Question

A solid metallic cube having total surface area \( 24 \, m^2 \) is uniformly heated. If its temperature is increased by \( 10^\circ C \), calculate the increase in volume of the cube.
Given: \( \alpha = 5.0 \times 10^{-4} \, C^{-1} \)

• A. \( 2.4 \times 10^6 \, cm^3 \)
• B. \( 1.2 \times 10^5 \, cm^3 \)
• C. \( 6.0 \times 10^4 \, cm^3 \)
• D. \( 4.8 \times 10^5 \, cm^3 \)

Hint:

For thermal expansion in solids, volume change is calculated using \( \Delta V = V_0 \gamma \Delta T \), where \( \gamma = 3\alpha \).

Answer 1

The Correct Option is B

Step 1: {Formula for volume expansion}
\[ \Delta V = V_0 \gamma \Delta T \] Step 2: {Volume relation with side length}
\[ \Delta V = a^3 (3\alpha) \Delta T \] Step 3: {Finding cube side length}
\[ 6a^2 = 24 \quad \Rightarrow \quad a^2 = 4 \quad \Rightarrow \quad a = 2 \] Step 4: {Substituting values}
\[ \Delta V = 2^3 (3 \times 5 \times 10^{-4}) \times 10 = 1200 \times 10^{-4} \, m^3 \] \[ = 1200 \times 10^2 \, cm^3 = 1.2 \times 10^5 \, cm^3 \] Thus, the correct answer is \( 1.2 \times 10^5 \, cm^3 \).

Answer 2

Step 1: Given total surface area of the cube is \( 24 \, m^2 \).
For a cube, total surface area = \( 6a^2 \), where \( a \) is the side length.
So, \( 6a^2 = 24 \Rightarrow a^2 = 4 \Rightarrow a = 2 \, m \)

Step 2: Volume of cube = \( a^3 = (2)^3 = 8 \, m^3 \)

Step 3: Use the formula for volumetric expansion:
\( \Delta V = 3\alpha V \Delta T \)
Given: \( \alpha = 5.0 \times 10^{-4} \, C^{-1} \), \( V = 8 \, m^3 \), \( \Delta T = 10^\circ C \)

Step 4: Substitute the values:
\( \Delta V = 3 \times 5.0 \times 10^{-4} \times 8 \times 10 = 0.12 \, m^3 \)

Step 5: Convert \( m^3 \) to \( cm^3 \):
\( 0.12 \, m^3 = 0.12 \times 10^6 = 1.2 \times 10^5 \, cm^3 \)

Final Answer: \( 1.2 \times 10^5 \, cm^3 \)