Question

Two rods \(A\) and \(B\) of equal dimensions having thermal conductivities \(K\) and \(2K\) respectively are joined in series. Under steady-state conditions, if the temperature difference between the rods \(A\) and \(B\) is \(12^\circ\mathrm{C}\), then the temperature difference between the open ends of the rods \(A\) and \(B\) is

• A. \(36^\circ\mathrm{C}\)
• B. \(6^\circ\mathrm{C}\)
• C. \(18^\circ\mathrm{C}\)
• D. \(24^\circ\mathrm{C}\)

Hint:

For rods connected in series, \[ \boxed{ \Delta T \propto R=\frac{L}{KA} } \] where \(R\) is the thermal resistance.

Answer 1

The Correct Option is C

Step 1: Relate temperature drops across the rods. In steady state, \[ \frac{\Delta T_A}{\Delta T_B} = \frac{R_A}{R_B}. \] Since the rods have equal length and cross-sectional area, \[ R=\frac{L}{KA}. \] Therefore, \[ R_A:R_B = \frac1K:\frac1{2K} = 2:1. \] Hence, \[ \Delta T_A:\Delta T_B = 2:1. \]

Step 2: Use the given information. The temperature difference across rod \(A\) is \[ 12^\circ\mathrm{C}. \] Thus, \[ \Delta T_B = 6^\circ\mathrm{C}. \]

Step 3: Find the total temperature difference. The temperature difference between the two open ends is \[ \Delta T = \Delta T_A+\Delta T_B = 12+6 = 18^\circ\mathrm{C}. \] Hence, \[ \boxed{18^\circ\mathrm{C}} \] Therefore, \[ \boxed{(C)} \] is the correct answer.