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(C) From the figure,the distance between two consecutive nodes or a node and an adjacent antinode is $\frac{\lambda}{4}$. Thus,$ab = \frac{\lambda}{4}

Vedclass · Accepted Answer

$I$ and $III$

Vedclass · Answer

(C) From the figure,the distance between two consecutive nodes or a node and an adjacent antinode is $\frac{\lambda}{4}$. Thus,$ab = \frac{\lambda}{4}$,which implies $\lambda = 4ab$.Statement $I$: The speed of the wave $v = n\lambda = n(4ab) = 4n 	imes ab$. This statement is correct.Statement $II$: The point $a$ is at a crest and $d$ is at a node. For $a$ to be in the same phase as $d$,the wave must travel a distance such that $d$ reaches the crest position. The distance between $a$ and $d$ is $\frac{3\lambda}{4}$. The time taken is $t = \frac{	ext{distance}}{	ext{speed}} = \frac{3\lambda/4}{n\lambda} = \frac{3}{4n} 	ext{ s}$. The statement says $\frac{4}{3n} 	ext{ s}$,which is incorrect.Statement $III$: The path difference between $b$ (node) and $e$ (crest) is $\frac{3\lambda}{4}$. The phase difference $\Delta\phi = \frac{2\pi}{\lambda} 	imes \Delta x = \frac{2\pi}{\lambda} 	imes \frac{3\lambda}{4} = \frac{3\pi}{2}$. This statement is correct.Therefore,statements $I$ and $III$ are correct.

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