In steady state, heat gained by middle plate $=$ heat lost by it.
$\Rightarrow$ Heat radiations from plate $A+$ Heat radiations from plate $B=$ Heat radiations emitted from both surfaces
$\Rightarrow A \varepsilon \sigma T^{4}+A \varepsilon \sigma(2 T)^{4}=2 A \varepsilon \sigma\left(T_{1}\right)^{4}$
$\Rightarrow \quad T^{4}+2^{4} T^{4}=2 T_{1}^{4} \Rightarrow \frac{17}{2} T^{4}=T_{1}^{4}$
$\Rightarrow \quad T_{1}^{4}=\left(\frac{17}{2}\right)^{\frac{1}{4}} \cdot T \text { or } T_{1}=1.7 T$
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$1.$ If the radius of the opening of the dropper is $\mathrm{r}$, the vertical force due to the surface tension on the drop of radius $R$ (assuming $\mathrm{r} \ll R$ ) is
$(A)$ $2 \pi r T$ $(B)$ $2 \pi R T$ $(C)$ $\frac{2 \pi r^2 T}{R}$ $(D)$ $\frac{2 \pi R^2 T}{r}$
$2.$ If $\mathrm{r}=5 \times 10^{-4} \mathrm{~m}, \rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2, \mathrm{~T}=0.11 \mathrm{Nm}^{-1}$, the radius of the drop when it detaches from the dropper is approximately
$(A)$ $1.4 \times 10^{-3} \mathrm{~m}$ $(B)$ $3.3 \times 10^{-3} \mathrm{~m}$
$(C)$ $2.0 \times 10^{-3} \mathrm{~m}$ $(D)$ $4.1 \times 10^{-3} \mathrm{~m}$
$3.$ After the drop detaches, its surface energy is
$(A)$ $1.4 \times 10^{-6} \mathrm{~J}$ $(B)$ $2.7 \times 10^{-6} \mathrm{~J}$
$(C)$ $5.4 \times 10^{-6} \mathrm{~J}$ $(D)$ $8.1 \times 10^{-6} \mathrm{~J}$
Give the answer question $1,2$ and $3.$