Answer:
Young's modulus(Y) = 2.4 × 10⁸ (Approx)
Explanation:
Given:
Length of rope (l) = 55 m
Diameter of rope = 1.3 cm
Radius of rope (r) = 1.3 / 2 = 0.65 cm = 0.0065 m
Weight of climber (m) = 83 kg
Rope elongate(Δl) = 1.4 m
Find:
Young's modulus(Y).
Computation:
Force (F) = Weight of climber (m) × g
Force (F) = 83 × 9.8
Force (F) = 813.4 N
Area of rope (A) = πr²
Area of rope (A) = 0.000132665
Young's modulus(Y) =[tex]\frac{(F)(l)}{(\delta l)(A)}[/tex]
Young's modulus(Y) = 44,737 / 0.000185731
Young's modulus(Y) = 2.4 × 10⁸ (Approx)
The Young's modulus of the rope is [tex]1.55 \times 10^{5} \;\rm N/m^{2}[/tex].
Given data:
The length of rope is, L = 55 m.
The diameter of rope is, d = 1.3 cm = 0.013 m.
The mass of climber is, m = 83 kg.
The elongation of rope is, [tex]\Delta L = 1.4 \;\rm m[/tex].
The ratio of stress and strain produced in a material is known as Young's Modulus of material . Then,
[tex]E=\dfrac{\sigma }{\epsilon}\\\\E=\dfrac{W/A }{L/ \Delta L}\\\\\\E = \dfrac{W \times \Delta L}{AL}[/tex]
Here,
W is the weight of climber and its value is, W = mg.
A is the cross-sectional area of rope and its value is, [tex]A =\dfrac{\pi}{4}d^{2}[/tex].
Solving as,
[tex]E = \dfrac{mg \times \Delta L}{\dfrac{\pi}{4} \times d^{2} \times L}\\\\\\E = \dfrac{83 \times 9.8 \times 1.4}{\dfrac{\pi}{4} \times (0.013)^{2} \times 55}\\\\\\E=1138.76/7.30 \times 10^{-3}\\\\E = 1.55 \times 10^{5} \;\rm N/m^{2}[/tex]
Thus, we can conclude that the Young's modulus of the rope is [tex]1.55 \times 10^{5} \;\rm N/m^{2}[/tex].
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