Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity 2K while that in the middle has thermal conductivity K. The left end of the combination is maintained at temperature 3T and the right end at T. The rods are thermally insulated from outside. In steady state, temperature at the left junction is T₁ and that at the right junction is T₂. The ratio T₁/T₂ is

The correct option is 5/3.

Step-by-step Explanation:
Let us analyze the system of three identical heat conducting rods connected in series as shown in the diagram:
- The left rod has a thermal conductivity of $2K$ and its left end is maintained at temperature $3T$.
- The middle rod has a thermal conductivity of $K$.
- The right rod has a thermal conductivity of $2K$ and its right end is maintained at temperature $T$.
- The temperature at the left junction is $T_1$ and the temperature at the right junction is $T_2$.

Since the rods are identical, they all have the same length ($L$) and cross-sectional area ($A$).
The thermal resistance ($R$) of a conducting rod is given by the formula:

$R=\frac{L}{kA}$

Let the thermal resistance of the middle rod (with conductivity $K$) be:

$R_{\mathrm{middle}}=R_2=\frac{L}{KA}=R_0$

Then, the resistances of the left and right rods (each with conductivity $2K$) are:

$R_{\mathrm{left}}=R_1=\frac{L}{2KA}=\frac{R_0}{2}$

$R_{\mathrm{right}}=R_3=\frac{L}{2KA}=\frac{R_0}{2}$

In the steady state, the rate of heat flow ($H$) through all three series-connected rods must be equal:

$H=\frac{3T-T_1}{R_1}=\frac{T_1-T_2}{R_2}=\frac{T_2-T}{R_3}$

Substituting the values of $R_1$, $R_2$, and $R_3$ in terms of $R_0$:

$\frac{3T-T_1}{R_0/2}=\frac{T_1-T_2}{R_0}=\frac{T_2-T}{R_0/2}$

Multiplying the entire relation by $R_0$ gives:

$2(3T-T_1)=T_1-T_2=2(T_2-T)$

We can establish two equations from these expressions. First, equating the left and right terms:

$2(3T-T_1)=2(T_2-T)$

Dividing by 2 on both sides:

$3T-T_1=T_2-T$

Rearranging the terms:

$T_1+T_2=4T$ — (Equation 1)

Next, equating the middle and right terms:

$T_1-T_2=2(T_2-T)$

Expanding the right-hand side:

$T_1-T_2=2T_2-2T$

Rearranging the terms:

$T_1-3T_2=-2T$ — (Equation 2)

Now, substitute $T_1=4T-T_2$ (from Equation 1) into Equation 2:

$(4T-T_2)-3T_2=-2T$

Simplify to solve for $T_2$:

$4T-4T_2=-2T$

$6T=4T_2$

$T_2=\frac{6}{4}T=\frac{3}{2}T$

Substitute the value of $T_2$ back into Equation 1 to find $T_1$:

$T_1=4T-\frac{3}{2}T=\frac{5}{2}T$

Finally, calculate the ratio of $T_1$ to $T_2$:

$\frac{T_1}{T_2}=\frac{\frac{5}{2}T}{\frac{3}{2}T}=\frac{5}{3}$