Question

A body of mass 10 kg is attached to one end of a wire of length 0.3 m and area of cross-section \(10^{-6} \, \text{m}^2\). If the maximum stress the wire can withstand is \(2.7 \times 10^7 \, \text{N/m}^2\), then the maximum angular velocity with which the wire-body system can be rotated in a horizontal circle about the other end of the wire is

• A. \( 4 \, \text{rad/s} \)
• B. \( 5 \, \text{rad/s} \)
• C. \( 9 \, \text{rad/s} \)
• D. \( 3 \, \text{rad/s} \)

Hint:

Relate maximum stress to tension using \( \sigma = \frac{T}{A} \). Then use \( T = m\omega^2 r \) to find angular velocity.

Answer 1

The Correct Option is D

Step 1: Use centripetal force and maximum tension from stress: \[ T = \sigma \cdot A = 2.7 \times 10^7 \cdot 10^{-6} = 27 \, \text{N} \] Step 2: Centripetal force: \( T = m \omega^2 r \) \[ 27 = 10 \cdot \omega^2 \cdot 0.3 \Rightarrow \omega^2 = \frac{27}{3} = 9 \Rightarrow \omega = 3 \, \text{rad/s} \]