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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 \(340\, {ms}^{-1}\) )

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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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