Question

$A$ stone of mass $m$ is 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

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

Answer 1

Answer: (C)

The situation is as shown in the figure. For vertical equilibrium of stone
$T \,cos \,\theta=mg$ $\quad$ or $\quad$ $T=\frac{mg}{cos \,\theta}$ $\quad\ldots\left(i\right)$
As$\quad$ $Y$ $=\frac{T}{A}$ $\frac{L}{\Delta L}$
$\therefore \quad$ $\Delta L$ $=\frac{TL}{AY}$ $\quad\left(Using \left(i\right)\right)$
$=\frac{mgL}{cos \,\theta\left(\pi D^{2} /4\right)Y}$ $=\frac{4mgL}{\pi D^{2}Y cos\, \theta}$