Question

If two metallic plates of equal thickness and thermal conductivities ${K_1}$ and ${K_2}$ are put together face to face and a common plate is constructed, then the equivalent thermal conductivity of this plate will be –
A. $\dfrac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}$
B. $\dfrac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}$
C. $\dfrac{{{{({K_1}^2 + {K_2}^2)}^{3/2}}}}{{{K_1}{K_1}}}$
D. $\dfrac{{{{({K_1}^2 + {K_2}^2)}^{3/2}}}}{{2{K_1}{K_1}}}$

Answer 1

Hint: Generally when temperature difference is maintained then heat transfers from higher temperature body to the lower temperature body. The amount of heat transferred depends upon the various factors like the temperature difference, thermal resistance of the material.

Formula used:
$R = \dfrac{L}{{KA}}$

Complete step-by-step answer:
Flow of a quantity with time is known as a current. It can be fluid current or heat current or electric current.
In case of fluid current a pressure difference is maintained and that drives the flow of fluid and fluid always flows from high pressure region to the low pressure region and fluid current is governed by fluid resistance too.
Whereas in electric current the electric charge flows with time and the voltage difference and electric resistance combined will govern the electric current. Charge flows from higher voltage to lower voltage naturally.
Similarly in thermal current it is governed by temperature difference and thermal resistance.
Thermal resistance of a thermal conductor of length ‘L’ and cross sectional area ‘A’ and thermal conductivity ‘K’ is given by $R = \dfrac{L}{{KA}}$
Thermal resistance of first conductor is ${R_1} = \dfrac{L}{{{K_1}A}}$
Thermal resistance of the second conductor is ${R_2} = \dfrac{L}{{{K_2}A}}$
When both are put together face to face then they will be in series and series effective thermal resistance will be ‘R’
$R = {R_1} + {R_2}$
Effective length will be ‘2L’ and effective cross section will be ‘A’ and effective thermal conductivity be ‘K’
$R = {R_1} + {R_2}$
$\eqalign{
  & \Rightarrow \dfrac{{2L}}{{KA}} = \dfrac{L}{{{K_1}A}} + \dfrac{L}{{{K_2}A}} \cr
  & \Rightarrow \dfrac{2}{K} = \dfrac{1}{{{K_1}}} + \dfrac{1}{{{K_2}}} \cr
  & \Rightarrow K = \dfrac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}} \cr} $
Hence the answer will be option B

So, the correct answer is “Option B”.

Note: If the metallic plates are connected in parallel rather than series then the effective cross-sectional area will be doubled and the effective length will be the same. Then ultimately the effective resistance will be something different. We can see some similarities in finding effective resistance for electric current and thermal current.