Question
A large tank, open at the top, has two small holes in the vertical wall. One is a square hole of side 's' at a depth h below the top and the other is a circular hole of radius r at a depth 4h below the top (given that s ≪ h; r ≪ h). When the tank is completely filled up to the brim with water, the quantity of water flowing out per second from each hole is the same, then r is equal to :
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Solution
[Key Concept: Efflux speed from a small hole at depth d is √(2gd), and volume flow is area times speed.]
$$v_{s} = \sqrt{2gh}$$
$$v_{c} = \sqrt{2g(4h)}$$
$$v_{c} = \sqrt{8gh}$$
$$Q_{s} = \text{~Area~~×~~velocity~}$$
$$Q_{s} = s^{2}\sqrt{2gh}$$
$$Q_{c} = \left( \pi r^{2} \right)\sqrt{8gh}$$
According to given condition
$$\begin{matrix} & Q_{s} = Q_{c} \\ & s^{2}\sqrt{2gh} = \pi r^{2}\sqrt{8gh} \\ & s^{4}(2gh) = \pi^{2}r^{4}(8gh) \\ & s^{4} = 4\pi^{2}r^{4} \\ & r^{4} = \frac{s^{4}}{4\pi^{2}} \\ & r = \frac{s}{\left( 4\pi^{2} \right)^{1/4}} \\ & r = \frac{s}{\sqrt{2\pi}} \end{matrix}$$
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