In a resonance tube experiment when the tube is filled with water up to a height of 17 . 0cm, from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning

Correct answer: A. 2200Hz (λ)/(2) = 24 . 5 - 17 = 7 . 5cm f = (V)/(λ) = (330 × 100)/(15) = 2200Hz

JEE Main 20205 SeptemberMorning Shift

JEE Main 2020 · 5 September · Morning Shift · Physics · Waves and Sound

In a resonance tube experiment when the tube is filled with water...

In a resonance tube experiment when the tube is filled with water up to a height of

17 . 0cm\text{,}

from bottom, it resonates with a given tuning fork. When the water level is raised the next resonance with the same tuning fork occurs at a height of

24 . 5cm\text{.}

If the velocity of sound in air is

330m{s}^{- 1}\text{,}

the tuning fork frequency is :

A. $2200Hz$

$\dfrac{\lambda}{2} = 24 . 5 - 17 = 7 . 5cm$

$f = \dfrac{V}{\lambda}$

$= \dfrac{330 \times 100}{15} = 2200Hz$