Name: _____________________________________________Date: _______________________ Class: ____________________
Archimedes’ Principle, Pascal’s Law and Bernoulli’s Principle Lesson—Practice Problems Worksheet Answer Key 1
Practice Problems Worksheet Answer Key
Show complete solutions to the following problems and box final answers with units.
1. A sample of an unknown material weighs 300 N in air and 200 N when submerged in an alcohol
solution with a density of 0.70 x 10
3
kg/m
3
. What is the density of the material?
Given:
F
g(air)
= 300 N
F
g(alcohol)
= 200 N
ρ
alcohol
= 0.7 x 10
3
kg/m
3
Unknown:
ρ
material
or ρ
o
Solution:
F
B
= F
g(air)
– F
g(alcohol)
= 300 N – 200N
F
B
= 100 N
F
g(air)
/ F
B
= ρ
o
/ ρ
alcohol
ρ
o
= F
g(air)
/ F
B
* ρ
alcohol
= (300 N / 100 N) * 0.7 x 10
3
kg/m
3
ρ
o
= 2.1 x 10
3
kg/m
3
2. A 40-cm tall glass is filled with water to a depth of 30 cm.
a. What is the gauge pressure at the bottom of the glass?
b. What is the absolute pressure at the bottom of the glass?
Given:
h = 30 cm = 0.3 m
g = 9.81 m/s
2
ρ
water
= 1.0 x 10
3
kg/m
3
Uknown:
a) P
gauge
b) P
absolute
Solution:
a) P
gauge
= ρgh = (1.0 x 10
3
kg/m
3
) (9.81 m/s
2
) (0.3 m)
P
gauge
= 2.9 x 10
3
kg/m
3
Pa
b) P
absolute
= P
atm
+ P
gauge
P
absolute
= 1.01 x 10
5
Pa + 2.9 x 10
3
kg/m
3
Pa
P
absolute
= 1.04 x 10
5
Pa
Name: _____________________________________________Date: _______________________ Class: ____________________
Archimedes’ Principle, Pascal’s Law and Bernoulli’s Principle Lesson— Practice Problems Worksheet Answer Key 2
3. Water circulates throughout a house in a hot water heating system. If the water is pumped at a
speed of 0.50 m/s through a 4.0-cm diameter pipe in the basement under a pressure of 3.03x10
5
Pa, what will be the velocity and pressure in a 2.6-cm diameter pipe on the second floor 5.0 m
above?
Given:
v
1
= 0.50 m/s v
2
= ?
h
1
= 0 m (basement) h
2
= 5.0 m
d
1
= 0.04 m d
2
= 0.026 m
A
1
= π (d
1
/ 2)
2
= 0.0004π A
2
= π (d
2
/ 2)
2
= 1.69 x 10
-4
π
P
1
= 3.03 x 10
5
Pa P
2
= ?
Unknown:
v
2
P
2
Solution:
A
1
v
1
= A
2
v
2
v
2
= A
1
v
1
/ A
2
= (0.0004π * 0.50 m/s) / 1.69 x 10
-4
π
v
2
= 11.83 m/s
P
1
+ ½ ρ v
1
2
+ ρgh
1
= P
2
+ ½ ρ v
2
2
+ ρgh
2
P
2
= P
1
+ ½ ρ (v
1
2
- v
2
2
) - ρgh
2
P
2
= (3.03 x10
5
Pa) + ½ (1.0 x 10
3
kg/m
3
) [(0.50 m/s)
2
– (11.83 m/s)
2
] – (1.0 x 10
3
kg/m
3
) (9.81 m/s
2
)
(5.0 m)
P
2
= 1.84 x 10
5
Pa
4. The small piston of a hydraulic lift has an area of 0.20 m
2
. A car weighing 1.2 x 10
4
N sits on a rack
mounted on the large piston. The large piston has an area of 0.90 m
2
. How large force must be
applied to the small piston to support the car?
Given:
A
1
= 0.20 m
2
A
2
= 0.90 m
2
F
1
= ? F
2
= 1.2 x 10
4
N
Unknown:
F
1
Solution:
F
1
/ A
1
= F
2
/ A
2
F
1
= F
2
/ A
2
(A
1
) = (1.2 x 10
4
N / 0.90 m
2
) * 0.20 m
2
F
1
= 2.7 x 10
3
N
Name: _____________________________________________Date: _______________________ Class: ____________________
Archimedes’ Principle, Pascal’s Law and Bernoulli’s Principle Lesson— Practice Problems Worksheet Answer Key 3
5. Calculate the absolute pressure at an ocean depth of 1.0 x 10
3
m. Assume that the density of the
water is 1.025 x 10
3
kg/m
3
and that P
0
= 1.01 x 10
5
Pa.
Given:
h = 1.0 x 10
3
m
ρ = 1.025 x 10
3
kg/m
3
P
atm
or P
o
= 1.01 x 10
5
Pa
Unknown:
P
absolute
Solution:
P
absolute
= P
atm
+ P
gauge
P
absolute
= P
atm
+ ρgh = 1.01 x 10
5
Pa + (1.025 x 10
3
kg/m
3
) (9.81 m/s
2
) (1.0 x 10
3
m)
P
absolute
= 1.01 x 10
7
Pa
6. A water tank has a spigot near its bottom. If the top of the tank is open to the atmosphere,
determine the speed at which the water leaves the spigot when the water level is 0.5 m above the
spigot.
Given:
P
1
= Patm = 1.01 x 105 Pa = P
2
(both are open to atmosphere)
v
1
= 0 (negligible)
h
1
= 0.5 m
h
2
= 0 m
Unknown:
v
2
Solution:
P
1
+ ½ ρ v
1
2
+ ρgh
1
= P
2
+ ½ ρ v
2
2
+ ρgh
2
P
1
+ ρgh
1
= P
2
+ ρgh
2
v
2
= sqrt(2gh
1
)
v
2
= sqrt(2 (9.81 m/s
2
) (0.5m))
v
2
= 3.13 m/s