Question

A body of mass 10 kg is attached to a wire 0.3m long. Its breaking stress is \[4.8 \times {10^7}\,N{m^2}\]. The area of cross-section of the wire is \[{10^{ - 6}}{m^2}\]. Find the maximum angular velocity with which it can be rotated in a horizontal circle without breaking.
A. \[2\,rad{s^{ - 1}}\]
B. \[4\,rad{s^{ - 1}}\]
C. \[6\,rad{s^{ - 1}}\]
D. \[8\,rad{s^{ - 1}}\]

Answer 1

Hint:Before we start addressing the problem, we need to know about the stress. The stress of an object is defined as the force applied per unit area. Here, we find the force using the equation of stress and thereby calculate the maximum angular velocity.

Formula Used:
To find the stress the formula is,
\[Stress = \dfrac{F}{A}\]
Where, F is force and A is cross-sectional area.

Complete step by step solution:
Consider a body of mass 10 kg that is attached to a wire of 0.3m long. Its breaking stress is \[4.8 \times {10^7}N{m^2}\] and the area of cross-section of the wire is \[{10^{ - 6}}{m^2}\]. We need to find the maximum angular velocity with which it can be rotated in a horizontal circle without breaking.To find the stress,
\[\text{Stress} = \dfrac{\text{Force}}{\text{Area}}\]
\[\text{Force} = \text{Stress} \times \text{Area}\]
Substitute the value of stress and area, we get,
\[F = 4.8 \times {10^7} \times {10^{ - 6}}\]
\[\Rightarrow F = 48\,N\]

Now, Force=Tension in the wire. That is,
\[F = m{\omega ^2}r\]
Rearrange the above equation for \[\omega \], we get,
\[{\omega ^2} = \dfrac{F}{{mr}}\]
Substitute the value of \[m = 10kg\] , \[r = 0.3m\] and \[F = 48N\] in above equation then, we get,
\[{\omega ^2} = \dfrac{{48}}{{10 \times 0.3}}\]
\[\Rightarrow {\omega ^2} = 16\]
\[\therefore \omega = 4rad{s^{ - 1}}\]
Therefore, the maximum angular velocity with which it can be rotated in a horizontal circle without breaking is \[4\,rad{s^{ - 1}}\].

Hence, option B is the correct answer.

Note: In this problem it is important to remember the equation for stress and the relation between the force and the tension in the wire then it will be easy to solve this problem.