A) 16 Q
B) 4Q
C) 8Q
D) 4.5 Q
Correct Answer: B
Solution :
By Stefens law \[=\frac{3\times 2\times 5000\times {{10}^{-10}}}{0.2\times {{10}^{-3}}}\] The rate of dissipation of heat by a body emissivity e is given by \[=\frac{6\times 5\times {{10}^{-7}}}{2\times {{10}^{-4}}}\] Here \[=15mm=1.5cm\] and \[C=\frac{{{\varepsilon }_{0}}A}{d}\] \[=\frac{8.854\times {{10}^{-12}}\times 5\times {{10}^{-4}}}{0.885\times {{10}^{-2}}}=0.5pF\] \[=8C=8\times 0.5=4pF\] Q = 0.25 x o(2 T)4
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