12 Fluids
Exam practice questions
Pages 184-187 Exam practice questions
1 a) At the instant of release, the viscous force is zero; the ball will begin to fall at the rate of the
−2
acceleration due to gravity, 9.8 m s – the answer is D.
[Total 1 Mark]
b) At point 3 the sphere is travelling at constant velocity; the acceleration is
zero – the answer is A.
[Total 1 Mark]
c) Between points 1 and 2 the sphere is still accelerating downwards; there must be a resultant
downward force, so the weight is greater than the sum of the two upward forces, U and F –
the answer is D.
[Total 1 Mark]
d) At point 3 the sphere is moving at the terminal velocity; the resultant force is zero, so the
weight must equal the sum of the two upward forces – the answer is C.
2 Stokes’ law force does not depend on the mass – the answer is B.
−3
−2
[Total 1 Mark]
[Total 1 Mark]
5
3 h × 13 600 kg m × 9.8 m s = 1.01 × 10 Pa
[1]
→ h = 0.76 m of mercury
[1]
[Total 2 Marks]
4 a) ρ =
mass
50 × 10−3 kg
= 4
= 1.5 kg m−3
3
volume
π
(0.20
m)
3
[2]
b) The air is compressed so its density inside the balloon is greater than that at normal
[1]
atmospheric pressure.
[Total 3 Marks]
5 a)
V × ρ × g = 600 N
600 N
V =
= 5.1 × 10−2 m3
1200 kg m −3 × 9.8 m s −2
[2]
b) The above volume is less than the volume displaced by the less dense water in the pool.
[1]
[Total 3 Marks]
6 a) The temperature [1] and the place of origin [1] will both affect the viscosity of the oil. The
rate of flow is inversely proportional to the viscosity – stickier fluids move more slowly
through the pipeline, so the rate of flow is greater at higher temperatures. [1]
Increasing the diameter of the pipe will greatly increase the rate at which the oil flows
through it (at the same pressure). [1]
b) If the flow rate is too fast, turbulence occurs [1] and much more energy is needed to
transport the oil.
[Total 5 Marks]
7 a) Laminar at A and B [1]; turbulent at C [1]
3
2
5
b) Uplift, F = ΔP × A = 4.0 × 10 Pa × 120 m [1] = 4.8 × 10 N [1]
© Hodder & Stoughton Limited 2015
12 Fluids
Exam practice questions
c) For level flight: uplift = weight
5
−2
4.8 × 10 N = m × 9.8 m s → m = 49 000 kg (49 tonnes)
[1]
[Total 5 Marks]
8 a) mg = 6πηrv
[1]
4 × 10−12 kg × 9.8 m s −2
−1
≈ 1 mm s
6 π (2 × 10−5 N s m −2 )(0.1 × 10−3 m)
[2]
b) The larger raindrops have a much bigger terminal velocity.
[1]
v=
[Total 4 Marks]
9 a) U + F upward; W or mg down
[2]
−3
b) W = 34 π r 3 ρ v = 34 π (1.5 × 10−3 m)3 (7800 kg m −3 )(9.8 m s −2 ) = 1.08 × 10 N
[2]
c) U + F = W
[1]
−3
−3
d) 0.20 × 10 N + 6πηrv = 1.1 × 10 N
v=
[1]
0.9 × 10−3 N
= 9.1 × 10−2 m s −1
6π × 0.35 Pas × 1.5 × 10−3 m
[2]
−1
e) i)
[1]
0.16 m s
ii)
A rise in the terminal velocity indicates a reduction in the viscosity of the liquid. [1] As
the temperature rises, the viscosity decreases [1] until, after a temperature of about 70
°C, the viscosity stays the same. [1]
iii)
[1]
Viscostatic means that the viscosity does not vary with temperature.
[Total 13 Marks]
10 a) Larger particles have a higher terminal velocity than the smaller ones [1]. The smaller
particles reach their terminal velocity much more quickly [1] than the larger ones; in such a
short distance the pebbles will not reach their terminal velocity [1] and will fall quickly to the
bottom.
The reason for this is that the viscous drag is proportional to the radius whereas the weight is
proportional to the radius cubed [1]. This leads to the expression:
v=
2( ρs − ρf ) g r 2
[1]
9η
Showing that terminal velocity is proportional to the radius squared. The layers will therefore
have the largest particles at the bottom with those of decreasing size settling in order on top.[1]
b) For the fine sand particles, v =
=
2 × (1200 kg m
−6
−1
= 5.4 × 10 m s
−3
2( ρs − ρf ) g r 2
9η
−3
) × 9.8 m s
−2
9 × 0.8 N s m
- 1000 kg m
−2
× (1.0 × 10
−4
m)
2
[1]
[1]
© Hodder & Stoughton Limited 2015
12 Fluids
Exam practice questions
Assuming terminal velocity is reached instantaneously,
time to settle =
distance
=
terminal velocity
0.22 m
−6
−1 = 40 000 s = 11 hours
5.44 × 10 m s
[2]
[Total 10 Marks]
11 a) The non-streamlined shape of the obstruction means that there will be abrupt changes in the
velocity of the fluid as it moves past the object [1]. This causes turbulence [1] and the
formation of vortices (eddies) in the fluid beyond the obstruction. The vortices move along at
the speed of the fluid [1], and, if the diameter of the tube is known, the rate of flow can be
calculated.
b) Rate of flow =
volume
Speed of flow =
2
=
π r l
time
= π r
t
60
12 × 10
−3
60 s
m
3
−4
3 −1
[1] = 2.0 × 10 m s
3 −1
3 −1
m s = 0.65 m s [1] → v =
Frequency of pulses =
[1]
−4
2
39
v=
3 −1
m s
−1
[1] = 0.53 m s
−2
2
π (1.1 × 10 m)
2.0 × 10
c) Area of pipe = π × (0.30 m) = 0.283 m
Av =
2
2.3 m s
[1]
2
3 −1
0.65 m s
−1
= 2.3 m s
2
0.283 m
[1]
−1
0.53 m s
−1
[1]
× 20 Hz [1] = 87 Hz
d) Assume that the flow is laminar [1] and the velocity of the flow is constant across the crosssection of the pipe. [1]
e) If the diameter is halved, the pressure across the ends will need to be eight times bigger for
the same rate of flow. [1]
The area of cross-section is four times less than that of the 60 cm pipe. This means that the
−1
speed of flow will be four times as big as that of the wider pipe [1] (9.2 m s ).
Although the higher pressure could cause problems, it is likely that the flow will become
turbulent at such high speeds of flow. [1]
[Total 16 Marks]
© Hodder & Stoughton Limited 2015