The bob of a simple pendulum at rest is given minimum velocity in horizontal direction, so that it describes a vertical circle of radius equal to the length of the simple pendulum. If velocity of the bob at the highest point is V, then velocity is at an angular displacement $60^o$ from lowest point:
(A) $\sqrt 2 $ V
(B) $\sqrt 3 $ V
(C) 2 V
(D) 3 V

Hint: Calculate the work done to use it in the work energy theorem and the velocity from the kinetic energy at the required state is the velocity at the given angular displacement.

Complete step – by - step solution:
First we need to calculate work done; in this case we know that,
$W = FS(1 - \cos \theta )$
F is the force, F=mg N
S is the displacement, S = 2R (twice radius at the top)
 θ = 60 is angle subtended.
$
  \therefore W = mg \times 2R(1 - \cos 60) \\
  W = \dfrac{3}{2}mgR \\
  $
 Now using work energy theorem as work equal to difference in kinetic energy,
$
  W = \Delta K \\
  \dfrac{3}{2}mgR = \dfrac{{mV_0^2}}{2} - \dfrac{{m{{\left( {\sqrt {gR} } \right)}^2}}}{2} \\
  {V_0} = 2\sqrt {gR} \\
  {V_0} = 2V \\
  $
Since $V = \sqrt {gR} $ and $V_0$ =2 V is the velocity at the angular displacement 60 from lowest point.

The correct option is C.

Note: The radius is taken as 2R because the radius at the top of the circle is twice the length of the pendulum so read as the diameter.