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Response Details:
a) water pressure acting on the piston P = Patm + ρgh
= Patm + 1000 kg/m3*10 m/s2*(0.6-0.4)m
= 101325 N/m2 + 2000 N/m2 = 103325 N/m2,
in perpendicular upward direction
b) there is also atmospheric pressure acting on the piston in perpendicular
downward direction
so, net pressure in upward direction = 2000 N/m2
so, net pressure force in upward direction = 2000 N/m2 * 0.01m2
= 20N
As the piston is in equilibrium this force will be equal and opposite to weight
of the load
or, 20N = mg
or, m = 20N/(10m/s2) = 2 kg
c) As the system is in equilibrium i.e. pressure force is balanced by weight of
the load
always at the piston there should be a hydrostatic pressure of Patm+ 2000 N/m2
or, gauge pressure of 2000 N/m2
or, height from the zero gauge pressure level(from the main tank) = (2000N/m2)/ρg
= (2000N/m2)/(1000kg/m3*10m/s2)
= 0.2 m = 20cm
Now as load and piston moves up by 10 cm, water in the main tank will also
move up by 10 cm to keep the height of the piston at 20 cm.
d) Suppose after adding water in the main tank, load would not have moved up.
The pressure at the piston will increase and this will give fluid a flow energy
(flow energy = P/ρ). Thus this flow energy will move the fluid up and doing so pressure
will decrease and thus the flow energy will decrease, Which gets accumulated as
load's potential energy.
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