Question

A rod of length $L$ and uniform cross-sectional area has varying thermal conductivity which changes linearly from $2K$ at end $B$ to $K$ at the other end $A$. The ends $A$ and $B$ of the rod are maintained at constant temperature ${100}^{\circ }C$ and $0^{\circ }C$, respectively. At steady state, the graph of temperature : $T=T\left(x\right)$ where $x=$ distance from end $A$ will be

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$K\left(x\right)=K+\frac{Kx}{L}$
$\frac{dQ}{dt}=I=-\frac{K\left(x\right)dT}{dx}A$
$I=-(K+\frac{Kx}{L})\frac{dT}{dx}A$
$\underset{0}{\overset{x}{\int }}\frac{ILdx}{-KA(x+L)}=\underset{T_1}{\overset{T}{\int }}dT$
$\ln (x+L{)}_0^x=-(T-T_1)\times \frac{KA}{IL}$
$T=T_1-\ln \left(\frac{x+L}{L}\right)\times \frac{IL}{KA}$ ... (1)
at $x=L$ and $T=T_2$
$T_2=T_1-\ln \left(2\right)\frac{IL}{KA}$
$\frac{T_1-T_2}{\ln 2}=\frac{IL}{KA}$
Put it in equ. (1)
$T=T_1-\ln \left(\frac{x+L}{L}\right)\left(\frac{T_1-T_2}{\ln 2}\right)$
$T=A-C\ln \left(Bx\right)$