The rectangular surface of area 8 cm $\times$ 4cm of a black body at a temperature of $127^\circ \mathrm{C}$ emits energy at the rate of E per second. If the length and breadth of the surface are each reduced to half of the initial value and the temperature is raised to $327^\circ \mathrm{C}$ , the rate of emission of energy will become
$(a) \frac{3}{8} E (b) \frac{81}{16} E (c) \frac{9}{16} E (d) \frac{81}{64} E$
Text Solution
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$(Q)_{\text{Blackbody}} = A \sigma T^{4} t \Rightarrow \frac{Q}{t} \propto P = A \sigma T^{4}$ Breadth are halved so area becomes one fourth. $\Rightarrow \frac{P_1}{P_2} = \frac{A_1}{A_2} \times \left(\frac{T_1}{T_2}\right)^4 \Rightarrow \frac{A_1}{(A_1/4)} \times \left(\frac{273+327}{273+127}\right)$ $\Rightarrow P_2 = \frac{81}{64} E$
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