A) \[{{K}_{1}}+{{K}_{2}}\]
B) \[{{K}_{1}}{{d}_{1}}+{{K}_{2}}{{d}_{2}}\]
C) \[\frac{{{d}_{1}}{{K}_{1}}+{{d}_{2}}{{K}_{2}}}{{{d}_{1}}+{{d}_{2}}}\]
D) \[\frac{{{d}_{1}}+{{d}_{2}}}{({{d}_{1}}/{{K}_{1}}+{{d}_{2}}/{{K}_{2}})}\]
Correct Answer: D
Solution :
When two rods are connected in series \[Q=\frac{A({{T}_{1}}-{{T}_{2}})t}{\frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{1}}}}=\frac{A({{T}_{1}}-{{T}_{2}})t}{({{d}_{1}}+{{d}_{2}})/K}\] \[\therefore \] \[\frac{{{d}_{1}}+{{d}_{2}}}{K}=\frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}};\] \[\therefore \] \[K=\frac{({{d}_{1}}+{{d}_{2}})}{\frac{{{d}_{1}}}{{{K}_{1}}}+\frac{{{d}_{2}}}{{{K}_{2}}}}\]You need to login to perform this action.
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