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A U - shaped smooth wire has a semicircular bending between A and B as shown in the figure. A bead of mass m moving with uniform speed v through the wire enters the semicircular bend at A and leaves at B. the average force, exerted by the bead on the part AB of the wire is

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$\frac{4 {mv}^{2}}{πd}$

$\frac{2 {mv}^{2}}{πd}$

$\frac{{mv}^{2}}{πd}$

answer is B.

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Detailed Solution

$F = \frac{2 mv}{t} = \frac{2 mv}{(\frac{πd}{2 v})} = \frac{4 {mv}^{2}}{πd}$

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