As per the given figure, two plates A and B o | vedclass

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Question

$\frac{\Delta Q }{\Delta t }=\left(\frac{1}{ R }ight) \Delta T$

$R$ : Thermal resistivity

$	herefore R _{1}=\frac{ L _{1}}{ K _{1} A }=\frac{

Vedclass · Accepted Answer

$21$

Vedclass · Answer

$\frac{\Delta Q }{\Delta t }=\left(\frac{1}{ R }ight) \Delta T$

$R$ : Thermal resistivity

$	herefore 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, $R_{	ext {eq }}$ of this series combination

$R _{	ext {eq }}= R _{1}+ R _{2}$

where $L _{ eq }=4+2.5=6.5$

$\frac{ L _{ eq }}{ K _{ eq }( A )}=\frac{4}{ K (120)}+\frac{5}{\frac{2}{2 K (120)}}$

$\frac{6.5}{ K _{ eq }(120)}=\frac{4}{ K (120)}+\frac{5}{4 K (120)}$

$\frac{6.5}{ K _{	ext {eq }}}=\frac{21}{4 K }$

$K _{ eq }=\frac{26}{21} K =\left(1+\frac{5}{21}ight) K$

$	herefore a =21$

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