Q38 (2026)
Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points A and F are maintained at $100^{circ}\text{C}$ and $40^{circ}\text{C}$ respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points B and E are (close to):
{{IMG1}}
- $89^{circ}\text{C}$ and $73^{circ}\text{C}$ respectively
- $80^{circ}\text{C}$ and $60^{circ}\text{C}$ respectively
- $80^{circ}\text{C}$ and $70^{circ}\text{C}$ respectively
- $60^{circ}\text{C}$ and $45^{circ}\text{C}$ respectively
### Related Formula
$$H = \frac{\Delta T}{R_{\text{eq}}}, \quad R = \frac{l}{KA}$$
### Core Logic
{{SOL_IMG1}}
Let thermal resistance of rod y be $3R$ and rod x be $R$ (since $K_x = 3K_y$).
Using thermal current equations:
$$H = \frac{100 - 40}{\frac{11R}{2}}$$
$$H = \frac{100 - T_B}{R}, \quad H = \frac{T_E - 40}{3R}$$
Solving yields $T_B = 89^{circ}\text{C}$ and $T_E = 73^{circ}\text{C}$.
### Pattern Recognition
Sees: Composite thermal conduction network with series/parallel segments.
Shortcut: Calculate equivalent thermal resistances and equate heat currents across junctions.
Check: Matches option (1). ✓
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
