A) \[\frac{2{{K}_{1}}+3{{K}_{2}}}{5}\]
B) \[\frac{{{K}_{1}}+{{K}_{2}}}{2}\]
C) \[{{K}_{1}}+{{K}_{2}}\]
D) \[\frac{{{K}_{1}}+3{{K}_{2}}}{4}\]
Correct Answer: D
Solution :
| [d] We know that termal resistance \[R=\frac{I}{KA}\] |
| For inner cylinder \[{{R}_{1}}=\frac{l}{{{K}_{1}}\pi {{R}^{2}}}\] |
| For outer cylinder\[{{R}_{2}}=\frac{l}{{{K}_{2}}\pi [{{(2R)}^{2}}-{{(R)}^{2}}]}\] |
| \[{{R}_{2}}=\frac{l}{{{K}_{2}}\pi 3{{R}^{2}}}\] |
| \[{{R}_{eq}}\]is the equivalent thermal resistance of the cylinder |
| \[\frac{1}{{{R}_{eq}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] |
| \[\frac{{{K}_{eq}}4\pi {{R}^{2}}}{l}=\frac{{{K}_{l}}\pi {{R}^{2}}}{l}+\frac{3{{K}_{2}}\pi {{R}^{2}}}{l}\] |
| \[4{{K}_{eq}}={{K}_{1}}+3{{K}_{2}}\] |
| \[{{K}_{eq}}=\frac{{{K}_{1}}+3{{K}_{2}}}{4}\] |
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