K.E. of a particle moving along a circle of radius ‘R’ depends on the distance as K.E. = $C S^{2}$ , then the force acting on the particle can be expressed as

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$2 C S {(1 + \frac{S^{2}}{R^{2}})}^{1 / 2}$

2 CS

$\frac{2 C S}{R}$

$\frac{2 S}{R}$

answer is C.

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Detailed Solution

$g i v e n K E = C S^{2} \Rightarrow \frac{1}{2} m V^{2} = C S^{2} m V^{2} = 2 C S^{2}$ ______(1)

$c e n t r i p e t a l f o r c e i s F_{c} = \frac{m V^{2}}{R} ⇒ F_{c} = \frac{2 C S^{2}}{R} ---(2)$

$F r o m e q n (1) \to V = \sqrt{\frac{2 C}{m}} \cdot S - - - (3) d i f f e r e n t i a t e w i t h r e s p e c t t o t i m e t, t o f i n d a c c e l e r a t i o n$
$∴ a = \frac{d v}{d t} a = \sqrt{\frac{2 C}{m}} \cdot \frac{d s}{d t} a = \sqrt{\frac{2 C}{m}} \cdot V s u b s t i t u t e e q n (3) a = \sqrt{\frac{2 C}{m}} \sqrt{\frac{2 C}{m}} . S a = \frac{2 C S}{m} ---(4)$
$∴ \tan g e n t i a l f o r c e i s F_{T} = m a s u b s t i t u t e e q n (4) F_{T} = 2 C S$

$∴ the Net force is given by F = \sqrt{{F_{C}}^{2} + {F_{T}}^{2}} s u b s t i t u t e t h e o b t a i n e d v a l u e s i n a b o v e e q u a t i o n$

$F = \sqrt{{(2 C S)}^{2} + {(\frac{2 C S^{2}}{R})}^{2}} = 2 C S {(1 + \frac{S^{2}}{R^{2}})}^{1 / 2}$