Answer:
[tex] 1.06 [/tex] mm/min
Explanation:
h = height of the water level above the hole = 1 m
[tex]d_{h}[/tex] = diameter of hole = 4 mm = 0.004 m
[tex]A_{h}[/tex] = Area of cross-section of hole = [tex](0.25) \pi d_{h}^{2}[/tex]
[tex]d_{t}[/tex] = diameter of tank = 2 m
[tex]A_{t}[/tex] = Area of cross-section of tank = [tex](0.25) \pi d_{t}^{2}[/tex]
R = rate at which the water level drop
Rate at which the water level drop is given as
[tex]R = \frac{A_{h} (\sqrt{2gh}))}{A_{t}}[/tex]
[tex]R = \frac{((0.25) \pi d_{h}^{2}) (\sqrt{2gh}))}{((0.25) \pi d_{t}^{2})}[/tex]
[tex]R = \frac{(d_{h}^{2}) (\sqrt{2gh}))}{(d_{t}^{2})}[/tex]
[tex]R = \frac{(0.004)^{2} \sqrt{2(9.8)(1)}}{(2)^{2}}[/tex]
[tex]R = 17.7\times 10^{-6}[/tex] m/s
[tex]R = 0.0177[/tex] mm/sec
[tex]R = (0.0177) (60)[/tex] mm/min
[tex]R = 1.06 [/tex] mm/min
