Question

The temperature of a metallic sphere of radius \(R\) is increased by a small amount \(\Delta T\). If the linear coefficient of thermal expansion of the metal is \(\alpha\), the approximate increase in the volume of the sphere is:

• A. \(6\pi R^3\alpha\Delta T\)
• B. \(2\pi R^3\alpha\Delta T\)
• C. \(3\pi R^3\alpha\Delta T\)
• D. \(4\pi R^3\alpha\Delta T\)

Hint:

Volume coefficient equals \(3\alpha\). For isotropic solids use \(\gamma=3\alpha\). Remember volume of sphere \(=\frac43\pi R^3\). Use approximation for small temperature changes.

Answer 1

The Correct Option is D

Step 1: State the data.
A metallic sphere of radius $R$ is heated by a small amount $\Delta T$. The linear expansion coefficient is $\alpha$. We want the change in volume.
Step 2: Initial volume.
The volume of a sphere is $V = \tfrac{4}{3}\pi R^3$.
Step 3: Link linear and volume expansion.
For isotropic solids the volume expansion coefficient is $\gamma = 3\alpha$.
Step 4: Write the expansion formula.
The increase in volume is $\Delta V = \gamma V \Delta T = 3\alpha V \Delta T$.
Step 5: Substitute the volume.
\[ \Delta V = 3\alpha \left(\tfrac{4}{3}\pi R^3\right)\Delta T \]
Step 6: Simplify.
The factor of $3$ cancels with the $\tfrac{1}{3}$.
\[ \Delta V = 4\pi R^3 \alpha \Delta T \]
\[ \boxed{4\pi R^3 \alpha \Delta T} \]