20 worked answers · short steps first, explanations when you need them
Chapter 9 Exercises
Textbook pages 200–201
Question 1
(a) Blood pressure is greater at the feet than at the brain. (b) Atmospheric pressure at about 6 km is nearly half its sea-level value. (c) Explain why hydrostatic pressure is a scalar even though pressure is force divided by area.
Steps only
Pressure in a liquid increases with depth: p = ρgh.
Air is compressed by the weight of the air above it, so air is densest near sea level. A large part of the atmosphere's mass lies in the first few kilometres; the air above 6 km is much thinner.
Pressure is the normal force per unit area at a point. Its value has magnitude only. The direction of the normal force depends on the orientation of the surface.
Answer
(a) The feet are lower, so the blood column above them is taller. (b) Much of the atmosphere’s mass lies near sea level, so less air lies above a point at 6 km. (c) Pressure has magnitude only; the normal force direction changes with the surface.
Question 2
Answer: (a) Why is the angle of contact obtuse for mercury in glass but acute for water? (b) Why does water wet glass but mercury does not? (c) Is surface tension dependent on surface area? (d) Why does detergent help water clean? (e) Why are liquid drops spherical?
Steps only
(a) Mercury's attraction to itself is stronger than its attraction to glass, while water is attracted strongly to glass.
Answer
(a) Mercury: obtuse; water: acute. (b) Water wets glass; mercury does not. (c) No. (d) Detergent lowers surface tension and improves spreading. (e) A sphere minimises surface area.
Question 3
Fill in the blanks: (a) Surface tension of a liquid generally ___ as temperature rises. (b) Viscosity of gases ___ with temperature, while viscosity of liquids ___. (c) In a solid, shearing force is proportional to ___; in a fluid, it is proportional to ___. (d) In steady flow, the increase in speed at a constriction follows conservation of mass or Bernoulli’s principle. (e) A small plane model in a wind tunnel reaches the turbulence threshold at a ___ speed than an actual plane.
Steps only
(a) Surface tension falls as temperature rises.
Answer
(a) decreases; (b) increases, decreases; (c) shear strain, rate of shear strain; (d) conservation of mass (continuity); (e) greater.
Question 4
Explain: (a) A sheet of paper rises when we blow over it. (b) Water gushes faster from a tap when it is partly closed. (c) A needle controls liquid flow better than thumb pressure. (d) A water-jet can push a container backwards. (e) A spinning ball does not follow a parabolic path.
Steps only
(a) Faster air above the paper has lower pressure; the higher pressure below pushes the paper up.
Answer
(a) Lower pressure above lifts it. (b) Smaller opening gives a faster jet. (c) Flow is very sensitive to needle radius. (d) The jet's momentum gives the container an opposite push. (e) Spin creates a Magnus force.
Question 5
A 50 kg girl stands on one heel of diameter 1.0 cm. Find the pressure on the floor. Take g=9.8 m/s².
Steps only
The force is her weight: F=mg=50×9.8=490 N.
Answer
The pressure is about 6.2×10⁶ Pa.
Question 6
What height of wine (density 984 kg/m³) balances atmospheric pressure 1.013×10⁵ Pa? Take g=9.8 m/s².
Steps only
For a liquid column, pressure p=ρgh.
Answer
A wine column about 10.5 m high is needed.
Question 7
A submarine is at a depth of 3.0 km. Is it safe if its shell can withstand an additional pressure of 1.0×10⁹ Pa? Use seawater density 1.03×10³ kg/m³.
Steps only
The extra water pressure is ρgh=1.03×10³×9.8×3000≈3.03×10⁷ Pa.
Answer
Yes, under these assumptions: the water pressure is about 3.0×10⁷ Pa, below the shell limit.
Question 8
A hydraulic lift raises a 3000 kg car. The large piston has area 425 cm². What pressure must the oil provide? Take g=9.8 m/s².
Steps only
Convert area: 425 cm²=0.0425 m². The car's weight is 3000×9.8=29,400 N.
Answer
The oil must provide about 6.9×10⁵ Pa of gauge pressure.
Question 9
In a U-tube manometer, 9.0 cm of water balances 12.5 cm of an unknown spirit above mercury. Find the spirit's relative density.
Steps only
The mercury levels are equal, so the pressures from the two liquid columns are equal.
Answer
The spirit's relative density is 0.72.
Question 10
Water and a spirit of relative density 0.72 are poured into the two arms of a U-tube over mercury. If each liquid column is 15 cm high, find the difference in mercury levels. Mercury's relative density is 13.6.
Steps only
The pressure difference of the two columns is equivalent to 15(1−0.72)=4.2 cm of water.
Answer
The mercury levels differ by about 0.31 cm (3.1 mm).
Question 11
Can Bernoulli's equation be applied to a fast, turbulent river? Explain.
Steps only
Bernoulli's equation assumes steady flow along a streamline and negligible viscosity.
Answer
Not reliably; turbulent, rapidly changing river flow does not meet Bernoulli's assumptions.
Question 12
In Bernoulli's equation, can pressure be gauge pressure instead of absolute pressure?
Steps only
Gauge pressure equals absolute pressure minus atmospheric pressure.
Answer
Yes. Use gauge pressure consistently; the common atmospheric-pressure offset cancels.
Question 13
Glycerine (density 1.3×10³ kg/m³, viscosity 0.83 Pa·s) flows through a 1.5 m long tube of radius 1.0 cm at 4.0 g/s. Find the pressure difference and check whether the flow is laminar.
Steps only
Mass flow is 0.004 kg/s, so volume flow Q=ṁ/ρ=3.08×10⁻⁶ m³/s.
Answer
Pressure difference ≈9.8×10² Pa; Re≈0.31, so the flow is laminar.
Question 14
An aeroplane wing has area 2.5 m². Air speeds above and below it are 70 m/s and 63 m/s. Find the lift, using air density 1.3 kg/m³.
Steps only
Bernoulli's equation gives higher pressure below the wing: Δp=½ρ(vabove²−vbelow²).
Answer
The lift is about 1.5×10³ N upward.
Question 15
A diagram shows streamlines in a pipe that narrows and widens. Identify the incorrect streamline pattern and explain.
Steps only
For steady incompressible flow, streamlines crowd where the pipe narrows because speed increases there.
Answer
The pattern showing fewer or more widely spaced streamlines at the constriction is incorrect; they should crowd together there.
Question 16
Water enters a spray pump tube of area 8.0 cm² at 1.5 m/min and leaves through 40 holes, each 1.0 mm in diameter. Find the exit speed.
Steps only
Inlet speed is 1.5/60=0.025 m/s. Inlet area=8.0×10⁻⁴ m².
Answer
The water exits at about 0.64 m/s.
Question 17
A soap film on a U-shaped wire supports a weight of 1.5×10⁻² N. The movable side is 30 cm long. Find the film’s surface tension.
Steps only
A soap film has two surfaces, so S=W/(2l)=1.5×10⁻²/(2×0.30).
Answer
Surface tension=2.5×10⁻² N/m.
Question 18
Figure 9.21(a) shows a liquid film supporting 4.5×10⁻² N. What weight can the same liquid film support in figures (b) and (c) at the same temperature? Each movable side is 40 cm long.
Steps only
The force is 2Sl. Since S and the 40 cm movable-side length are the same in all three figures, the supported weight is unchanged.
Answer
Figures (b) and (c) each support 4.5×10⁻² N.
Question 19
Find the pressure inside a mercury drop of radius 3.0 mm. Surface tension is 0.465 N/m and atmospheric pressure is 1.01×10⁵ Pa.
Steps only
For one liquid surface, excess pressure Δp=2T/r.
Answer
The inside pressure is about 1.013×10⁵ Pa (310 Pa above atmospheric pressure).
Question 20
(a) Find the pressure inside a soap bubble of radius 5.0 mm when T=0.025 N/m. (b) Find the pressure inside an air bubble of the same radius, 40 cm below a soap solution of relative density 1.2. Atmospheric pressure is 1.01×10⁵ Pa.
Steps only
(a) A soap bubble has two surfaces, so excess pressure is 4T/r=4×0.025/0.005=20 Pa. Its inside pressure is 1.01×10⁵+20=1.0102×10⁵ Pa.