A force $\dfrac{{m{v^2}}}{r}$ is acting on a body of mass m moving with a speed v in a circle of radius r. What is the work done by the force in moving the body over half the circumference of the circle?
${\text{A}}{\text{. }}\dfrac{{m{v^2}}}{r} \times \pi r$
${\text{B}}{\text{.}}$ Zero
${\text{C}}{\text{. }}\dfrac{{m{v^2}}}{{{r^2}}}$
\[{\text{D}}{\text{. }}\dfrac{{\pi {r^2}}}{{m{v^2}}}\]

Hint – As the body is in circular motion, so the centripetal force, i.e., $\dfrac{{m{v^2}}}{r}$ is directed towards the centre and at every point it is perpendicular to the displacement. Keep this point in mind to solve the question.
Formula used - $W = FS\cos \theta $

Complete step-by-step answer:
We have been given that a body of mass m is moving with a speed v in a circle of radius r.
As, we know when a body moves in a circular path then centripetal force acts on it which is given by $\dfrac{{m{v^2}}}{r}$
This centripetal force is perpendicular to the displacement of the body at every point.
So, as we know the work done is given by $W = FS\cos \theta $ , where F is the force acting on the body and S is the displacement and theta is the angle between them and W is the work done.
Now, here as mentioned above the angle between the centripetal force and the displacement is 90 degrees.
Keeping the value of theta, we get-
$
  W = FS\cos {90^ \circ } \\
   \Rightarrow W = 0\{ \because \cos {90^ \circ } = 0\} \\
$
Therefore, the work done is Zero.
Hence, the correct option is B.

Note – Whenever such types of questions appear, then first write all the things given in the question and then by using the formula, $W = FS\cos \theta $ and also by knowing the angle between the force and displacement find the work done. In this case the work done is zero, as the force is perpendicular to displacement.