question_answer

The figure shows a system of two concentric spheres of radii \[{{r}_{1}}\] and \[{{r}_{2}}\] are kept at temperatures \[{{T}_{1}}\]and \[{{T}_{2}}\], respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to

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         \[\frac{dQ}{dt}=\frac{-4\pi K{{r}_{1}}{{r}_{2}}({{T}_{2}}-{{T}_{1}})}{({{r}_{2}}-{{r}_{1}})}\]

\[\therefore \,\,\frac{dQ}{dt}\propto \frac{{{r}_{1}}\,{{r}_{2}}}{({{r}_{2}}-{{r}_{1}})}\]