Question:medium

Two bodies A and B at temperatures ' $\text{T}_1$ ' $\text{K}$ and ' $\text{T}_2$ ' $\text{K}$ respectively have the same dimensions. Their emissivities are in the ratio $16 : 1$. At $\text{T}_1 = x\text{T}_2$, they radiate the same amount of heat per unit area per unit time. The value of $x$ is

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If emissivity increases, temperature must decrease to maintain the same power output.
Updated On: May 7, 2026
  • 8
  • 4
  • 2
  • 0.5
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
The amount of heat radiated per unit area per unit time by a body is its emissive power.
According to the Stefan-Boltzmann law, the emissive power depends on the emissivity of the surface and the fourth power of its absolute temperature.
Step 2: Key Formula or Approach:
The Stefan-Boltzmann law states: \[ E = \epsilon \sigma T^4 \] where $E$ is the emissive power (heat radiated per unit area per unit time), $\epsilon$ is the emissivity, $\sigma$ is the Stefan-Boltzmann constant, and $T$ is the absolute temperature.
We are given that $E_A = E_B$.
Step 3: Detailed Explanation:
For body A: \[ E_A = \epsilon_A \sigma T_1^4 \] For body B: \[ E_B = \epsilon_B \sigma T_2^4 \] Since they radiate the same amount of heat per unit area per unit time: \[ E_A = E_B \] \[ \epsilon_A \sigma T_1^4 = \epsilon_B \sigma T_2^4 \] We can cancel the Stefan-Boltzmann constant $\sigma$: \[ \epsilon_A T_1^4 = \epsilon_B T_2^4 \] Rearrange to group the temperatures and emissivities: \[ \frac{T_1^4}{T_2^4} = \frac{\epsilon_B}{\epsilon_A} \] \[ \left( \frac{T_1}{T_2} \right)^4 = \frac{\epsilon_B}{\epsilon_A} \] We are given the ratio of emissivities $\epsilon_A : \epsilon_B = 16 : 1$. Therefore, $\frac{\epsilon_B}{\epsilon_A} = \frac{1}{16}$.
Substitute this into the equation: \[ \left( \frac{T_1}{T_2} \right)^4 = \frac{1}{16} \] Taking the fourth root of both sides: \[ \frac{T_1}{T_2} = \left( \frac{1}{16} \right)^{1/4} = \frac{1}{2} \] We are given the relation $\text{T}_1 = x\text{T}_2$, which means $x = \frac{T_1}{T_2}$.
Therefore, $x = \frac{1}{2} = 0.5$.
Step 4: Final Answer:
The value of $x$ is 0.5.
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