As per the given figure,two plates $A$ and $B$ of thermal conductivity $K$ and $2K$ are joined together to form a compound plate. The thickness of the plates are $4.0 \,cm$ and $2.5 \,cm$ respectively,and the area of cross-section is $120 \,cm^{2}$ for each plate. If 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
    $22$

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