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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
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B
1 : 16
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C
4 : 1
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D
16 : 1
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Solution

The correct option is B 4 : 1
Let there be two start, so STAR1 has following emission power E1 radius r1 STAR2 has following emission power E2 radius r2.
Now, it is given that r2=8×r1 and T2=T14
So, we know that radiation emitted by body is: E=σϵAT4
E1E2=(r1r2)2×(T1T2)4
E1E2=(18)2×(4×T1T1)4=4

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