Question

Two perfectly black sphere $A$ and $B$ having radii $8cm$ and $2cm$ are maintained at temperatures ${127}^{\circ }C$ and ${527}^{\circ }C$, respectively. The ratio of the energy radiated by $A$ to that by $B$ is

• A. $1:2$
• B. $1:1$
• C. $2:1$
• D. $1:4$
• E. $1:16$

Answer 1

Answer: (B)

$1:1$

Key Concept Total energy emitted per second by a unit area of a body is proportional to the fourth power of its absolute temperature.
i.e. $E=\sigma T^4$
where, $\sigma$ is a universal constant called Stefan Boltzmann constant.
Given, $T_1={127}^{\circ }C$
$={127}^{\circ }+273=400K$
$T_2={527}^{\circ }C=800K$
and $r_1=8cm\Rightarrow r_2=2cm$
We know that total energy,
$E=\sigma A{T}^4$
In the first condition
$E_1={\sigma }_1{A}_1{T}_1^4....\left(i\right)$
In the second condition,
$E_2={\sigma }_2{A}_2{T}_2^4....\left(ii\right)$
On dividing Eq. (i) by Eq. (ii), we get
$\frac{E_1}{E_2}=\frac{A_1{T}_1^4}{A_2{T}_2}[\therefore {\sigma }_1={\sigma }_2]$
$=\frac{8_2}{2^2}\times \frac{(400{)}^4}{(800{)}^4}$
$\Rightarrow \frac{E_1}{E_2}=4^2\times \frac{1}{2^4}$
$\Rightarrow \frac{E_1}{E_2}=1:1$.