Question

A cylindrical can 0.90m high with a diameter of 0.30m is filled with water to a depth of 0.70m from the bottom. A small hole with a cross sectional area of 0.004m^2 is in the side of the can at the bottom. Density of water is 1000kg/m^3 What is the speed of water flowing out of the bottom of the can?

Answer 1

I have attached a rough figure of the cylinder as asked in the question.

Let P_A be the atmospheric pressure. v₁ be the initial speed of water in the cylindrical can and let v₂ be the speed of water from the hole.

Let us apply equation of continuity

Area of cylinder * v₁ = Area of hole *v₂ ...........(1)

Diameter of cylinder = 0.30 m

Radius = 0.15 m

Area o cross-section of the cylindrical can =

Area of cross-section of the hole = 0.004 m²

From equation (1),

0.07 * v1 = 0.004 * v2

v₂ = 17.5 v₁ .......(2)

In order to find the speed of water, we need to consider Bernoulli's theorem.

.....(3)

(Since, the hole is at the bottom, its height is 0. d is the density of water.)

Rearranging equation (3), we can write

......(4)

h=0.70 m given. Let, g= 10 ms^-2.

From equations (2) and (4), we get

This is the initial speed of water in the cylindrical can.

Using equation (2) ,

This v₂ is the speed of water flowing out of the bottom of the can.