Question

A large, cylindrical water tank with diameter 2.40 mm is on a platform 2.00 mm above the ground. The vertical tank is open to the air and the depth of the water in the tank is 2.00 mm. There is a hole with diameter 0.600 cmcm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole.
A) When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?
B) How long does it take you to collect 1.00 gal of water in the bucket?

Answer 1

Answer:

Explanation:

A ) radius of tank r = 1.2 m

depth of water in the tank = 2 m

1 gal of water =  1 / 264.17 m³

= 3.785 x 10⁻³ m³

Let h be the change in height of water in the tank .

volume of water flowing out

= π r² x h = 3.785 x 10⁻³

3.14 x 1.2² x h = 3.785 x 10⁻³

h = 83.71 x 10⁻⁵ m

= .84 mm .

B )

change in height is negligible .

velocity of efflux of water from the hole at the bottom

v = √ 2 gh

h is height of water level which is 2 m

v = √ (2 x 9.8 x 2 )

= 6.26 m / s

radius of hole = .3 x 10⁻² m

area of cross section

= π r²

= 3.14 x ( .3 x 10⁻² )²

= 28.26 x 10⁻⁶ m²

volume of water flowing through the hole per unit area

= area of cross section x velocity of efflux

= 28.26  x 10⁻⁶ x 6.26

If t be the time required ,

28.26  x 10⁻⁶ x 6.26 x t   = 3.785 x 10⁻³

t = 21.4 s