Q 12-03-178JEE MainJEE Main 2022 (27 Jun, Shift 2)Medium
In the given circuit $'a'$ is an arbitrary constant. The value of $m$ for which the equivalent circuit resistance is minimum, will be $\sqrt{\dfrac x2}$. The value of $x$ is ______ .
Numerical value type. Enter your answer.
Answer: 3
$$R_{eq} = \frac{ma}{3} + \frac{a/m}{2} = a\left(\frac m3 + \frac{1}{2m}\right)$$
$$\frac{dR_{eq}}{dm} = a\left(\frac13 - \frac{1}{2m^2}\right) = 0 \Rightarrow m^2 = \frac32$$
$m = \sqrt{\dfrac32}$, so $x = 3$.
Solution by Sreeraj P, M.Sc Physics