Question

A body of mass 3.14 kg is suspended from one end of a wire of length 10 m. The radius of cross-section of the wire is changing uniformly from $5\times {10}^{-4}$ m at the top (i.e. point of suspension) to $9.8\times {10}^{-4}$ m at the bottom. Young's modulus of elasticity is $2\times {10}^{11}N/m^2$. The change in length of the wire is

• A. $4\times {10}^{-3}$ m
• B. $3\times {10}^{-3}$ m
• C. ${10}^{-3}$ m
• D. $2\times {10}^{-3}$ m

Answer 1

The correct option is C ${10}^{-3}$ m

$\begin{array}{c}\text{Let us consider a element at distanice}x\text{from top. of}\\ textthicknessdxandradiusr\text{.}\\ \text{det}{r}_1,r_{2}b{e}^{}\text{radii of top and bottom respectively.}\end{array}$

$\begin{array}{c}\text{Given length}=10m\\ \text{Young's modulus =}y=2\times {10}^{11}{Nm}^{-2}\\ r_1=5\times {10}^{-6}\text{and}{r}_2=9.8\times {10}^{-4}\end{array}$

$\begin{array}{c}\text{Let}\delta l\text{be change in lengthof element}\\ \delta l=\frac{\left(3.14\right)gdx}{\pi r^2\times y}(\text{from formula}\Delta l=\frac{FL}{Ay})\end{array}$

$\text{radius varies as}r=r_1+\frac{(r_2-r_1)}{L}\times x\text{with distance}$

$\begin{array}{c}\delta l=\frac{3.149}{\pi y}\frac{dx}{{(r_1+\frac{r_2-r_1}{L}\times x)}^2}\\ \text{Integrating bothsides}\\ \int dL=\frac{3.149}{\pi y}{\int }_0^L\frac{dx}{{(r_1+\left(\frac{r_2-r_1}{L}\right)x)}^2}\end{array}$

$=\frac{3.149}{\pi y}{[\frac{-1}{(r_1+\frac{r_2-r_1}{L}x)}\times \frac{L}{r_2-r_1}]}_0^2$$={10}^{-3}m$