question_answer

One end of a thermally insulated rod is kept at a temperature\[m{{y}^{2}}+(1-{{m}^{2}})xy-m{{x}^{2}}=0\]and the other at\[xy=0,\]. The rod is composed of two sections of lengths\[-\frac{1}{2}\]and\[-2\] and thermal conductivities \[{{K}_{1}}\] and \[{{K}_{2}}\] respectively. The temperature at the interface of the two sections is       AIEEE  Solved  Paper-2007

A)  \[f(x)=\int_{1}^{x}{\frac{\log \,t}{1+t}}dt\]

B)  \[\frac{1}{2}\]

C)  \[f:R\to R\]

D)  \[f(x)=\]

Correct Answer: C

Solution :

Let temperature at the interface be T. For part AB, Rate of heat transmission, \[\pm 1\] For part BC, Rate of heat transmission, \[F(x)=f(x)+f\left( \frac{1}{x} \right),\] Here, A is area of cross-sections. At equilibrium, \[f(x)=\int_{1}^{x}{\frac{\log \,t}{1+t}}dt\] \[\frac{1}{2}\] \[f:R\to R\] \[f(x)=\] \[min\{x+1,|~x|+1\}\]