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(B) The force acting on the body is the centripetal force,which is always directed towards the center of the circular path.At any point on the circula

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Zero

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(B) The force acting on the body is the centripetal force,which is always directed towards the center of the circular path.At any point on the circular path,the direction of the velocity (and thus the instantaneous displacement) is along the tangent to the circle.The radius vector (direction of force) and the tangent (direction of displacement) are always perpendicular to each other,meaning the angle $	heta$ between the force $F$ and the displacement $s$ is $90^o$.The work done is given by the formula $W = F \cdot s \cdot \cos 	heta$.Since $\cos 90^o = 0$,the work done $W = F \cdot s \cdot 0 = 0$.Therefore,the work done by the centripetal force in moving the body over any part of the circular path is zero.

$A$ body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{mv^2}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle?

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