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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A cylinder of radius r and thermal conductivity $\;K_{1}\;$ is surrounded by cylindrical shell of inner radius r and outer radius 2r and thermal conductivity $\;K_{2}\;$ . The effective thermal conductivity when ends of combined system $\;t_{1}\;$ and $\;t_{2}\;$ under steady state is

$(a)\;(\large\frac{K_{2}+3K_{1}}{4})\qquad(b)\;(\large\frac{3K_{2}+K_{1}}{4})\qquad(c)\;\large\frac{K_{1}}{4}\qquad(d)\;\large\frac{K_{2}}{4}$

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Answer : $\;(\large\frac{K_{1}+3K_{2}}{4})$
Explanation :
$\large\frac{dQ}{dt} = \large\frac{K_{1} (\pi r^2) (t_{1}-t_{2}) }{L} + \large\frac{K_{2} (\pi (2r)^2 - \pi r^2)(t_{1} - t_{2})}{L}$
$= \large\frac{K_{1} (\pi r^2) (t_{1}-t_{2}) }{L} + \large\frac{3K_{2} \pi r^2 (t_{1} - t_{2}) }{L}$
$=\large\frac{(K_{1}+3K_{2} )\pi r^2 (t_{1}-t_{2}) }{L}$
$K_{eff} = \; $ Thermal conductivity of combined system
Then ,
$\large\frac{dQ}{dt} = \large\frac{K_{eff}(\pi (2r)^2 (t_{1} - t_{2}) )}{L}$
Therefore , $\;K_{eff} = \large\frac{K_{1} + 3K_{2}}{4}$
answered Mar 31, 2014 by yamini.v
 

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