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A particle starts at ‘A’ with initial speed $V_{0}$ . It moves along a circular path under the action of a variable force which is always directed towards ‘B”. A and B are end points of a diameter. When it reaches the point ‘P’ as shown , the speed is (Neglect the effect of gravity)

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$2 V_{0}$

$V_{0}$

$3 V_{0}$

$4 V_{0}$

answer is D.

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Detailed Solution

$\tan \frac{θ}{2} = \frac{F_{t}}{F_{r}} = \frac{m (\frac{d v}{d t})}{(\frac{m v^{2}}{R})} = V \frac{d v}{d s} . \frac{R}{v^{2}}$

$\int_{0}^{2 π / 3} \tan \frac{θ}{2} d θ = \int_{v_{0}}^{v} \frac{d v}{v} [∴ d s = R d θ] ∴ V = 4 V_{0}$

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