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 following through a slab in time \[t,\] \[\theta =\frac{KA\Delta \theta }{l}\] For same heat flow through each slab and composite slab, we have \[\frac{{{K}_{1}}A(\Delta {{\theta }_{1}})}{l}=\frac{{{K}_{2}}A(\Delta {{\theta }_{2}})}{l}\] \[=\frac{KA(\Delta {{\theta }_{1}}+\Delta {{\theta }_{2}})}{2l}\] Or \[{{K}_{1}}\Delta {{\theta }_{1}}={{K}_{2}}\Delta {{\theta }_{2}}\] \[=\frac{K}{2}(\Delta {{\theta }_{1}}+\Delta {{\theta }_{2}})=C\] 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 \[\frac{C}{{{K}_{1}}}+\frac{C}{{{K}_{2}}}=-\frac{2C}{K}\] \[\therefore \] \[K=\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}\] Given, \[{{K}_{1}}=K,\]\[{{K}_{2}}=2K\] So \[K=\frac{2K\times 2K}{K+2K}=\frac{4}{3}\]
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