Question

Three rods of same length and area of cross-section but of thermal conductivities $K,2K$ and $3K$ are connected in series in the same order. Free end of the first rod is at $0^{\circ }\text{C}$ and free end of third rod is at ${55}^{\circ }\text{C}$. In steady state, temperature difference across the middle rod is

• A. ${15}^{\circ }\text{C}$
• B. ${30}^{\circ }\text{C}$
• C. ${10}^{\circ }\text{C}$
• D. ${25}^{\circ }\text{C}$

Answer 1

The correct option is A ${15}^{\circ }\text{C}$


Given that,
Let the length and area of each rod is $l$ and $A$ respectively and temperature of the junction is $t_1$ and $t_2$.
Rate of heat transfer by first rod = Rate of heat transfer by second rod = Rate of heat transfer by third rod.
$H_1=H_2=H_3$
$\Rightarrow \frac{t_1-0}{\frac{l}{AK}}=\frac{t_2-t_1}{\frac{l}{2KA}}=\frac{{55}^{\circ }\text{C}-t_2}{\frac{l}{3KA}}$

$\Rightarrow 3K(55-t_2)=2K(t_2-t_1)=K{t}_1$
$\Rightarrow 3(55-t_2)=2(t_2-t_1)=t_1\dots \dots \left(1\right)$

From equation ($1$)
$2t_2-2t_1=t_1$
$t_2=\frac{3}{2}{t}_1\dots \dots \left(2\right)$

Again from equation ($1$)
$3(55-t_2)=t_1$
$\Rightarrow 165-3t_2=t_1$
$\Rightarrow 165=t_1+3t_2$
$\Rightarrow 165=t_1+(\frac{3}{2}\times 3)t_1$
$\Rightarrow 330=11t_1$
$\Rightarrow t_1={30}^{\circ }\text{C}$
Now
$t_2-t_1=\frac{t_1}{2}\dots \dots \left(3\right)$
$\therefore t_2-t_1=\frac{30}{2}={15}^{\circ }\text{C}$
Hence, option (s) is correct.

(OR)

Given that,
Let the length and area of each rod is $l$ and $A$ respectively and temperature of the junction is $t_1$ and $t_2$.
Thermal resistance of first rod is $R_1=\frac{l}{KA}$
Thermal resistance of second rod is $R_2=\frac{l}{2KA}$
Thermal resistance of third rod is $R_3=\frac{l}{3KA}$

As these rods are in series heat flow through them can be considered as
$H=\frac{\Delta T}{R_1+R_2+R_3}=\frac{55-0}{\frac{l}{KA}+\frac{l}{2KA}+\frac{l}{3KA}}....i$
Heat transfer through second rod is $H=\frac{\Delta T_2}{R_2}=\frac{\Delta T_2}{\frac{l}{2KA}}....ii$
Equating $i\&ii$
$\frac{55-0}{\frac{l}{KA}+\frac{l}{2KA}+\frac{l}{3KA}}=\frac{\Delta T_2}{\frac{l}{2KA}}$

$\frac{55-0}{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}}=\frac{\Delta T_2}{\frac{1}{2}}$
$\Delta T_2=55\times \frac{3}{11}={15}^{\circ }$