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A uniformly tapering conical wire is made from a material of Young's modulus $Y$ and has a normal, unextended length $L.$ The radii, at the upper and lower ends of this conical wire, have values $R$ and $3R,$ respectively. The upper end of the wire is fixed to a rigid support and a mass $M$ is suspended from its lower end. The equilibrium extended length, of this wire, would equal
$L\left( {1 + \frac{2}{9}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$
$L\left( {1 + \frac{1}{9}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$
$L\left( {1 + \frac{1}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$
$L\left( {1 + \frac{2}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$
Solution
Consider a small element $dx$ of radius $r$,
$r = \frac{{2R}}{L}x + R$
At equilibrium change in length of the wire
$\int\limits_0^l {dL = \int {\frac{{Mgdx}}{{\pi {{\left[ {\frac{{2R}}{L}x + R} \right]}^2}y}}} } $
Taking limit from $0$ to $L$
$\Delta L = \frac{{Mg}}{{\pi y}} – \frac{1}{{\left[ {\frac{{2Rx}}{L} + R} \right]_0^L}} \times \frac{L}{{2R}} = \frac{{MgL}}{{3\pi {R^2}y}}$
The equilibrium extended length of wire
$ = L + \Delta L$
$ = L + \frac{{MgL}}{{3\pi {R^2}Y}} = L\left( {1 + \frac{1}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$
