Question

In a steel plate of length \(100\) cm and breadth \(50\) cm, there is a hole in the shape of an equilateral triangle of side \(10\) cm. The coefficient of linear expansion of steel is \(1.2\times10^{-5}\,{}^{\circ}\mathrm{C}^{-1}\). The temperature of the steel plate is increased by \(100^\circ\mathrm{C}\). What will be the percentage change in the area of the triangular hole?

• A. \(0.24\%\)
• B. \(-0.24\%\)
• C. \(0.12\%\)
• D. \(-1.2\%\)

Hint:

A hole expands just like the surrounding material. For area expansion, \[ \beta=2\alpha. \] Percentage change: \[ \beta\Delta T\times100. \]

Answer 1

The Correct Option is A

Step 1: The key idea about a hole.
A hole in a heated plate grows just like a solid piece of the same metal would. So we only need the area-expansion rule, and the plate dimensions and triangle side are not needed at all.

Step 2: Area expansion factor.
For any area, the fractional change is \[ \frac{\Delta A}{A}=\beta\,\Delta T, \] where $\beta=2\alpha$ is the area-expansion coefficient.

Step 3: Find $\beta$.
Given $\alpha=1.2\times10^{-5}\,^\circ\mathrm{C}^{-1}$, so \[ \beta=2\times1.2\times10^{-5}=2.4\times10^{-5}\,^\circ\mathrm{C}^{-1}. \]

Step 4: Put in the temperature rise.
Here $\Delta T=100^\circ\mathrm{C}$. So \[ \frac{\Delta A}{A}=2.4\times10^{-5}\times100=2.4\times10^{-3}. \]

Step 5: Convert to percentage.
Multiply by $100$: \[ \frac{\Delta A}{A}\times100=0.24\%. \]

Step 6: Sign of the change.
The temperature went up, so the hole got bigger, meaning the change is positive.
\[ \boxed{+0.24\%} \]