Animated Solution for Physics - Properties of Solids and Liquids: Two rods $A$ and $B$ of identical dimensions are at temperature $30^{\circ }\text{C}$. If $A$ is heated upto $180^{\circ }\text{C}$ and $B$ upto $T^{\circ }\text{C}$, then new lengths are the same. If the ratio of the coefficients of linear expansion of $A$ and $B$ is $4:3$, then the value of $T$ is

Have you ever wondered how engineers design bridges or railway tracks so they don't buckle under the scorching summer sun? The secret lies in understanding thermal expansion. Today, we are going to tackle a brilliant JEE problem that tests exactly this concept. It’s not just about plugging numbers into a formula; it’s about visualizing the physical reality of two metal rods stretching out as they heat up.

Let's dive into the story of Rod A and Rod B.

Two Rods, One Goal

Imagine you have two metal rods, A and B, lying side by side. The problem tells us they have "identical dimensions" and are both chilling at a comfortable room temperature of $30^{\circ }\text{C}$.

This is our starting line. Because their dimensions are identical, their initial lengths are exactly the same. Let's call this initial length $l_0$.

Now, we turn up the heat. Rod A is thrown into a furnace and heated up to $180^{\circ }\text{C}$. Rod B is also heated, but to an unknown temperature $T^{\circ }\text{C}$.

Here is the magical constraint of the problem: after heating, their new lengths are exactly the same.

Before we write down any equations, let's think about what this constraint means physically.

If both rods started at the same length ($l_0$) and ended up at the same final length ($l_{\text{new}}$), what does that tell us about how much they grew?

It means the change in their lengths must be identical!

Mathematically, if $l_{\text{new}}=l_0+\Delta l_A$ and $l_{\text{new}}=l_0+\Delta l_B$, then it must be true that:

This simple realization is the master key to unlocking the entire problem.

Now, let's bring in our trusty tool for linear thermal expansion:

Let's apply this to Rod A. It started at $30^{\circ }\text{C}$ and went up to $180^{\circ }\text{C}$. So, its change in temperature is $\Delta T_A=180-30=150^{\circ }\text{C}$. The expansion for Rod A is:

Next, let's look at Rod B. It started at $30^{\circ }\text{C}$ and went up to $T^{\circ }\text{C}$. Its change in temperature is $\Delta T_B=T-30$. The expansion for Rod B is:

We already established our master key: $\Delta l_A=\Delta l_B$. Let's equate our two expressions:

Notice how beautifully $l_0$ cancels out from both sides? This is why the problem didn't need to give us the actual length of the rods. The physics holds true regardless of whether the rods are 1 meter long or 100 meters long!

Let's rearrange the equation to group the $\alpha$ terms together, because the problem gave us their ratio:

The problem states that the ratio of the coefficients of linear expansion, ${\alpha }_1:{\alpha }_2$, is $4:3$.

Let's substitute this into our equation:

Now, it's just a matter of simple algebra. Let's isolate $T$:

Divide $150$ by $3$ to get $50$, and multiply by $4$:

Finally, add $30$ to both sides:

And there we have it! Rod B must be heated to exactly $230^{\circ }\text{C}$ for it to match the length of Rod A.

This problem is a beautiful example of how physical constraints (identical initial and final lengths) translate into elegant mathematical symmetries (equating $\Delta l$). Always look for these symmetries—they are the hallmarks of great physics problems!