Q.Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass $M$ is attached to each of the free ends at the centre of the rods.

The key idea is Young’s modulus: steel has a very high modulus (≈ $2 \times 10^{11}$ Pa), while rubber has a very low modulus (≈ $10^{6}$ Pa). Under the same load, steel deforms negligibly in shape, but rubber undergoes large lateral contraction (Poisson effect) in addition to elongation.

1. Both rods experience the same tensile stress $\sigma = \frac{Mg}{A}$.
2. For steel, the strain $\epsilon = \frac{\sigma}{Y}$ is tiny — elongation is small and lateral contraction is imperceptible. …

1. Understanding the setup and the concept.

   Both rods are identical in dimensions — same length $L$, same cross-sectional area $A$. Each is fixed at the top and has a mass $M$ hanging from the free end. The mass applies a tensile force $F = Mg$ along the axis of each rod.

   The question is about two effects: elongation (axial strain) and lateral contraction (change in shape of the cross-section). The first is governed by Young’s modulus $Y$, the second by Poisson’s ratio $\nu$. But here the key difference is the magnitude of the axial strain, which determines whether the lateral contraction is perceptible.

2. Elongation: how much does each rod stretch?

   For a rod under axial tension, the elongation $\Delta L$ is given by

Steel has $Y \approx 2 \times 10^{11} \, \text{Pa}$, rubber has $Y \approx 5 \times 10^6 \, \text{Pa}$ — a factor of about $40\,000$ difference. For the same $F$, $L$, and $A$, the rubber rod stretches roughly $40\,000$ times more than the steel rod.

So both elongate, but the rubber rod’s elongation is enormous and easily visible, while the steel rod’s elongation is microscopic.

3. Lateral contraction: the Poisson effect. When a rod stretches axially, it contracts laterally (gets thinner). The lateral strain $\epsilon_\text{lat}$ is related to the axial strain $\epsilon_\text{axial}$ by Poisson’s ratio $\nu$:

For steel, $\nu \approx 0.3$; for rubber, $\nu \approx 0.5$ (nearly incompressible).

The absolute lateral contraction is $\Delta d = -\nu \, \epsilon_\text{axial} \, d$, where $d$ is the original diameter. Since $\epsilon_\text{axial} = \Delta L / L$, and $\Delta L$ is tiny for steel, $\Delta d$ is utterly imperceptible. For rubber, $\Delta L$ is huge, so $\Delta d$ is also large — the rod becomes noticeably thinner.

4. Shape of the bottom edge: why does it change? The mass $M$ is attached at the centre of the free end, not uniformly across the cross-section. This means the load is concentrated at a point. For steel: the rod is so stiff that the point load is transmitted almost uniformly across the cross-section within a very short distance (Saint-Venant’s principle). The bottom face remains essentially flat — no perceptible shape change. …

Step 1: deltaL=FL/(AY); steel's Y is ~40000x rubber's, so rubber elongates far more for same F. Step 2: lateral Poisson contraction scales with axial strain -- negligible for steel, large for rubber. Step 3: point load at centre of free end spreads uniformly for stiff steel ( …