Two conducting slabs having same shape and size of heat conductivities $$k1$$ and $$k2$$ joined as shown in figure. The temperature at ends of the slabs are θ$$1$$ and θ$$2$$ $$($$θ$$1$$>θ$$2)$$. The final temperature of junction $$($$θm $$)$$ is:

Correct option is 1k1θ1+k2θ2k1+k2

Questions

Two conducting slabs having same shape and size of heat conductivities k₁ and k₂ joined as shown in figure. The temperature at ends of the slabs are $θ_{1}$ and $θ_{2}$ ( $θ_{1} > θ_{2}$ ). The final temperature of junction ( $θ_{m}$ ) is:

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k1θ1+k2θ2k1+k2

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k1θ2-k2θ1k1+k2

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detailed solution

Correct option is A

Assuming both slabs have identical thicknesses, ∆Q∆t=k1A(θ1-θm)l=k2A(θm-θ2)l⇒θm=k1θ1+k2θ2k1+k2

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Two conducting slabs having same shape and size of heat conductivities k1 and k2 joined as shown in figure. The temperature at ends of the slabs are θ1 and θ2 (θ1>θ2). The final temperature of junction (θm ) is:

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Ready to Test Your Skills?

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Two conducting slabs having same shape and size of heat conductivities k₁ and k₂ joined as shown in figure. The temperature at ends of the slabs are $θ_{1}$ and $θ_{2}$ ( $θ_{1} > θ_{2}$ ). The final temperature of junction ( $θ_{m}$ ) is: