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:

We have a conical wire whose upper, thinner end of radius $$R$$ is rigidly fixed and whose lower, thicker end of radius $$3R$$ supports a mass $$M$$. Because the mass hangs at the very end, every cross-section of the wire carries the same tensile force, namely the weight $$Mg$$ of the mass. The weight of the wire itself is not mentioned, so we neglect it.

The extension of a small element will be obtained from the basic definition of Young’s modulus. First we state the formula:

For a small element of length $$dx$$, cross-sectional area $$A$$ and tension $$F$$, the incremental extension is

$$d\ell=\frac{F}{YA}\;dx.$$

To find the total extension we integrate this expression along the full length of the wire, taking into account that the area varies with position.

Choose the coordinate $$x$$ measured from the fixed upper end (so $$x=0$$ at the top and $$x=L$$ at the bottom). Because the radius increases linearly from $$R$$ to $$3R$$, the radius at a distance $$x$$ from the top is

$$r(x)=R+\left(\frac{3R-R}{L}\right)x=R+\frac{2R}{L}x=R\left(1+\frac{2x}{L}\right).$$

Hence the cross-sectional area is

$$A(x)=\pi r(x)^2=\pi R^2\left(1+\frac{2x}{L}\right)^2.$$

The tensile force in every element is simply $$F=Mg$$. Substituting these expressions into the incremental extension formula, we get

$$d\ell=\frac{Mg}{Y\;\pi R^2\left(1+\dfrac{2x}{L}\right)^2}\;dx.$$

Integrating from the top ($$x=0$$) to the bottom ($$x=L$$) gives the total extension $$\Delta L$$:

$$\Delta L=\int_{0}^{L}\frac{Mg}{Y\;\pi R^2\left(1+\dfrac{2x}{L}\right)^2}\;dx.$$

Now we perform the integral. Put

$$u=1+\frac{2x}{L}\quad\Rightarrow\quad du=\frac{2}{L}dx\quad\Rightarrow\quad dx=\frac{L}{2}\,du.$$

When $$x=0$$, $$u=1$$; when $$x=L$$, $$u=3$$. Substituting, we obtain

$$\Delta L=\frac{Mg}{Y\pi R^2}\int_{u=1}^{3}\frac{L}{2}\,\frac{du}{u^{2}}$$ $$=\frac{MgL}{2Y\pi R^2}\int_{1}^{3}u^{-2}\,du.$$

We evaluate the integral:

$$\int u^{-2}\,du=-\frac{1}{u}\quad\text{so}\quad\int_{1}^{3}u^{-2}\,du=-\frac{1}{3}+1=\frac{2}{3}.$$

Substituting this result gives

$$\Delta L=\frac{MgL}{2Y\pi R^2}\cdot\frac{2}{3}=\frac{MgL}{3Y\pi R^2}.$$

The equilibrium (extended) length $$L_{\text{eq}}$$ is therefore

$$L_{\text{eq}}=L+\Delta L=L\left(1+\frac{Mg}{3Y\pi R^{2}}\right).$$

This expression exactly matches Option C.

Hence, the correct answer is Option C.