JEE Mains · Physics · STD 11 - 10.2 transmission of heat
If two metallic plates of equal thicknesses 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 \(\frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}\)
- B \(\frac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}\)
- C \(\frac{ K _{1}+ K _{2}}{2 K _{1} K _{2}}\)
- D \(\frac{ K _{1}+ K _{2}}{ K _{1} K _{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}\)
Step-by-step Solution
Detailed explanation
In series \({R_{eq}} = {R_1} + {R_2}\) ==> \(\frac{{2l}}{{{K_{eq}}A}} = \frac{l}{{{K_1}A}} + \frac{l}{{{K_2}A}}\) ==> \(\frac{2}{{{K_{eq}}}} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}}\) ==> \({K_{eq}} = \frac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}\)
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