MHT CET · Physics · Thermal Properties of Matter
Two spherical black bodies of radii ' \(r_1\) ' and ' \(r_2\) ' at temperature ' \(T_1\) ' and ' \(T_2\) ' respectively radiate power in the ratio \(1: 2\) Then \(r_1: r_2\) is
- A \(\frac{1}{2}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^4\)
- B \(\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2\)
- C \(2\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^4\)
- D \(2\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)^2\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2\)
Step-by-step Solution
Detailed explanation
Power Radiated by black body, \(\mathrm{P}=\sigma \mathrm{AT}^4\)
\(\therefore \quad\) For first black body:
\(\mathrm{P}_1=\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4\)
\(\therefore \quad\) For second black body:
\(\mathrm{P}_2=\sigma 4 \pi \mathrm{r}_2{ }^2 \mathrm{~T}_2{ }^4\)
\(\therefore \quad\) The ratio will be:
\(
\begin{aligned}
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4}{\sigma 4 \pi \mathrm{r}_2^2 \mathrm{~T}_2^4} \\
& \frac{1}{2}=\frac{\mathrm{r}_1^2 \mathrm{~T}_1^4}{\mathrm{r}_2^2 \mathrm{~T}_2^4} \\
& \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{1}{2} \frac{\mathrm{T}_2^4}{\mathrm{~T}_1^4} \\
& \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2
\end{aligned}
\)
\(\therefore \quad\) For first black body:
\(\mathrm{P}_1=\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4\)
\(\therefore \quad\) For second black body:
\(\mathrm{P}_2=\sigma 4 \pi \mathrm{r}_2{ }^2 \mathrm{~T}_2{ }^4\)
\(\therefore \quad\) The ratio will be:
\(
\begin{aligned}
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\sigma 4 \pi \mathrm{r}_1^2 \mathrm{~T}_1^4}{\sigma 4 \pi \mathrm{r}_2^2 \mathrm{~T}_2^4} \\
& \frac{1}{2}=\frac{\mathrm{r}_1^2 \mathrm{~T}_1^4}{\mathrm{r}_2^2 \mathrm{~T}_2^4} \\
& \frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{1}{2} \frac{\mathrm{T}_2^4}{\mathrm{~T}_1^4} \\
& \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{\sqrt{2}}\left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)^2
\end{aligned}
\)
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