Question

A stone of mass m tied to one end of a thread of length l. The diameter of the thread is d and it is suspended vertically. The stone is now rotated in a horizontal plane and makes an angle $\theta$ with the vertical. Find the increase in length of the wire. Young's modulus of the wire is Y.

• A. $\frac{4mgl}{\pi d^{2}Y\cos \theta }$
• B. $\frac{4mgl}{\pi d^{2}Y\sin \theta }$
• C. $\frac{4mgl}{\pi d^{2}Y}$
• D. $\frac{4mgl}{\pi d^{2}Y\sec \theta }$

Answer 1

The correct option is A $\frac{4mgl}{\pi d^{2}Y\cos \theta }$

The Tension generated in string can be calculated as $Tcos\theta =mg$
$ThusT=\frac{mg}{cos\theta }$
Now, $\Delta l=\frac{Tl}{AY}=\frac{mg\times l}{cos\theta \times \frac{\pi d^2}{4}\times Y}=\frac{4mgl}{\pi d^{2}Ycos\theta }$