Two materials having coefficients of thermal conductivity $3K$ and $K$ and thickness $d$ and $3\text{d}$ respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ${\theta }_{2}$ and ${\theta }_{1}$ respectively, $({\theta }_{2}>{\theta }_{1}).$ The temperature at the interface is ![JEE Main 2019 Physics, Thermal Properties of Matter — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Thermal_Properties_of_Matter/648b5a66417cc3fb48d664d1/question_1__q_648b5a66417cc3fb48d664d1__cdn-question-pool.getmarks.app__06babf8a-a101-4a42-bfbc-b1ded828777e-image6__20ee6bb8f4_final_ppt_sync.png)

![JEE Main 2019 Physics, Thermal Properties of Matter — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Thermal_Properties_of_Matter/648b5a66417cc3fb48d664d1/solution_1__q_648b5a66417cc3fb48d664d1__cdn-question-pool.getmarks.app__8d810774-fb4a-49e3-8761-94a69a79128f-image7__c5d2f8e605_final_ppt_sync.png) Let the temperature of the junction $T ^{\circ}C$ . Rate of heat flow in Rod $1$ = rate of heat flow in Rod $2$ $\frac{3\mathrm{kA}}{d}({\theta }_{2}-T)=\frac{\mathrm{kA}}{3d}(T-{\theta }_{1})$ $\Rightarrow 9({\theta }_{2}-T)=(T-{\theta }_{1})$ $\Rightarrow 10T=9{\theta }_{2}+{\theta }_{1}$ $\Rightarrow T=\frac{9{\theta }_{2}+{\theta }_{1}}{10}=\frac{{\theta }_{1}}{10}+\frac{9{\theta }_{2}}{10}$

JEE Main 2019 — Physics Thermodynamics

2019mcqmedium

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

JEE Main 2019 Physics, Thermal Properties of Matter — question figure

Official previous-year question

Held on 9 Apr 2019 · Verified 6 Jul 2026.

Options

A
$\frac{{\theta }_{2}+{\theta }_{1}}{2}$
B
$\frac{{\theta }_{1}}{6}+\frac{5{\theta }_{2}}{6}$
C
$\frac{{\theta }_{1}}{3}+\frac{2{\theta }_{2}}{3}$
D
$\frac{{\theta }_{1}}{10}+\frac{9{\theta }_{2}}{10}$

Did you get this right?

Solution

JEE Main 2019 Physics, Thermal Properties of Matter — solution figure

Let the temperature of the junction $T ^{\circ}C$ .

Rate of heat flow in Rod $1$ = rate of heat flow in Rod $2$

$\frac{3\mathrm{kA}}{d}({\theta }_{2}-T)=\frac{\mathrm{kA}}{3d}(T-{\theta }_{1})$

$\Rightarrow 9({\theta }_{2}-T)=(T-{\theta }_{1})$

$\Rightarrow 10T=9{\theta }_{2}+{\theta }_{1}$

$\Rightarrow T=\frac{9{\theta }_{2}+{\theta }_{1}}{10}=\frac{{\theta }_{1}}{10}+\frac{9{\theta }_{2}}{10}$

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Options

$\frac{{\theta }_{2}+{\theta }_{1}}{2}$

$\frac{{\theta }_{1}}{6}+\frac{5{\theta }_{2}}{6}$

$\frac{{\theta }_{1}}{3}+\frac{2{\theta }_{2}}{3}$

$\frac{{\theta }_{1}}{10}+\frac{9{\theta }_{2}}{10}$

Solution

JEE Main 2019 Physics, Thermal Properties of Matter — solution figure

Let the temperature of the junction $T ^{\circ}C$ .

Rate of heat flow in Rod $1$ = rate of heat flow in Rod $2$

$\frac{3\mathrm{kA}}{d}({\theta }_{2}-T)=\frac{\mathrm{kA}}{3d}(T-{\theta }_{1})$

$\Rightarrow 9({\theta }_{2}-T)=(T-{\theta }_{1})$

$\Rightarrow 10T=9{\theta }_{2}+{\theta }_{1}$

$\Rightarrow T=\frac{9{\theta }_{2}+{\theta }_{1}}{10}=\frac{{\theta }_{1}}{10}+\frac{9{\theta }_{2}}{10}$

JEE Main 2019 — Physics Thermodynamics

Options

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Solution

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JEE Main 2019 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics