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

Options

A. `sqrt(2)`
B. 3 K
C. `4/(3)K`
D. `2/(3)K`

Answer

C

Explanation

Equivalent thermal circuit
`1/(R_("eq"))=1/(R_(1))+1/(R_(2))`
`rArr(K_("eq")xx2A)/(l)=(KA)/(l)+(2KA)/(l)`
`rArrK_("eq")=3/(2)K`
In parallel
`K_("eq")=(2K_(1)K_(2))/(K_(1)+K_(2))`
`thereforeK_("eq")=(4K)/(3)`