A) \[90{}^\circ C\]
B) \[85{}^\circ C\]
C) \[70{}^\circ C\]
D) \[50{}^\circ C\]
Correct Answer: C
Solution :
Key Idea: In the steady state, the rate of flow of heat in both the conductors will be the same. Taking the two conductors with same cross-section area A, joined in series, with same length d Let H be heat flow through this combination and \[\theta \] is temperature of common surface. In steady state rate of flow of heat is same hence, \[H=\frac{Q}{t}=\frac{{{K}_{1}}A({{\theta }_{1}}-\theta )}{d}=\frac{{{K}_{2}}A(\theta -{{\theta }_{2}})}{d}\] Given \[{{\theta }_{1}}=100{{\,}^{o}}C,\,{{\theta }_{2}}={{20}^{o}}C\] and \[{{K}_{1}}:{{K}_{2}}=\frac{5}{3}\] \[\therefore \] \[{{K}_{1}}({{100}^{o}}-\theta )={{K}_{2}}(\theta -{{20}^{o}})\] \[\Rightarrow \] \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{\theta -{{20}^{o}}}{{{100}^{o}}-\theta }\] \[\Rightarrow \] \[\frac{5}{3}=\frac{\theta -{{20}^{o}}}{{{100}^{o}}-\theta }\] \[\Rightarrow \] \[8\theta ={{560}^{o}}\] \[\Rightarrow \] \[\theta ={{70}^{o}}\]You need to login to perform this action.
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