JEE Main 2022 · 27 Jul · Shift 1 · Physics Q37

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PhysicsSingle correct+4 / -1Question 37 of 90

If $K_1$ 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 $\dfrac{K_2}{K_1}=9$ , $\dfrac{A_1}{A_2}=2$ , $\dfrac{L_1}{L_2}=2$ . 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 :

A steel rod from a furnace at 450 °C joined at temperature T to a thinner copper rod leading to an ice box at 0 °C, both surrounded by insulating material.

18°C

14°C

45°C

150°C

Answer and worked solution

Official answer: C

Thermal resistances $R=\frac L{KA}$ : $\frac{R_s}{R_c}=\frac{L_1}{L_2}\cdot\frac{K_2}{K_1}\cdot\frac{A_2}{A_1}=2\times9\times\frac12=9$ .

Junction. $T=450\times\frac{R_c}{R_s+R_c}=45$ °C.

(C)

Topics:Conduction

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JEE Main 2022 · 27 Jul · Shift 1, Physics question 37