1.

Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities K and 2 K respectively. The equivalent thermal conductivity of the slab is

Answer»

`2/3K`
`3K`
`SQRT(2)K`
`(4)/(3)K`.

Solution :In series equivqlent conductivity is
`K_(s)=(2K_(1)K_(2))/(K_(1)+K_(2))=(2K.2K)/(K+2K)=(4K^(2))/(3K)=(4)/(3)K`.
Correct choice is (d).


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