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: Understanding the Concept:
This question deals with thermal expansion. When a solid object is heated, it expands in all dimensions: length, area, and volume. We need to compare the percentage change in different geometrical properties of a sphere (radius, diameter, surface area, and volume) for a given change in temperature. The mass of the sphere does not change upon heating.
Step 2: Key Formula or Approach:
Let the initial radius be \(r_0\) and the temperature change be \( \Delta T \). Let \( \alpha \) be the coefficient of linear expansion.
The new radius \(r\) after heating is given by \( r = r_0(1 + \alpha \Delta T) \).
The change in radius is \( \Delta r = r - r_0 = r_0 \alpha \Delta T \).
The fractional change in radius is \( \frac{\Delta r}{r_0} = \alpha \Delta T \).
The percentage change in radius is \( % \Delta r = \alpha \Delta T \times 100 \).
We can derive the relationships for surface area and volume expansion from the linear expansion. The coefficient of superficial (area) expansion is \( \beta \approx 2\alpha \), and the coefficient of cubical (volume) expansion is \( \gamma \approx 3\alpha \).
Step 3: Detailed Explanation:
Let's calculate the percentage change for each quantity.

Radius (r) and Diameter (d):
The fractional change in radius is \( \frac{\Delta r}{r_0} = \alpha \Delta T \).
Since diameter \( d = 2r \), the fractional change in diameter is \( \frac{\Delta d}{d_0} = \frac{2\Delta r}{2r_0} = \frac{\Delta r}{r_0} = \alpha \Delta T \).
The percentage change in both radius and diameter is \( \alpha \Delta T \times 100 \).

Surface Area (A):
The surface area of a sphere is \( A = 4\pi r^2 \).
The fractional change in area is \( \frac{\Delta A}{A_0} = \beta \Delta T \approx 2\alpha \Delta T \).
The percentage change in surface area is approximately \( 2\alpha \Delta T \times 100 \).

Volume (V):
The volume of a sphere is \( V = \frac{4}{3}\pi r^3 \).
The fractional change in volume is \( \frac{\Delta V}{V_0} = \gamma \Delta T \approx 3\alpha \Delta T \).
The percentage change in volume is approximately \( 3\alpha \Delta T \times 100 \).

Mass (m):
Heating does not add or remove matter. The mass of the sphere remains constant. The percentage change in mass is 0.

Comparison:
Percentage change in radius/diameter \( \propto \alpha \).
Percentage change in surface area \( \propto 2\alpha \).
Percentage change in volume \( \propto 3\alpha \).
Since \( \alpha \) is a positive value, we have \( 3\alpha>2\alpha>\alpha \). Therefore, the percentage change is maximum for the volume.
Step 4: Final Answer:
The maximum percentage change will be observed in the volume of the sphere. This corresponds to option (A).