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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A copper rod AB of length 100 cm is fixed with a steel rod BC of length 20 cm at B . If the ends A and C are maintained at temperature $\;100^{0}C\;$ and $\;0^{0}C\;$ , then the temperature of the function B is : [thermal conductivities of copper and steel are 0.9 and 0.18 CGS units respectively ]

$(a)\;40^{0}C\qquad(b)\;60^{0}C\qquad(c)\;80^{0}C\qquad(d)\;50^{0}C$

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Answer : $\;50^{0}C$
Explanation :
Rate of heat flow is the same
$\large\frac{K_{1}A(T_{1}-\theta)}{L_{1}} = \large\frac{K_{2}A (\theta - T_{2})}{L_{2}}$
$\large\frac{0.9 A (100 - theta)}{100} = \large\frac{0.18 A (\theta - 0^{0})}{20}$
$90 - 0.9 \theta = 0.9 \theta$
$1.8 \theta = 90$
$\theta = \large\frac{90}{1.8} = 50^{0}C$
answered Apr 8, 2014 by yamini.v
 

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