A) \[In\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)\]
B) \[\frac{({{r}_{2}}-{{r}_{1}})}{({{r}_{1}}\,{{r}_{2}})}\]
C) \[({{r}_{2}}-{{r}_{1}})\]
D) \[\frac{{{r}_{1}}\,{{r}_{2}}}{({{r}_{2}}-{{r}_{1}})}\]
Correct Answer: D
Solution :
Consider a shell of thickness (dr) and of radii (r) and the temperature of inner and outer surfaces of this shell be \[T,\,(T-dT)\] |
\[\frac{dQ}{dt}=\] rate of flow of heat through it g \[=\frac{KA[(T-dT)-T]}{dr}=\frac{-KAdT}{dr}\]\[=-4\pi K{{r}^{2}}\frac{dT}{dr}\] \[(\because \,A=4\pi {{r}^{2}})\] |
To measure the radial rate of heat flow, integration technique is used, since the area of the surface through which heat will flow is not constant. |
Then, \[\left( \frac{dQ}{dt} \right)\int\limits_{{{r}_{1}}}^{{{r}_{2}}}{\frac{1}{{{r}^{2}}}}\,dr=-4\pi K\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{dT}\] |
\[\frac{dQ}{dt}\,\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]=-4\pi K\,\left[ {{T}_{2}}-{{T}_{1}} \right]\] |
or \[\frac{dQ}{dt}=\frac{-4\pi K{{r}_{1}}{{r}_{2}}({{T}_{2}}-{{T}_{1}})}{({{r}_{2}}-{{r}_{1}})}\] |
\[\therefore \,\,\frac{dQ}{dt}\propto \frac{{{r}_{1}}\,{{r}_{2}}}{({{r}_{2}}-{{r}_{1}})}\] |
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