JEE PYQ: Thermodynamics Question 25
Question 25 - 2020 (06 Sep Shift 2)
Three rods of identical cross-section and lengths are made of three different materials of thermal conductivity $K_1$, $K_2$ and $K_3$, respectively. They are joined together at their ends to make a long rod (see figure). One end of the long rod is maintained at 100$°$C and the other at 0$°$C (see figure). If the joints of the rod are at 70$°$C and 20$°$C in steady state and there is no loss of energy from the surface of the rod, the correct relationship between $K_1$, $K_2$ and $K_3$ is :
(1) $K_1 : K_3 = 2 : 3$, $K_1 < K_3 = 2 : 5$
(2) $K_1 < K_2 < K_3$
(3) $K_1 : K_2 = 5 : 2$, $K_1 : K_3 = 3 : 5$
(4) $K_1 > K_2 > K_3$
Show Answer
Answer: (1)
Solution
As the rods are identical, so they have same length ($l$) and area of cross-section ($A$). They are connected in series. So, heat current will be same for all rods.
$\frac{(100 - 70)K_1 A}{l} = \frac{(70 - 20)K_2 A}{l} = \frac{(20 - 0)K_3 A}{l}$
$K_1(100 - 70) = K_2(70 - 20) = K_3(20 - 0)$
$K_1(30) = K_2(50) = K_3(20)$
$\frac{K_1}{10} = \frac{K_2}{6} = \frac{K_3}{15} \Rightarrow K_1 : K_2 : K_3 = 10 : 6 : 15$
$K_1 : K_3 = 2 : 3$
BETA
