Q161 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
The rectangular surface area $(12\ \text{cm}\times6\ \text{cm})$ of the black body, at a temperature $127^\circ\text{C}$, emits heat energy at the rate of Q units per second. If the length and breadth of the surface area are each reduced to half of its initial value and the temperature is increased to $327^\circ\text{C}$, then the emission of heat energy will be
- A$\dfrac{9Q}{4}$
- B$\dfrac{81Q}{16}$
- C$\dfrac{81Q}{32}$
- Correct answerD$\dfrac{81Q}{64}$
Solution
According to Stefan-Boltzmann law, the rate of heat emission is given by $E = \sigma A T^4$.
Initial area $A_1 = 12 \times 6 = 72\ \text{cm}^2$
Initial temperature $T_1 = 127^\circ\text{C} = 400\ \text{K}$
Initial rate of heat emission $Q_1 = Q$
Final area $A_2 = \dfrac{12}{2} \times \dfrac{6}{2} = 18\ \text{cm}^2 = \dfrac{A_1}{4}$
Final temperature $T_2 = 327^\circ\text{C} = 600\ \text{K}$
Using the relation $Q \propto A T^4$, we get:
$\dfrac{Q_2}{Q_1} = \dfrac{A_2}{A_1} \left(\dfrac{T_2}{T_1}\right)^4$
$\dfrac{Q_2}{Q} = \dfrac{1}{4} \left(\dfrac{600}{400}\right)^4$
$\dfrac{Q_2}{Q} = \dfrac{1}{4} \left(\dfrac{3}{2}\right)^4 = \dfrac{1}{4} \times \dfrac{81}{16} = \dfrac{81}{64}$
$Q_2 = \dfrac{81Q}{64}$
Answer: $\dfrac{81Q}{64}$
Initial area $A_1 = 12 \times 6 = 72\ \text{cm}^2$
Initial temperature $T_1 = 127^\circ\text{C} = 400\ \text{K}$
Initial rate of heat emission $Q_1 = Q$
Final area $A_2 = \dfrac{12}{2} \times \dfrac{6}{2} = 18\ \text{cm}^2 = \dfrac{A_1}{4}$
Final temperature $T_2 = 327^\circ\text{C} = 600\ \text{K}$
Using the relation $Q \propto A T^4$, we get:
$\dfrac{Q_2}{Q_1} = \dfrac{A_2}{A_1} \left(\dfrac{T_2}{T_1}\right)^4$
$\dfrac{Q_2}{Q} = \dfrac{1}{4} \left(\dfrac{600}{400}\right)^4$
$\dfrac{Q_2}{Q} = \dfrac{1}{4} \left(\dfrac{3}{2}\right)^4 = \dfrac{1}{4} \times \dfrac{81}{16} = \dfrac{81}{64}$
$Q_2 = \dfrac{81Q}{64}$
Answer: $\dfrac{81Q}{64}$
Hence, the correct answer is option: 'D'.