Two rods of length `d_(1)` and `d_(2)` and coefficients of thermal conductivites `K_(1)` and `K_(2)` are kept touching each other. Both have the same area of cross-section. The equivalent thermal conductivity.

Options

A. `K_(1)d_(1) + K_(2)d_(2)`
B. `K_(1)+K_(2)`
C. `(K_(1)d_(1)+K_(2)d_(2))/(d_(1)+d_(2))`
D. `(d_(1)+d_(2))/(d_(1)/K_(1))+(d_(2)/K_(2))`

Answer

D

Explanation

Consider the diagram where two rods are connected end to end (series).

Thermal resistance (R ) of a rod of length l, thermal conductivity K and area of cross-section A is given by
`R = l/(KA)`

When rods are connected in series
`R_(eq)` = equivalent thermal resistance = `R_(1) + R_(2)`..............(i)
where, `R_(1)`= Thermal resistance of first rod
`R_(2)` = thermal resistance of second rod
For fig. (a).
Thermal resistance of first rod, `R_(1)`= (d_(1)/(K_(1)A)`
Where, A is area of cross-section of each rod. Similarly, thermal resistance of second rod, `R_(2)=d_(1)/(K_(2)A)`
`Now, From fig. a) `R_(eq) = (d_(1))/(K_(1)A) + (d_(2))/(K_(2)A)`
`rArr (L_(eq)/(K_(eq)A_(eq))` = `(d_(1))/(K_(1)A_) + d_(2)/(K_(2)A)`..........(ii)
where, `L_(eq)` = equivalent length of rod
`E_(eq)` = equivalent are of cross-section .
For series connection,
`L_(eq)` = equivalent length =`d_(1)+d_(2)`
`A_(eq)` = equivalent area of cross-section area = A
Now, from fig (b),
`(d_(1)+d_(2))/(K_(eq)A)=d_(1)/(K_(1)A) + d_(2)/(K_(2)A)rArr (d_(1)+d_(2))/K_(eq) = d_(1)/K_(1) + d_(2)/K_(2)`
`rArr K_(eq) = ((K_(1)K_(2))/(K_(2)d_(1)+K_(2)d_(2)))(d_(1) + d_(2)) = (d_(1)+d_(2))/((d_(1)/K_(1)+(d_(2)/K_(2))))`
[Dividing numerator and denomintator by `K_(1)K_(2)]`
Equivalent thermal conductivity, `K_(eq) = (d_(1)+d_(2))/((d_(1)/K_(1) + d_(2)/K_(2))`