MCQ
Two bodies $A$ and $B$ having equal surface areas are maintained at temperatures $10^\circ C$ and $20^\circ C$. The thermal radiation emitted in a given time by $A$ and $B$ are in the ratio :
- ✓$1 : 1.15$
- B$1 : 2$
- C$1 : 4$
- D$1 : 16$
✓
Answer
Correct option: A.
$1 : 1.15$
From Stefan $-$ Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area $A$ is given by,
$\text{u}=\sigma\text{AT}^4$
Here, $\sigma$ is Stefan $-$ Boltzmann constant.
The thermal radiation emitted in a given time by $A$ and $B$ will be in the ratio.
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{\text{T}_\text{A}^4}{\text{T}_\text{n}^4}$
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{(273+10)^4}{(273+20)^4}$
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{1}{1.15}$
$\text{u}=\sigma\text{AT}^4$
Here, $\sigma$ is Stefan $-$ Boltzmann constant.
The thermal radiation emitted in a given time by $A$ and $B$ will be in the ratio.
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{\text{T}_\text{A}^4}{\text{T}_\text{n}^4}$
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{(273+10)^4}{(273+20)^4}$
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{1}{1.15}$
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