Q

# Help me answer: - Properties of Solids and Liquids - JEE Main

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 $'\theta_2 '\:\:and\:\:'\theta_{1}'$  respectively , $(\theta_{2}>\theta_{1})$  .The temperature at the interface is :

• Option 1)

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

• Option 2)

$\frac{\theta_{2}+\theta_{1}}{2}$

• Option 3)

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

• Option 4)

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

Views

$\Rightarrow Conservation \:\:of \:\:heat$

$\frac{T-\Theta_{2}}{d/3kA}+\frac{T-\Theta_{1}}{3d/kA}=0$

$\Rightarrow37-3\theta_{2}-\frac{T}{3}-\frac{\theta_{1}}{3}=0$

$\Rightarrow97-9\theta_{2}+T-\theta_{1}=0$

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

Option 1)

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

Option 2)

$\frac{\theta_{2}+\theta_{1}}{2}$

Option 3)

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

Option 4)

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

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