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(B) The wavelength $\lambda$ of the wave is given by the formula $\lambda = \frac{v}{f}$.Given $v = 300\, m/s$ and $f = 500\, Hz$,we have $\lambda = \

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$0.1$

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(B) The wavelength $\lambda$ of the wave is given by the formula $\lambda = \frac{v}{f}$.Given $v = 300\, m/s$ and $f = 500\, Hz$,we have $\lambda = \frac{300}{500} = 0.6\, m$.$A$ phase difference of $360^{\circ}$ corresponds to a path difference of one full wavelength $\lambda$.Therefore,a phase difference of $\Delta\phi = 60^{\circ}$ corresponds to a path difference $\Delta x$ given by $\Delta x = \frac{\Delta\phi}{360^{\circ}} 	imes \lambda$.Substituting the values,$\Delta x = \frac{60^{\circ}}{360^{\circ}} 	imes 0.6\, m = \frac{1}{6} 	imes 0.6\, m = 0.1\, m$.

Two points on a travelling wave having frequency $500\, Hz$ and velocity $300\, m/s$ are $60^{\circ}$ out of phase,then the minimum distance between the two points is ..... $m$.

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