Question

Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length 3L/2
,
is connected across AB (See figure). In steady state, temperature difference between P and Q will be close to:
(A) 45°C
(B) 75°C
(C) 60°C
(D) 35°C [JEE Main 2019]

Answer 1

The temperature difference between P and Q will be close to 45°C.

Option (A) is correct.

Explanation:

Heat Current -

IH = θ1 -θ2 / Rth

Wherein

θ1 - θ2 = Temperature difference between two ends of the conductor.  

Thermal conductor = θA - θB /Req

Thermal conductor =  θA - θB / 8R / 5

Thermal conductor = 5 x 120  / 8 x R

Thermal conductor = 75 / R

θp - θq = 75 / R . 3 R / 5

θp - θq = 45°C

Thus the temperature difference between P and Q will be close to 45°C