Question 801: Thermodynamics & KTG

Question

Two spherical black bodies of radii ' $R_1$ ' and ' $\mathrm{R}_2$ ' and with surface temperature ' $\mathrm{T}_1$ ' and ' $\mathrm{T}_2$ ' respectively radiate the same power. The ratio of ' $R_1$ ' to ' $R_2$ ' will be

Accepted Answer

The power radiated by a black body is given by Stefan-Boltzmann law:
$$ P = \sigma A T^4 $$
where:
$ \sigma $ is the Stefan-Boltzmann constant,
$ A $ is the surface area of the black body,
$ T $ is the absolute temperature.
For a spherical black body with radius $ R $, the surface area $ A $ is $ 4\pi R^2 $. Therefore, the power $ P $ radiated by a spherical black body can be written as:
$$ P = \sigma (4\pi R^2) T^4 $$
For two spherical black bodies with radii $ R_1 $ and $ R_2 $ and temperatures $ T_1 $ and $ T_2 $ respectively, radiating the same power, we have:
$$ \sigma (4\pi R_1^2) T_1^4 = \sigma (4\pi R_2^2) T_2^4 $$
Simplifying this equation:
$$ R_1^2 T_1^4 = R_2^2 T_2^4 $$
To find the ratio of $ R_1 $ to $ R_2 $, rearrange the above equation:
$$ \left(\frac{R_1}{R_2}\right)^2 = \frac{T_2^4}{T_1^4} $$
Taking the square root of both sides:
$$ \frac{R_1}{R_2} = \left(\frac{T_2}{T_1}\right)^2 $$
Thus, the ratio of $ R_1 $ to $ R_2 $ is:
Option B:
$ \left(\frac{T_2}{T_1}\right)^2 $

Answer

$\left(\frac{T_2}{T_1}\right)^4$

Answer

$\left(\frac{T_2}{T_1}\right)^2$

Answer

$\left(\frac{T_1}{T_2}\right)^4$

Answer

$\left(\frac{T_1}{T_2}\right)^2$

Thermodynamics & KTG

Two spherical black bodies of radii ' R_1 ' and ' \mathrm{R}_2 ' and with surface temperature ' \mathrm{T}_1 ' and ' \mathrm{T}_2 ' respectively radiate the same power. The ratio of ' R_1 ' to ' R_2 ' will be

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