A composite block is made of slabs A,B,C,D and E of different thermal conductivities (given in terms of a constant K and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat 'Q' flows only from left to right through the blocks. Then in steady state

Options

A. (i),(ii)
B. (i), (ii), (iii)
C. (i), (iii),(iv)
D. (ii), (iv)

Answer

C

Explanation

(3) `A` and `E` are in series, hence heat current is same, (i) is `O.K.`
Thermal resistances
Let `b` is width of slabs
`R_(A) = (L)/(2 K (4 Lb)) = (R )/(8), R_(B) = (4 L)/(3 K (Lb)) = (4 R)/(3)`
`R_(C ) = (4 L)/(4K (2 Lb)) = (R )/(2) , R_(D) = (4L)/(5 K(Lb)) = (4 R)/(5)`
`R_(E) = (L)/(6 K (4 Lb)) = (R )/(24)`
Temperature difference = Heat current `xx` resistance
Since resistance of `E` is samllest, hence, temperature difference across it is minimum, (iii), is `O.K`.
`i_(C ) = (Delta theta)/(R_(C )) = (Delta theta)/(R//2) = (2 Delta theta)/(R )`
`i_(B) = (Delta theta)/(R_(B)) = (Delta theta)/(4 R//3) = (3 Delta theta)/(4R)`
`i_(D) = (Delta theta)/(R_(D)) = (Delta theta)/(4 R//5) = (5 Delta theta)/(4 R)`
`i_(C ) = i_(B) + i_(D)`, (iv) is `O.K`.