Question

A stone of mass $m$ tied to one end of an elastic 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 horizontal. If Young's modulus is $Y$ then increase in length of thread is

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

Answer 1

Answer: (C)

$\frac{4mgl}{\pi d^{2}Y\sin \theta }$

Horizontal Force ef mass $m$.

$T\cos \theta =m\cdot {\omega }^2\cdot l\cos \theta$

There is vertical equilibrium $T\sin \theta =mg$

$\begin{array}{cc}\Delta l=& \frac{Fl}{AY}=\frac{Fl}{AY}\\ & =\frac{mgl}{\sin \theta \times \frac{\pi d^{2}Y}{4}}\\ =& \frac{4mgl}{Y\pi d^2\sin \theta }.\end{array}$