A composite slab consists of two slabs A and B of different materials but of the same thickness placed one on top of the other. 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 will be:

Options

A. `4^(@)C`
B. `8^(@)C`
C. `12^(@)C`
D. `16^(@)C`

Answer

B

Explanation

Let temperature of junction be `T_(0)^(@)C`
Temperature difference across slab`A = (T_(1)- 7'_(0))^(@)C`
Temperature difference across slab `B = (T_(0) - T_(2)^(@)C`
Since both slabs are connected in series, heat current through both will be equal
`(k_(1)A(T_(1)-T_(0))/(t/2) =(k_(2)A(T_(0)-T_(2))/(t//2)`
As `k_(1) =(k_(2))/(2)`
`rArr (k_(2))/(2) (T_(1)-T_(0)) =(T_(0)-T_(2))k_(2)`
`T_(1)-T_(0) = 2T_(0) - 2T_(2)`
`T_(1) + 2T_(2) = 3T_(0)`...(1)
As `T_(1)-T_(2)= 12^(@)C`
`rArr T_(2) =T_(1) -12` ...(2)
From (I) and (2), we get
`T_(1)+2(T_(1)-12) = 3T_(0)`
`3T_(1) - 24 = 3T_(0)`
`T_(1) -T_(0) = 8^(@)C`