The temperature $\theta$ at the junction of two insulating sheets, having thermal resistances $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ as well as top and bottom temperatures $\theta_{1}$ and $\theta_{2}$ (as shown in figure) is given by :

(A) $\frac{\theta_{1} R_{2}+\theta_{2} R_{1}}{R_{1}+R_{2}}$

(B) $\frac{\theta_{1} R_{1}+\theta_{2} R_{2}}{R_{1}+R_{2}}$

(C) $\frac{\theta_{2} R_{2}-\theta_{1} R_{1}}{R_{2}-R_{1}}$

(D) $\frac{\theta_{1} R_{2}-\theta_{2} R_{1}}{R_{2}-R_{1}}$

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Heat & Thermodynamics

The temperature

θ

at the junction of two insulating sheets, having thermal resistances

R_{1}

and

R_{2}

as well as top and bottom temperatures

θ_{1}

and

θ_{2}

(as shown in figure) is given by :

Answer: (A)

In steady state, the heat current through both resistances is the same:

\frac{θ _{1} - θ}{R _{1}} = \frac{θ - θ _{2}}{R _{2}}

. Cross-multiplying:

R_{2} (θ_{1} - θ) = R_{1} (θ - θ_{2})

, which gives

R_{2} θ_{1} - R_{2} θ = R_{1} θ - R_{1} θ_{2}

. Solving for

θ

θ (R_{1} + R_{2}) = R_{2} θ_{1} + R_{1} θ_{2}

, so

θ = \frac{θ _{1} R _{2} + θ _{2} R _{1}}{R _{1} + R _{2}}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option