A) \[3\,K\]
B) \[\frac{4}{3}K\]
C) \[\frac{2}{3}K\]
D) \[\sqrt{2}\,K\]
Correct Answer: B
Solution :
| The quantity of heat flowing across 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 \[{{K}_{1}}\Delta \,{{\theta }_{1}}={{K}_{2}}\Delta {{\theta }_{2}}=\frac{K'}{2}\,(\Delta {{\theta }_{1}}+\Delta {{\theta }_{2}})=C\,\,\,(say)\] |
| 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 \[C\left( \frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}\,{{K}_{2}}} \right)=\frac{2C}{K'}\] |
| \[\therefore \] \[K'=K,\,{{K}_{2}}=2K\] |
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