Question

A very powerful vacuum cleaner which has a hose of circular cross section can lift a brick of mass 12 kg when the hose is placed perpendicularly on a horizontal flat smooth face of a brick. What is the minimum radius the hose opening should have to achieve this?

A). 1.9 cm
B). 2.6
C). 1.4 cm
D). 3.2 cm

Answer 1

A). 1.9 cm

Step-by-step explanation:

m = Mass of brick = 12 kg

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of hose

A = Area =
`\pi r^2`

F = Force =
`mg`

Let us assume that the pressure required to lift the brick would be atmospheric pressure

`P=(F)/(A)\\\Rightarrow P=(F)/(\pi r^2)\\\Rightarrow r=\sqrt{(F)/(\pi P)}\\\Rightarrow r=\sqrt{(12* 9.81)/(\pi* 101325)}\\\Rightarrow r=0.01923\ m=1.9\ cm`

The radius of the hose should be 1.9 cm