Question

One end of a nylon rope of length 4.5m and diameter 6mm is fixed to a stem of a tree. A monkey weighing 100N jumps to catch the free end and stays there. What will be the change in the diameter of the rope? (Given Young’s modulus of nylon = \[4.8 \times {10^{11}}\,{\text{N}}\,{{\text{m}}^{ - 2}}\] and Poisson’s ratio of nylon = 0.2)
A. \[8.8 \times {10^{ - 9}}\,{\text{m}}\]
B. \[7.4 \times {10^{ - 9}}\,{\text{m}}\]
C. \[6.4 \times {10^{ - 8}}\,{\text{m}}\]
D. \[5.6 \times {10^{ - 9}}\,{\text{m}}\]

Answer 1

Hint:The Poisson’s ratio is the ratio of transverse train to lateral strain. The transverse strain is the ratio of change in diameter to the original diameter while the lateral strain is the ratio of change in length to the original length. Calculate the change in length using the formula for the Young’s modulus of a material.

Formula used:
Poisson’s ratio, \[\nu = \dfrac{{\Delta D/D}}{{\Delta L/L}}\]
where, \[\Delta D\] is the change in the diameter, D is the original diameter, \[\Delta L\] is the change in length and L is the original length of the rope.
Young’s modulus, \[Y = \dfrac{{F\,L}}{{A\,\Delta L}}\]
where, F is the force, A is the area of cross-section of the rope.

Complete step by step answer:
We know that Poisson’s ratio is the ratio of transverse strain to the lateral strain. That is, it is the ratio of change in diameter of the rope to the change in length of the rope. Therefore, we can express the Poisson’s ratio as,
\[\nu = \dfrac{{\Delta D/D}}{{\Delta L/L}}\] …… (1)
Here, \[\Delta D\] is the change in the diameter, D is the original diameter, \[\Delta L\] is the change in length and L is the original length of the rope.
Now, we have to calculate the change in the length of the rope as the monkey jumps to catch the free end. We have the expression for the Young’s modulus,
\[Y = \dfrac{{F\,L}}{{A\,\Delta L}}\]
Here, F is the force, A is the area of cross-section of the rope.
Here the force on the other end of the rope is the weight of the monkey. Therefore, we can substitute W for F in the above equation.
\[Y = \dfrac{{W\,L}}{{A\,\Delta L}}\]
\[ \Rightarrow \Delta L = \dfrac{{W\,L}}{{\pi {r^2}\,Y}}\]
Here, r is the radius of the rope.
Substituting \[W = 100\,{\text{N}}\], \[L = 4.5\,{\text{m}}\], \[r = 3\,{\text{mm}}\] and \[Y = 4.8 \times {10^{11}}\,{\text{N}}\,{{\text{m}}^{ - 2}}\] in the above equation, we get,
\[\Delta L = \dfrac{{\left( {100} \right)\left( {4.5} \right)}}{{\left( {3.14} \right){{\left( {3 \times {{10}^{ - 3}}} \right)}^2}\,\left( {4.8 \times {{10}^{11}}} \right)}}\]
\[ \Rightarrow \Delta L = \dfrac{{450}}{{1.356 \times {{10}^7}}}\]
\[ \Rightarrow \Delta L = 3.32 \times {10^{ - 5}}\,{\text{m}}\]
We have the expression for the Poisson’s ratio,
\[\nu = \dfrac{{\Delta D/D}}{{\Delta L/L}}\]
\[ \Rightarrow \Delta D = \dfrac{{\Delta L}}{L} \times \nu D\]
Substituting \[\Delta L = 3.32 \times {10^{ - 5}}\,{\text{m}}\], \[L = 4.5\,{\text{m}}\], \[\nu = 0.2\] and \[D = 6\,{\text{mm}}\] in the above equation, we get,
\[\Delta D = \dfrac{{3.32 \times {{10}^{ - 5}}}}{{4.5}} \times \left( {0.2} \right)\left( {6 \times {{10}^{ - 3}}} \right)\]
\[ \therefore \Delta D = 8.8 \times {10^{ - 9}}\,{\text{m}}\]
Therefore, the change in diameter of the rope is \[8.8 \times {10^{ - 9}}\,{\text{m}}\].

So, the correct answer is option A.

Note: Poisson’s ratio is unit less quantity and thus it has no dimensions. It is actually a negative ratio of transverse strain to the longitudinal strain. We have assumed that the monkey catches the free end firmly while jumping. If the monkey jumped from a certain height to catch the free end, the force acted on the free end would have been greater than the weight of the monkey.