A) 1 cal/s
B) 2 cal/s
C) 8 cal/s
D) 16 cal/s
Correct Answer: A
Solution :
Rate of flow of heat is equal \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{\frac{{{K}_{1}}{{A}_{1}}({{\theta }_{1}}-{{\theta }_{2}})}{{{l}_{1}}}}{\frac{{{K}_{2}}{{A}_{1}}({{\theta }_{1}}-{{\theta }_{2}})}{{{l}_{2}}}}\] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{\frac{{{K}_{1}}{{A}_{1}}({{\theta }_{1}}-{{\theta }_{2}})}{l}}{\frac{{{K}_{2}}{{A}_{2}}({{\theta }_{1}}-{{\theta }_{2}})}{{{l}_{2}}}}\] But \[({{\theta }_{1}}-{{\theta }_{2}})\] is same. \[\therefore \] \[K>1\] \[\therefore \] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{\frac{{{K}_{1}}{{A}_{1}}}{{{l}_{1}}}}{\frac{{{K}_{2}}{{A}_{2}}}{{{l}_{2}}}}=\left( \frac{{{K}_{1}}}{{{K}_{2}}} \right)\,\,\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)\,\,\left( \frac{{{A}_{1}}}{{{A}_{2}}} \right)\] \[\therefore \] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\left( \frac{{{K}_{1}}}{{{K}_{2}}} \right)\,\,\left( \frac{{{l}_{2}}}{{{l}_{1}}} \right)\,\,{{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}=\frac{1}{2}\times \frac{2}{1}\times {{\left( \frac{1}{2} \right)}^{2}}\] \[\therefore \] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{1}{4}\] \[\Rightarrow \] \[{{H}_{1}}=\frac{{{H}_{2}}}{4}=\frac{4}{4}=1\,\,cal/s\]
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