JEE Mains · Physics · STD 11 - 14. waves and sound
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 \(340\, {ms}^{-1}\) )
- A \(340\)
- B \(34\)
- C \(17\)
- D \(3.4\)
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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