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10-2.Transmission of Heat
normal
Radiated energy at $TK$ temperature is $E$ for a body of diameter $'d'$. If temperature becomes $(2T)$ and diameter becomes $\frac{d}{4}$ then radiated energy will be :-
A
$4\, E$
B
$16\, E$
C
$E$
D
$E/16$
Solution
$E=\sigma A_{1} T^{4}$
Surface area $\mathrm{A}_{1}=\pi \mathrm{r}_{1}^{2}=\pi\left(\frac{\mathrm{d}_{1}}{2}\right)^{2}$
if diameter is $=\frac{d_{1}}{4}$
surface area $\mathrm{A}_{2}=\pi\left(\frac{\mathrm{d}_{1}}{2 \times 4}\right)^{2}=\frac{\mathrm{A}_{1}}{16}$
$\mathrm{E}_{2}=\sigma \mathrm{A}_{2} \mathrm{T}_{2}^{4}=\sigma \frac{\mathrm{A}_{1}}{16}(2 \mathrm{T})^{4}=\sigma \mathrm{A}_{1} \mathrm{T}^{4}=\mathrm{E}$
Standard 11
Physics
