Q98 / 100 · BITSAT 2024 (Memory Based Paper 2) · Physics
A solid metallic cube having total surface area $24 \mathrm{~m}^2$ is uniformly heated. If its temperature is increased by $10^{\circ} \mathrm{C}$, calculate the increase in volume of the cube.
$\left(\right.$ Given : $\left.\alpha=5.0 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\right)$
- A$2.4 \times 10^6 \mathrm{~cm}^3$
- Correct answerB$1.2 \times 10^5 \mathrm{~cm}^3$
- C$6.0 \times 10^4 \mathrm{~cm}^3$
- D$4.8 \times 10^5 \mathrm{~cm}^3$
Solution
We have $\Delta V=V_0 \gamma \Delta T$
$\Delta V=a^3,(3 \alpha) \Delta T$
Now, $6 a^2=24 \quad[\because$ Total surface area of cube $\left.=6 a^2\right]$
$\Rightarrow a^2=4 \Rightarrow a=2$
So, $\Delta V=2^3\left(3 \times 5 \times 10^{-4}\right) \times 10=1200 \times 10^{-4} \mathrm{~m}^3$
$=1200 \times 10^2 \mathrm{~cm}^3=1.2 \times 10^5 \mathrm{~cm}^3$
Hence, the correct answer is option: 'B'.