OAT Physics - Problem Drill 14: Fluids
Question No. 1 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
1. Suppose that a person’s systolic blood pressure (measured while arm is level
Question #01 with the heart) is 100 mm Hg. Given that the density of blood is 1050 kg/m3, what
would happen to this pressure if that person’s arm was raised 40 cm above the
heart? (1 N/m2 = 7.5×10-3 mm Hg)
(A) 68 mm Hg
(B) 100 mm Hg
(C) 132 mm Hg
(D) 150 mm Hg
(E) none of the above
A. Correct.
Apply Bernoulli’s law at the level of the heart and at the level of the raised arm.
The pressure at the level of the arm will be reduced by pgh compared to the normal
systolic pressure, where p is density of blood, and h is the positive height the arm
is raised. See the solution below for a more detailed explanation. Ignore the effect
of blood velocity in this problem.
B. Incorrect.
By Bernoulli’s law, the pressure will change when the arm is elevated.
Feedback on
Each Answer
Choice
C. Incorrect.
By Bernoulli’s law, the pressure will decrease when the arm is elevated.
D. Incorrect.
By Bernoulli’s law, the pressure will decrease when the arm is elevated.
E. Incorrect.
One of the above choices is correct; consider the pressure variation with height of
the fluid.
Let p0 be the pressure at the level of the heart, pa be the pressure at the level of
the arm, and let h be the height of the arm. Then, using Bernoulli’s equation, we
obtain:
p0 =pa +ρgh
which can be put into the form:
pa =p0 -ρgh .
Thus pa is less than p0 by the term pgh.
ρ gh= (1050 kg/m3 ) × (10 m/s2 ) × ( 0.4 N/m2 )
Solution
=4200 N/m2
= ( 4200 N/m2 ) ×
7.5×10-3 mm Hg
1 N/m2
=31.5 mm Hg
=32 mm Hg
Subtract 32 mm Hg from 100 mm Hg, and a value of 68 mm Hg is obtained.
Thus the correct answer choice is (A).
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 2 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
2. A garden hose with an inner diameter of 2.0 cm is connected to a lawn sprinkler
Question #02 with 20 uniform holes of diameter 2 mm. If the flow velocity at the end of the
hose is 100 cm/s, what is the flow velocity at each of the holes?
(A) 20 cm/s
(B) 100 cm/s
(C) 500 cm/s
(D) 200 cm/s
(E) none of the above
A. Incorrect.
Since the combined area of the sprinkler holes is less than the area of the garden
hose, the velocity will increase, not decrease. Use the equation of continuity for an
incompressible fluid.
B. Incorrect.
Since the combined area of the sprinkler holes is less than the area of the garden
hose, the velocity at the sprinkler will change. Use the equation of continuity for an
incompressible fluid.
Feedback on
Each Answer
Choice
C. Correct!
This answer can be obtained by using the equation of continuity for an
incompressible fluid.
D. Incorrect.
Since the combined area of the sprinkler holes is less than the area of the garden
hose, the velocity at the sprinkler will change. Use the equation of continuity for an
incompressible fluid.
E. Incorrect.
One of the above answer choices is correct.
Use the equation of continuity for an incompressible fluid:
A1v1=A 2 v 2 ,
Where A1 is the area at the end of the garden hose, v1 is the velocity at the end of
the garden hose, A2 is the total area of the sprinklers, n is the number of sprinklers,
and v2 is the quantity to be found.
2
2
⎛d ⎞
⎛d ⎞
π ⎜ 1 ⎟ v1=nπ ⎜ 2 ⎟ v 2
⎝2⎠
⎝ 2⎠
Eliminate π and factors of 2:
d12 v1=nd22 v 2
Solution
then solve for v2:
d12
v1=v 2 .
nd22
Plugging in numbers, we obtain:
( 2 cm )
2
20(0.2 cm)2
(100 cm/s ) =v 2
⎛ 100
⎞
100 ⎟ cm/s=v 2
⎜
⎝ 20
⎠
500 cm/s=v 2
So the correct answer choice is (C).
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 3 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
3. If a fluid is unwettable to an open-ended tube and the tube is inserted to the
Question #03 fluid, what the relative position between the surface of the outside fluid and the
surface of the inside fluid of the tube?
(A) Inside is lower
(B) Inside is higher
(C) Same
(D) Uncertain
(E) None of the above
A. Correct!
If the fluid is unwettable to the tube, the fluid in the tube will keep a convex shape.
So the contacting angle is a sharp angle. As result, the surface tension will pull the
fluid inside the tube down till the surface tension is balanced by the pressure drop
between the outside fluid and the inside fluid.
Feedback on
Each Answer
Choice
B. Incorrect.
If the fluid is wettable to the tube, the fluid in the tube will keep a concave shape.
So the contacting angle is a blunt angle. As a result, the surface tension will lift the
fluid inside the tube till the surface tension is balanced by the pressure drop
between the inside fluid and the outside fluid.
C. Incorrect.
Same height means there is no pressure drop between the inside fluid and outside
fluid. So, this mean there is no surface tension.
D. Incorrect.
The answer should be certain with a correct answer above.
E. Incorrect.
One of the outcomes must be correct.
Define “Surface tension”
The force on the surface of fluids to keep the fluid drop has the minimum surface
area. If the fluid is unwettable to the tube, the fluid in the tube will keep a convex
shape. So the contacting angle is a sharp angle. As a result, the surface tension will
pull the fluid inside the tube down till the surface tension is balanced by the
pressure drop between the outside fluid and the inside fluid.
The correct answer choice is (A).
Solution
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 4 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
4. If you dive down to the bottom of a 10.0m pool of water, what pressure will
you feel? The mass density of water is 1000 kg/m3. The existing air pressure is
1.01x105 Pa.
Question #04
(A) 1.01x105Pa
(B) 2x105 Pa
(C) 2.5x105Pa
(D) 5x105 Pa
(E) 0
A. Incorrect!
This is the original pressure exerted on the surface of the water. There will be
additional pressure due to the water above the diver.
Feedback on
Each Answer
Choice
B. Correct!
Use the formula that relates pressure variation to depth. The original pressure and
mass density of water is given. Substitute and calculate. Your final answer will be
in N/m2, or Pascals.
C. Incorrect!
Be sure to use the acceleration from gravity, 9.8m/s/s in the term for the additional
pressure due to the water.
D. Incorrect!
Use the formula that relates pressure variation to depth.
E. Incorrect!
The pressure cannot be zero. Be sure to factor in the additional pressure due to the
10.0 depth of water.
Use the formula that relates pressure variation to depth.
P = Po + ρgh
The original pressure is the air pressure that is exerted on the surface of the water.
This is given in the problem.
The additional pressure due to the 10.0m of water above the diver is given by the
second term. The “g” is the acceleration from gravity.
Solution
Substitute and calculate carefully.
P = (1.01x10 5 Pa) + (1000kg/m 3 )(9.8m/s 2 )(10.0m)
P = 1.01x10 5 Pa + 9.8x10 4 N/m2
P = 2x105 Pa
The correct answer is (B).
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 5 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
5. Why is the curvature of the upper edge of an aircraft’s wing higher than that of
the lower edge?
Question #05
(A) Beauty
(B) Force balance consideration
(C) To generate lifting aerodynamic force
(D) To mimic birds
(E) No apparent reason
A. Incorrect!
That wouldn’t be a very scientific reason to design a wing.
B. Incorrect!
Force balance is assured by the tail. So, force balance is not the reason for the
shape of wings.
Feedback on
Each Answer
Choice
C. Correct!
This statement satisfies the principles for fixed wing aircraft. You may recall this
question was discussed in the core tutorial. Consider the fluid motion of the air
over the top side of the wing.
D. Incorrect!
The wing of birds is also with such a shape, which is used to induce lifting
aerodynamic forces. But, the wing of aircraft is not designed just to mimic the wing
of birds but has its physical properties.
E. Incorrect!
The wing of birds is also with such a shape, which is used to induce lifting
aerodynamic forces. But, the wing of aircraft is not designed just to mimic the wing
of birds but has its physical properties.
Principles for fixed wing aircraft:
The curvature of the upper edge is always higher than that of lower edge.
According to the continuity law, the flow speed should be higher near the upper
edge (travel distance is long). According to Bernoulli’s law, the pressure at the
upper edge should be lower than that at the lower edge. Therefore, a lifting
aerodynamic force is generated.
The correct answer choice is (C).
Solution
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 6 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
6. What is the lifting force on a wing that has a top side velocity of 500 m/s and a
bottom side velocity of 400 m/s if the wing has an area of 100 m2? Assume the
density of air is 1.3 kg/m3.
Question #06
(A) 6×104 N
(B) 6×104 N
(C) 6×106 N
(D) 0
(E) None of the above
A. Incorrect.
To find the force on the wing, multiply the relative pressure by 100 m2, not divided
by 100 m2.
B. Incorrect.
This is the value of the relative pressure of the bottom of the wing compared to the
top of the wing.
Feedback on
Each Answer
Choice
C. Correct!
First, solve for the relative pressure at the bottom of the wing using Bernoulli’s law.
Next, multiply this pressure by the area of the wing to find the upward force on the
wing.
D. Incorrect.
The answer cannot be zero. First, solve for the relative pressure at the bottom of
the wing using Bernoulli’s law. Next, multiply this pressure by the area of the wing
to find the upward force on the wing.
E. Incorrect.
One of the above answer choices is correct.
By applying Bernoulli’s law to the top and to the bottom of the wing, we obtain:
1
1
p1+ ρair v12 = ρair v 22 ,
2
2
Where p1 denotes relative pressure at the bottom of the wing, v1 denotes velocity
on the bottom side of the wing, and v2 denotes velocity at the top side of the wing,
and
ρair denotes the density of air.
This equation can be solved for p1:
1
p1= ρair ( v 22 -v12 ) .
2
Solution
Using the definition of pressure, p = F/A, find the force using p1 and the area, A =
100 m2.
1
F=p1A= ρair ( v 22 -v12 ) A .
2
In the final step, substitute all numerical quantities and solve:
F=0.5×1.3× ( 5002 -4002 ) ×100 N=0.65×90000×100 N»6×106 N
The correct answer choice is (C).
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 7 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
7. Suppose a U-tube which is open to the atmosphere on both ends is filled on
both sides: on one side with water (of density 1×103 kg/m3), and on the other side
with an oil of density 2×103 kg/m3. If the height of the water column is 60 cm,
what is the height of the column containing the oil?
Question #07
(A) 30 cm
(B) 60 cm
(C) 120 cm
(D) 100 cm
(E) none of the above
A. Correct!
Because the oil has twice the density of water, the height of the column of oil
required is half the height of the column of water.
B. Incorrect.
Because the oil has a different density than water, the heights of the oil and water
columns will not be the same.
Feedback on
Each Answer
Choice
C. Incorrect.
The height of the column of oil will be a factor of two smaller than the height of the
column of water, not two times larger.
D. Incorrect.
Because the oil has twice the density of water, the height of the column of oil
required is half the height of the column of water.
E. Incorrect.
One of the above answer choices is correct.
By applying Bernoulli’s law to both sides of the U-tube, we obtain:
patm +ρw ghw =patm +ρo gho ,
Where patm denotes the atmospheric pressure, pw denotes the density of water, hw
denotes the height of the water column, ho denotes the density of oil, and ho
denotes the height of the oil column. Cancel the term patm from both sides to
obtain the following relation:
ρw ghw =ρo gho .
Canceling g from both sides, then solving for h0, obtain:
Solution
ρw hw
=ho .
ρo
In the final step, substitute all numerical quantities and solve:
ho =
1.0×103 kg/m3
60 cm=30 cm
2×103 kg/m3
The correct answer choice is (A).
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 8 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
8. A rectangular section of lumber measures 2.0m x .05m x .08m. How much
weight can it support if it is entirely submerged under water? This lumber has a
mass density of 550 kg/m3.
Question #08
(A)
(B)
(C)
(D)
(E)
43 N
0.008 N
550 N
4.4 N
0N
A. Correct!
Use the buoyancy formula B= mass density times the acceleration from gravity
times the volume.
B. Incorrect!
This is only the volume of the lumber section. The buoyant force is the weight that
this amount of water would displace.
Feedback on
Each Answer
Choice
C. Incorrect!
This is the mass density of the wood. This describes the amount of mass per
volume of the substance. Use the buoyancy formula to find the force that the wood
will support.
D. Incorrect!
You forgot to include the acceleration of gravity that is in the buoyancy formula.
E. Incorrect!
It is not zero. The buoyant force is the weight that this amount of water would
displace.
If the entire wood section is submerged underwater, it would have a buoyant force
equal to the weight of water that is displaced by that wood.
B = ρgV
The mass density of water is given, the acceleration of gravity is the typical 9.8
m/s/s, and the volume of the wood can be calculated.
Since the section of wood is rectangular, Volume = Base x height x width= Bhw
V=(2.0m)(.05m)(.08m)=.008m3
Solution
Substitute this value into the buoyancy force formula along with the other known
values.
B = (550kg/m 3 )(9.8m/s/s)(.008m 3 ) = 43N
This represents the maximum weight the miniature wood raft could support without
going under. Any weight less than this, and the mini wood raft would be at least
partially above the water line.
The correct answer choice is (A).
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 9 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
9. A block of wood floats with 1/3rd of its volume submerged in a large pool of
water. Find the density of the wood. (The density of water is 1.0×103 kg/m3.)
Question #09
Feedback on
Each Answer
Choice
(A) 330 kg/m3
(B) 670 kg/m3
(C) 1000 kg/m3
(D) 100 kg/m3
(E) none of the above
A. Correct!
The highest velocity is in the center, the lowest velocity is near the wall of the pipe.
B. Incorrect.
The force of weight is equal to the buoyant force; the buoyant force depends on the
weight of the displaced seawater, and has nothing to do with the unsubmerged
volume.
C. Incorrect.
This would be true if the block were totally submerged.
D. Incorrect.
The highest velocity is in the center, the lowest velocity is near the wall of the pipe.
E. Incorrect.
One of the above answer choices is correct.
The force of weight of the wooden block can be written in terms of the density of
the block,
ρb, the volume of the block, Vb, and g:
FW =ρb Vb g .
On the other hand, the weight of the displaced water, which is equivalent to the
buoyant force, is given by:
Fb =ρw Vw g
Where w is the density of water, and VW is the volume of displaced water.
Because we are given the following information
VW =Vb /3 ,
(the block is 1/3rd submerged)
and since we know:
FW =Fb ,
Solution
the density of water can be solved (with a little algebra) :
1
1000
ρb = ρW =
kg/m3 »330 kg/m3
3
3
Thus choice (A) is correct.
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved
Question No. 10 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
10. Two wooden balls of equal volume but different density are held beneath the
surface of a container of water. Ball A has one-half the density of ball B. Which of
the following statements is true when the balls are released?
Question #10
(A) Ball A has a smaller acceleration than ball B.
(B) Ball A has a greater acceleration than ball B.
(C) Ball A has the same acceleration as ball B.
(D) It cannot be determined.
(E) None of the above.
A. Incorrect.
For ball B, it’s smaller mass and lower density make it easier to move than ball B
B. Correct!
Since ball A is less dense, it has a smaller mass to move. Each ball has the same
volume, thus the same buoyant force. With an equal force, but a smaller mass,
ball A accelerates more.
Feedback on
Each Answer
Choice
C. Incorrect.
Consider the effect of the different densities.
D. Incorrect.
Yes, it can be determined from the given information. Consider how the masses
and buoyant forces differ for each ball.
E. Incorrect.
There is one correct answer above.
The less dense ball, ball A, has less weight than ball B but has the same buoyant
force as ball B. Thus the net force on ball A will be higher, so ball A will have a
higher acceleration. This conclusion can also be derived quantitatively.
The force of weight of the wooden block can be written in terms of the density of
water,
ρW, the density of the ball, ρb, the volume of the block, V, and g:
∑ F=F -F
b
W
=ρw Vg-ρb Vg .
On the other hand, the net force experienced by each ball depends on the density
of each ball
∑ F=m a=ρ Va
b
Solution
b
The two preceding equations can be combined to obtain the acceleration of each
ball:
⎛ρ
⎞
a= ⎜ W -1⎟ g .
⎝ ρb ⎠
Since we know that the density of ball A is less than that of ball B:
ρb,a <ρb,b
The formula for acceleration can be used to show:
aa >ab
Thus choice (B) is correct.
RapidLearningCenter.com
©Rapid Learning Inc. All Rights Reserved