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 :-

\theta _1 \over 10 + 9\theta _2 \over 10

\theta _1 \over 6 + 5\theta _2 \over 6

\theta _1 \over 3 + 2\theta _2 \over 3

Heat and Thermodynamics PYQ — JEE Main 2019 · 9 Apr · Shift 2 · Q48 with Solution

View written solutionFree

Correct answer: A

Concept used: steady-state heat conduction through composite slab

In steady state, the same heat current flows through both slabs.

For a slab,

H = \frac{kA(\Delta \theta)}{L}

where:

$k$ = thermal conductivity
$A$ = cross-sectional area
$L$ = thickness

Let the interface temperature be $\theta$ .

Data for the two slabs

First slab: conductivity $3K$ , thickness $d$
Second slab: conductivity $K$ , thickness $3d$
Outer temperatures are $\theta_2$ and $\theta_1$ , with $\theta_2 > \theta_1$

Assume heat flows from the side at $\theta_2$ through the first slab, then through the second slab to $\theta_1$ .

So,

Temperature drop across first slab = $\theta_2 - \theta$
Temperature drop across second slab = $\theta - \theta_1$

Equate heat current through both slabs

For first slab:

H = \frac{3KA(\theta_2-\theta)}{d}

For second slab:

H = \frac{KA(\theta-\theta_1)}{3d}

At steady state,

\frac{3KA(\theta_2-\theta)}{d} = \frac{KA(\theta-\theta_1)}{3d}

Cancel $KA/d$ :

3(\theta_2-\theta) = \frac{\theta-\theta_1}{3}

Multiply by $3$ :

9(\theta_2-\theta) = \theta-\theta_1

Expand:

9\theta_2 - 9\theta = \theta - \theta_1

9\theta_2 + \theta_1 = 10\theta

Hence,

\theta = \frac{9\theta_2 + \theta_1}{10}

Match with options

\theta = \frac{\theta_1}{10} + \frac{9\theta_2}{10}

This matches Option A.

Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

So, the derived answer agrees with the stored correct answer.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: A

Concept used: steady-state heat conduction through composite slab

In steady state, the same heat current flows through both slabs.

For a slab,

H = \frac{kA(\Delta \theta)}{L}

where:

$k$ = thermal conductivity
$A$ = cross-sectional area
$L$ = thickness

Let the interface temperature be $\theta$ .

Data for the two slabs

First slab: conductivity $3K$ , thickness $d$
Second slab: conductivity $K$ , thickness $3d$
Outer temperatures are $\theta_2$ and $\theta_1$ , with $\theta_2 > \theta_1$

Assume heat flows from the side at $\theta_2$ through the first slab, then through the second slab to $\theta_1$ .

So,

Temperature drop across first slab = $\theta_2 - \theta$
Temperature drop across second slab = $\theta - \theta_1$

Equate heat current through both slabs

For first slab:

H = \frac{3KA(\theta_2-\theta)}{d}

For second slab:

H = \frac{KA(\theta-\theta_1)}{3d}

At steady state,

\frac{3KA(\theta_2-\theta)}{d} = \frac{KA(\theta-\theta_1)}{3d}

Cancel $KA/d$ :

3(\theta_2-\theta) = \frac{\theta-\theta_1}{3}

Multiply by $3$ :

9(\theta_2-\theta) = \theta-\theta_1

Expand:

9\theta_2 - 9\theta = \theta - \theta_1

9\theta_2 + \theta_1 = 10\theta

Hence,

\theta = \frac{9\theta_2 + \theta_1}{10}

Match with options

\theta = \frac{\theta_1}{10} + \frac{9\theta_2}{10}

This matches Option A.

Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

So, the derived answer agrees with the stored correct answer.

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