Two slabs A and B of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of A and B are $k_{1}$ and $k_{2}$ respectively. A steady temperature difference of $12^{\circ} C$ is maintained across the composite slab. If $k_{1} = \frac{k_{2}}{2}$ , the temperature difference across slab A is:

4^{\circ} C

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6^{\circ} C

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8^{\circ} C

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10^{\circ} C

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Solution

The correct option is C $8^{\circ} C$
Assumption: Temperature at Slab A is lower.
Let temperature at junction be $T_{o}$ .

Given, $T_{B} - T_{A} = 12$ ; $k_{2} = 2 k_{1}$
Then, $k_{1} \frac{T_{o} - T_{A}}{R_{A}} = k_{2} \frac{T_{B} - T_{o}}{R_{B}}$

or, $k_{1} (T_{o} - T_{A}) = k_{2} (T_{B} - T_{o})$

Solving, $T_{o} - T_{A} = 8^{o} C$

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The correct option is C 8∘CAssumption: Temperature at Slab A is lower.Let temperature at junction be To.Given, TB−TA=12 ; k2=2k1Then, k1To−TARA=k2TB−ToRBor, k1(To−TA)=k2(TB−To)Solving, To−TA=8oC