The Kinetic energy 'K' of a particle moving along a circle of radius R depends on the distance covered 'S' as $K = a S^{2}$ . The force acting on the particle is

2 a \frac{S^{2}}{R}

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2 a S {(1 + \frac{S^{2}}{R^{2}})}^{\frac{1}{2}}

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2 a S

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2 a \frac{R^{2}}{S}

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Solution

The correct option is B $2 a S {(1 + \frac{S^{2}}{R^{2}})}^{\frac{1}{2}}$

$K = a S^{2}; K = \frac{1}{2} m v^{2} F_{c} = \frac{m v^{2}}{R} = \frac{2 K}{R} = \frac{2 a S^{2}}{R} F_{t} . d S = d K \Rightarrow F_{t} = \frac{d K}{d s} = 2 a S F = \sqrt{F_{c}^{2} + F_{t}^{2}} = \sqrt{{(\frac{2 a S^{2}}{R})}^{2} + (2 a S)^{2}} = 2 a S \sqrt{1 + {(\frac{S}{R})}^{2}}$

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