A uniformly tapering conical wire is made fro | vedclass

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(C) Let the upper end be at $x=0$ and the lower end at $x=L$. The radius $r(x)$ at a distance $x$ from the top is given by $r(x) = R + \frac{3R-R}{L}x

Vedclass · Accepted Answer

$L\left( {1 + \frac{1}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$

Vedclass · Answer

(C) Let the upper end be at $x=0$ and the lower end at $x=L$. The radius $r(x)$ at a distance $x$ from the top is given by $r(x) = R + \frac{3R-R}{L}x = R(1 + \frac{2x}{L})$.The extension $d\Delta L$ of a small element of length $dx$ is given by $d\Delta L = \frac{Mg dx}{Y A(x)}$,where $A(x) = \pi r(x)^2 = \pi R^2 (1 + \frac{2x}{L})^2$.Integrating from $x=0$ to $x=L$:$\Delta L = \int_0^L \frac{Mg dx}{\pi Y R^2 (1 + \frac{2x}{L})^2} = \frac{Mg}{\pi Y R^2} \int_0^L (1 + \frac{2x}{L})^{-2} dx$.Let $u = 1 + \frac{2x}{L}$,then $du = \frac{2}{L} dx$,or $dx = \frac{L}{2} du$.When $x=0, u=1$; when $x=L, u=3$.$\Delta L = \frac{Mg}{\pi Y R^2} \cdot \frac{L}{2} \int_1^3 u^{-2} du = \frac{MgL}{2 \pi Y R^2} \left[ -\frac{1}{u} \right]_1^3 = \frac{MgL}{2 \pi Y R^2} \left( 1 - \frac{1}{3} \right) = \frac{MgL}{2 \pi Y R^2} \cdot \frac{2}{3} = \frac{MgL}{3 \pi Y R^2}$.The total extended length is $L + \Delta L = L + \frac{MgL}{3 \pi Y R^2} = L \left( 1 + \frac{1}{3} \frac{Mg}{\pi Y R^2} \right)$.

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