Consider two cylindrical rods of indentical dimesnions, one of rubber and the other of steel. Both the rods are fixed rigidiy at one end to the roof. `A` mass `M` is attached to each of the free ends at the centre of the rods.

Options

A. Both the rods will elongate but there shall be no perceptible change in shape
B. The steel rod will elongate and change shape but  the rubber rod will only elongate.
C. The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.
D. The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.

Answer

To determine which rod will elongate more when a mass \( M \) is attached to the free ends of two cylindrical rods (one made of rubber and the other made of steel), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have two cylindrical rods of identical dimensions: one made of rubber and the other made of steel. Both rods are fixed at one end, and a mass \( M \) is attached to the free ends.

2. **Identify the Relevant Properties**: The elongation of a rod under a load can be described using Young's modulus \( Y \), which is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0} \]
where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length (elongation), and \( L_0 \) is the original length.

3. **Determine the Forces**: The force \( F \) acting on each rod due to the mass \( M \) is given by:
\[ F = M \cdot g \]
where \( g \) is the acceleration due to gravity.

4. **Calculate the Elongation**: The elongation \( \Delta L \) of each rod can be expressed as:
\[ \Delta L = \frac{F L_0}{A Y} \]
Since the dimensions of both rods are identical, the cross-sectional area \( A \) and the original length \( L_0 \) are the same for both rods.

5. **Compare Young's Modulus**: The Young's modulus for rubber is significantly lower than that for steel. For example:
- Young's modulus for rubber \( (Y_{\text{rubber}}) \) is approximately \( 0.01 - 0.1 \, \text{GPa} \).
- Young's modulus for steel \( (Y_{\text{steel}}) \) is approximately \( 200 \, \text{GPa} \).

6. **Conclusion on Elongation**: Since the elongation \( \Delta L \) is inversely proportional to Young's modulus \( Y \), the rod with the lower Young's modulus (rubber) will elongate more than the rod with the higher Young's modulus (steel). Therefore:
\[ \Delta L_{\text{rubber}} > \Delta L_{\text{steel}} \]

7. **Final Statement**: The rubber rod will elongate more than the steel rod when the same mass \( M \) is applied to both.