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