Question

One end of a nylon rope of length \( 1 \) and diameter \( 10 \, \text{mm} \) is fixed to free limb. A monkey weighing \( 100 \, \text{N} \) 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 \( \sigma \))

• A. \( \frac{4000\sigma}{\pi Y} \)
• B. \( \frac{400\sigma}{\pi Y} \)
• C. \( \frac{40000\sigma}{\pi Y} \)
• D. \( \frac{40000\pi}{\sigma Y} \)

Hint:

Poisson ratio connects lateral strain with longitudinal strain. First calculate stress, then longitudinal strain, and finally change in diameter.

Answer 1

The Correct Option is C

Step 1: Write given values.
Force applied by monkey is:
\[ F = 100 \, N \]
Diameter of rope is:
\[ d = 10 \, \text{mm} = 0.01 \, m \]

Step 2: Find radius of rope.
\[ r = \frac{d}{2} = 0.005 \, m \]

Step 3: Calculate area of cross-section.
\[ A = \pi r^2 \]
\[ A = \pi (0.005)^2 \]
\[ A = 25\pi \times 10^{-6} \, m^2 \]

Step 4: Write longitudinal stress.
\[ \text{Stress} = \frac{F}{A} \]
\[ \text{Stress} = \frac{100}{25\pi \times 10^{-6}} \]
\[ \text{Stress} = \frac{4 \times 10^6}{\pi} \]

Step 5: Write longitudinal strain.
\[ \text{Longitudinal strain} = \frac{\text{Stress}}{Y} \]
\[ = \frac{4 \times 10^6}{\pi Y} \]

Step 6: Use Poisson ratio relation.
Lateral strain is:
\[ \frac{\Delta d}{d} = \sigma \times \frac{4 \times 10^6}{\pi Y} \]

Step 7: Calculate change in diameter.
\[ \Delta d = d \times \sigma \times \frac{4 \times 10^6}{\pi Y} \]
\[ \Delta d = 0.01 \times \sigma \times \frac{4 \times 10^6}{\pi Y} \]
\[ \Delta d = \frac{40000\sigma}{\pi Y} \]