A) \[\frac{72}{7}\frac{r\rho c}{\sigma }\]
B) \[\frac{7}{72}\frac{r\rho c}{\sigma }\]
C) \[\frac{27}{7}\frac{r\rho c}{\sigma }\]
D) \[\frac{7}{27}\frac{r\rho c}{\sigma }\]
Correct Answer: B
Solution :
\[\frac{dT}{dt}=\frac{\sigma \,A}{mcJ}\,\,({{T}^{4}}-T_{0}^{4})\] [In the given problem fall in temperature of body \[dT=(200-100)=100K\], temp. of surrounding T0 = 0K, Initial temperature of body \[T=200K].\] \[\frac{100}{dt}=\frac{\sigma 4\pi {{r}^{2}}}{\frac{4}{3}\pi {{r}^{3}}\rho \,c\,J}({{200}^{4}}-{{0}^{4}})\] Þ \[dt=\frac{r\rho \,c\,J}{48\sigma }\times {{10}^{-6}}s=\frac{r\rho \,c}{\sigma }.\frac{4.2}{48}\times {{10}^{-6}}\] \[=\frac{7}{80}\frac{r\rho \,c}{\sigma }\mu \,s\tilde{}\frac{7}{72}\frac{r\rho \,c}{\sigma }\mu \,s\] [As J = 4.2]You need to login to perform this action.
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