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_______
JEE MAIN 2022, Diffcult
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$\frac{\Delta Q }{\Delta t }=\left(\frac{1}{ R }\right) \Delta T$
$R$ : 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 series combination
$R _{\text {eq }}= R _{1}+ R _{2}$
where $L _{ eq }=4+2.5=6.5$
$\frac{ L _{ eq }}{ K _{ eq }( A )}=\frac{4}{ K (120)}+\frac{5}{\frac{2}{2 K (120)}}$
$\frac{6.5}{ K _{ eq }(120)}=\frac{4}{ K (120)}+\frac{5}{4 K (120)}$
$\frac{6.5}{ K _{\text {eq }}}=\frac{21}{4 K }$
$K _{ eq }=\frac{26}{21} K =\left(1+\frac{5}{21}\right) K$
$\therefore a =21$
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