Question

A star behaves like a perfectly black body emitting radiant energy. The ratio of radiant energy emitted per second by this star to that emitted by another star having 8 times the radius of former, but having temperature one-fourth of that of the former in kelvin is :

• A. 1 : 4
• B. 1 : 16
• C. 4 : 1
• D. 16 : 1

Answer 1

Answer: (C)

4 : 1

Let there be two start, so ${STAR}_1$ has following emission power $E_1$ radius $r_1$ ${STAR}_2$ has following emission power $E_2$ radius $r_2$.

Now, it is given that $r_2=8\times r_1$ and $T_2=\frac{T_1}{4}$

So, we know that radiation emitted by body is: $E=\sigma ϵA{T}^4$

$\frac{E_1}{E_2}=(\frac{r_1}{r_2}{)}^2\times (\frac{T_1}{T_2}{)}^4$

$\frac{E_1}{E_2}=({\frac{1}{8})}^2\times {\left(\frac{4\times T_1}{T_1}\right)}^4=4$