Q82 / 90 · JEE Main 2021 (27 Aug Shift 1) · Physics

A rod $C D$ of thermal resistance $10.0 {KW}^{- 1}$ is joined at the middle of an identical rod $A B$ as shown in figure. The ends $A, B$ and $D$ are maintained at $200 ° C, 100 ° C$ and $125 ° C$ respectively. The heat current in $C D$ is $P W$ . The value of $P$ is

Correct answer: 2

Solution

$Q_{A C} + Q_{D C} = Q_{C B}$

$\frac{K A (200 - t)}{\frac{ℓ}{2}} + \frac{K A (125 - t)}{ℓ} = \frac{(t - 100) K A}{\frac{ℓ}{2}}$

$400 - 2 t + 125 - t = 2 t - 200$

$725 = 5 t$

$t = 145 º C$

$Q_{C D} = \frac{K A (145 - 125)}{ℓ} = \frac{20}{\frac{ℓ}{K A}} = 2 W$

Hence, the correct answer is 2.