A vessel is completely filled with a liquid of density 1.1×10^3 kg m^-3 calculate the pressure at a depth of 5cm from free surface
Answers
Answer:
Calculate the pressure due to a water–column of height 120 m. (Take g = 10 m s–2 and density of water = 103 kg m–3)?
Sol:
Given height of water–column = 120 m, g = 10 m s–2
And density of water = 103 kg m–3
Pressure = h ρ g
= 120 × 103 × 10
= 12 × 105 N m–2
2) At what depth below the surface of water will the pressure be equal to the atmospheric pressure ? The atmospheric pressure is 10 N cm–2. Density of water = 103 kg m–3, g = 9.8 m s–2?
Sol:
Pressure at a depth h below the surface of water = atmospheric pressure + pressure due to water column of height h.
If atmospheric pressure is P, then 2P = P + hρg or hρg = P
But P = 10 N cm–2
= l0N/(10–2 m)2
= 105 N m–2
Hence h = P/ρg
= 105/(103 × 9.8)
= 10.2 m
Thus a water column of height 10.2 m exerts a pressure equal to the atmospheric pressure.
3) A square plate of side 10 m is placed horizontally 2 m below the surface of water. The atmospheric pressure is 1.013 × 105 N m–2. Calculate the total thrust on the plate.(Density of water = 103 kg m–3, g = 9–8 m s–2)?
Sol:
The pressure at a point 2 m below the surface of water = atmospheric pressure = (1.013 × 105 N m2) + pressure due to a column of water of height 2 m.
Pressure due to 2 m column of water = h ρ g
= 2 × 103 × 9.8
= 0.196 × 105 N m–2
∴ Total pressure = (1.013 × 105) + (0.196 × 105)
= 1.209 × 105 N m–2
Area of the plate = 10 × 10
= 102 m2
∴ Total thrust on the plate = Pressure × Area
= (1.209 × 105) × 102
= 1.209 × 107 N
4) When an air bubble rises from the bottom of a lake to the surface, its volume is doubled. Find the depth of the lake. (Atmospheric pressure = 76 cm of Hg, density of Hg = 13.6 g cm–3 and density of water = 1 g cm–3)?
Sol:
Let the depth of lake be h cm
Pressure at the bottom of lake,
P1 = Atmospheric pressure + pressure due to water column in the lake
= 76 × 13.6 × g + hρg
= (76 × 13.6 + h × 1) g
Volume of bubble at the bottom, V1 = V
Volume of bubble at the surface, V2 = 2V
Pressure at the surface P2 = Atmospheric pressure
= 76 cm of Hg
= 76 × 13.6 × g
From Boyle's law, P1V1 = P2V2
(76 × 13.6 + h × 1) g × V = 76 × 13.6 × g× 2V
or h = (2 ×76 ×13.6) – (76 × 13.6)
= 76 × 13.6
= 1034 cm
= 10.34 m
5) Figure shows a rectangular object immersed in water of density 103 kg m–3 Calculate the pressure at the top and bottom of the object, the resultant pressure on the object. Take atmospheric pressure = 105 Pa and g = 9.8 N kg–1?
Sol:
Atmospheric pressure P0 = 105 Pa,
g = 9.8 N kg–1
Depth of top of the object from water surface h1 = 20 cm
= 0.2 m
Depth of bottom of object from water surface h2 = 20 + 15 = 35 cm =0.35 m
(i) Pressure at the top surface of object P1 = Po + h1 ρg
= 105 + (0.20 × 103 × 9.8)
= 1.0196 × 105 Pa
Pressure at the bottom surface of object P2 = Po + h2 ρg
= 105 + (0.35 × 103× 9.8)
= 1.0343 × 105 Pa
(ii) Resultant pressure on the object = P2 – P1
= (1.0343 × 105) – (1.0196 × 105)
= 0.0147 × 105 Pa (or 1.47 × 103 Pa)
6) A U tube is partially filled with mercury. If water is added in one arm and oil is added in the other arm, calculate ratio of water and oil columns so that the mercury level is the same in the two arms. Given density of water = 103 kg m–3, density of oil = 900 kg m–3?
Sol:
Since the level of mercury is the same in the two arms of the U tube,
Pressure of water column on the surface of mercury in one arm = Pressure of oil column on the surface of mercury in the other arm.
i.e., h1ρ1g = h2ρ2 g
Here h1 = height of water column,
ρ1 = density of water
= 103 kg m–3,
h2 = height of oil column,
ρ2 = density of oil
= 900 kg m–3.
∴ h1/h2 = ρ1/ρ2
= 900/103
= 9/10
7) A rectangular vessel of dimensions 120 cm × 80 cm × 200 cm is completely filled with a liquid of density 1.1 × 103 kg m–3. Neglecting the atmospheric pressure, find (i) the thrust at the bottom of the vessel, (ii) the pressure at the bottom of the vessel, (iii) the pressure at a depth of 5 cm from the free surface, (iv) the force experienced by a metal foil placed at a depth 5 cm from the free surface and (v) the thrust at the bottom of the vessel if the atmospheric pressure (= 1 × 105 N m–2) is taken into account? [Take g = 9.8 m s–2]
Sol:
(i) Volume of vessel = (120/100 m )× (80/100m) ×(200/100 m)
= 1.92 m3
Thrust at the bottom of the vessel = Weight of liquid in the vessel.
= Volume × density × g
= 1.92 × (1.1 × 103) × 9.8
= 2.069 × 104 N
(ii) Area of the bottom of vessel = (120/100 m)×(80/100m)
= 0.96 m2
Pressure at the bottom of vessel = Thrust/Area
= (2.069 × 104 N)/(0.96 m2)
= 2.15 × 104 N m–2
(iii) Pressure at a depth of 5 cm from the free surface = hρg
= (5/100) (× (1.11 × 103) × 9.8)
= 539 N m–2
(iv) Net force on the metal foil will be zero because the force exerted on each of the two faces of the foil, by the liquid will be equal and opposite.
(v) Pressure at the bottom of vessel = Atmospheric pressure + pressure due to liquid column
= (1.0 × 105) + (0.215× 105)
= 1.215 × 105 N m–2
∴ Thrust at the bottom = Pressure × Area
= (1.215× 105) × 0.96
= 11.664 × 104 N
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