The rods AB,BC & CA made of same material & have equal cross - section . So , the temperature at C = ?

$(a)\;90^{0}C\qquad(b)\;60^{0}C\qquad(c)\;30^{0}C\qquad(d)\;Data\;not\;appropriate$

Answer : Data not appropriate
Explanation :
Let temperature of B pt . be $\;T_{B}$
$\large\frac{(T_{A}-T_{B})}{l_{0}}=\large\frac{(T_{B}-T)}{2l_{0}}$
$=90-T_{B}=\large\frac{T_{B}}{\alpha}-\large\frac{T}{\alpha}$
$\large\frac{3T_{B}}{2}=90+\large\frac{T}{2}----(1)$
Also , $\;\large\frac{(T_{B}-T_{0})}{2l_{0}}=\large\frac{(T_{C}-90)}{3l_{0}}$
$\large\frac{T_{B}}{2}-\large\frac{T}{2}=\large\frac{T}{3}-30$
$\large\frac{T_{B}}{2}=\large\frac{5T}{6}-30-----(2)$
From (1) & (2) - $\;30+\large\frac{T}{6}=\large\frac{5T}{6}-30$
$T=90^{0}C$
Not possible
answered Mar 13, 2014 by