Question

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 the rope (Given Young’s modulus of nylon=$4.8\times10^{11} N$ $m^{-2}$ and Poisson’s ratio of nylon = $0.2$.), what will be the change in the diameter of the rope?

• A. $8.8\times10^{-9} m$
• B. $7.4\times10^{-9} m$
• C. $6.4\times10^{-8} m$
• D. $5.6\times10^{-9} m$

Answer 1

Answer: (A)

Poisson’s ratio, $\sigma$ $=\frac{\Delta D/ D}{\Delta l /l}$ $=\frac{\Delta D}{D}\cdot$ $\frac{l}{\Delta l}$
$\therefore $ $\quad$ $\Delta D=$ $\frac{\sigma D \Delta l}{l}$ $=\frac{0.2\times6\times10 ^{-3}\times3.32\times10^{-5}}{4.5}$
$=8.8\times10^{-9}\,m$