One end of a nylon rope of length 1 and diameter 10 mm is fixed to free limb. A monkey weighing 100 N jumps to catch the free end and stays there. The change in diameter of the rope is (Young's modulus of the wire is Y and Poisson Ratio is $\sigma$ )

The Young modulus $Y$ is defined as the ratio of longitudinal stress to longitudinal strain. The longitudinal stress is $\sigma_{stress} = \dfrac{F}{A} = \dfrac{F}{\pi r^2} = \dfrac{F}{\pi (d/2)^2} = \dfrac{4F}{\pi d^2}$, where $F = 100$ N and $d = 10$ mm = $0.01$ m.

The longitudinal strain is $\epsilon_L = \dfrac{\Delta l}{l} = \dfrac{\sigma_{stress}}{Y} = \dfrac{4F}{\pi d^2 Y}$.

The Poisson ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = - \dfrac{\epsilon_D}{\epsilon_L}$, where $\epsilon_D = \dfrac{\Delta d}{d}$ is the lateral strain.

Therefore, the change in diameter $\Delta d$ is given by $\Delta d = - \sigma \epsilon_L d = - \sigma \left( \dfrac{4F}{\pi d^2 Y} \right) d = - \dfrac{4F \sigma}{\pi d Y}$.

Substituting the given values $F = 100$ N and $d = 0.01$ m:

$\Delta d = - \dfrac{4 \times 100 \times \sigma}{\pi \times 0.01 \times Y} = - \dfrac{400 \sigma}{0.01 \pi Y} = - \dfrac{40000 \sigma}{\pi Y}$.

The magnitude of the change in diameter is $\dfrac{40000 \sigma}{\pi Y}$.

Answer: $\dfrac{40000 \sigma}{\pi Y}$