A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is K/π rev./min. The value of K is :
(Assume the string is massless and unstretchable)

Solution

Physics sir · 4 steps

The string's tension alone supplies the centripetal force, so set mω² r equal to 80 N.

1. Tension is centripetal. The stone moves in a horizontal circle of radius r = 2 m. Treating the string as horizontal and ignoring gravity, the tension is the only force toward the centre. The maximum speed is reached when the tension hits its limit of 80 N.

   Tₘₐₓ = m ωₘₐₓ² r

2. Find omega. With m = 0.1 kg: 80 = 0.1 × ω² × 2, so ω² = 400 and ω = 20 rad s⁻¹.
3. Convert to rev/min. One revolution is 2π radians, so 20/2π = 10/π rev per second. Multiply by 60 to get 600/π rev per minute.
4. So. Matching 600/π with K/π gives K = 600, option (c).

The idea to keep. Maximum tension gives maximum ω through T = mω² r. Always convert grams to kg first, then turn rad/s into rev/min by dividing by 2π and multiplying by 60.

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