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

• 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

Answer: (C)

$\frac{4mgL}{\pi D^{2}Y\cos\theta}$

Answer 2

: The situation is as shown in the figure.

Refer figure

For vertical equilibrium of stone

$T\cos\theta = mgorT = \frac{mg}{\cos\theta}$ ….(i)

As $Y = \frac{T}{A}\frac{L}{\Delta L}$

$\therefore\Delta L = \frac{TL}{AY}$ (Using (i))

$= \frac{mgl}{\cos\theta(\pi D^{2}/4)Y} = \frac{4mgL}{\pi D^{2}YCos\theta}$