Radiated energy at TK temperature is E for&nb | vedclass

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Question

$E=\sigma A_{1} T^{4}$

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

Vedclass · Accepted Answer

$E$

Vedclass · Answer

$E=\sigma A_{1} T^{4}$

Surface area $\mathrm{A}_{1}=\pi \mathrm{r}_{1}^{2}=\pi\left(\frac{\mathrm{d}_{1}}{2}ight)^{2}$

if diameter is $=\frac{d_{1}}{4}$

surface area $\mathrm{A}_{2}=\pi\left(\frac{\mathrm{d}_{1}}{2 	imes 4}ight)^{2}=\frac{\mathrm{A}_{1}}{16}$

$\mathrm{E}_{2}=\sigma \mathrm{A}_{2} \mathrm{T}_{2}^{4}=\sigma \frac{\mathrm{A}_{1}}{16}(2 \mathrm{T})^{4}=\sigma \mathrm{A}_{1} \mathrm{T}^{4}=\mathrm{E}$

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 :-

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