Q. A body of mass $m = 10$ kg is attached to a wire of length 0.3 m. The maximum angular velocity with which it can be rotated in a horizontal circle is (Breaking stress of wire =$4.8 \times 10^{7}Nm^{- 2}\ $and area of cross-section of a wire = $10^{- 2}m^{2})$
A) $4\ rads^{- 1}$
B) $8\ rads^{- 1}$
C) $1\ rads^{- 1}$
D) $2\ rads^{- 1}$
Explanation
(a) Breaking strength = tension in the wire = $mr\omega^{2}$
4.8 $\times 10^{7} \times 10^{- 6} = 10 \times 0.3\ \times \omega^{2}$
$\omega^{2} = \frac{48}{0.3\ \times 10} = 16$
$\omega = 4\ rads^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.