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QJEEM19Phe3b2: JEE Main 2019 Physics - Class 11 JEEM19Phe3b2 Solution: Two materials having coefficients of thermal conductivity '3K'…

Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ' θ\theta 2' and ' θ\theta 1' respectively, ( θ\theta 2 > θ\theta 1). The temperature at the interface is :-

JEE Main 2019 (Online) 9th April Evening Slot Physics - Heat and Thermodynamics Question 310 English

Multiple Choice Options

  • A θ110+9θ210{{{\theta _1}} \over {10}} + {{9{\theta _2}} \over {10}}
  • B θ2+θ12{{{\theta _2} + {\theta _1}} \over 2}
  • C θ16+5θ26{{{\theta _1}} \over {6}} + {{5{\theta _2}} \over {6}}
  • D θ13+2θ23{{{\theta _1}} \over {3}} + {{2{\theta _2}} \over {3}}

Step-by-Step Verified Solution

Let interface temperature in steady state conduction is θ\theta , then assuming no heat loss through sides;

( Rate of heat  flow through  first slab )=( Rate of heat  flow through  second slab )\left(\begin{array}{l}\text { Rate of heat } \\\text { flow through } \\\text { first slab }\end{array}\right)=\left(\begin{array}{l}\text { Rate of heat } \\\text { flow through } \\\text { second slab }\end{array}\right)

amp;⇒(3K)A(θ2−θ)d=KA(θ−θ1)3damp;⇒9(θ2−θ)=θ−θ1amp;⇒9θ2+θ1=10θamp;⇒θ=910θ2+110θ1\begin{aligned}& \Rightarrow \frac{(3 K) A\left(\theta_2-\theta\right)}{d}=\frac{K A\left(\theta-\theta_1\right)}{3 d} \\\\& \Rightarrow \quad 9\left(\theta_2-\theta\right)=\theta-\theta_1 \\\\& \Rightarrow \quad 9 \theta_2+\theta_1=10 \theta \\\\& \Rightarrow \quad \theta=\frac{9}{10} \theta_2+\frac{1}{10} \theta_1\end{aligned}

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