A) \[{{t}_{2}}:{{t}_{1}}\]
B) \[{{t}_{1}}:{{t}_{2}}\]
C) \[{{t}_{1}}^{2}:{{t}_{2}}^{2}\]
D) \[{{t}_{2}}^{2}:{{t}_{1}}^{2}\]
Correct Answer: A
Solution :
Relation between temperature gradient (TG)and thermal conductivity (K) So, \[d\frac{\theta }{dt}=-KAd\frac{\theta }{dx}\,=-KA\,\times (TG)\] i.e., \[t\propto \,\frac{1}{K}\,\,\,\,\,\,or\,\,\,\,K\propto \,\frac{1}{t}\] \[{{K}_{1}}\propto \,\frac{1}{{{t}_{1}}}\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,{{K}_{2}}\propto \,\frac{1}{{{t}_{2}}}\] From eqs. (i) and (ii), we get \[\frac{{{K}_{1}}}{{{K}_{2}}}\,=\frac{1/{{t}_{1}}}{1/{{t}_{2}}}\] \[{{K}_{1}}:{{K}_{2}}\,={{t}_{2}}:{{t}_{1}}\]
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