1. The document contains solutions to 10 physics problems involving fluid flow and pressure. Problems involve calculating flow rates, speeds, and pressures using equations like continuity, Bernoulli's principle, and relationships between pressure, density, and height. 2. Key calculations include determining the speed of water flowing over Niagara Falls, the velocity of water through a changing diameter pipe, and the pressure difference between a giraffe's heart and brain due to its height. 3. Questions are solved by setting up and manipulating the relevant equations to find the unknown values, like using continuity to relate velocities at different pipe diameters.

1. 1
Solved problems on chapter 5
Q1: Each second, 5 525 𝑚3
of water flows over the 670𝑚 wide cliff of the Horseshoe Falls
portion of Niagara Falls. The water is approximately 2 m deep as it reaches the cliff. What is its
speed at that instant?
Solution
The cross section area of the water reaching the cliff is
𝐴 = 670𝑚 × 2𝑚 = 1340𝑚2
The flow rate is,
𝑄 = 𝐴𝑣 = 5525
𝑚3
𝑠𝑒𝑐
→ 𝑣 =
𝑄
𝐴
=
5525
𝑚3
𝑠𝑒𝑐
1340𝑚2
= 4.123 𝑚/𝑠𝑒𝑐
Q2: A water hose 2.00 cm in diameter is used to fill a 20.0-L bucket. If it takes 1.00 min to fill
the bucket, what is the speed 𝑣 at which water moves through the hose?
Solution
The flow rate is
𝑄 = 𝐴𝑣 =
𝑉
𝑡
=
20 × 10−3
𝑚3
60𝑠𝑒𝑐
= 3.33 × 10−4
𝑚3
/𝑠𝑒𝑐
Thus,
𝑣 =
3.33 × 10−4
𝑚3
/𝑠𝑒𝑐
𝐴
=
3.33 × 10−4
𝑚3
/𝑠𝑒𝑐
𝜋(1 × 10−2)2 𝑚2
= 1.06 𝑚/𝑠𝑒𝑐
Q3: A horizontal pipe 10.0 cm in diameter has a smooth reduction to a pipe 5.00 cm in diameter.
If the pressure of the water in the larger pipe is 8.0 × 104
𝑃𝑎 and the pressure in the smaller pipe
is 6.0 × 104
𝑃𝑎, at what rate does water flow through the pipes?
Solution
Applying Bernoulli’s equations,
𝑃1 +
1
2
𝜌𝑣1
2
= 𝑃2 +
1
2
𝜌𝑣2
2
8.0 × 104
𝑃𝑎 +
1
2
1000𝑣1
2
= 6.0 × 104
𝑃𝑎 +
1
2
1000𝑣2
2
We need to find a relation between 𝑣1 and 𝑣2, thus apply the continuity equation,
𝐴1 𝑣1 = 𝐴2 𝑣2 → 𝑣2 =
𝐴1 𝑣1
𝐴2
=
𝑟1
2
𝑟2
2 𝑣1 =
100
25
𝑣1 = 4𝑣1
𝑣2 = 4𝑣1
Thus,
8.0 × 104
𝑃𝑎 +
1
2
1000𝑣1
2
= 6.0 × 104
𝑃𝑎 +
1
2
1000 × 16𝑣1
2
Solve for 𝑣1, we have 𝑣1 = 1.63 𝑚/𝑠𝑒𝑐
𝑄 = 𝐴𝑣 = 𝜋(0.05)2
𝑚2
× 1.63
𝑚
𝑠𝑒𝑐
= 0.256𝑚3
/𝑠𝑒𝑐

2. 2
Q4: Water flows through a horizontal pipe that gradually tapers. At a point where the radius is 𝑅
the velocity is 2𝑚𝑠−1
and the pressure is 3 atm. At a point where the radius is 𝑅/2, find a. the
velocity; b. the pressure.
Solution:
a. From the continuity equation, we can find the velocity
11
12212
2
1
2
1
2
2
1
2
1
2211
8244
4
1
4
2










smsmvvvvv
R
vR
v
R
vR
vAvA


b. Use Bernoulli's equation where the potential energy per unit volume is constant
   
    
atmatmatmP
mNmsmkgmsmkgatmP
msmkgPmsmkgatmvPvP
b
b
bba
7.2296.03
)/10013.1/(/1032/1023
8/10
2
1
2/10
2
1
3
2
1
2
1
2521332133
213321332
2
2
1






Q5: A 2.0 N force pushes an A = 25 𝑚𝑚² syringe plunger and forces water out an area a = 0.010
𝑚𝑚² needle. What is the speed of the exiting water?
Solution
The force cause a pressure at cross sectional area A = 25 𝑚𝑚² is
𝑃𝑝𝑙𝑢𝑛𝑔𝑒𝑟 =
𝐹
𝐴
=
2𝑁
25(10−3 𝑚)2
= 80,000 𝑁/𝑚2
Use Bernoulli’s equation
𝑃1 +
1
2
𝜌𝑣1
2
= 𝑃2 +
1
2
𝜌𝑣2
2
Use the continuity equation to find a relation between the velocities
𝐴1 𝑣1 = 𝐴2 𝑣2 → 𝑣1 =
𝐴2 𝑣2
𝐴1
=
0.01𝑚𝑚2
25𝑚𝑚2
𝑣2 = 4 × 10−4
𝑣2
𝑃1 = 𝑃𝑎𝑡𝑚 + 𝑃𝑝𝑙𝑢𝑛𝑔𝑒𝑟 and 𝑃2 = 𝑃𝑎𝑡𝑚
𝑃𝑎𝑡𝑚 + 𝑃𝑝𝑙𝑢𝑛𝑔𝑒𝑟 +
1
2
𝜌𝑣1
2
= 𝑃𝑎𝑡𝑚 +
1
2
𝜌𝑣2
2

3. 3
𝑃𝑝𝑙𝑢𝑛𝑔𝑒𝑟 +
1
2
𝜌𝑣1
2
=
1
2
𝜌𝑣2
2
Solve for 𝑣2
You will find
80000𝑁/𝑚2
+
1
2
𝜌(4 × 10−4
𝑣2)
2
=
1
2
𝜌𝑣2
2
80000𝑁
𝑚2
=
1
2
𝜌𝑣2
2
−
1
2
𝜌(4 × 10−4
𝑣2)
2
=
1
2
𝜌(𝑣2
2
− 16 × 10−8
𝑣2
2)
80000𝑁
𝑚2
=
1
2
1000(1 − 16 × 10−8)𝑣2
2
𝑣2 = 12.6 𝑚/𝑠𝑒𝑐
Q6: How tall must a barometer be if it uses red wine (ρ = 984 kg/m3) as its working liquid?
Solution
𝑃 = 𝜌𝑔𝐻 → 𝐻 =
𝑃
𝜌𝑔
=
1.013 × 105
𝑃𝑎
1000𝑘𝑔/𝑚3 × 10𝑚/𝑠2
= 10.5 𝑚
Q7: A Venturi tube may be used as a fluid flow meter. If the difference in pressure is𝑃1 − 𝑃2 =
21𝑘𝑃𝑎, find the fluid flow rate in cubic meters per second, given that the radius of the outlet tube
is 1.00 𝑐𝑚, the radius of the inlet tube is 2.00 𝑐𝑚, and the fluid is gasoline, 𝜌 = 700 𝑘𝑔/𝑚3
.
Solution:
The flow rate is given by 𝑄 = 𝐴1 𝑣1
Thus we need to find 𝑣1 to able to find the flow rate, that is
𝑣1 = √
2 (𝑝1 − 𝑝2)
𝜌 [(
𝐴1
𝐴2
⁄ )
2
− 1]
→ √
2 × 21 × 103 𝑁/𝑚2
700 𝑘𝑔/𝑚3 [( 𝜋22
𝜋12⁄ )
2
− 1]
= 6.324 𝑚/𝑠𝑒𝑐
𝑄 = 𝐴1 𝑣1 = 𝜋(2 × 10−2)2
𝑚2
× 6.324
𝑚
𝑠𝑒𝑐
= 0.00795
𝑚3
𝑠𝑒𝑐
≅ 8 𝑙𝑖𝑡/𝑠𝑒𝑐
Q8: If the difference in pressure on the two sides of a closed door of area 2 𝑚2
is 0.01 atm, what
is the net force on the door? Do you think you can open the door by hand?
Solution
The net force is NmmNAP 2026210013.101.0 225
 
!!!!
Do you think you have the enough force to open this door! Good luck 

4. 4
Q9: What is the pressure difference between the heart and brain of a giraffe if the brain is 2 m
above the heart? (Assume that the velocity of the blood is the same in both locations).
33
100595.1 
 kgmblood .
Assume the velocity of the blood is the same, this means we can use
   
22
2
21233
39.2078739.20787
39.20787281.9100595.1
m
N
m
kgms
P
skgmmmskgmhgP
yygPP
blood
abba







Q10: Water is flowing in a circular pipe of varying cross-section. a. At one place the radius of the
pipe is 0.2 m. what the water velocity must be at this point if the volume flow rate in the pipe is
3 1
0.8m s
 b. At the second point in the pipe the water velocity is
1
3.8m s
 . What is the radius
of the pipe at this point?
Solution:
a. 3 1
1 0.2 , 0.8 , ?
V
r m and m s what isv
t

  

 
3 1
3 1 11
1 1 1 2 2
1
0.8
0.8 6.3662
0.2
A lV m s
Av m s v m s
t t A m

  
       
  
b.  
221
1 1 2 2 2 1 2 2
2
6.3662
0.2 0.26
3.8
v
Av A v A A r r m
v
        