a 0.3 kg ball is attached to a string that is 50 cm in length, and completes 2 circles every second.Find the tension required to whirl the ball with the given constriants and also find the angle theta?

Sounds like you're missing a diagram. Can you describe what you mean by the angle theta?

Consider the sum of forces acting in the vertical direction:


T \cos \theta - mg = 0
.

where \theta = angle of the rope to the vertical.

For the horizontal direction the horizontal component of the rope's tension must equal the ball's mass times its centripedal acceleration:


T \sin \theta = m\omega^2 R


R is the radius of the circle traced by the ball: R= L \sin \theta. You can find \omega (rotational velocity) from the period of rotation:


\omega = \frac 1 {2 \pi (0.5s)} = \frac 1 {\pi} s^{-1}


So now you have two equation in two unknowns: \theta and T. Let us know what you get for an answer and we'll check it for you.