Question
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 :
A.
$$L\left( {1 + \frac{2}{9}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$$
B.
$$L\left( {1 + \frac{1}{9}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$$
C.
$$L\left( {1 + \frac{1}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$$
D.
$$L\left( {1 + \frac{2}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$$
Answer :
$$L\left( {1 + \frac{1}{3}\frac{{Mg}}{{\pi Y{R^2}}}} \right)$$
Solution :
Consider a small element $$dx$$ of radius $$r,$$
$$r = \frac{{2R}}{L}x + R$$
At equilibrium change in length of the wire
$$\int\limits_0^1 {dL} = \int {\frac{{Mg\,dx}}{{\pi {{\left[ {\frac{{2R}}{L}x + R} \right]}^2}y}}} $$
Taking limit from $$0$$ to $$L$$
$$\Delta L = \frac{{Mg}}{{\pi y}}\left[ { - \frac{1}{{\left[ {\frac{{2Rx}}{L} + R} \right]_0^L}} \times \frac{L}{{2R}}} \right[ = \frac{{MgL}}{{3\pi {R^2}y}}$$
The equilibrium extended length of wire $$ = L + \Delta L$$
$$\eqalign{ & = L + \frac{{MgL}}{{3\pi {R^2}Y}} \cr & = L\left( {1 + \frac{1}{3}\frac{{Mg}}{{3\pi Y{R^2}}}} \right) \cr} $$
Consider a small element $$dx$$ of radius $$r,$$
$$r = \frac{{2R}}{L}x + R$$
At equilibrium change in length of the wire
$$\int\limits_0^1 {dL} = \int {\frac{{Mg\,dx}}{{\pi {{\left[ {\frac{{2R}}{L}x + R} \right]}^2}y}}} $$
Taking limit from $$0$$ to $$L$$
$$\Delta L = \frac{{Mg}}{{\pi y}}\left[ { - \frac{1}{{\left[ {\frac{{2Rx}}{L} + R} \right]_0^L}} \times \frac{L}{{2R}}} \right[ = \frac{{MgL}}{{3\pi {R^2}y}}$$
The equilibrium extended length of wire $$ = L + \Delta L$$
$$\eqalign{ & = L + \frac{{MgL}}{{3\pi {R^2}Y}} \cr & = L\left( {1 + \frac{1}{3}\frac{{Mg}}{{3\pi Y{R^2}}}} \right) \cr} $$
