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

Step 1: Understanding the Concept:
For a heavy rod hanging vertically, the tension is not uniform. It is maximum at the support and zero at the free end. The effective force causing elongation is half the weight of the rod.
Step 2: Key Formula or Approach: \[ \Delta L = \frac{WL}{2AY} \] Where \( W \) is weight, \( L \) is length, \( A \) is area, and \( Y \) is Young's modulus.
Step 3: Detailed Explanation:
Given:
\( W = 10\,\text{N} \) (Weight unit \(\text{kg}\,\text{ms}^{-2}\) is equivalent to Newton).
\( L = 20\,\text{cm} = 0.2\,\text{m} \).
\( A = 100\,\text{cm}^2 = 100 \times 10^{-4}\,\text{m}^2 = 10^{-2}\,\text{m}^2 \).
\( Y = 2 \times 10^{11}\,\text{Nm}^{-2} \).
Calculation:
\[ \Delta L = \frac{10 \times 0.2}{2 \times 10^{-2} \times 2 \times 10^{11}} \] \[ \Delta L = \frac{2}{4 \times 10^9} \] \[ \Delta L = 0.5 \times 10^{-9} = 5 \times 10^{-10}\,\text{m} \] Step 4: Final Answer:
The elongation of the rod is \(5 \times 10^{-10}\,\text{m}\).