A Cylindrical container rotates with the constant angular speed. The radius of the cylinder is R . Find the height has shown at which water is in equilibrium with respect to the container.
Answers
A cylindrical container of radius 'R' and height 'h' is completely filled with a liquid. Two horizontal L-shaped pipes of small cross-sectional area 'a' are connected to the cylinder as shown in the figure. Now the two pipes are opened and fluid starts coming out of the pipes horizontally in opposite directions.
The side wall of a wide vertical cylindrical vessel of height
h=
75cm
has a narrow vertical slit running all the way down to the bottom of the vessel. The length of the slit is
l=
50cm
and the width
b=
1.0mm
. With the slit closed, the vessel is filled with water. Find the resultant force of reaction (in
N) of the water flowing out of the vessel immediately after the slit is opened.
Study later
View Answer
A jet of water having velocity
=
10m/s
and stream cross-section
=
2
cm
2
hits a flat plate perpendicularly, with the water splashing out parallel to plate.
I HOPE IT HELPS. ☺FOLLOW ME ✌
A cylindrical container rotates with the constant angular speed, 10 rad/s. The radius of the cylinder is R = 5cm
To find : The height has shown at which water is in equilibrium with respect to the container.
solution : consider a fluid particle of mass m is at position (x, y).
Pseudo force acting on the particle, F = mω²x
weight of particle , W = mg
at equilibrium, net force should be perpendicular to the free surface.
so, tanθ = mω²x/mg [ where θ is angle between net force and mg ]
now dy/dx = tanθ = mω²x/mg = ω²x/g
⇒g∫dy = ω²∫ x dx
⇒gy = ω²x²/2
⇒y = ω²x²/g
here , ω = 10 rad/s , x = R = 5 × 10¯² m and g = 10m/s²
so, y = {10² × (5 × 10¯²)²}/2(10)
= (100 × 25 × 10¯⁴)/(20)
= 125 × 10¯⁴
= 0.0125 m
= 1.25 cm
Therefore the height at which water is in equilibrium with respect to the container.