A) 2 : 3
B) 3 : 5
C) 5 : 2
D) 1 : 1
Correct Answer: C
Solution :
The quantity of heat flowing in the vessel is same, hence \[Q=\frac{{{K}_{1}}\,A\,({{\theta }_{1}}-{{\theta }_{2}})\,{{t}_{1}}}{d}\] \[=\frac{{{K}_{2}}\,A\,({{\theta }_{1}}-{{\theta }_{2}})\,{{t}_{2}}}{d}\] Where \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] are temperatures of the outer and the inner walls of the two vessels. From this, we have \[{{K}_{1}}{{t}_{1}}={{K}_{2}}{{t}_{2}}\] \[\Rightarrow \] \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{25}{20}=\frac{5}{2}\]
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