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_______

\(R\) : 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 series combination

\(R _{\text {eq }}= R _{1}+ R _{2}\)

where \(L _{ eq }=4+2.5=6.5\)

\(\frac{ L _{ eq }}{ K _{ eq }( A )}=\frac{4}{ K (120)}+\frac{5}{\frac{2}{2 K (120)}}\)

\(\frac{6.5}{ K _{ eq }(120)}=\frac{4}{ K (120)}+\frac{5}{4 K (120)}\)

\(\frac{6.5}{ K _{\text {eq }}}=\frac{21}{4 K }\)

\(K _{ eq }=\frac{26}{21} K =\left(1+\frac{5}{21}\right) K\)

\(\therefore a =21\)