91Ó°ÊÓ

$L=15\text{ }m$

$l=2.8cm\left(\frac{1\text{ }m}{100\text{ }cm}\right)=0.028\text{ }m$

Diameter of the rope$d=9.6\text{ }mm\left(\frac{1\text{ }m}{100\text{ }cm}\right)=9.6×{10}^{−3}\text{ }m$

Acceleration due to gravity$\text{g}=9.8\text{ }m/s^2$

$\text{σ=}\frac{\text{F}}{\text{A}}$We have been given all the required values to calculate the stress, strain, and Young’s Modulus. We convert them into equivalent units. By plugging these values into the formula:

Strain localid="1661252112148" $\mathit{Ï\mu }\mathbf{\text{=}}\frac{\mathbf{l}}{\mathbf{L}}$

Stress

Young’s Modulus $E\text{=}\frac{\text{σ}}{\mathrm{Ï\mu }}$.

By usingthe formula for strain,

$\begin{array}{c}\mathrm{Ï\mu }=\frac{l}{L}\\ =\frac{0.028\text{ }m}{15\text{ }m}\\ =1.9×{10}^{−3}\end{array}$

Strain,$\mathrm{Ï\mu }=1.9×{10}^{−3}$

We have the formula for stress,

$\mathrm{σ}=\frac{\text{F}}{\text{A}}$

We calculate the area of rope,

$\begin{array}{c}A=\frac{\mathrm{π}}{4}×{\text{d}}^2\\ =\frac{\mathrm{π}}{4}×{(9.6×{10}^{−3})}^2\\ =7.2×{10}^{−5}{m}^2\end{array}$

The force will be the weight of the rock climber,

$\begin{array}{c}\text{F}=\text{mg}\\ =95\text{ }kg×9.8\text{ }m/s^2\\ =931\text{ }N\end{array}$

Using the above-mentioned stress formula, we get

$\begin{array}{c}\mathrm{σ}=\frac{931\text{ }N}{7.2×{10}^{−5}\text{ }{m}^2}\\ =1.29×{10}^7\text{ }N/m^2\end{array}$

Stress $\mathrm{σ}=1.29×{10}^7\text{ }N/m^2$,

We have the formula for E,

$E=\frac{\mathrm{σ}}{\mathrm{Ï\mu }}$

$\begin{array}{c}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}=\frac{1.3×{10}^7\text{ }N/m^2}{1.9×{10}^{−3}}\\ =6.84×{10}^9\text{ }N/m^2\end{array}$

Young’s modulus, E =$6.84×{10}^9\text{ }N/m^2$