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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V 0

3 V 0

2 V 0

4 V 0

answer is D.

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Detailed Solution

$\tan \frac{θ}{2} = \frac{F_{t}}{F_{r}} = \frac{m (\frac{dv}{dt})}{(\frac{{mv}^{2}}{R})} = v \frac{dv}{ds} \cdot \frac{R}{v^{2}}$

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

solving this, we get, $v = 4 V_{0}$

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