Question

A bob of mass 10 kg is attached to wire 0.3 m long. Its breaking stress is $4.8\times {10}^{7}N/m2$. 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

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