Search any question & find its solution
Question:
Answered & Verified by Expert
A perfectly black body emits a radiation at temperature ' $\mathrm{T}_1$ ' $\mathrm{K}$. If it is to radiate at 16 times this power, its temperature ' $\mathrm{T}_2$ ' $\mathrm{K}$ should be
Options:
Solution:
1758 Upvotes
Verified Answer
The correct answer is:
$2 \mathrm{~T}_1$
$$
\begin{aligned}
& \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^4 \\
& \therefore 16=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^4 \\
& \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=2
\end{aligned}
$$
\begin{aligned}
& \frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^4 \\
& \therefore 16=\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^4 \\
& \therefore \frac{\mathrm{T}_2}{\mathrm{~T}_1}=2
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.