MHT CET · Physics · Thermal Properties of Matter
If the temperature of a hot body is increased by \(50 \%\), then the increase in the quantity of emitted heat radiation will be approximately
- A \(125 \%\)
- B \(200 \%\)
- C \(300\%\)
- D \(400 \%\)
Answer & Solution
Correct Answer
(D) \(400 \%\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Given: } \mathrm{T}_2=\mathrm{T}_1+\frac{50}{100} \mathrm{~T}_1 \\ & \mathrm{~T}_2=1.5 \mathrm{~T}_1\end{aligned}\)
According to Stefan's law,
\(
\frac{\mathrm{Q}}{\mathrm{At}}=\sigma \mathrm{T}^4
\)
Percentage change in radiation is,
\(
\begin{aligned}
& \Delta \mathrm{E} \%=\frac{\mathrm{T}_2^4-\mathrm{T}_1^4}{\mathrm{~T}_1^4} \times 100 \\
& \Delta \mathrm{E} \%=\frac{(1.5)^4 \mathrm{~T}_1^4-\mathrm{T}_1^4}{\mathrm{~T}_1^4} \times 100 \\
& \Delta \mathrm{E} \% \approx 400 \%
\end{aligned}
\)
According to Stefan's law,
\(
\frac{\mathrm{Q}}{\mathrm{At}}=\sigma \mathrm{T}^4
\)
Percentage change in radiation is,
\(
\begin{aligned}
& \Delta \mathrm{E} \%=\frac{\mathrm{T}_2^4-\mathrm{T}_1^4}{\mathrm{~T}_1^4} \times 100 \\
& \Delta \mathrm{E} \%=\frac{(1.5)^4 \mathrm{~T}_1^4-\mathrm{T}_1^4}{\mathrm{~T}_1^4} \times 100 \\
& \Delta \mathrm{E} \% \approx 400 \%
\end{aligned}
\)
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