9.3 Solutions
407
P =
8ηlV
πr 4 =
8 × 0.001 × 4000 × 0.002
3.14 × (0.04) 4
= 0.0796 × 10
5 Pa
Pressure head h =
P
ρg
=
0.0796 × 10 5
1000 × 9.8
= 0.8 m
9.26 P A − P B =
8ηl 1 Q
π r 4 =
8ηQ
π
(0.16)
(2 × 10 −3 ) 4 =
8ηQ (0.01)
π × 10 −12
(1)
P B − P 0 =
8ηQ × (0.04)
π × 10 −12
(2)
Adding (1) and (2)
P A − P 0 =
8ηQ × 0.05
π × 10 −12
(3)
Dividing (2) by (3)
P B − P 0
P A − P 0
= 0.8
∴ P B − P 0 = 0.8 × (P A − P 0 ) = 0.8 × 3 = 2.4 cm of water.
9.27 The terminal velocity v T is given by
v T =
2
9
r
2 g
(ρ 1 − ρ 2 )
η
(1)
where r is the radius of the drop, ρ 1 and ρ 2 are the densities of the drop and
air, respectively, g is the gravity and η is the coefficient of viscosity. If the new
radius is r and the new terminal velocity v
T , then
v
T
v T
=
r 2
r 2
(2)
Under the assumption that the drops are incompressible, the volume remains
constant:
4π
3
(r
)
3
= 2 ×
4π
3
r
3
∴ r
= 2
1/3 r
(3)
Using (3) in (2)
v
T = 2
2/3
v T = 4
1/3
v T
407
P =
8ηlV
πr 4 =
8 × 0.001 × 4000 × 0.002
3.14 × (0.04) 4
= 0.0796 × 10
5 Pa
Pressure head h =
P
ρg
=
0.0796 × 10 5
1000 × 9.8
= 0.8 m
9.26 P A − P B =
8ηl 1 Q
π r 4 =
8ηQ
π
(0.16)
(2 × 10 −3 ) 4 =
8ηQ (0.01)
π × 10 −12
(1)
P B − P 0 =
8ηQ × (0.04)
π × 10 −12
(2)
Adding (1) and (2)
P A − P 0 =
8ηQ × 0.05
π × 10 −12
(3)
Dividing (2) by (3)
P B − P 0
P A − P 0
= 0.8
∴ P B − P 0 = 0.8 × (P A − P 0 ) = 0.8 × 3 = 2.4 cm of water.
9.27 The terminal velocity v T is given by
v T =
2
9
r
2 g
(ρ 1 − ρ 2 )
η
(1)
where r is the radius of the drop, ρ 1 and ρ 2 are the densities of the drop and
air, respectively, g is the gravity and η is the coefficient of viscosity. If the new
radius is r and the new terminal velocity v
T , then
v
T
v T
=
r 2
r 2
(2)
Under the assumption that the drops are incompressible, the volume remains
constant:
4π
3
(r
)
3
= 2 ×
4π
3
r
3
∴ r
= 2
1/3 r
(3)
Using (3) in (2)
v
T = 2
2/3
v T = 4
1/3
v T