Question

Only one end of a nylon rope of length 4.5 m and diameter 6 mm is fixed to a free limb. A monkey weighing 100 N jumps to catch the free end and stays there. Find the elongation of rope, (Given Young’s modulus of nylon = $4.8\times {{10}^{11}}N{{m}^{-2}}$ and Poisson’s ratio of nylon = 0.2)
A. $33.2\mu m$
B. $0.151\mu m$
C. $0.625\mu m$
D. $0.425\mu m$

Answer 1

Hint: First we will make some assumptions that will make the question solvable by the given data in the question. Then we will use the formula for Young’s modulus and using that we will derive the expression for the expansion in length and calculate the elongation in the length of the rope.

Formula used: Young’s Modulus:
\[Y=\dfrac{FL}{A\Delta L}\]

Complete step by step answer:
The only force acting on the rope will be due to the weight of the monkey holding it. We will assume the monkey has no velocity when he grabbed the rope and that the rope is a light rope that does not have any mass. So, we will use the formula for Young’s modulus and derive the expression for \[\Delta L\]i.e. change in length.
\[Y=\dfrac{FL}{A\Delta L}\Rightarrow \Delta L=\dfrac{FL}{AY}\]
Here F is the force on the string, L is the length of the rope, Y is the Young’s modulus, A is the cross-sectional area of the rope. Putting in values in the formula, we get
\[\Delta L=\dfrac{100\times 4.5}{(\pi \times {{0.003}^{2}})(4.8\times {{10}^{11}})}=3.32\times {{10}^{-5}}=33.2\mu m\]

So, the correct answer is “Option A”.

Note: The assumptions are necessary for the correct approach to solve the question and physically visualize what is given in the question. We need to consider a steady state when solving the question i.e. none of the bodies in the question must be in motion as that may give rise to additional forces.