91Ó°ÊÓ

The initial length of the rod,$\text{L}=0.8000\text{″¾}$

The initial radius of the rod,$\text{r}=1000.0\text{‰ӾÀ}\left(\frac{{10}^{−6}\text{″¾}}{1\text{‰ӾÀ}}\right)=1000.0×{10}^{−6}\text{m}$

The final radius of the rod, $=999.9\text{‰ӾÀ}\left(\frac{{10}^{−6}\text{″¾}}{1\text{‰ӾÀ}}\right)=999.9×{10}^{−6}\text{″¾}$

You can use the concept of modulus of elasticity, and expression of density and volume of the rod. The formulas used are given below.

$\begin{array}{c}\mathbf{ÒÏ}\mathbf{=}\frac{\mathbf{M}}{\mathbf{V}}\\ \mathbf{V}\mathbf{=}{\mathbf{Ï€°ù}}^{\mathbf{2}}\mathbf{h}\\ \frac{\mathbf{F}}{\mathbf{A}}\mathbf{=}\mathbf{E}\frac{\mathbf{Δ³¢}}{\mathbf{L}}\end{array}$

One end of the rod is clamped and the other end is stretched by the machine in such a way that the rod’s density remains the same after elongation of the rod.

The expression of the density of the rod is

$\begin{array}{l}\mathrm{ÒÏ}=\frac{\mathrm{M}}{\mathrm{V}}\\ \mathrm{V}=\frac{\mathrm{M}}{\mathrm{ÒÏ}}=\mathrm{constant}\\ \mathrm{V}={\mathrm{Ï€°ù}}^2\mathrm{L}=\mathrm{constant}\end{array}$

The expression of the volume of the rod before and after elongation is:

$\begin{array}{c}{\mathrm{Ï€°ù}}^2\mathrm{L}=\mathrm{Ï€}{{\mathrm{r}}^{\text{'}}}^2\mathrm{L}\text{'}\\ {\mathrm{r}}^2\mathrm{L}={{\mathrm{r}}^{\text{'}}}^2\mathrm{L}\text{'}\\ {\mathrm{L}}^{\text{'}}=\frac{{\mathrm{r}}^2\mathrm{L}}{{{\mathrm{r}}^{\text{'}}}^2}\end{array}$

Change in length of the rod due to the applied force by the machine is:

$\begin{array}{c}\mathrm{Δ³¢}={\mathrm{L}}^{\text{'}}−\mathrm{L}\\ =\frac{{\mathrm{r}}^2\mathrm{L}}{{{\mathrm{r}}^{\text{'}}}^2}−\mathrm{L}\\ =\mathrm{L}\left(\frac{{\mathrm{r}}^2}{{{\mathrm{r}}^{\text{'}}}^2}−1\right)\end{array}$

The expression of Young’s modulus of elasticity is:

$\begin{array}{c}\frac{\mathrm{F}}{\mathrm{A}}=\mathrm{E}\frac{\mathrm{Δ³¢}}{\mathrm{L}}\\ \mathrm{F}=\mathrm{AE}\frac{\mathrm{Δ³¢}}{\mathrm{L}}\\ \mathrm{F}={\mathrm{Ï€°ù}}^2\mathrm{E}\frac{\mathrm{L}\left(\frac{{\mathrm{r}}^2}{{\mathrm{r}}^{\text{'}2}}−1\right)}{\mathrm{L}}\\ \mathrm{F}=3.14×{(1000×{10}^{−6}\text{ }\mathrm{m})}^2×70×{10}^9{\text{ N/³¾}}^{\text{2}}×\left[\frac{{(1000×{10}^{−6}\text{ }\mathrm{m})}^2}{{(999.9×{10}^{−6}\mathrm{m})}^2}−1\right]\\ \mathrm{F}=44\text{ N}\end{array}$