Question 617: Thermodynamics & KTG

Question

Two spherical black bodies have radii ' $R_1$ ' and ' $R_2$ '. Their surface temperatures are $T_1 K$ and $T_2 K$ respectively. If they radiate the same power, the ratio $\frac{R_1}{R_2}$ is

Accepted Answer

Step 1: Power radiated by a black body
By Stefan-Boltzmann law:
$ P = \sigma A T^4 $
where $A$ is surface area of sphere, $A = 4\pi R^2$.
So,
$ P = \sigma \cdot (4\pi R^2) \cdot T^4 $
$ P = 4\pi \sigma R^2 T^4 $
Step 2: Equating powers
Given both radiate same power:
$ 4\pi \sigma R_1^2 T_1^4 = 4\pi \sigma R_2^2 T_2^4 $
Canceling constants:
$ R_1^2 T_1^4 = R_2^2 T_2^4 $
Step 3: Solve for ratio
$ \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{T_2}{T_1}\right)^4 $
Taking square root:
$ \frac{R_1}{R_2} = \left(\frac{T_2}{T_1}\right)^2 $
Final Answer:
$ \frac{R_1}{R_2} = \left(\frac{T_2}{T_1}\right)^2 $
Ã¢ÂÂ Correct Option:
D

Answer

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

Answer

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

Answer

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

Answer

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

Thermodynamics & KTG

Two spherical black bodies have radii ' R_1 ' and ' R_2 '. Their surface temperatures are T_1 K and T_2 K respectively. If they radiate the same power, the ratio \frac{R_1}{R_2} is

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