathrm{cm}

, so

e=17-15=2\ \mathrm{cm}=20\ \mathrm{mm}

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake