Explanation:
Given that,
Mass of the rock climber, m = 90 kg
Original length of the rock, L = 16 m
Diameter of the rope, d = 7.8 mm
Stretched length of the rope, [tex]\Delta L=3.1\ cm[/tex]
(a) The change in length per unit original length is called strain. So,
[tex]\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193[/tex]
(b) The force acting per unit area is called stress.
[tex]\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa[/tex]
(c) The ratio of stress to the strain is called Young's modulus. So,
[tex]Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2[/tex]
Hence, this is the required solution.
