JEE MAIN - Physics (2022 - 29th June Morning Shift - No. 20)
As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm2 for each plate. The equivalent thermal conductivity of the compound plate is $$\left( {1 + {5 \over \alpha }} \right)$$ K, then the value of $$\alpha$$ will be ______________.
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Explanation
$${{{L_1}} \over {{K_1}{A_1}}} + {{{L_2}} \over {{K_2}{A_2}}} = {{{L_1} + {L_2}} \over {{K_{eff}}{A_{eff}}}}$$
$$ \Rightarrow {4 \over K} + {{2.5} \over {2K}} = {{6.5} \over {{K_{eff}}}}$$
$$ \Rightarrow {{10.5} \over {2K}} = {{6.5} \over {{K_{eff}}}}$$
$$ \Rightarrow {K_{eff}} = {{13K} \over {10.5}} = \left( {1 + {5 \over {21}}} \right)K$$
$$ \Rightarrow \alpha = 21$$
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