A) \[\frac{A{{k}_{0}}}{bl}\left[ \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right]\]
B) \[\frac{A{{k}_{0}}}{bl}\left[ \frac{a+b{{T}_{2}}}{a+b{{T}_{1}}} \right]\]
C) \[\frac{A{{k}_{0}}}{bl}\ln \left[ \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right]\]
D) \[\frac{A{{k}_{0}}}{al}\ln \left[ \frac{a+b{{T}_{2}}}{a+b{{T}_{1}}} \right]\]
Correct Answer: C
Solution :
| [c] Idea Rate of flow of heat is given by \[\frac{dQ}{dt}=-KA\frac{dT}{dx}\]. |
| As we know that, \[\frac{dQ}{dt}=-KA\frac{dT}{dx}\] |
| \[\frac{dQ}{dt}=-\frac{{{k}_{0}}A}{a+bT}\frac{dT}{dx}\] |
| On integrating both sides within the proper limits. |
| \[\frac{dQ}{dt}\int_{0}^{l}{dx}=-{{k}_{0}}A\int_{{{T}_{1}}}^{{{T}_{2}}}{\frac{dT}{dx}}\] |
| This gives \[\frac{dQ}{dt}=\frac{A{{k}_{0}}}{bl}\ln \left[ \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right]\] |
| TEST Edge Question related to equivalent thermal conductivity of two or more rods in series and parallel at various temperature can be asked. In series equivalent conductivity is given by \[{{K}_{eq}}=\frac{{{K}_{1}}{{K}_{2}}({{L}_{1}}+{{L}_{2}})}{({{L}_{1}}{{K}_{1}}+{{L}_{2}}{{K}_{2}})}\] |
| In parallel equivalent conductivity is given by \[{{K}_{eq}}=\left( \frac{{{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}})}{{{A}_{1}}+{{A}_{2}}} \right)\] |
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