JEE Mains · Physics · STD 11 - 10.2 transmission of heat
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 \(120 \,cm ^{2}\) for each plate. The equivalent thermal conductivity of the compound plate is \(\left(1+\frac{5}{\alpha}\right) K\), then the value of \(\alpha\) will be_______
- A \(20\)
- B \(21\)
- C \(23\)
- D \(22\)
Answer & Solution
Correct Answer
(B) \(21\)
Step-by-step Solution
Detailed explanation
\(\frac{\Delta Q }{\Delta t }=\left(\frac{1}{ R }\right) \Delta T\) \(R\) : Thermal resistivity \(\therefore R _{1}=\frac{ L _{1}}{ K _{1} A }=\frac{ L _{1}}{ K (120)}\) \(L _{1}=4 \,cm\) \(A =120 \,cm ^{2}\) \(R _{2}=\frac{2.5}{(2 K )(120)}\) Now, \(R_{\text {eq }}\) of this…
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