JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2021
- Medium
- Numerical
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 \(340\, {ms}^{-1}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(34\)
Step-by-step Solution
Detailed explanation
\(\frac{\lambda}{4}=\ell \Rightarrow \lambda=4 \ell\)
\({f}=\frac{{V}}{\lambda}=\frac{{V}}{4 \ell}\)
\(\Rightarrow 250=\frac{340}{4 \ell}\)
\(\Rightarrow \ell=\frac{34}{4 \times 25}=0.34\, {m}\)
\(\ell=34\, {cm}\)
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