Animated Solution for Physics - Properties of Solids and Liquids: A uniform heavy rod of weight $10{\text{ kg ms}}^{-2}$, cross-sectional area $100{\text{ cm}}^2$ and length $20\text{ cm}$ is hanging from a fixed support. Young's 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.

When we first learn about Hooke's Law, we usually deal with massless strings or rods being pulled by a heavy block at the end. In those ideal cases, the tension is uniform throughout the material. But what happens when the rod itself is heavy?

Imagine you are standing at the bottom of a human pyramid versus standing at the top. The person at the bottom feels the weight of everyone above them, while the person at the top feels nothing. Similarly, for a heavy rod hanging from a ceiling, the tension is not uniform. It is zero at the free bottom end and reaches its maximum at the top support, where it must hold the entire weight of the rod.

The Differential Element

Because the tension varies, we cannot simply plug the total weight $W$ into the standard elongation formula $\Delta l=\frac{WL}{AY}$. Doing so would assume the entire rod is experiencing the maximum tension, which is false.

To solve this, we must slice the rod into infinitesimally small elements. Let's consider a tiny element of length $dx$ located at a distance $x$ from the bottom free end.

What is the tension $T(x)$ acting on this specific element? It is exactly equal to the weight of the portion of the rod hanging strictly below it. Since the rod is uniform, the weight per unit length is $\frac{W}{L}$. Therefore, the weight of the bottom portion of length $x$ is:

Now that we know the exact tension acting on our tiny element $dx$, we can safely apply Hooke's Law just to this microscopic slice. The tiny elongation $d(\Delta l)$ of this element is:

To find the total elongation of the entire rod, we must add up the elongations of all these tiny slices from the bottom ($x=0$) to the top ($x=L$). This is where the magic of integration comes in:

Pulling the constants out of the integral, we get:

Evaluating the integral $\int xdx=\frac{x^2}{2}$ from $0$ to $L$ yields:

This is a beautiful and profound result! It tells us that a heavy rod stretches exactly half as much under its own weight as it would if its entire weight was concentrated and hung at the very bottom.

Now, we just need to substitute the given values into our derived master equation.

Given: - Weight, $W=10\text{ N}$ - Length, $L=20\text{ cm}=0.2\text{ m}$ - Area, $A=100{\text{ cm}}^2=10^{-2}{\text{ m}}^2$ - Young's Modulus, $Y=2\times 10^{11}{\text{ N/m}}^2$

Plugging these in:

Converting this to standard scientific notation, we get our final answer:

This perfectly matches option (d). The physics is elegant, and the math confirms it!