Question

A uniform heavy rod of weight \(10\,\text{kg}\,\text{ms}^{-2}\), cross-sectional area \(100\,\text{cm}^2\) and length \(20\,\text{cm}\) is hanging from a fixed support. Young modulus of the material of the rod is \(2 \times 10^{11}\,\text{Nm}^{-2}\). Neglecting the lateral contraction, find the elongation of rod due to its own weight :

• A. \(2 \times 10^{-9}\,\text{m}\)
• B. \(5 \times 10^{-10}\,\text{m}\)
• C. \(5 \times 10^{-8}\,\text{m}\)
• D. \(4 \times 10^{-8}\,\text{m}\)

Hint:

For problems involving own weight, you can assume the entire weight acts at the center of mass, which is \(L/2\) from the support, leading to the factor of 1/2 in the formula.

Answer 1

The Correct Option is B

To find the elongation of the rod due to its own weight, we use the concept of Young's modulus and the formula for elongation due to self-weight. The relevant formula is:

\(\Delta L = \frac{\rho g L^2}{2Y}\)

where:

• \( \Delta L \): Elongation
• \( \rho \): Density of the rod material
• \( g \): Acceleration due to gravity (\approx 9.8 \, \text{ms}^{-2})
• \( L \): Length of the rod
• \( Y \): Young's modulus

Given:

• Weight of the rod = 10 \, \text{kg}\, \text{ms}^{-2} \Rightarrow \text{mass} = 1 \, \text{kg} (since weight = mass × gravity, and here gravity is taken as 1).
• Cross-sectional area \(= 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2\)
• Length \(L = 20 \, \text{cm} = 0.2 \, \text{m}\)
• Young's modulus \(Y = 2 \times 10^{11} \, \text{Nm}^{-2}\)

First, calculate the density:

\( \rho = \frac{\text{mass}}{\text{volume}} = \frac{1}{0.2 \times 100 \times 10^{-4}} \, \text{kg/m}^3\)

\( \rho = \frac{1}{0.002} = 500 \, \text{kg/m}^3 \)

Now, substituting the values into the elongation formula:

\(\Delta L = \frac{500 \times 9.8 \times (0.2)^2}{2 \times 2 \times 10^{11}}\)

\(\Delta L = \frac{500 \times 9.8 \times 0.04}{4 \times 10^{11}}\)

\(\Delta L = \frac{196}{4 \times 10^{9}}\)

\(\Delta L = 49 \times 10^{-9} \, \text{m}\)

\(\Delta L = 4.9 \times 10^{-10} \, \text{m} \approx 5 \times 10^{-10} \, \text{m}\)

Thus, the elongation of the rod due to its own weight is approximately 5 \times 10^{-10} \, \text{m}.