Question

Three rods of same dimensions have thermal conductivities \(3K\), \(2K\) and \(K\). They are arranged as shown in the figure. The ends are maintained at \(100^\circ C\), \(50^\circ C\) and \(0^\circ C\). Then, the temperature of the junction in steady state is

• A. \(\dfrac{200}{3}^\circ C\)
• B. \(\dfrac{100}{3}^\circ C\)
• C. \(75^\circ C\)
• D. \(\dfrac{50}{3}^\circ C\)

Hint:

At steady state junction: \(\sum k(T_{hot}-T)=\sum k(T-T_{cold})\). Use proportional form when rods have same length and area.

Answer 1

The Correct Option is A

Step 1: Let junction temperature be \(T\).
At steady state, net heat flow into junction = net heat flow out.
Step 2: Use heat current formula.
\[ H = \frac{kA}{L}(T_1 - T) \]
Since rods have same \(A\) and \(L\),
\[ H \propto k(T_1 - T) \]
Step 3: Write heat currents.
From \(100^\circ C\) through \(3K\):
\[ H_1 = 3K(100 - T) \]
From \(50^\circ C\) through \(2K\):
\[ H_2 = 2K(50 - T) \]
To \(0^\circ C\) through \(K\):
\[ H_3 = K(T - 0) = KT \]
Step 4: Apply steady state condition.
Heat entering = heat leaving:
\[ H_1 + H_2 = H_3 \]
\[ 3(100 - T) + 2(50 - T) = T \]
\[ 300 - 3T + 100 - 2T = T \]
\[ 400 - 5T = T \Rightarrow 400 = 6T \Rightarrow T = \frac{400}{6} = \frac{200}{3} \]
Final Answer: \[ \boxed{\dfrac{200}{3}^\circ C} \]