Question

When a metallic sphere is heated, maximum percentage change will be observed in its

• A. volume
• B. radius
• C. diameter
• D. surface area
• E. mass

Hint:

For small thermal expansion: linear change is proportional to \(1\), area change to \(2\), and volume change to \(3\). So volume always shows the maximum percentage change.

Answer 1

The Correct Option is A

Step 1: Recall thermal expansion of solids.
When a metallic sphere is heated, its linear dimensions increase. That means radius, diameter, surface area, and volume all change, while mass remains unchanged.

Step 2: Let the fractional linear expansion be small.
Suppose the radius increases from \(r\) to \(r+\Delta r\). Then the fractional change in radius is:
\[ \frac{\Delta r}{r} \] This is the linear percentage change.

Step 3: Find percentage change in diameter.
Since diameter \(d=2r\), its fractional change is the same as that of radius:
\[ \frac{\Delta d}{d}=\frac{\Delta r}{r} \] So radius and diameter have the same percentage change.

Step 4: Find percentage change in surface area.
Surface area of a sphere is:
\[ S=4\pi r^2 \] For small changes, \[ \frac{\Delta S}{S}=2\frac{\Delta r}{r} \] So the percentage change in surface area is twice the percentage change in radius.

Step 5: Find percentage change in volume.
Volume of a sphere is:
\[ V=\frac{4}{3}\pi r^3 \] For small changes, \[ \frac{\Delta V}{V}=3\frac{\Delta r}{r} \] So the percentage change in volume is three times the percentage change in radius.

Step 6: Compare all percentage changes.
Thus: \[ \text{radius change} \sim 1\frac{\Delta r}{r} \] \[ \text{diameter change} \sim 1\frac{\Delta r}{r} \] \[ \text{surface area change} \sim 2\frac{\Delta r}{r} \] \[ \text{volume change} \sim 3\frac{\Delta r}{r} \] and mass does not change. Therefore, the maximum percentage change is in the volume.

Step 7: State the final answer.
Hence, when a metallic sphere is heated, the maximum percentage change is observed in its:
\[ \boxed{\text{volume}} \] So the correct option is \((1)\).