Question

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

• A. $0.332$ $\mu m$
• B. $0.151$ $\mu m$
• C. $0.625$ $\mu m$
• D. $0.425$ $\mu m$

Answer 1

Answer: (A)

Here $l=4.5m$, $D=6mm$ $=6\times 10^{-3}m$, $F=100N$,
$Y=4.8\times 10^{11}N{m}^{-2}$, $\sigma =0.2$
As $Y=\frac{F}{A}\cdot$ $\frac{l}{\Delta l}$
$\therefore$ $\Delta l$ $=\frac{F}{A}\cdot$ $\frac{l}{Y}$ $=\frac{100\times 4.5}{3.14\times {\left(3\times 10^{-3}\right)}^2\times 4.8\times 10^{11}}$
$=3.32\times 10^{-5}m$