Q. The top of a water tank is open to air and its water level is maintained. It is giving out $0.74 m^3$ water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to :

Solution:

In flow volume = outflow volume
$\Rightarrow \frac{0.74}{60} = \left(\pi \times4 \times10^{-4}\right) \times\sqrt{2gh} $
$ \Rightarrow \sqrt{2gh} = \frac{74 \times100}{240 \pi} $
$ \Rightarrow \sqrt{2gh} \frac{740}{24\pi} $
$ \Rightarrow 2gh = \frac{740 \times740 }{24 \times24\times10} \left(\pi^{2} = 10\right)$
$ \Rightarrow h = \frac{74 \times74}{2 \times24 \times24} $
$ \Rightarrow h \approx 4.8 m $

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