JEE PYQ: Thermodynamics Question 38
Question 38 - 2019 (09 Jan Shift 1)
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 $\frac{3L}{2}$, is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to:
(1) 45$°$C
(2) 75$°$C
(3) 60$°$C
(4) 35$°$C
Show Answer
Answer: (1)
Solution
$\frac{\Delta T_{AB}}{R_{AB}} = \frac{120}{8R} = \frac{120 \times 5}{8R}$
In steady state temperature difference between P and Q,
$\Delta T_{PQ} = \frac{120 \times 5}{8R} \times \frac{3}{5} \times R = \frac{360}{8} = 45°$C
BETA
