A slightly conical wire of length L and end radii r₁ and r₂ is stretched by two forces F, F applied parallel to length in opposite directions and normal to end faces. If Y denotes the Young’s modulus, then extension produced is

The correct answer is FL/πr₁r₂Y.

To find the extension produced in a slightly conical wire when subjected to a stretching force, we can analyze the extension of an infinitesimal element of the wire and integrate it over the entire length.

Let the wire have a length $L$, with the radius at one end $(x=0)$ being $r_1$ and the radius at the other end $(x=L)$ being $r_2$.

Assuming a linear variation of the radius along the length of the wire, the radius $r$ at a distance $x$ from the end of radius $r_1$ is given by:
$r=r_1+\left(\frac{r_2-r_1}{L}\right)x$

The cross-sectional area $A$ of the wire at this distance $x$ is:
$A=\pi r^2=\pi {[r_1+\left(\frac{r_2-r_1}{L}\right)x]}^2$

Let us consider a small element of length $dx$ at a distance $x$. The extension $d\delta$ of this small element under the tension force $F$ is given by Hooke's Law:
$d\delta =\frac{Fdx}{AY}$
where $Y$ is the Young's modulus of the wire.

Substituting the expression for the cross-sectional area $A$ into the equation:
$d\delta =\frac{Fdx}{\pi {[r_1+\left(\frac{r_2-r_1}{L}\right)x]}^{2}Y}$

To find the total extension $\delta$ of the wire, we integrate this expression from $x=0$ to $x=L$:
$\delta ={\int }_0^L\frac{Fdx}{\pi Y{[r_1+\left(\frac{r_2-r_1}{L}\right)x]}^2}$

Let $u=r_1+\left(\frac{r_2-r_1}{L}\right)x$. Then, the differential is:
$du=\left(\frac{r_2-r_1}{L}\right)dx\Rightarrow dx=\frac{L}{r_2-r_1}du$
The limits of integration change from $x=0$ to $u=r_1$, and from $x=L$ to $u=r_2$.

Substituting these variables into the integral:
$\delta =\frac{F}{\pi Y}\left(\frac{L}{r_2-r_1}\right){\int }_{r_1}^{r_2}\frac{du}{u^2}$

Integrating $u^{-2}$:
$\delta =\frac{FL}{\pi Y(r_2-r_1)}{[-\frac{1}{u}]}_{r_1}^{r_2}$

$\delta =\frac{FL}{\pi Y(r_2-r_1)}[\frac{1}{r_1}-\frac{1}{r_2}]$

$\delta =\frac{FL}{\pi Y(r_2-r_1)}\left[\frac{r_2-r_1}{r_1{r}_2}\right]$

Cancelling out the term $(r_2-r_1)$ from both the numerator and the denominator, we get:
$\delta =\frac{FL}{\pi r_1{r}_{2}Y}$

Thus, the extension produced in the wire is indeed $\frac{FL}{\pi r_1{r}_{2}Y}$.