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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 6 with the vertical. If Young's modulus of the wire is Y, then the increase in the length of the wire is

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$\frac{4 mgL}{{πD}^{2} Y}$

$\frac{4 mgL}{{πD}^{2} Ysinθ}$

$\frac{4 mgL}{{πD}^{2} Ycosθ}$

$\frac{4 mgL}{{πD}^{2} Ytan θ}$

answer is C.

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Detailed Solution

The situation is as shown in the figure.

Refer figure
For vertical equilibrium of stone
$T cosθ = mg or T = \frac{mg}{\cos θ}$ --------(i)
As $Y = \frac{T}{A} \frac{L}{∆ L}$ (Using (i))

$∴ ∆ L = \frac{TL}{AY}$

= $\frac{mgL}{\cos θ (\frac{{πD}^{2}}{4}) Y} = \frac{4 mgL}{{πD}^{2} Ycosθ}$

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