MCQ
A black body is heated from ${27^o}C$ to ${127^o}C$. The ratio of their energies of radiations emitted will be
- A$3:4$
- B$9:16$
- C$27 : 64$
- ✓$81: 256$
✓
Answer
Correct option: D.
$81: 256$
d
(d) $\frac{{{Q_1}}}{{{Q_2}}} = \frac{{T_1^4}}{{T_2^4}} = {\left( {\frac{{273 + 27}}{{273 + 127}}} \right)^4} = {\left( {\frac{{300}}{{400}}} \right)^4} = \frac{{81}}{{256}}$
(d) $\frac{{{Q_1}}}{{{Q_2}}} = \frac{{T_1^4}}{{T_2^4}} = {\left( {\frac{{273 + 27}}{{273 + 127}}} \right)^4} = {\left( {\frac{{300}}{{400}}} \right)^4} = \frac{{81}}{{256}}$
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