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(B) The circumference of the circular tube is $2\pi R$.The two paths from $S$ to $D$ are along the arcs of the circle. Since $S$ and $D$ are at right

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$2\pi R$

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(B) The circumference of the circular tube is $2\pi R$.The two paths from $S$ to $D$ are along the arcs of the circle. Since $S$ and $D$ are at right angles,the shorter path length is $L_1 = \frac{1}{4}(2\pi R) = \frac{\pi R}{2}$.The longer path length is $L_2 = \frac{3}{4}(2\pi R) = \frac{3\pi R}{2}$.The path difference between the two waves is $\Delta x = L_2 - L_1 = \frac{3\pi R}{2} - \frac{\pi R}{2} = \pi R$.For destructive interference (minima) to occur at $D$,the path difference must be an odd multiple of half-wavelengths:$\Delta x = (2n + 1) \frac{\lambda}{2}$,where $n = 0, 1, 2, \dots$Equating the path difference:$\pi R = (2n + 1) \frac{\lambda}{2}$$\lambda = \frac{2\pi R}{2n + 1}$To find the maximum value of $\lambda$,we choose the smallest non-negative integer $n = 0$:$\lambda_{	ext{max}} = \frac{2\pi R}{2(0) + 1} = 2\pi R$.

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