Advertisements
Advertisements
Question
The area of base of a cylindrical vessel is 300 cm2. Water (density= 1000 kg m-3) is poured into it up to a depth of 6 cm. Calculate : (a) the pressure and (b) the thrust of water on the base. (g = 10m s-2.
Solution
Given Area of base of vessel , a = 300 cm2 = 300 × 10-4 m2
Density of water , ρ = 1000 kg m-3
Depth , h = 6 cm = 0.06 m
Acceleration due to gravity , g = 10 ms-2
(a) Pressure = hρg = 0.06 × 1000 × 10 = 600 pascal
(b) Thrust , T = pressure × area = 600 × 300 × 10-4 = 18 N
APPEARS IN
RELATED QUESTIONS
The density of brass is 8.4 g cm-3. What do you mean by this statement?
Define the term relative density of a substance.
For a floating body, how is its weight related to the buoyant force?
How does the pressure exerted by thrust depend on the area of surface on which it acts? Explain with a suitable example.
Explain the following statement:
Sleepers are laid below the rails.
The pressure inside a liquid of density p at a depth h is :
A vessel contains water up to a height of 1.5 m. Taking the density of water 103 kg m-3, acceleration due to gravity 9.8 m s-2 and area of base of vessel 100 cm2, calculate: (a) the pressure and (b) the thrust at the base of vessel.
What is meant by up thrust?
The magnitude of buoyant force acting on an object immersed in a liquid depends on ______ of the liquid.
The upward force that is caused due to the pressure difference in liquid (or fluid) is called ______.