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Q.

A particle having a uniform speed V moves along a circle of radius r. The average acceleration of the particle during the time in which the particle moves over the quadrant of the circle is

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a

$\frac{2 \sqrt{2} V^{2}}{π r}$

b

$\frac{2 V^{2}}{π r}$

c

$\frac{V^{2}}{π r}$

d

$\frac{2 \sqrt{2} V^{2}}{π r^{2}}$

answer is B.

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Detailed Solution

$a = \frac{Δ V}{(\frac{T}{4})}$

$Δ V = 2 V \sin \frac{θ}{2}; T = \frac{2 π r}{V}$

$a = \frac{4 \times 2 V \sin 45^{0}}{2 π r} \times V$

$= \frac{2 \sqrt{2} V^{2}}{π r}$

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