Case i: Three vessel have
different shapes, but the same base area and the same weight when empty. They
are filled with water to the same level.
a)
Are the water pressure at the bottom of the each vessel are same? If not, which
is larger
and
which is smaller?
Solution: As,
P = Patm + Pgh
PA = PB = PC
b)
If the three vessels containing water are weighed on a scale, do they give the
same reading? If not, which weighs most and which weight the least?
Solution: The
weight of water is more in A. Therefore, A weighs most and B weighs the least, (Assuming
weight of each vessel is same).
c)
Is the force excreted by the water on the base of vessel is equal to the weight
of water in the vessel?
Solution:
From
(a) we know that upward force exerted by base of each vessel is same. Let this
force be F. If we draw the diagram for the liquid in each of the vessel we have:
(NH and NV are horizontal and vertical components of the
normal force by side walls, W is weight of fluid in vessel)
Case ii: Is the pressure
at point 2, P₂ is the same as the pressure at point 1, P₁ even though the column
of water above point 2 is less than the column of water above point 1?
Consider
a horizontal cylindrical element between 1 and 2
Net
force acting on the cylindrical element is P₁A - P₂A = 0 as the liquid is non-acceleration.
P₁
= P₂
(i.e.)
for same liquid which is not accelerating horizontally, pressure will be same
at all Points which are at the same level.
Case iii: Pressure at two
points at the same level of the same liquid which is not accelerating horizontally
when the liquid elements at those points are in “communication” must be same. i.e.;
in following situation.
Since,
lengths are equal, PA = PB = PC = PD
Example: Water is filled in a flask up to a
height of 20 cm. The bottom of the flask is circular with radius 10 cm. If the
atmospheric pressure is 1.01 × 10⁵
Pa, find the force exerted by the water on the bottom. Take g = 10 m/s² and density of water =
1000 kg/m³.
Solution: The
pressure at the surface of the water is equal to the atmospheric pressure P₀.
The pressure at the base P = P₀ + hrg
= 1.01
x 10₆ Pa + (0.20m) (1000 kg/m³) (10 m/s²)
=
1.01 x 10⁵ Pa + 0.20 x 10⁵ Pa = 1.03 x 10⁵ Pa
The area of the bottom = πr²
= 3.14 x (0.1m)
= 0.0314 m²
The force on the bottom is, therefore,
F = Pπr² = (1.03 x 10⁵) x (0.0314m²) = 3230N.
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