Question

Breaking stress for steel is $F/A$, where $A$ is the area of cross-section of steel wire. A body of mass $M$ is tied at the end of the steel wire of length $L$ and whirled in a horizontal circle. The maximum number of revolution it can make per second is:

• A. $\sqrt{\frac{FA}{{4\pi }^{2}ML}}$
• B. $\sqrt{\frac{F}{{4\pi }^{2}MLA}}$
• C. $\sqrt{\frac{F}{{4\pi }^{2}ML}}$
• D. $\sqrt{\frac{{4\pi }^2}{MLA}}$

Answer 1

The correct option is C $\sqrt{\frac{F}{{4\pi }^{2}ML}}$

Breaking stress for steel is $\frac{F}{A}$. On rotating the mass $M$, $F=m{\omega }^{2}L\Rightarrow w=\sqrt{\frac{F}{ML}}$ $\therefore$ Revolutions per second $=f=\frac{w}{2\pi }=\sqrt{\frac{F}{4{\pi }^{2}ML}}$