Q. A mass of 2 kg is whirle‘d in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute. Keeping the radius constant, the tension in the string is double. The new speed is nearly
A) 2.25 rpm
B) 7 rpm
C) 10 rpm
D) 14 rpm
Explanation
(b) $F = mr\omega^{2} = mr(2\pi v)^{2}\ ie,F \propto v^{2}$
$\frac{2F}{F} = \left( \frac{v^{'}}{v} \right)^{2}$ or $v^{'} = v\sqrt{2} = 5\sqrt{2} = 7$rpm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.