A parallel plate capacitor having its lower end fixed and upper end is attached with spring having spring constant K. Upper plate is in equilibrium before switch is closed. After switch is closed, the condition on the potential of battery so that the system can acquire new equilibrium position is $V \leq \sqrt{{(\frac{p}{q})}^{r} \frac{L^{3} K}{ε_{0} A}}$ . Then $p + q + r$ is? [p & q are smallest possible integers and V is potential difference across the battery ]

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Detailed Solution

Let in new equilibrium the upper plate is displaced by x units. If we get x less than L then only new equilibrium is achieved :
$K x = \frac{1}{2 A \in_{0}} {[\frac{A \in_{0}}{L - x} V]}^{2} K x {(L - x)}^{2} = \frac{A \in_{0} V^{2}}{2}$
Maximum value of $K x {(L - x)}^{2} = ?$
$K {(L - x)}^{2} + 2 K x (L - x) (- 1) = 0 x = L / 3$
Maximum value of $K x {(L - x)}^{2} = K \frac{4 L^{3}}{27}$
So for real $x K \frac{4 L^{2}}{27} \geq \frac{A \in_{0}}{2} V^{2}$
$\Rightarrow V \leq \sqrt{\frac{8}{27} \frac{L^{3} K}{A \in_{0}}} \Rightarrow V \leq \sqrt{{(\frac{2}{3})}^{3} \frac{L^{3} K}{A \in_{0}}}$
or maximum kinetic energy of string in its first overtone $\frac{a_{0}^{2} π^{2} T}{l}$