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^(@)C` is maintained across the composite slab . If `K_(1) =(K_(2))/(2)` , the temperature difference across slab A is

Options

A. `4^(@)C`
B. `6^(@)C`
C. `8^(@)C `
D. `10^(@)C`

Answer

C

Explanation

Rate of flow of heat across A= Rate of flow of heat across B
`(K_(1)A(T_(1)-T))/(l) = (K_(2)A(T-T_(2)))/(l)`
`K_(1)(T_(1)-T)=K_(2)(T-T_(2))`
`implies T_(1)-T = 2(T - T_(2)` `( because K_(2) = 2K_(1))`
`T_(1) + 2T_(2) - 3T = 0`
Given temperature difference across the slab,
`T_(1) - T_(2) = 12^(@)C`
`implies T_(1) + 2(T_(1) - 12) - 3T = 0`
`implies 3T_(1) - 24 - 3T = 0`
`implies T_(1) - T = 8^(@)C`
`therefore` Temperature difference acress slab, `A = 8^(@)C`.