A ball of mass 0.5 kg is attached to a string of length 50 cm. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400 N. The maximum possible value of angular velocity of the ball in rad / s is, :

Solution

Physics sir · 3 steps

The tension supplies the centripetal force, so set mω² L equal to the 400 N limit.

1. What pulls inward. The string tension is the only horizontal force on the ball, so it provides the centripetal force. If the string tilts at angle θ from the vertical, the radius is L sin θ and the inward pull is T sin θ.
2. Angle cancels. T sin θ = mω² (L sin θ) gives T = mω² L at any angle. A faster spin means more tension, so ω is largest when T reaches 400 N.

   ωₘₐₓ = √(Tₘₐₓ/mL) = √(400/(0.5 × 0.5)) = √1600

3. So. With L = 50 cm = 0.5 m, ωₘₐₓ = 40 rad s⁻¹, which is option (c).

The idea to keep. For a ball whirled on a string, tension equals mω² L, with L the string length. The breaking tension caps ω. Convert cm to m and take the square root at the end.

Answer from the NTA final answer key. Solution by PrepWiser.