A black metal foil is warmed by radiation fro | vedclass

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Question

As we know that emissive power of body is given as: $E=\sigma T^{4}$

and they are kept at some known distance so $E \propto \frac{1}{r^{2}}$

now

Vedclass · Accepted Answer

$4 P$

Vedclass · Answer

As we know that emissive power of body is given as: $E=\sigma T^{4}$

and they are kept at some known distance so $E \propto \frac{1}{r^{2}}$

now if the temperature and distances are $T$ and $r$ the power received is $P$ and when both temperature and distance increased twice, we get the mathematical relation

$\frac{T_{1}^{4}}{r_{1}^{2}}=\frac{T_{2}^{4}}{r_{2}^{2}}$

So, $\frac{T_{1}^{4}}{r_{1}^{2}}=P$

Then $\frac{\left(2 T_{1}\right)^{4}}{\left(2 r_{1}\right)^{2}}=4 P$

A black metal foil is warmed by radiation from a small sphere at temperature $' T'$ and at a distance $' d'$ . It is found that the power received by the foil is $P$ . If both the temperature and distance are doubled, the power received by the foil will be :

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