Question

A wire of conical shape of length $L$ and end radii $r_1$, and $r_2$ is stretched by two forces F applied parallel to the length in opposite directions normal to the end faces as shown. If $Y$ denotes the Young's modulus elasticity then extension produced is

• A. $\frac{FL}{\pi r_1{r}_{2}Y}$
• B. $\frac{FL}{2\pi r_1{r}_{2}Y}$
• C. $\frac{FL}{\pi r_1^{2}Y}$
• D. $\frac{2FL}{\pi r_1{r}_{2}Y}$

Answer 1

The correct option is A $\frac{FL}{\pi r_1{r}_{2}Y}$

$\text{Let}{r}_1,r_2\text{be top and bottom radii}$

$\begin{array}{c}\text{L be length}\\ \text{F be Force}\end{array}$

$\begin{array}{c}\text{Consider an clement of radius r}\text{and thickness}dx\\ \text{at distance}x\text{from left.}\\ \text{elongation of element}=\delta l=\frac{Fdx}{\pi r^{2}Y}\end{array}$

$\begin{array}{c}r\text{varies as}r=r_1+\left(\frac{r_2-r_1}{2}\right)x\text{w.r.t}x.\\ \int \delta L=\frac{F}{\pi y}{\int }_0^L\frac{dx}{{(r_1+\left(\frac{r_2-r_1}{L}\right)x)}^2}\\ =\frac{F}{\pi y}{\frac{-L}{(r_1+\frac{(r_2-r_1)x}{L}x)(r_2-r_1)\mid }}_0^L\end{array}$

$\begin{array}{c}=\frac{F}{\pi Y}\times [\frac{-L}{(r_1+\frac{r_2-x_1}{L}{L}^{})(r_2-r_1)}+\frac{L}{r_1(r_2-r_1)}]\\ =\frac{FL}{\pi Y}\times \frac{1}{(r_2-r_1)}(\frac{1}{r_1}-\frac{1}{r_2})\\ =\frac{FL}{\pi Y}\times \frac{1}{r_2-r_1}\times \frac{r_2-r_1}{r_1{r}_2}\end{array}$

$\text{elongation}=\frac{FL}{\pi r_1{r}_{2}Y}$