In the given circuit ' $a$ ' is an arbitrary constant. The value of $m$ for which the equivalent circuit resistance is minimum, will be $\sqrt{\frac{ x }{2}}$. The value of $x$ is ...........
JEE MAIN 2022, Diffcult
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$R =\left(\frac{ ma }{3}\right)+\left(\frac{ a }{2 m }\right)$
$\frac{ dR }{ dm }=\frac{ a }{3}-\frac{ a }{2 m ^{2}}=0$
$\frac{ a }{3}=\frac{ a }{2 m ^{2}}$
$m ^{2}=\frac{3}{2}$
$m =\sqrt{\frac{3}{2}}$
$x =3$
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