Question

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

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

Answer 1

The correct option is C $\frac{4mgL}{\pi D^{2}Y\cos \theta }$

The situation is as shown in the figure. For vertical equilibrium of stone

$T\cos \theta =mgorT=\frac{mg}{\cos \theta }....\left(i\right)$
As $Y=\frac{T}{A}\frac{L}{\Delta L}$
$\therefore \Delta L=\frac{TL}{AY}$ [Using (i)]
$=\frac{mgL}{\cos \theta \left(\pi D^2\right)Y}=\frac{4mgL}{\pi D^{2}Y\cos \theta }$