Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
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
संबंधित प्रश्न
What happens to the buoyant force as more and more volume of a solid object is immersed in a liquid ? When does the buoyant force become maximum ?
What is the law of floatation?
State whether pressure is a scalar or vector?
A metal solid cylinder tied to a thread is hanging from the hook of a spring balance. The cylinder is gradually immersed into the water contained in a jar. What changes do you expect in the readings of the spring balance? Explain your answer.
Figure shows the same block of wood floating in three different liquids A, B and C of densities ρ1, ρ2 and ρ3 respectively. Which of the liquid has the highest density? Give a reason for your answer.
Select the correct option:
When an object sinks in a liquid, its:
Select the correct option:
When a body is wholly or partially immersed in a liquid, it experiences a buoyant force which is equal to:
How is thrust different from pressure?
The upward force that is caused due to the pressure difference in liquid (or fluid) is called ______.
If the buoyant force is less, the object will float.
