A body of mass 10 kg is attached to a wire 0.3 m long. Its breaking stress is 4.8 x 10⁷ N /m² . The area of cross-section of the wire is 10⁻⁶ m² . What is the maximum angular velocity with which it can be rotated in the horizontal circle

The correct option is 4 rad/sec.

Step-by-step Explanation:

We are given the following parameters for the system:
Mass of the body, m = 10 kg
Length of the wire (which acts as the radius of the circular path), r = 0.3 m
Breaking stress of the wire, $\sigma =4.8\times {10}^7{\text{N/m}}^2$
Area of cross-section of the wire, $A={10}^{-6}{\text{m}}^2$

Step 1: Calculate the maximum tension ($T_{\text{max}}$) the wire can withstand
The maximum tension is related to the breaking stress and the cross-sectional area of the wire by the formula:
$T_{\text{max}}=\text{Breaking Stress}\times \text{Area of Cross-section}$
$T_{\text{max}}=\sigma \times A$
Substituting the given values:
$T_{\text{max}}=(4.8\times {10}^7{\text{N/m}}^2)\times {10}^{-6}{\text{m}}^2$
$T_{\text{max}}=48\text{N}$

Step 2: Relate tension to the centripetal force
When the body is rotated in a horizontal circle, the tension in the wire provides the necessary centripetal force to keep it in circular motion:
$T=m{\omega }^{2}r$
where $\omega$ is the angular velocity.

Step 3: Solve for the maximum angular velocity (${\omega }_{\text{max}}$)
To find the maximum angular velocity, we use the maximum tension that the wire can handle:
$T_{\text{max}}=m{\omega }_{\text{max}}^{2}r$
Substitute the values into the equation:
$48=10\times {\omega }_{\text{max}}^2\times 0.3$
Simplifying the expression:
$48=3{\omega }_{\text{max}}^2$
${\omega }_{\text{max}}^2=\frac{48}{3}$
${\omega }_{\text{max}}^2=16$
Taking the square root on both sides:
${\omega }_{\text{max}}=4\text{rad/sec}$

Therefore, the maximum angular velocity with which the body can be rotated is 4 rad/sec.