Animated Solution for Physics - Properties of Solids and Liquids: A steel wire having a radius of $2.0\text{ mm}$, carrying a load of $4\text{ kg}$, is hanging from a ceiling. Given that $g=3.1\pi {\text{ ms}}^{-2}$, what will be the tensile stress that would be developed in the wire?

The journey of understanding materials begins with a simple question: how do they respond when we pull them? In this problem, we are exploring the very essence of material strength by calculating the tensile stress developed in a steel wire under a load.

Imagine you are an engineer designing a suspension bridge or a simple elevator. The cables supporting the weight are under immense tension. If the stress exceeds the material's limit, it snaps. This is why calculating stress is not just an academic exercise; it is the foundation of structural integrity!

Let's visualize the physical reality of our problem. We have a steel wire hanging vertically from a rigid ceiling. Attached to its lower end is a block of mass $m=4\text{ kg}$.

Gravity is relentlessly pulling this mass downwards with a force equal to its weight, $W=mg$. To prevent the mass from falling, the wire must pull upwards with an equal and opposite force. This internal restoring force is what we call Tension ($T$). In a state of equilibrium, the tension in the wire is exactly equal to the weight of the hanging mass.

We are also given the dimensions of the wire. It has a radius $r=2.0\text{ mm}$. In physics, we must always be vigilant about units. Before we proceed, we must convert this radius into standard SI units (meters) to ensure our final answer is in Pascals (${\text{N/m}}^2$). Therefore, $r=2.0\times 10^{-3}\text{ m}$.

Now, what exactly is stress? Tensile stress is defined as the internal restoring force acting per unit cross-sectional area of the deformed body.

Mathematically, it is expressed as:

In our scenario, the restoring force $F$ is the tension $T$, which equals $mg$. The wire is cylindrical, so its cross-section is a perfect circle. The area of this circle is given by $A=\pi r^2$.

Substituting these into our master equation, we get:

This is where the magic of well-crafted physics problems shines. Let's substitute our known values into the equation. We have $m=4\text{ kg}$, $g=3.1\pi {\text{ ms}}^{-2}$, and $r=2\times 10^{-3}\text{ m}$.

I know this expression might look a bit dense, but let's take a breath and simplify the denominator first. Squaring the radius gives us $4\times 10^{-6}{\text{ m}}^2$.

Look closely at this beautiful symmetry! The examiner intentionally gave us $g=3.1\pi$ to make our lives easier. The $\pi$ in the numerator and the $\pi$ in the denominator cancel each other out perfectly. Even better, the $4$ from the mass and the $4$ from the squared radius also cancel out!

We are left with a remarkably simple fraction:

Finally, bringing the $10^{-6}$ from the denominator to the numerator changes the sign of its exponent, giving us our final answer:

And there we have it! The tensile stress developed in the steel wire is $3.1\times 10^6{\text{ Nm}}^{-2}$. This elegant cancellation is a classic hallmark of JEE problems, rewarding students who set up the full equation before rushing to calculate intermediate values.