JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

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 °C and 40 °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) :

A
89 °C and 73 °C respectively
B
80 °C and 60 °C respectively
C
80 °C and 70 °C respectively
D
60 °C and 45 °C respectively

Solution

Use the thermal resistance analogy. For a rod of length $ℓ$ , area $A$ and conductivity $k$ : $R_{t h} = \frac{ℓ}{k A}$ .

Let $R = \frac{ℓ}{3 K A}$ be the resistance of an x-rod (since $k_{x} = 3 K$ ). Then a y-rod has resistance $3 R$ (since $k_{y} = K$ ).

Resistances along the path A→B→{C,D in parallel branches}→E→F:

$R_{A B} = R$ (x-rod), $R_{E F} = 3 R$ (y-rod).

Upper branch B→C→E (two y-rods in series): $3 R + 3 R = 6 R$ .

Lower branch B→D→E (two x-rods in series): $R + R = 2 R$ .

Parallel combination from B to E: $\frac{( 6 R ) ( 2 R )}{6 R + 2 R} = \frac{12 R ^{2}}{8 R} = \frac{3 R}{2}$ .

Total: $R + \frac{3 R}{2} + 3 R = \frac{11 R}{2}$ .

Steady-state heat current: $H = \frac{100 - 40}{11 R /2} = \frac{120}{11 R}$ .

Across $R_{A B}$ : $100 - T_{B} = H R = \frac{120}{11}$ , so $T_{B} = 100 - \frac{120}{11} \approx 89.1 d e g; C$ .

Across $R_{E F}$ : $T_{E} - 40 = H (3 R) = \frac{360}{11}$ , so $T_{E} = 40 + \frac{360}{11} \approx 72.7 d e g; C$ .

So $T_{B} \approx 89 d e g; C$ and $T_{E} \approx 73 d e g; C$ .

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