Question

A slightly tapering wire of length / and radii $r_1$ and $r_2(r_2>r_1)$ subjected to twoequal and opposite stretching force $F$ . If young's modulus of materials of wire is$Y,$ then the elongation in the wire is

• A. $\frac{Fl}{\pi Y{r}_1/2}$
• B. $\frac{2Fl}{\pi Y{r}_1/2}$
• C. $\frac{3Fl}{\pi Y{r}_1/2}$
• D. $\frac{2FI}{Y{r}_1{r}_2}$

Answer 1

The correct option is D $\frac{2FI}{Y{r}_1{r}_2}$

$\begin{array}{c}\text{By Hookes law}\\ \text{We know}\\ \Delta l=\frac{Fl}{AY}\end{array}$

$\text{We have taken a small.}$

$\text{length}l=dr\text{at which (area)}A=\pi y^2$

$\text{Elongation}={\int }_0^l\frac{Fdr}{\pi y^{2}Y}$

$\begin{array}{c}\text{putting}y\text{in terme of}{r}_1,r_2\text{and}l\text{, and}\\ \text{after integrating,}\end{array}$

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

$\begin{array}{c}\text{For such type of question we can take effective area}\\ \text{Effective area =}\pi r_1{r}_2\text{in place of}\pi rr\text{as}\\ \text{radius is varying.}\\ \text{Elongation}=\frac{Fl}{A_{\text{effective}}Y}=\frac{Fl}{\pi r_1{r}_{2}Y}\end{array}$