A)
B)
C)
D)
Correct Answer: D
Solution :
[d] \[-ms\frac{dT}{dt}=e\sigma A\left( {{T}^{4}}-T_{0}^{4} \right)\] |
\[-\frac{dT}{dt}=\frac{e\sigma A}{ms}\left( {{T}^{4}}-T_{0}^{4} \right)\] |
\[-\frac{dT}{dt}=\frac{4e\sigma AT_{0}^{3}}{ms}\left( \Delta T \right)\] |
\[T={{T}_{0}}+({{T}_{i}}-{{T}_{0}}){{e}^{-kt}}\] |
where\[k=\frac{4e\sigma AT_{0}^{3}}{ms}\] |
\[k=\frac{4e\sigma AT_{0}^{3}}{\rho \text{v}s}\] |
\[\left| \frac{dT}{dt} \right|\propto k\] |
\[\therefore \]\[\left| \frac{dT}{dt} \right|\propto \frac{1}{\rho s}\] |
\[{{\rho }_{A}}{{s}_{A}}=2000\times 8\times {{10}^{2}}=16\times {{10}^{5}}\] |
\[{{\rho }_{B}}{{s}_{B}}=4000\times {{10}^{3}}=4\times {{10}^{6}}\] |
\[{{\rho }_{A}}{{s}_{A}}<{{\rho }_{B}}{{s}_{B}}\] |
\[{{\left| \frac{dT}{dt} \right|}_{A}}>{{\left| \frac{dT}{dt} \right|}_{B}}\] |
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