Question

A bob of mass 10 kg is attached to wire 0.3 m long. Its breaking stress is 4.8 $\times$ ${10}^7$ $\frac{N}{m^2}$. The area of cross section of the wire is ${10}^{-6}$$m^2$. The maximum angular velocity with which it can be rotated in a horizontal circle

• A. 8 rad/sec
• B. 4 rad/sec
• C. 2 rad/sec
• D. 1 rad/sec

Answer 1

The correct option is B 4 rad/sec

Centripental force = breaking force
$\Rightarrow$ m ${\omega }^2$ r = Breaking stress $\times$ cross sectional area
$\Rightarrow$ m ${\omega }^2$ r = p $\times$ A $\Rightarrow$ $\omega$ = $\sqrt{\frac{p\times A}{mr}}$ = $\sqrt{\frac{4.8\times {10}^7\times {10}^{-6}}{10\times 0.3}}$
$\therefore$ $\omega$ = 4 rad/sec