A) \[116{}^\circ C,74{}^\circ C\]
B) \[120{}^\circ C,180{}^\circ C\]
C) \[125{}^\circ C,50{}^\circ C~\]
D) \[130{}^\circ C,40{}^\circ C\]
Correct Answer: A
Solution :
Suppose \[{{\theta }_{1}}\]and \[{{\theta }_{2}}\]be the temperature of junctions \[B\]and\[C\], respectively. |
In the steady state, the rate of flow of heat through each bar will be same. |
\[\frac{Q}{t}=\frac{2K\times A\left( 200-{{\theta }_{1}} \right)}{X}\]\[=\frac{4K\times A({{\theta }_{1}}-{{\theta }_{2}})}{x}\] |
\[=\frac{3K\times A\left( {{\theta }_{2}}-18 \right)}{X}\] \[2\left( 200-{{\theta }_{1}} \right)=4\left( {{\theta }_{1}}-{{\theta }_{2}} \right)=3\left( {{\theta }_{2}}-18 \right)\] \[200-{{\theta }_{1}}=2{{\theta }_{1}}-2{{\theta }_{2}}\operatorname{and}4{{\theta }_{1}}-4{{\theta }_{2}}=3{{\theta }_{2}}-54\] |
\[\Rightarrow 3{{\theta }_{1}}-2{{\theta }_{2}}=200\,\operatorname{and}4{{\theta }_{1}}-7{{\theta }_{2}}=-54\]\[\Rightarrow \left( -8+21 \right){{\theta }_{2}}=\left( 800+162 \right)\]\[\Rightarrow {{\theta }_{2}}=\frac{962}{13}=74{}^\circ \operatorname{C}\] |
\[{{\theta }_{1}}=\frac{200+2\times 74}{3}=116{}^\circ \operatorname{C}\] |
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