JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Heat & Thermodynamics

K_{1}

and

K_{2}

are the thermal conductivities,

L_{1}

and

L_{2}

are the lengths and

A_{1}

and

A_{2}

are the cross sectional areas of steel and copper rods respectively such that

\frac{K _{2}}{K _{1}} = 9, \frac{A _{1}}{A _{2}} = 2, \frac{L _{1}}{L _{2}} = 2

. Then, for the arrangement as shown in the figure, the value of temperature T of the steel-copper junction in the steady state will be :

Answer: (C)

Solution

In steady state, the heat flow rate through both rods is equal:

\frac{K _{1} A _{1}}{L _{1}} (450 - T) = \frac{K _{2} A _{2}}{L _{2}} (T - 0)

. Rearranging:

\frac{450 - T}{T} = \frac{K _{2} A _{2} L _{1}}{K _{1} A _{1} L _{2}} = 9 \times \frac{1}{2} \times 2 = 9

. Thus

450 - T = 9 T

, giving

T = 4 5^{\circ} C

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option