Question

A slightly conical were of length l and end radii $r_1$ and $r_2$ is stretched by two forces $F$ and $F$ applied parallel to length in opposite direction and normal to end faces, If Y denotes the Young's modulus. Then extension produced is-

• A. $\frac{FL}{\pi {r_1}^{2}Y}$
• B. $\frac{FL}{\pi r_{1}Y}$
• C. $\frac{FL}{\pi r_1{r}_{2}Y}$
• D. $\frac{FL}{\pi r_1{r}_2}$

Answer 1

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

$\begin{array}{c}\text{Let}{r}_1,r_2\text{be top and bottom radii}\\ \begin{array}{cc}& L\text{be length}\\ & F\text{be Force}\end{array}\end{array}$

$\begin{array}{c}\text{consider and element of radius}r\text{and thickness}dx\\ \text{at distance}x\text{from top.}\\ \text{elongaten of element}\end{array}$

$=\delta L=\frac{Fdx}{\pi r^{2}Y}$

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

${=\frac{F}{\pi y}\frac{-L}{(r_1+\frac{(r_2-r_1)x}{L})(r_2-r_1)}]}_0^L$

$\begin{array}{c}=\frac{FL}{\pi y}\times \frac{1}{(r_2-r_1)}(\frac{1}{r_1}-\frac{1}{r_2})\\ =\frac{FL}{\pi y}\times \frac{1}{r_2-r_1}\times \frac{r_2-r_1}{r_1{r}_2}\\ \text{elongation}=\frac{FL}{\pi r_1{r}_{2}y}\end{array}$