Q 11-14-095JEE MainJEE Main 2021 (27 Aug, Shift 2)Easy
A tuning fork is vibrating at $250$ Hz. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ______ cm. (Take speed of sound in air as $340$ m s$^{-1}$)
Numerical value type. Enter your answer.
Answer: 34
The shortest closed pipe resonates in its fundamental mode, $L = \dfrac{\lambda}{4} = \dfrac{v}{4f}$:
$$L = \frac{340}{4\times250} = 0.34\ \text{m} = 34\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics