Get Answers to all your Questions

header-bg qa

Q11   A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm ^2 . Calculate the elongation of the wire
when the mass is at the lowest point of its path.

Answers (1)


Mass of the body = 14.5 kg

Angular velocity, \omega = 2 rev/s 

\omega =4\pi \ rad/s

The radius of the circle, r = 1.0 m

Tension in the wire when the body is at the lowest point is T

\\T=mg+m\omega ^{2}r\\ T=14.5\times 9.8+14.5\times (4\pi )^{2}\\ T=2431.84N

Cross-Sectional Area of wire, A = 0.065 cm2

Young's Modulus of steel, Y=2\times 10^{11}Nm^{-2}

\\\Delta l=\frac{Fl}{AY}\\ \Delta l=\frac{2431.84\times1}{0.065\times 10^{-4}\times 2\times 10^{11}}\\ \Delta l=1.87 mm

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support