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 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 \[\frac{C}{{{K}_{1}}}+\frac{C}{{{K}_{2}}}=\frac{2C}{K'}\] \[orC\left( \frac{{{K}_{1}}+{{K}_{2}}}{{{K}_{1}}\,{{K}_{2}}} \right)=\frac{2C}{K'}\] \[\therefore K'=K,\,{{K}_{2}}=2K\] Given, \[{{K}_{1}}=K,\,{{K}_{2}}=2K\] So, \[K'=\frac{2K\times 2K}{K+2K}=\frac{4}{3}\]
You need to login to perform this action.
You will be redirected in
3 sec
