JEE Mains · Physics · STD 11 - 14. waves and sound
The distance between two consecutive points with phase difference of \(60^{\circ}\) in a wave of frequency \(500\,Hz\) is \(6.0\,m\). The velocity with which wave is traveling is \(.........km / s\)
- A \(17\)
- B \(16\)
- C \(15\)
- D \(18\)
Answer & Solution
Correct Answer
(D) \(18\)
Step-by-step Solution
Detailed explanation
\(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x\) \(\frac{\pi}{3}=\frac{2 \pi}{\lambda}(6 m )\) \(\lambda=36\,m\) \(V = f \lambda=(500\,Hz )(36\,m )\) \(=18000\,m / s\) \(=18\,km / s\)
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