A) 3K
B) \[\frac{4}{3}K\]
C) \[\frac{2}{3}K\]
D) \[\sqrt{2}K\]
Correct Answer: B
Solution :
The quantity of heat flowing through a slab in time t, \[Q=\frac{KA\Delta \theta }{l}\] For same heat flow through each slab and composite slab, we have \[\frac{{{K}_{1}}A(\Delta {{\theta }_{1}})}{l}=\frac{{{K}_{2}}A(\Delta {{\theta }_{2}})}{l}\] \[=\frac{K'A(\Delta {{\theta }_{1}}+\Delta {{\theta }_{2}})}{2l}\] Or \[{{K}_{1}}\Delta {{\theta }_{1}}={{K}_{2}}\Delta {{\theta }_{2}}\] \[=\frac{K'}{2}(\Delta {{\theta }_{1}}+{{\theta }_{2}})=C\] (say) So, \[=\Delta {{\theta }_{1}}=\frac{C}{{{K}_{1}}}\Delta {{\theta }_{2}}=\frac{C}{{{K}_{2}}}\] And \[(\Delta {{\theta }_{1}}+\Delta {{\theta }_{2}})=\frac{2C}{K'}\] Or \[\frac{C}{{{K}_{1}}}+\frac{C}{{{K}_{2}}}=\frac{2C}{K'}\] Or \[C\left( \frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}{{K}_{2}}} \right)=\frac{2C}{K'}\] \[\therefore \] \[K'=\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}\] Given, \[{{K}_{1}}=K,{{K}_{2}}=2K\] So \[K'=\frac{2K\times 2K}{K+2K}=\frac{4}{3}\]
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