Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book

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388 8Waves

8.3.9 Beat Frequency

8.81 When the man moves towards the source

(v + v

) f

When the man moves away from the source



(v − v

) f



∴ f

− f



( f

+ f



) + f

− f



1.5

330

(548 + 552) + 548 − 552 = 1s

∴ Beat frequency = 1/s.

8.82 Suppose the frequency of the unknown fork (unloaded) is n. Then n = 300±2

Case (i) Suppose n = 300 − 2 = 298

Let the frequency of the loaded unknown fork be n

and the loaded known

fork be n

300 − n

= 5(1)

298 − n

= 9(2)

Also frequency changes in both the forks are the same

∴ 300 −n

= 298 −n

or n

− n

= 2(3)

Subtracting (2) from (1)

− n

=−6(4)

Obviously (3) and (4) are inconsistent.

Case (ii) Suppose n = 300 +2 = 302

300 − n

= 5(5)

302 − n

= 9(6)

also 300 − n

= 302 −n

or n

− n

= 2(7)

Subtracting (5) from (6)

− n

= 2(8)

Thus (7) and (8) are consistent. Therefore, correct solution is n = 302.

8.3 Solutions 389

8.3.10 Waves in Pipes

8.83

L =

2 f

342

2 × 439

= 0.3895 m

The new frequency with the changed length L

= 439 +2 = 441

= 439 −2 = 437

2 f

342

2 × 441

= 0.3877 m

2 f

342

2 × 437

= 0.3913 m

The pipe must be shortened by 0.3895 − 0.3877 = 0.0018 m or 1.8 mm or

lengthened by 0.3913 − 0.3895 = 0.0018 m = 1.8 mm, so that 2 beats/s may

be heard when it is sounded with the fork.

8.84 Time taken for the plate to fall is

t =





2 × 10

980

Time period

T =

7 × 8

Frequency, f =

= 56 Hz

8.85

v = v

= v



t + 273

273

= 33150



20 + 273

273

= 34343 cm/s

L = (2N + 1)

(2N + 1)v

4 f

(2N + 1) × 34, 343

4 × 210

(N = 0, 1, 2)... (1)

Also L = (2M +1)

4 f

(2M + 1) ×34343

4 × 350

(M = 0, 1, 2 ...) (2)

Equating right-hand sides of (1) and (2) and simplifying

(2N + 1)

(2M + 1)

(3)

The choice of N = 1 and M = 2 satisﬁes (3). Using N = 1in(1)orM in (2)

gives L = 122.65 cm.

390 8Waves

Resonance with tuning forks of different frequencies is possible because

resonance for f = 210 occurs with the ﬁrst overtone of the tube and for

f = 350 it occurs with the second overtone.

8.86

f =

(open pipe, fundamental)

(open pipe, ﬁrst overtone)

(closed pipe, second overtone)

− f

= 100 (by problem)

−

= 100

∴ Fundamental frequency of open pipe, f =

= 200 Hz.

Chapter 9

Fluid Dynamics

Abstract Chapter 9 is concerned with ﬂuid dynamics comprising equation of

continuity, Bernoulli’s equation, Torricelli’s theorem, Reynolds number, viscosity

and terminal velocity.

9.1 Basic Concepts and Formulae

Steady ﬂow (laminar ﬂow): In this type of ﬂow, the velocity of the ﬂuid (liquid or

gas) at a point is always the same although the velocity of the ﬂuid may be different

at different points along the line of ﬂow.

Irrotational ﬂow: In this type of ﬂow the element of ﬂuid at each point has no net

angular velocity about that point. It implies the absence of eddies and vortices.

Incompressible ﬂuid ﬂow: A liquid is incompressible if its density is constant.

Stream lines are imaginary curves drawn through a ﬂuid to indicate the direction

of motion of the ﬂow of the ﬂuid.

Tube of ﬂow is a bundle of stream lines which deﬁne the boundary of the ﬂuid.

Turbulent motion and Reynold’s number: For any liquid through a pipe there

exists a critical velocity at which the laminar ﬂow suddenly changes into turbulent

type of ﬂow. The stability of ﬂuid ﬂow is described by a dimensionless quantity

called Reynold’s number (R).

R = ρvd/η (9.1)

where η and ρ are the viscosity and density of the ﬂuid, respectively, d is the diam-

eter of the pipe and v is the velocity. If

R < 2200, the ﬂow is steady

R = 2200, the ﬂow is unstable

R > 2200, the ﬂow is usually turbulent

The Equation of Continuity

The principle of conservation of mass leads to the equation of continuity. For steady

ﬂow, the mass of ﬂuid passing all sections in a stream of ﬂuid per unit time is

391

392 9 Fluid Dynamics

constant. At two points 1 and 2,

= ρ

= constant (one-dimensional ﬂow) (9.2)

For incompressible ﬂuids, ρ

= ρ

, and (9.2) is reduced to

Q = A

= A

= constant (9.3)

∇.ρv +

∂ρ

∂t

= 0 (equation of continuity in three dimensions) (9.4)

For steady incompressible ﬂow, ρ = constant and (9.4) r educes to

∂v

∂x

∂v

∂y

∂v

∂z

= 0 (9.5)

Bernoulli’s equation for steady, non-viscous, incompressible ﬂow

+ gh +

= constant (9.6)

or P + ρgh +

ρv

= constant (9.7)

In (9.6) the ﬁrst term P/ρ is the pressure head or the potential energy per unit mass

of the liquid due to the pressure, the second term hg is the elevation head or the

potential energy per unit mass of the liquid due to gravity and the third term v

is the kinetic energy per unit mass of the liquid. Thus, Bernoulli’s equation results

from the conservation of energy.

Torricelli’s theorem: A tank is ﬁlled with a liquid. An oriﬁce is located at the

side of the tank, at a depth h below the surface of the liquid. Then the velocity of

emergence of the liquid from the oriﬁce is given by

v =



2gh (9.8)

The venturi meter: It works as a gauge to measure the ﬂow speed of a liquid. Let a

liquid of density ρ ﬂow through a pipe of cross-sectional area A with velocity v.At

the constriction, called throat, the area is reduced to a (Fig. 9.1). A manometer con-

taining a suitable liquid of density serves to register the pressure difference between

points 1 and 2. In Bernoulli’s equation the gravitational energy term will be absent

as the centre of the cross-sectional areas A and a is at the same horizontal level.

Finally, we obtain

v = a



2(ρ



− ρ)gh

ρ(A

− a

)

(9.9)

9.1 Basic Concepts and Formulae 393

Fig. 9.1

The pitot tube is a device to measure the ﬂow speed of a gas. If ρ



is the density of

the liquid in the manometer, ρ the density of the ﬂowing gas and h the difference in

height in the limbs of the manometer, then

v =



2ghρ



(9.10)

Viscosity: The coefﬁcient of viscosity of a liquid is the tangential force per unit

area per unit velocity gradient. The backward tangential force acting on any liquid

layer is

F =−η A

(9.11)

where dv/dy is the velocity gradient which is identical with v/y for constant gra-

dient. The negative sign shows that the viscous drag acts opposite to the velocity of

the liquid.

Poiseuille’s Method for Viscosity Determination

Volume V ﬂowing per second through a tube of radius a and length L under pressure

head P is given by

V =

π Pa

8ηL

(9.12)

Terminal velocity

Drag force F = 6πηrv (Stokes law) (9.13)

When a sphere of radius r and of density ρ

is dropped in an extensive ﬂuid of

density ρ its speed increases linearly as for a free fall (v = gt ). However, due to

viscous drag v approaches asymptotically to a constant value v

given by

2gr

9η

(ρ

− ρ) (9.14)

394 9 Fluid Dynamics

9.2 Problems

9.2.1 Bernoulli’s Equation

9.1 The radius of a water pipe decreases from 10 to 5 cm. If the average velocity in

the wider portion is 4 m/s, ﬁnd the average velocity in the narrower region.

9.2 Verify if the continuity equation for steady incompressible ﬂow is satisﬁed for

the following velocity components:

= 3x

− xy + 2z

= 2x

− 6xy + y

=−2xy − yz +2y

9.3 Air streams horizontally across an aeroplane wing of area 4 m

, weighing

300 kg. The air speed is 70 and 55 m/s over the top surface and the bottom

surface, respectively. Find (a) the lift on the wing; (b) the net force on it.

9.4 A venturi meter has a pipe diameter of 20 cm and a throat diameter of 10 cm.

If the water pressure in the pipe is 60,000 Pa and in the throat is 45,000 Pa,

calculate the rate of ﬂow of water in m

/s.

9.5 A pitot tube which is used to determine the speed of an aircraft relative to

air is mounted on the wing of a plane. The tube contains alcohol of density

810 kg/m

and registers a level difference of 15.0 cm. Assuming that the den-

sity of air at NTP is 1.293 kg/m

, ﬁnd the plane’s speed in km/h relative to the

air.

9.6 A garden sprinkler has 80 small holes each 2.5mm

in area. If water is supplied

at the rate of 2 × 10

−3

/s, ﬁnd the average velocity of the spray.

9.7 For steady, incompressible ﬂow which of the following values of velocity com-

ponents are possible?

(a) v

= 3xy + y

= 5xy + 2x

(b) v

= 3x

+ y

=−6xy

9.8 If the speed of ﬂow past the lower surface of the wing of an aeroplane is

100 m/s, what speed of ﬂow over the upper surface would give a pressure dif-

ference of 1000 Pa? Assume an air density of 1.293 kg/m

9.9 A venturi meter has a pipe diameter of 4 cm and a throat diameter of 2 cm. The

velocity of water in the pipe section is 10 cm/s. Find (a) the pressure drop; (b)

the velocity in the throat.

9.10 Water is observed to ﬂow through a capillary of diameter 1.0 mm with a speed

of 3 m/s. Viscosity of water in CGS units is

(a) 0.018 at 0

◦

(b) 0.008 at 30

◦

9.2 Problems 395

Calculate the Reynold’s number and test at which of these three temperatures

is the ﬂow likely to be streamlined. Assume that for Reynold’s number R <

2200 ﬂow is steady.

9.11 A horizontal tube AB of length L, open at A and closed at B, is ﬁlled with an

ideal ﬂuid. The end B has a small oriﬁce. The tube is set in rotation in the hor-

izontal plane with angular velocity ω about a vertical axis passing through A,

Fig. 9.2. Show that the efﬂux velocity of the ﬂuid is given by v = ωl



− 1

where l is the length of the ﬂuid.

Fig. 9.2

9.12 A pitot tube, Fig. 9.3, is mounted along the axis of a gas pipeline of cross-

sectional area A. Calculate the rate of ﬂow of the gas across the section of the

pipe if h is the difference in the liquid column and ρ

and ρ

are the densities

of the liquid and the gas, respectively.

Fig. 9.3

9.13 Water ﬂows in a horizontal pipe of varying cross-section. Two manometer

tubes ﬁxed on the pipe, Fig. 9.4, at sections A

and A

indicate a difference

h in the water columns. Calculate the rate of ﬂow of water in the pipe.

9.14 A cylinder ﬁlled with water of volume V is ﬁtted with a piston and is placed

horizontally. There is a small hole of cross-sectional area s at the other end of

396 9 Fluid Dynamics

Fig. 9.4

the cylinder, s being much smaller than the cross-sectional area of the piston

(Fig. 9.5). Show that the work to be done by a constant force acting on the

piston to squeeze all water from the cylinder in time t is given by

W =

ρV

where ρ is the density of water. Neglect friction and viscosity.

Fig. 9.5

9.15 A cylindrical vessel with water is rotated about its vertical axis with a constant

angular velocity ω. Show that

(a) the water pressure distribution along its radius is given by P = P

ρω

, where ρ is the density of water and P

is the pressure at the

central point.

(b) Show that the ﬁgure of revolution of water is a paraboloid.

9.16 A manometer is ﬁxed to a water tap. When the valve is closed the manometer

shows the reading of 3.5×10

Pa. When the valve is open the reading becomes

3.1 × 10

Pa. Find the speed of water.

9.2.2 Torricelli’s Theorem

9.17 A water container is ﬁlled up to a height H. A small hole is punched at the

side wall at a depth h below the water surface. Show (a) that the distance from

the foot of the wall at which the stream strikes the ﬂoor is 2

√

h(H −h); (b)

the second hole through which the second stream has the same range must be

punched at a depth H − h.

9.2 Problems 397

9.18 In prob. (9.17) show that the hole must be punched at a depth h = H/2for

maximum range and that this maximum distance is H.

9.19 A large tank is ﬁlled with water at the rate of 70 cm

/s. A hole of cross-section

0.25 cm

is punched at the bottom of the tank. Find the maximum height to

which the tank can be ﬁlled.

9.20 A tank of cross-sectional area A is ﬁlled with water up to a height h

. Water

leaks out from a small hole of area ‘a’ at the bottom. Find the time taken for

the water level to decrease from h

to h

9.21 A l arge tank is ﬁlled with water. The total pressure at the bottom is 3.0 atm. If

a small hole is punched at the bottom what is the velocity of efﬂux?

9.22 Two tanks with a large opening are ﬁlled with a liquid. A hole of cross-

sectional area A

is punched in tank 1 and another of cross-sectional area

in tank 2 at depths h

and h

, respectively. If A

= 2A

and the volume

ﬂux is identical, then what should be the ratio h

/ h

9.23 A wide container with a small oriﬁce in the bottom is ﬁlled with water and

kerosene. If the water column measures 60 cm and kerosene column 40 cm,

calculate the efﬂux velocity of water. Take the speciﬁc gravity of water as 1.0

and kerosene as 0.8 and neglect viscosity.

9.24 A wide vessel ﬁlled with water is punched with two holes on the opposite side

each with cross-sectional area of 1.0cm

. If the difference in height of the

holes is 51 cm, calculate the resultant force of reaction of the water ﬂowing

out of the vessel.

9.2.3 Viscosity

9.25 Water is conveyed through a tube 8 cm in diameter and 4 km in length at the

rate of 120 l/min. Calculate the pressure required to maintain the ﬂow. Coefﬁ-

cient of viscosity of water, η = 0.001 SI units. 1 atm = 1.013 ×10

Pa.

9.26 Two capillary tubes AB and BC are joined end to end at B. AB is 16 cm long

and of diameter 0.4 cm. BC is 4 cm long and of diameter 0.2 cm. The compos-

ite tube is held horizontally as in Poiseuille’s experiment, with A connected to

a vessel of water giving a constant head of 3 cm and C open to air. Calculate

the pressure difference between B and C.

9.27 Two raindrops fall through air with terminal velocity of v

cm/s. If the drops

coalesce what will be the new terminal velocity?

9.28 Q cm

of water ﬂows per second through a horizontal tube of uniform bore

of radius r and of length l. Another tube of half the length but radius 2r is

connected in parallel to the same pressure head. What will be the total quantity

of water ﬂowing per second through these two tubes?