Fluid Mechanics With Engineering Applications 10th Edition By Finnemore, Joseph Franzini (Solutions Manual)

MEO
Physicstestbank Fo
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r mid-term Grade 10
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Second semester
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,1) At sea level, the average atmospheric pressure is 760 mmHg. How
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many
kilopascals is this? g g




a) 1.013 kPa g




b) 10.13 kPa g




c) 101.3 kPa g




d) 1013 kPa g




g Answer: C g




2) Which would be the least likely to sink into soft ground?
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a) a loaded lorry with six wheels
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b) a loaded lorry with four wheels
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c) an empty lorry with four wheels
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d) an empty lorry with six wheels
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Answer: D g




3) Water boils on Mount Everest at a temperature...
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a) less than 100 degrees Celsius because Mount Everest is always cold
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b) less than 100 degrees Celsius because there is less atmospheric p
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ressure
c) higher than 100 degrees Celsius because of the high altitude.
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d) water always boils at the same temperature no matter where it is on e
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arth

,Answer: B g




4) When the atmospheric pressure at the surface of a lake is 1 Atm, fi
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nd the lake depth when the pressure at its base is 3Atm? Knowing tha
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t : (P water =1000 kg/m3 , Pa=1,013 x10* N/m^2, g=10m/s^2)?
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a) 30.39 m g



b) 20.673 m g



c) 9.717 m g



d) 10.13
Answer: A g




Explanation : pressure at base = 3* 1.013* 10^5 = 303900
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𝜌𝑔ℎ= 30 g



3900
1000*10* h= 303900 g g




H= 30.39 m.
g g




5) When a barometer is put at a top of a mountain, the height of m
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ercury column in its tube
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a) 76 cm
g



b) 78 cm g



c) 74 cm
g



d) 80 cm A
g g




nswer : C g g

, 6) If a uniform body weighs 50N in air and 30 N in water, its specific gr
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avity equals ... g g




a) 1.5
b) 1.67
c) 2.5
d) 3 Answ g




er: B g




Explanation : 50 N = m x 9.81 m/s^2 m g g g g g g g g g




g = 5.1 kg
g g




Next, we can use the weight of the object in water to determine the bu
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oyant force acting on the object:g g g g g




weight in water = force due to gravity on the displaced water = buoy
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ant force g




30 N = (volume of object submerged in water) x (density of water) x g
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volume of object submerged in water = 30 N / (1000 kg/m^3 x 9.81 m/s
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^2) = 0.00305 m^3
g g g




Since the object is uniform, the volume of the submerged part is eq
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ual to its total volume, so the density of the object can be
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calculated as: g