GJIAPTEH.

l:J

FLUID MECHANICS

13.1 FLUIDS of the forces be F. We define the pressure of then .

at the point A as \lid
. Ma~te~ is broadly divided into three categories, . F
P= Llill AC'.'
sohd, hqwd and gas. The intermolecular forces are ,1$...0 = (13.1
strong in solids, so that the shape and size of solids homogeneous and nonviscous fluid
~o ?0 t _easily change. This force is comparatively less For a h . . , ti.:
quantity does not depend on t e o_nentat1on of 68 'Ill
m hqwds and so the shape is easily changed. Although
alk of pressure at a point. For such a ~
the sh~pe of a liquid can be easily changed, the volume h ence we t . h · nl 0\lid
of a given mass of a liquid is not so easy to change. pressure is a scalar quantity avmg o y magnitude '
I~ n~eds quite a good effort to change the density of
liqwds. In gases, the intermolecular forces are very Unit of Pressure
. -2
small and it is simple to change both the shape and The SI unit of pressure 1s N m called pascal and
the density of a gas. Liquids and gases together are
abbreviated as Pa.
called fluids, i.e., that which can flow.
In this chapter we shall largely deal with liquids. Variation of Pressure with Height
The equations derived may be applicable to gases in
many cases with some modifications. We shall assume
that the liquids we deal with are incompressible and r2~S2
T c::::: •B
nonuiscous. The first condition means that the density
dz
of the liquid is independent of the variations in
pressure and always remains constant. The second 1
condition means that parts of the liquid in contact do
not exert any tangential force on each other. The force
by one part of the liquid on the other part is
Figure 13.2
perpendicular to the surface of contact. Thus, there is
no friction between the adjacent layers of a liquid.
Let us consider two points A and B (figure 13.2
13.2 PRESSURE IN A FLUID
separated by a small vertical height dz. Imagine a
horizontal area 68 1 containing A and an identical
horizontal area 682 containing B. The area
68 1 = 682 = 68. Consider the fluid enclosed between
the two surfaces 68i, 682 and the vertical boundary
joining them. The vertical forces acting on this flwd
are
Figure 13.1 (a) F 11 vertically upward by the fluid below it
(bJ F2 , vertically downward by the fluid above it
Consider a point A in the fluid (figure 13.1). and
Imagine a small area 68 containing the point A. The (cJ weight W, vertically downward.
fluid on one side of the area presses the fluid on the Let the pressure at the surface A be P and the
other side and vice versa. Let the common magnitude pressure at B be P + dP. Then
 l'A < Al I.A~

l
I ii! II l h
I II nuld
\\ I
fl ( \." tel l
, t11rnl N:llllhb1mm,
t
J,'I 1,' + \\

p ~ (/> -+ cfP ) \S + p B(rlz) AS mu

d/> llig dz ( I3 2)
JI urc d1
gwc pomt.s nc
,J _ ti
,,A c \\ l' mo, C dup through cl ht•1ght ~ 1c pressure moy be stntcd as follow
1'11\'a:t ~ b) ~g z. "hl•n• 11 is tht• density of the nu1d If the prtssurc 111 a lu1u1d l8 cl,ang d at a
,bat l"' IJ\l, part1r11lar point, the change is transmitt d to the tnl1re
1
'd~r t \\O points
~ J\\ Nlll~l
· nt z = 0 and z = /I , If t he liquid ,, 1tlwut bc111g d1mu11shcd in magnitude
• .
,,al'l' nt z = 0 1s P1 and that at z = h is p th , I
· (132) 2' tn
rquat1on . ♦F
P2

f dP =J- pg dz

or.
0

If the density is same everywhere F1gun• 13 4

P2 -P1 =- pgz
As an example, suppose a flask fitted with a piston
or, . . . (13.3)
is filled with a liquid as shown in figure (13.4). Let an
Next consider two points A and B in the same external force F be applied on the piston. If the
horizontal line inside a fluid. Imagine a small vertical cross-sectional area of the piston is A. the pre,,:,ure
area 1$1 containing the point A and a similar vertical just below the piston is increased by F I A . By Pascal',
area 1.52 containing the point B. law, the pressure at any point B will also increase by
the same amount FI A. This is because the pressure
at B has to be pgz more than the pressure at the
piston, where z is the vertical distance of B below the
piston. By applying the force we do not appreciably
change z (as the liquid is supposed to be
Figure 13.3 incompressible) and hence the pressure difference
remains unchanged. As the pressure at the piston is
increased by FI A, the pressure at B also increases b,
The area /:JS 1 = /:JS 2 =68. Consider the liquid the same amount. ·
rontained in the horizontal cylinder bounded by /:JS 1
Pascal's law has several interesting applicanons
illld 1-5,. If the pressures at A and B are P1 and P2
Figure (13.5) shows the principle of a h\draulil :·?
re~pectively, the forces in the direction AB are used to raise heavy loads such as a car.·
{a) P16S towards right and
~F
(b) P.j).S towards left.
If the fluid remains in equilibrium, A A1

P1/:JS = P2/:JS.
or, P, = P2
Thus, the pressure is same at two points in the V
lame horizontal level.
r'ii:un• 13.5
 •
11 lll f/11 1(1 1)' I l11 1111 18 f
A nnd 11 or th t11hu f ill11 ,1,,w11 •
h)
l'OIIIII rt( (I
1111 d III both th•
f1 rm lixucl \\ 1th Il11
hquul I fil lu l 111 the
llld In lh1 ho11 zo11I 11 t11h1
J f 111111 1 l "hen pr 1 8 1·d
1

IC p1Rlm1 \ 1 ll'd h) 11 fon <'

l'H!nwht•rc by h gurt• J:J.fi
l th, hqu1d llll'
n F A Tlw , nh c \ 1s open and ·the liquid
flo,, th, c, hnde, B. It exert:-; an extra force Figure n3.6J shows schemati ~ally thr• itu;it,
F. on the larger piston in the upward direction Th upper part of the tube conta ins vacuurn
ecury goes down and no air isf allowed in Th as trie
,,h,ch ra1, c, the load upward. mer h us, Ui
pressure at the upper end A o t e mercury col l(:
The ad, nm a ge of this method is that if A is much · ·de the tube is P,.. = zero. u~
Let us consider a po·Int (
IDS! .
larger than A 1 • e,·en a small force F I is, able to on the mercury surface m the cup and another • Pei
generate a large force F, which can raise the load. It B in the tube at the same horizon~al level, ;:
may be noted that there is no gain in terms of work. pressure at c is equal to th~ atmosph enc pressure. Ai
The work done by F 1 is same as that by F 2 • The piston B and c are in the same honzonta l level, the pressures
A has to traverse a larger downward distance as at B and C are equal. Thus, the _p ressure at B is ~
compare d to the height raised by B . to the atmospheric pressure Po m the lab.
Suppose the point B is at a depth H below A If
13.4 ATMOSPHERIC PRESSURE AND BAROMETER . f p
be the density o mercury,
The atmosph ere of the earth is spread up to a PB =PA+ pgH
height of about 200 km. This atmosphere presses the or, Po= pgH · ... (13.4
bodies on the surface of the earth. The force exerted
by the air on any body is perpendicular to the surface The height H of the mercury column in the tube above
of the body. We define atmospheric pressure as follows. the surface in the cup is measure d. Knowing the
Consider a small surface .:!,S in contact with air. If the density of mercury and the acceleration due to gravitv
force exerted by the air on this part is F, the the atmospheric pressure can be calculated us~g
atmosph eric pressure is equation (13.4).

Po=~ :s · The atmospheric pressure is often given as the

length of mercury column in a barometer. Thus, a
pressure of 76 cm of mercury means
Atmospheric pressure at the top of the atmosphere 3 2
3
is zero as there is nothing above it to exert the force. P0 = (13·6 x 10 kg m ) (9·8 m s ) (0·76 m)
The pressure at a distance z below the top will be 5
z = 1·01 x 10 Pa.
j pg dz. Remember, neither p nor g can be treated as This pressure is written as 1 atm. If the tube i,
0
constant over large variations in heights. However, the insufficient in length, the mercury column will not fall
density of air is quite small and so the atmospheric down and no vacuum will be created. The inner surface
pressure does not vary appreciably over small of the tube will be in contact with the mercury at the
distances. Thus, we say that the atmospheric pressure top and will exert a pressure P,.. on it.
at Patna is 76 cm of mercury without specifying
Example 18.1
whether it is at Gandhi Maidan or at the top of
Golghar. Water is filled in a /7.ask up to a height of 20 cm. Tht
Torricelli devised an ingenious way to measure the bottom of the /7.ask is circular with radtus 10 cm If tht
atmosphe ric pressure. The instrume nt is known as atmospheric pressure 1s 1·01 x 10 ' Pa, find the fort1
I
barometer. exerted by the waler on the bottom. Take g 10 m s and
In this, a glass tube open at one end and having density of water - 1000 kg m .
a length of about a meter is filled with mercury. The
open end is temporar ily closed (by a thumb or Solution : The pressure at the surface of the water is equsl
otherwis e) and the tube is inverted in a cup of to the atmospheric pressure P . The pressure at iht
mercury. With the open end dipped into the cup, the bottom is
 A, h1111r.d111' ptin pl s tt l
I' ♦ lO (l
n1) ( IllO(l kg In ) '1 'l '"" twlly ,.,, fulfy dlJIII d lflto a fluid at ,
I" • • 111 I )
l x l0 1 ♦ 002,._ lO ' l'a cxutll ,,,, 111111 a11/ /u1cc of b1,oyo, cy tqual I
!Ix l(l
I
P11 ,,1,,,
11{ ti,, ,1, r!tl /711ul
111'\'ll af tl,,, b..111 nm It r • I
Anh1nlf'd,.s' pr111c1pl i n t n nd
' 14 )( l{) I 1111 I , 111ripll' und rnny b d• duccd from N wton I I
11
00;!) 1 Ill 1 motion .
(bl' fol'('I' ,,n tlw hot tom ,~. tlwrl-for~. · Consider the situut10n shown m figur 13.8 w
F P 1t r a body is shown dipped into a fluid Suppose th
l ()3 x IO'Pa) x 100314 m 'l- 3230 N dipped in the fluid is replaced by the me fluid of
equal volume. As the entire fluid now becom
)!in,,,..,etrr
.. homogeneous, all parts will remain in eqmhbrium The
part of the fluid substituting the ~dy als~ re~am m
,t:tnl1mcter is a simple device to equilibrium. Forces acting on this substJtutmg fluid
·' • 1~ d measure the
.ure 111 a c o:se vessel containing a g It .
f '· h · • as. consists are
I.• r-tube avmg some liquid. One end of the tube
(a) the weight mg of this part of the fluid, and
, ,,pen to the atmosphere and the other end is
:l)lllli>Cted to the vessel (figure 13.7). (b) the resultant B of the contact force:o by the
remaining fluid.
As the substituting fluid is in equilibrium. the.:,e
two should be equal and opposite. Thus,
B = mg (13.5)
and it acts in the vertically upward direction.
Now the substituting fluid just occupies the space
Figure 13.7
which was previously occupied by the body. Hence, the
shape of the boundary of the substituting fluid is same
The pressure of the gas is equal to the pressure as the boundary of the body. Thus, the magnitude and
at A direction of the force due to the pressure on any small
= pressure at B area of the boundary is same for the body as for the
= pressure at C + hpg substituting fluid. The force of buoyancy on the body
= P0 +hpg is, therefore, same as the force of buoyancy B on the
substituting fluid.
when P0 is the atmospheric pressure, h = BC is the
difference in levels of the liquid in the two arms and From equation (13.5) the force of buoyancy on a
pis the density of the liquid. dipped body is equal to the weight mg of the displaced
fluid and acts along the vertically upward direction.
The excess pressure P- P0 is called the quage This is Archimedes' principle.
pressure.
Note that in this derivation we have assumed that
13.5 ARCHIMEDES' PRINCIPLE the fluid is in equilibrium in an inertial frame. If 1t i$
not so, the force of buoyancy may be different from the
When a body is partially or fully dipped into a weight of the displaced fluid.
fluid, the fluid exerts forces on the body. At any small
~rtion of the surface of the body, the force by the fluid Floatation
G perpendicular to the surface and is equal to the
pressure at that point multiplied by the area (figure When a solid body is dipped into a fluid, the fluid
lJ.8). The resultant of all these contact forces is called exerts an upward force of buoyancy on the solid. If the
the force of buoyancy or buoyant force. force of buoyancy equals the weight of the solid. the
solid will remain in equilibrium. This is called
fioatation. When the overall density of the solid is
smaller than the density of the fluid, the solid floaG
with a part of it in the fluid. The fraction dippt>d is
such that the weight of the displaced fluid equals the
weight of the solid.
Figure 13.8
 ,~n ,p ll ,,I 1•J,y1 c• f ~w 8 , , thrc,• forc
es th, lie d
. 1
11dJ OJ
I II
with flf J fH •c,•lcrotwn a JI r
[Jndt •r, 1h o 1
, 1,pwurc
J
ll( Tl ')t 'fl ltl ll/,:
ccond uw, • c mg m a
, 8
() r,11 /71Hl f ,t
N( •wton " p t,S 1 2{. l,J
t11i J t I• /Ill f;)Zp(g + ao)
; t:Ji ,n(g + a 0 ) - (t,
,
" an rd i.•' l!{Ieug '
, r bi ha tin..., e ru be ,.~ "" llll ' •.. Oat,
H murl1 ml11 mr of th
~
or (P , I i)p pi n+ a o) Z.
' z- 16
• l 1)(l(l kg 111 • Pi
? 0..-IISJt) of 1m trr I cc•d
by . tlw or,
,,r th e n1 lic• IS hitJnn ig.ht . . .
~ •h M I Th
e W t'~
Th e
ht
bt111, an t fon
•
·,• 1!1 t•q•un
h
t °b 1·Rwe
th e
ins ide (b) Bu oy an t Fo rc
e
. .
s1de a liquid of
body 1s di pp ed m g up with
buornnt fon "C -. <' cu e • go in
• . II a , ,ilu mt• \' of t pposde ~n an el evat or ~
h<> WBl<'I d1<:pla,.,.,J
oft ccd wa ter = Vpg, Now supl e of buoyancy
ight of th e di~pla rc
e 1 fo
• ac t us ca lc ul at e th e
tht. water, the we density ~ a Le
r. Th ,
us le t us supJlOse
\\-hC'rtl f' ,, th
r de11~1ty of " ate n do· AB w as do ne ea rl ie r,
ac ce lera~ JO ui d by the sam
th is bo .['t
th e body in to th e liq ui d becomes e
B on liq
s Jume Th e en tir e
0·7 kg 7 10 • m ' -- 700 cm
'.
th at we su b ti 1u e th e su bs tit ut ed liquid -~
0· , kg X an d he nc e
\ •= - - = 1000 kg m vo f th e qw'd . Th us. the
,4 = IS
liq ui d of eq ua Ii
or , p 3
m as s
3
10 00 cm · ou s
homogen ·th es pe ct to th e · re st o I
cm ) = ·th
of the cu be = (1 0 r wi _d an acce eration
Th e to tal rn lu me at re st WIed liq uid is al so go mg h up li
th e wa ter is substitut e qw · . .
Th e rn lu me out.s ide 1
3 w ith th e re st o ft .
= 30 0 cm • a 0 toge th er liqwd are
1000 cm - 700 cm on th e su bs tit ut ed
2

The forces ac tin g

rce B an d
(a ) the buoyant fo liquid.
BUOYANT of th e su bs tit ut ed
PR ES SU RE DI FFERENCE AND (b) the weight m g
13.6 DS
LERATING FLUI cond la w
FORCE IN ACCE From Newton's se
were derived by B -m g= m a 0
Equations , 13 .3 ) an d (13.5 ) ation is in .. . 03.i/
at th e flu id under consider B = m (g + ao)
assum in g th not the case, or,
in an in er tial frame. If this is (1 3. 7) ar e si m ila
r to the
equilib rium discuss some an d
ua tio ns m us t be modified. We shall Equatio n (1 3. 6)
fo r un ac ce le ra te d liq
uid '1\-ith
the eq eq ua tio ns
celerating fluids. correspondi ng th e role of g.

I special cases of ac
A Li qu id Pl ac ed
in an El ev at or

es su re Differenc
(a l Pr
e
th e on ly differnce

Fr ee Su rf ac e of
th at g + a ta ke s

a Li qu id in H or
iz on ta l Acceleration

ac ed in a be ak er
which IS
d it is liq ui d pl
e a be ak er co nt ains some liquid an Cons id er a
w ith an accelera
tion a
Suppos th an on ta lly
at or wh ich is going up wi ints acceleratin g ho riz
in ts in the liquid
placed in an el ev B be two po A an d B be tw o po
re 13.9). Let A and (figure 13.10). Le t ta l lin e along the
acceleration a0 (figu at a vertical height z above A. tio n l in th e sa m e ho riz on
ing at a se pa ra th e pressure
in the liquid, B be nd A an d an sh al l fi rs t ob ta in
all ho riz ontal ar ea t1S arou ac ce le ra tio n a 0 • We
Cons tru ct a sm truct a vertical th e po in ts A an d
B.
tal ar ea around B. Cons difference between
eq ua l ho riz on . Consider th e
th e tw o areas as the faces j
cy lin de r "'i th is cylinder. Le t 1----J--
contained within th
motion of the liquid pressure at B .
at A and P 2 be the
P 1 be the pressure

Fig ure 13 .10

d
ct a JI ·ca l ar ea !}.S around A an
rti
Figure 13.9 Co ns tru 8 m a ve _J
d B · C ons1. der th e liq ui d conta.inf'II
an eqeuahl ar. ea ar o un the
d m th e th th e two ar eas as
the liquid containe in
a1 cylin de r w ith
. Force~ acting on cal direction, ar e : flat fa ce /7 ;o ntth be P 1 an d th e pres~ Lft
cylm der, m the verti at B be t e pr es su re at A
due to the liquid be
low it g th e lin e A B ar e
(a) P ,.l.S, upward P2• Th e forces al on ff
uid above it and ~s-ng · ht du e to th e liq ui d on the le
(b ) p ~ . down
ward due to the liq .
WP 1 M~ warw
wh an d
(c) weight m g =
(.l.SJ.zpg downward ' er e P the
18 · bt
density of the liqui
d. J ft
(b ) P:A,S ~Wards e du e to th e liq ui
d on the ng
 1 u IM It nl 1

11, n th,or fi111

1 1111 fiq II tel the 1,nrtlcl1 A
, lmdc1 lrN-1\
1 • t 1111111!:(li II, 1111 11 nil p
ll ,nd Im u 11i11lJ t11,\111d 1 II( Ji I I hr 011gh II ~Ill h fl n,,w ,,f fluid I ( fl
•
l'a \S I' \."i
1/1(10

(/'a J' \ \.\ I \." )l1l11 II

.
I\ /11 /1'<10.
( I:1 Hl
•
A
►

'1(1lC' t,Hl p,1mt:- Ill t lw ~anw hor·•inn t nI Ime do t

. no
t'\}ll,l I pn•~~m·t• if tlw lilt d . ~,gurc HI II
, II u1 is ncceleraled
ontn )
.\~ th~n' •~ no YCrt1cul acceleration the . In steady flow the velocity of fluid particles
j,., , ahd lf the atmospheric p ' . equation reaching a particular point is the same at all time
1.i.3 . ressure IS p the
:-e-::ure at A. IS P_ =Po+ h1pg and the pressureo,at B Thus, each particle follows the same path as taken by
P· a previous particle passing through that point.
p =Po .j. Ji;pg, where h, and h2 are the d epths of A
- '
.-d B from the free surface Substitut·mg m • (13.8) If the liquid is pushed in the tube at a rapid rate,
s:. · the flow may become turbulent. In this ca:,e, the
h,pg - h2pg = lpao velocities of different particles passing through the
h , -h2 ao same point may be different and change erratically
or. -r=- g with time. The motion of water in a high fall or a fast
flowing river is, in general, turbulent.
or. tane = ao
g
~here 9 is the inclination of the free surface with the
Steady flow is also called streamline fl.ow.

Line of Flow : Streamline

j
horizontal.

J3.7 FLOW OF FLUIDS

The path taken by a particle in flowing fluid is
called its line of flow. The tangent at any point on the
j
line of flow gives the direction of motion of that particle
The.flow of fluid is in general a complex branch of
mecharucs. If you look at the motion of water in a fall
like Rallah fall near Manali or Kemti fall near
at that point. In the case of steady flow, all the
particles passing through a given point follow the same
path and hence we have a unique line of flow passing
,
through a given point. In this case, the line of flow is
~foussurie) the view is very pleasant. The water falls
from a height and then proceeds on a flat bed or a
also called a streamline. Thus, the tangent to the
streamline at any point gives the direction of all the
1
slope with thumping, jumping and singing if you can
particles passing through that point. It is clear that •
appreciate the music. But if you try to analyse the
motion of each particle on the basis of laws of
two streamlines cannot intersect, othenvise the •
particle reaching at the intersection will hav~ two
mechanics, the task is tremendous ly difficult. Other different directions of motion.
examples of fluid flow are the sailing of clouds and the
motion of smoke when a traditional Chu.Iha using coal, Tube of Flow
wood or goitha (prepared from cowdung) in an Indian
1illage is lit. The motion of each smoke particle is Con~ider an area S in a fluid in steady flow. Draw
governed by the same Newton's laws but to predict the streamlines from all the points of the periphen· of S.
motion of a particular particle is not easy. The~e streamlines enclose a tube, of which S is ~ cross-
sect10n. Such a tube is called a tube of flou. As the
streamlines do not cross each other, fluid flowing
!3.8 STEADY AND TURBULENT FLOW through differnt tubes of flow cannot intermix
although there is no physical partition between th~
Consider a liquid passing through a glass tube
tubes. When a liquid is passed slowly through a pipe
!gure 13.11). Concentrate on a particular point A in
the pipe itself is one tube of flow. ·
e tube and look at the particles arriving at A. If the
ieloeity of the liquid is small, all the particles which
:rne
8
~ A will have same speed and will move in same
rection. As a particle goes from A to another point
p· •~s speed and direction may change, but all the
@ -§ --
al!articles reaching A will have the same speed at A and
at ~he Part!cles reaching B will have the same speed
-----------
Figure 13.12
· Also, 1f one particle passing through A has gone
 a
1 ,11 pt1 r 1'1, I•

1011" '1 1h1• I I II I fll I L11,11 flt ,, I A

muo, \11tl:-..A1 I I (I\\ 01 \N lltl' 111 Iw 111 ,. "
11111st hnvc
lt'."l O:\ll'IU ss1111 l '"I)
NON\ IS( 'Ol lH Fl.lJII>
,
1111' 11111 I I H l
r1<·ornflll' •11,11 w,,
A 1u,t1I A~n~t.l
' n h 1 11 f tlw 1111" ul 11 1111111 h1•rn11wH 11111ch A 1u1 Ai'i I'I
n~
Jhficd if ,,, n1n 11!1•1 tlw 111111I tu hp 1111·om111·t• 11111 hll- r
'/'h i' product of thl' area o <·ross ~l't'ittm ,1,111
nd nom I N.111 11111I !hut tlw flm'v ' " 11 rnl11lt01111I
.~Jll'l'cl rl'lnai 118 the same. at all points of a lulu, ,,{~I.
In mpn: 1h1ht, nwun 'I lh,11 tlw dt>nHtlv nf till' thud 'fl · J·8 coiled the equatwnf of continuity and <:x fir
1 me ot nil t lw Jl\llntK ,ind n•nH1111s t·nnHtnnl IIH timr 11 H . fl .
the law of conserva tion o mass m u1d dynam,ce
~ c This nssump1tn11 ,., quilt• good for ltqu1ds and
1" \'Uhd m u rtmn l'UM'H of flow of gases. Viscosity of ........._
Example 13.3
n tlu1d 1:; rr.l:ltt•d lo tlw mlcrnul friction when a layer
of fluid slips o, l'r another layer. Mechanic al energy is Figure (13.14) shows a liquid being pushed out of a 1 ~
piston. The area of cross sectzon 0f u
lo~t ngninst :-.uch v1:-.cous forces. The assumpti on of a by Pressing a 2 f 11i,
nom,scou :,. fluid will mean that we are neglectin g the ·ston is 1·0 cm and that o the tube at the Ou~~1,•.
pi
effect of :-.uch internal friction. Irrotatio nal flow means is pushed at a speed of 2
20 mm • If the piston
2

there ts no net angular velocity of fluid particles. When what is the speed of the outgoing liquid ? CJn
l ou put some washing powder in a bucket containin g I\
"ater and mix it by rotating your hand in circular
path along the wall of the bucket the water comes into
rotational motion. Quite often w~ter flowing in rivers
show small vortex formatio n where it goes in rotationa l Figure 13.14
motion about a centre. Now onwards we shall consider
only the irrotatio nal motion of an incompre ssible and Solution : From the equation of continuity
nonvisco us fluid. A 1v1 =A2v2
2 1 2
or, (1'0 cm ) (2 cm s- ) = (20 mm )v2
2
13.10 EQUATION OF CONTINUITY l·O cm -1
or, v2 = 2 x 2 cm s
20mm
We have seen that the fluid going through a tube 2
100 mm -1 -1
of flow does not intermix with fluid in other tubes. The = 2 x2cms =lOcms
.
20mm
total mass of fluid going into the tube through any
cross section should, therefore, be equal to the total
mass coming out of the same tube from any other cross 13.11 BERNOULLI EQUATION
section in the same time. This leads to the equation
Bernoulli equation relates the speed of a fluid at a
of continuit y.
point, the pressure at that point and the height of that
Let us consider two cross sections of a tube of flow point above a reference level. It is just the application of
at the points A and B (figure 13.13). Let the area of work-ene rgy theorem in the case of fluid flow.
cross section at A be Ai and that at B be A2, Let the
We shall consider the case of irrotational and
speed of the fluid be Vi at A and v2 at B . steady flow of an incompre ssible and nonviscous liqwd

-JC==lf
v1t.t
A1
Figure (13.15) shows such a flow of a liquid in a tube
of varying cross section and varying height. Consider
the liquid containe d between the cross sections A and
B of the tube. The heights of A and B are h 1 and h.
respectiv ely from a reference level. This liquid
advances into the tube and after a time t:J, is contained
Figure 13.13
between the cross sections A' and B' as shown in figure

How much fluid goes into the tube through the

cross section at A in a time interval Af ? Let us
construc t a cylinder of length v1Af at A as shown in B'

the figure. As the fluid at A has speed v 1, all the fluid

included in this cylinder will cross through A 1 in the
time interval Af. Thus, the volume of the fluid going
into the tube through the cross section at A is
Figure 13.15
A v Af . Similarly , the volume of the fluid going out of
1 1
 -
th \ll'fl l!r, l 1011 Ill \
11r ',,, !I 'I Ii• 1h 11w•· 111 kin t,c en rgy <K F
11f
\I
m:n '"' 11111111 11 1 H \ l11p111I Ill 11111 1 A / I
th pc,·,l ,,r Ilw liqu11! ,II , \ 1 K 1., of 1 /Ill I< I· 111 AA'//
tl11, ~,l<·,·d ,,1 t lw liquid 111 II ''a1 I< I~ of \'/I .,. KI•, 1>f /Ill' K f, of M ' KE of A'IJ
t)II' J'I ,•,~UH' ,It J\ /1 1
pr,•~~un' at H /\ K.E. of BH' K.E. of AA'
111,,
sud till'
"" ,
d,•n~it.' of tlw liquid fl.
di~t.inCl' and the d'1stance
. \rl' t• 1M
= zJ 2
(Ml ) l'o -
1
2 (/J.n1 JV1.
2 (111

,r ,11
Since the flow is assumed to be steady, the speed
\.'. 'fh,, Yolume between A and A' 1·s A U = A,
an d
at any point remains constant ~ ~iI?e and hence _the
tr 1 1

~ ,'l,lu
me ~tween B and B' is A u M
2 2 • By the t·
equa 10n KE. of the part A'B is same at m!t1al_and final time
j('llntinu1t_\ •
A1U1M:: A2U2Af . and cancels out when change in kinetic energy of the
system is considered.
fhe 01ass of this volume of liquid is
By the work-energy theorem, the total work. do~e
~n =pA1u,M =PA2u2Af. . .. (i) . equal to the change in its kinetic
on the system 1s
!,et us ca!culate t~e total work done on the part energy. Thus,
r the liquid Just considered.
The forces acting on this part of the liquid are P1 [ ~ )-P{~ )+ (!J.m)gh1- (6m'}gh2
ta1p 1A1, by the liquid on the left 1 2 1 2
ibl P2A2, by the liquid on the right = 2 (/J.m)U2 - 2 (6m )U1
ic) (j11t)g, the weight of the liquid considered and
or,
1 d) "'\', contact forces by the walls of the tube.
In time Af, the point of application of P1A 1 is 1 2 1 2
(13.10)
or, P1 + pgh1 + PU1 =P2 + pgh2 + 2 PU2
di;placed by AA'= U1M. Thus, the work done by P1A1 2
ID tiJlle M is 1 2
(13.11)
P + pgh +
or,
2 pu = constant
W1 = <P1A1) (u 1M) =P{~}
This is known as Bernoulli equation.
Similarly, the work done by P2 A2 in time !J.t is
Example13.4
W2 = - (P2 A2) (u 2M) = - P{~}
Figure (13.16) shows a liquid of density 1200 kg m..,
The work done by the weight is equal to the flowing steadily in a tube of varying cross section. The
2
negative of the change in gravitational potential cross section at a point A is l ·0 cm and that at B is
2
energy. 20 mm , the points A and B are in the same hon::ontal
1
The change in potential energy (P.E.) in time Af is plane. The speed of the liquid at A is 10 cm s- • Calculate
the difference in pressures at A and B.
P. E. of A' BB' -P. E. of AA'B
=P. E. of A'B +P. E. of BB'
-P. E. of AA' -P. E. of A'B A, •B

=P. E. of BB' -P. E. of AA'

Figure 13.16
= (!J.m)gh 2 - (!J.m)gh 1•
Thus, the work done by the weight in time Af is Solution : From equation of continuity, the speed t•1 at B
W3 = (!J.m)ghl - (!J.m)gh2, is given by,
The contact force .::::. \ does no work on the liquid A 1l'1 = A 2l '1
because it is perpendicular to the velocity. or, (l·0 cm
2
) (10 cm s
1
) = (20 mm 1)t•1
The total work done on the liquid considered, in
l·0 cm ' 1
tbe time interval Af is or, v = - --x l0cms =50cms 1
20mm·
W:::W, + W2+ W3 By Bernoulli equation,

==P{ ~ )- p ~ )+ (!J.m)gh
2[ 1 -(!J.m)gh 2 ... (ii) P, + pghl T ! pt•: pl + pgh. +½Pl':.
r ,,
c , rt• r , 1, 111 11

I I1• I' / I
I I I
,, ,,/ r ''"""ll
III l 111 r. 11
th • f, 1·1 111 10
1111/ Llu,,uglt a ,
,
I lt(,J-
t11• 1u1rnell,,"'
t //11,,11/fh tlw h, tHht h ~l ~
r,,,, /
\
11 tl,•11th /, lw ' ' '',
~ ,.,,. /11 11I f/ I ' ''/ fl , lt11,1
I n /J Ii i I 1 1 11 11 w11 11 1111111• ' " llu•,11 p,,,, . ..,~"
I 'l'h i H 114 " .,. 'h.
wmnty id comin~ ,,ut 18 coiled th, , 1~ •.,.
1
,\1 00k 111 )(.! , Pll1111' a '
1
10111•1n's > 141w<·d of t
lw I,qu ~d f
,'(flu.\
t,OQ kt; Ill X "1()1) 1 111' 8 1111'11,
E.-.:amplt• 1,'J.S
lz is constructed on th<' fop of 11 111,il,,
A water ta n ·-....u,
l~.l~ \Pl'Lll' \ l'lON~ OF BEHNOUI.Ll EQUATION eed will the water come out ,,[ ,1 I.rip P.
With w Iw t sp ? ' ._
ter level in the tank . Assume BU>ti,J., n_
(o) ll~ dn.,-.tntil·, below t h e w a ., w.
ressure above the water level ia er,u,z
and t h at the P ' ~ to
If the:- :,.pl' '-'d of t he fluid is zero everywhere, we get the atmospheric pressure.
the:- :,.ttu.tti,m ,,f h~ drost atics. Putting u1 == u2 == 0 in the e velocity is given by Torricelli's theorelll
Solution : Th
Bt.'rtHlUtli equ at ion (13.10) V =..J2gh
2
P 1 + pgh l == P 2 + pgh2 =-{2 x (9·8 m s- ) x (6·0 m ) "' 11 m s·'.
or. P 1 - P 2 == p g (h 2 - h 1)
a:-: expected from hydrostatics. (c) Ventury Tube
(b ) Speed of Efflux
tury tube is used to measure the flow Sl)eed
A ven . t f t ..
of a fluid in a tube. It cons1s s_ o a cons nct1on or a
Consider a liquid of density p filled in a tank of in the tube. As the flwd passes through tbE
throat . d .h
large cross-sectional area A 1 . There is a hole of constriction, its speed increases m accor ance wit the
cross-sectional area A 2 at the bottom and the liquid equation of continuity. The I_>ressure thus decreases as
flows ou t of the tank through the hole. The situation required by Bernoulli equation.
is shown in figure ( 13.17). Suppose A 2 <<A 1 .
T
h
l

Figure 13.17
Figure 13.18

Let u1 and u 2 be the speeds of the liquid at

A 1 and A 2 • As both the cross sections are open to the Figure (13.18) shows a ventury tube through which
atm osphere, the pressures there equals the a liquid of density p is flowing. The area of cross
atmospheric pressure P 0 • If the height of the free section is A 1 at the wider part and A 2 at the
surface a bove the hole is h , Bernoulli equation gives constriction. Let the speeds of the liquid at A1 and
1 2 1 2 A 2 be u 1 and u2 and the pressures at A 1 and A2 be
P 0 + pu 1 + pgh ==Po+ PU2 • . . . (i)
2 2 P 1 and P 2 respectively. By the equation of continuity
A 1U 1 == A 2V 2 ... (11
By th e equation of continuity
A1U1 == A 2U2 . and by Bernoulli equation,
P u ttin g u 1 in terms of u2 in (i), 1 2 1 2
pu2
P 1+
2 pu, == P 2+
2

½p[1:)'vi + pgh =½pui or, (P 1- P 2)==

1
2 p(u;-un.
... (ii

5
Figure (13.18) also shows two vertical tube
or, [1 - [ 1:]' ]vi = 2 gh
2
connected to the ventury tube at A 1 and A2- If ~
th
difference in heights of the liquid levels in these tube~
If A 2 << A i, this equation reduces to u2 == 2 gh is h, we have
or, v 2 .2 g h .
 -
7

my t
flr tlr II fr,,m If
I 1111 Ir I r,
IIUI
I lu
I II

'" I 111 1\', I

Ill W Ill Ii 1111 II 11111
Iii• 111111111\l r fl JII> 11 l
dt111llrtll 'llil' 111111111111 IS 1,, I
th
' " II II II ' " " I
I 111 l11111 11 11
I 1111111 \ II fill I pt, 1l111 pt1111p
11
II'S Ill II
' 11111111111 li I thlt1111/11l1 p 1111•
,,.(
I III f 1 11111 I II f II Ill III II
1'1lllll1•1 j
II ' ii
tlw hq11ul t11 Ii" 111111 II, I l'I It • 1111 Ill
111g
-a
\ 1:- pn:-Iw,l h1 tlw 111 ,t ,, t i 1Ill II II llllj•lt II
h1• .ur p.1:-:-1•~ th1 1111 ,,1i Iii 1, ' Figure 13.20
" IIIII 1111111111 /I
1:- l'1nt:-Hh-rahh 1111 1,,11 ,•d
•11 11 1 n111
'
, ·,111.-111 tv
,,:aUfl' dn1p:,; nu,, to 11•tl111·1•d 1' 11 ' 1"41111 • Ill 1111 •
'l'hf' plun<• of the figure represents honzontal plane.
cuonR. tlw hqnttl 1M m,1-1,•tl 111111 , II " ' \1'141 t• 1 IIIHI 'l'lw 111r 1.hnt, gocH from the A side of the ball in the
.,~ ,•d " 1th t lw l''qwlkd 111, . f1g11n· 1H dragged by the spin of the ball and its speed
_I._- _~L
--+-
A
1111T1•11H<'H. The air t,hat goes from the B side of the ball
111 1lw figure suffers an opposite drag and its speed
cl1·n1•11H<'H. The pressure of air is reduced on the A side
1111d iH increased on the B side as required by the
B1•rno11lli'H t,heorcm. As a result, a net force F acts on
1lw hull from the B side to the A side due to this
pn•1-1H11rc difference. This force causes the deviation of
el Change of Plane of Motion of n Hplnnlnw Bull
Iht- plunc of mot.ion.
Quite often when swing bo11 l,•rMof 1· , , 1,t, 1,t th•livt•r
the ball, the ball chnngt'S tl:4 plt1,11• 111' 11wl urn 111 111 r .

Worl,·1•d Out Examples

2
I. A beaker of circular cross St'ctio11 o/ md111s I 1·111 ,.~ f,lfrd (1 ·136 x 10 5 Nm ) x (3·14 x 0·04 m x 0·04 m)
u·ith mercury up to a hci!,iht o{ 10 rm Vind th,· /im·,• fi7 1 N.
exerted by the mercury 011 th,• bottom o{ th,• lwal.·,·r 'l'/11 •
6 1
atmospheric pressure = 10 N m , !>.·11.,,t,• 11/ 111,·rc111'\' 2. 'l'h<' density of air near earth's surface is 1 ·3 kg m -3 and
5 2
=13600kgm. Takeg =10m ~ thl' atmospheric pressure is l·0 x 10 N m - • If the
nt111osphere had uniform density, same as that obsened
Solution : The pressure nt lht• :-urliH·1•
nt th£• surface of the earth, what would be the height of
= atmospheric prl':.~un• th,• atmosphere to exert the same pressure ?

= 10 Nm . S11l11t'.<1" : Lcl the uniform density be p and atmospheric

lw1ghl be h. The pressure at the surface of the earth
The pressure at the bottom would be
6
= 10 N m + hp~ P pgh
8
= 10 N m • + lo· l m) l I ;!(i(){l h~ Ill )
1
(I() Ill H ) or, L·0 x l0 ~Nm ' (1·3kgm ,,)(9·8ms )h
l·0 x 10 ' Nm
= 10 Nm .- 1:1H00 Nm or, h
(1·3 kg m ' ) (9·8 m s-i) = 7850 m.
= 1'136 x JO N m . t•:wn Mount Ewresl (8848 m ) would have been outside
tlw ntmoi;pht'rt'.
The force exerted by the m,•n~llr) on tlw hnllt•m
' h t , nr "'' ' y
u ,t r If llr•
1 I Ill , /I/I f/

Joiuur1 ww:,

"librium the prf• Jeur1•11 1,t th

. . In equr ' t11
Sofut,oii · Jd be equal as they lH! 1n th!J
faces
sur_ shou .
J ]eve!. If the atmosphenc pressur1• ,, 1, ~Tl/.!
horrzoFnta_ pplied to maintain the equilibrium •~•
force 18 a ' ""'
pressures are F
F11{Un! l:l-\\ I p + ~ and P0 + cm 2 respectively.
10
0
1 cm
,otution : Suppo~e the atmospheric pressure = Po• This gives F = 50 N.
3
rn,,,ure at A = P + Ir (1000 kg m · ) g.
• e of mass 10 g is suspended by a ueri,,. 1
-3 6. A copper piec . ""'
Pre,-,-ure at B = P0 + (0·02 m)(l3600 kg m ) g • . "'h ring elongates 1 cm over its natural lennth
spring. ~- esp . .. . A b k _·-a
The:;:e pressures are equal as A and B are at the same iece in equilibrium. ea er containi••
to keep. the wP placed below . ""
horizontal level. Thus, the piece so as to immerse
water is no . h .
. mpletely in water. Find t e elongation of 1L.
h = (0·02 m) 13·6 the piece co -a •~
spring. Density of copper = 9000 kg m . Take g = 10 ms,
= 0·27 m = 27 cm.
. . Let the spring constant be k. When the piece 15
Solutwn. • . n;r the equilibnUDl
. conditi"on gives
.
4. A cylindrical uessel containing a liquid is closed by a h anging lD .,.,.. , 2
smooth piston of mass m as shown in the figure. The k(l cm) = (0·0l kg) (10 m s· )
area of cross section of the piston is A. If the atmospheric
. .. (iJ
pressure is P0 , find the pressure of the liquid just below k(l cm)= 0·l N .
or
the piston. The volume of the copper piece

=
0·0lkg _.!_ 10 -s m.
a
-3 - X
9000 kgm 9
This is also the volume of water displaced when the piece
is immersed in water. The force of buoyancy
= weight of the liquid displaced

Figure 13-W2
= ¼x 10 5 3
m 3 x (1000 kg m- ) x (10 ms
2
)

= 0·0llN.
Solution : Let the pressure of the liquid just below the
If the elongation of the spring is x when the piece i.
piston be P. The forces acting on the piston are
immersed in water, the equilibrium condition of the
(a ) its weight, mg (downward) piece gives,
(b) force due to the air above it, P0 A (downward)
kx = 0·l N - 0·0ll N = 0·089 N. •.. (iii
(c) force due to the liquid below it, PA (upward).
By (i) and (ii),
If the piston is in equilibrium,
PA=P0 A+mg 0·089
x =--cm
0·l
=0·89 cm.

or, p =Po+!!!.i. .
A
7. A cubical block of wood of edge 3 cm floats in water. Tht
0
5. The area of cross section of the two arms of a hydraulic lower surface of the cube just touches the free end of
tht
press are 1 cm 2 and 10 cm 2 respectively (figure 13-W3). uertical spring fixed at the bottom of the pol. Find
A force of 5 N is applied on the water in the thinner arm. maximum weight that can be put on the block u:ithoul
What force should be applied on the water in the thicker wetting it. Density of wood 800 kg m ' and sprr~
arm so that the water may remain in equilibrium ? conStant of the spring - 50 Nm . Take g - 10 ms ·
 I 1,1 I M, h11nl 1

rn
1111 \\l'il(hl Ill}/ '/./1,p

Thi• 11111RII of th, purl ()(' of Liu pl 1t1k (C(/ 1 )p

I / ~
'l'lw 11111H11 of w11l1, d1 pl11r1 rl () fi c1,slI p co&U
l.!.£.i.fl
Tlw huoynnl fon•f' J,' 1H, lhc•rc-fi,n, 1' COIi
~,,,,,. : '!'ht• ~f)\•r11it· gra, it} of the block 0·8 . 11ence the
..,,...~
i!ht 1n~1dl' wnh•r = 3 cm -,., 0·8 2 .4 cm. Tl1c height . Now, for cquilihrium, lh<· torqu1• r,f ml( nbout O hr,ulrl
, , ,
Supposc the balance the torque of F about 0
~,1,1d1' "all
,,.. . r - 3 cm - 2·4 0·6 cm ·
s_,1111 um
,H~1ghl that . .t .
can be put withoul weltmg So, mg(OB)sin8 F(OA) sin8
. . 1 18
·
The block m this case is completely immerse d.
(c~:e )( c~se)
11
'' m the
"~tt•r. The volume of the displaced water or, (Zlp)l 2
:: volume of the block= 27 x 10 am a 2 1
}{ence. the force of buoyancy or, cos e=2
= (27 x 10 a ma) x (1000 kg m-3) x (10 m s 2) 1
=0·27 N.
or,
2
cos0 = ✓ , or, 0 - 45°.

The spring is compressed by 0·6 cm and hence the 9. A cylindrical block of wood of mass Mis /7.oating in water
upward force exerted by the spring with its axis vertical. It is depressed a little and then
released. Show that the motion of the block is simple
= 50 N m ' x 0·6 cm= 0·3 N.
harmonic and find its frequency.
The force of buoyancy and the spring force taken
Solution : Suppose a height h of the block is dipped in the
together balance the weight of the block plus the weight
water in equilibrium position. If r be the radius of the
11 put on the block. The weight of the block is
6
cylindrical block, the volume of the waler displaced
W' = (27 x 10- m) x (800 kg m-3) x (10 ms 2)
=1tr 2h. For floating in equilibrium
=0·22 N. 1t r 2hpg = W ... (iJ
Thus, W = 0·27 N + 0·3 N - 0·22 N where p is the density of water and W the weight of the
=0·35 N. block.
Now suppose during the vertical motion, the block is
8, A wooden plank of length 1 m and uniform cross section further dipped through a distance x at some instant. The
2
1s hinged at one end to the bottom of a tank as shown volume of the displaced water is n r (h + x). The forces
m figure (13-W5). The tank is filled with water up to a acting on the block are, the weight W vertically
2
height of 0·5 m. The specific gravity of the plank is 0·5. downward and the buoyancy n r (h + x) pg vertically
Find the angle 0 that the plank makes with the vertical upward.
in the equilibrium position. (Exclude the case 0 = 0.J Net force on the block at displacement x from the
eqwlibrium position is
2
F - W - nr (h + x)pg
I 2
= W - nr hpg - 7tr pxg
2
g
0 Using (i),
2
F =- nr pgx = kx, where Ii= nr pg.
2
Figure 13-W5
Thus, the block executes SHM with frequency

1 - ik=_l - ~
Solution : The forces acting on the plank are shown in
the figure. The height of water level is l = 0·5 m. The V=
2n-\J M 2n -\J ~ ·
length of the plank is 1·0 = 2l. The weight of the plank
acti; through the centre B of the plank. We have
10. Water flows in a horizontal tube as shown in figure
OB .:cl. The buoyant force F acts through the point A
0 3-W6). The pressure of water changes by 600 N m 1
which is the middle point of the dipped part OC of the A
plank. ~
We have OA= OC =_I__ ~
2 2 cos0 Figure 13-W6
Let the mass per unit length of the plank be P·
 1-,,,/,1II"" I
i.--- -,A

1
1, md lh 11 ul JI •

I t
Figure 13-W7
l , l m ulh ,,qu 1IH111,
,' I I I
f ross sectio n of the tank is large cornpartd
I }'
A + 2 ♦'I" I + 2 J'I •
As the area c ening the speed of water in th,, I.at
O

or.
I
2 1'('>1
1
1 ) - - J11',
~ .. 2 n
1 a
=-
3
2 p11,
2
' th
to that of e opall a~ compa red to the speed at U:
will be very sm sure at the surfac e of water in the 1,.,
""II:
3 , a • g The pres
or, 600 ~ m =- tl000 kgm l 1•, opemn · th atmos phere plus due to the load .
2 is that due to e z
""'i'""7 3 I (20kg )(l0m s )=P +400 Nm·2
or. 1, \0·4 m s = 0·6 ms P1, =Po+ 0·5 m z o .
I 3 1
Pressure is that due to the
'l
Tht• r,lte of flo" - (30 cm ) (0·63 ms l = 1890 cm s • At the openin g, th e
2
atmosp here.
11. The area of cross section of a large tank is 0·5
m • It has Using Bernoulli equati on
an opening near the bottom hal'ing area of cross section 1 2 _! 2
the PA+ pgh +2pu1, =Ps + 2 PUs
1 cm 2• A load of 20 kg is applied on the water at
top. Find the l'elocit_v of the water coming out of the
is
Po+ 400 Nm z + (1000 kg m-a) (10 m s-2) (0·5 m)+O
opening at the time when the height of2 water level or,
1 -3 2

50 cm abol'e the bottom. Take g = 10 ms • =Po+ (1000 kg m )u8

2
2 k -1) 3
Us2
or, 5400N m- =(500 gm
-1
u8 "' 3·3 ms .
or,

QUESTIONS FOR SHORT ANSW ER

liquid is (a) at the equato r (b) at a pole (c) somewhere

I. Is it always true that the molecules of a dense liquid else.
are heavie r than the molecules of a lighter liquid ?
6. A barom eter tube reads 76 cm of mercu ry. If the
skin, tube
2. If someone presse s a pointed needle agains t your the open end immer sed in
if someo ne presse s a rod agains t your is gradua lly incline d keepin g
you are hurt. But
tolerat e. Explai n. the mercu ry reserv oir, will the length of mercu ry column
skin with the same force, you easily
that be 76 cm, more than 76 cm or less than 76 cm ?
3. In the derivation of P 1 - P2 = pgz, it was assumed
will this equati on not One
the liquid is incom pressib le. Why 7. A one meter long glass tube is open at both ends.
be strictly valid for a compressible liquid ? the tube is dipped into a mercu ry cup, the tube
end of
4. Suppose the density of air at Madra s is Po
and is kept vertica l and the air is pump ed out of the tube
atmospheric pressu re is P 0 • If we go up, the densit y and by connec ting the upper end to a suctio n pump. Can
•
the pressu re both decrease. Suppose we wish to calcula
te mercu ry be pulled up into the pump by this process
the pressu re at a height 10 km above Madra s. If we use
inside
8. A satelli te revolves round the earth. Air pressu re
the equat10n P - P = p, gz, will we get a pressu re more
the satelli te is maint ained at 76 cm of mercury. What
than the actual or less than the actual ? Neglect the ter
will be the height of mercu ry colum n in a barome
variati on m g. Does your answe r chang e if you also te ?
tube 1 m long placed in the satelli
con:.ider the vanati on in g ?
8
5. The free surface of a liquid resting in an inertia l frame 9 · Consid er the barom eter shown in figure (13-Ql) If
1s horizontal. Does the norma l to the free surfac
e pass tube,
s~all hole is made at a point P in the barom eter
Think separa tely if the
throug h the centre of the earth ? will the mercu ry come out from this hole ?
 11 I M In 1

A 1, rr"~ ho ,t Ii nd d wilt r k ~
I :I. t ..t t f
111 iilf.t I h rnnxlmum 1 11•
m111 u th in 1111 h, I ht of th lmd dt of
lt11la t11 fl• B unrl•, th lmdg• houl dd d?
h, •• moVl rl or 11111111 rnur• rock he ll
I I. Wnl••r 18 luv. ly r.orn111g r,ul from ll v rti ~ pipe Al,
w11t1•r rlf'!U'Prtde ufl• r l <Hn ing ,,ut, itA nr ~ er atarm of
rcd11n• . l~xplorn thrll rm thr bn J or t C qu
continuity.
, J,\l'\"h11n1•1h•i1' pnni·ipli• 'ulid 11\ nn l'l1'\'t1(or a1·celeratmg
If- • In 11 ,-.11 nr1•1•h•1-.1t m.: on n t1,,·l'l road ? I t
15. While watenng a dr stnnt p an ' o h fipartially
gardcnf.'1' er on
\\11.l 1~ ti ,•a~11'r to :-\\ im m Sl'a water than m fresh closes the exit hole of the pt~ by puttrng
it. Explain why this results m the water ea
stl:
~gg01ng
11 ,.stcr 1
\ gl.i,s 1,f " .all'r hns an icl' cube floating in water. The to a larger distance.
l'- .-:it,' r 1,,, l'I JU~! touches t_he rim of the glass. Will the . the car rull5 at
16. A Gipsy car has a canvass top. When
11111('r OH' rflll\\ when the tee melts ? high speed, the top bulges out. Explam.

OBJECTIVE I

2
I, A liquid can easily change its shape but a solid can not 2 (b) 3000 N m·
(a) 400 N m·
because 2 (d) zero.
(c) 1000 N m·
81the density of a liquid is smaller than that of a solid
bl the forces between the molecules is stronger in solid
than in liquids
cl the atoms combine to form bigger molecules in a solid T
d) the average separation between the molecules is 10cm
larger in solids. 1
2. Consider the equations
P = Lim :c, and P -P = pgz.
6B-+ 0 o.;,
1 2 Figure 13-Q3

In an elevator accelerating upward

(al both the equations are valid
tb) the first is valid but not the second 6. A beaker containing a liquid is kept inside a big closed
,cl the second is valid but not the first jar. If the air inside the jar is continuously pumpe~ o~t.
1d) both are invalid.
the pressure in the liquid near the bottom of the liqwd
will
3. The three vessels shown in figure (13-Q2) have same (a) increase (b) decrease (c) remain constant
base area. Equal volumes of a liquid are poured in the (d) first decrease and then increase.
three vessels. The force on the base will be
7. The pressure in a liquid at two points in the same
(a) maximum in vessel A (b) maximum in vessel B
(cl maximum in vessel C (d) equal in all the vessels.
horizontal plane are equal. Consider an elevator

u
accelerating upward and a car accelerating on a
horizontal road. The above statement is correct in

LJ LJ
Figure 13-Q2
(a) the car only
(b) the elevator only
(c) both of them
(d) neither of them.
8. Suppose the pressure at the surface of mercury in a
4· Equal mass of three liquids are kept in three _i~entical barometer tube is P, and the pressure at the surface of
mercury in the cup is P .
cylindrical vessels A B and C. The densities are
(a) P, =0, P2 =atmospheric pressure
P., Pa, Pc with PA< Ps; Pc· The force on the base will be
1
a, maximum in vessel A (b) maximum in vessel B (b) P1 = atmospheric pressure, P1 - 0
1 (c) P, = P2 = atmospheric pressure
cJ maximum in vessel c (d) equal in all the vessels.
(dl P, = P, = 0.
5, Figure U3-Q3) shows a siphon. The liquid shown is
Water. The pressure difference Ps - PA between the 9. A barometer kept in an elevator reads 76 cm when it is
l>Otnts A and B is at rest. If the elevator goes up with increasing ::-pt' t>d,
 I ,1 /1. I 111 11 1 I ul,I ,I l,1111 I rllllll'I• ti lyn11,
d With w
1
1111, 1 111 d h ,r1i11nl lly t,,w """ '1ght Ill(.,
1
11\ I• I Ill
1111 ,, I ht " llltnnl 11r.rrn ti f1,r , t'YU~l!
"'' I,, 1111 ,,1' ,,f 1lr• l,1,11
1 ,o,
'''"' II!"I'"
lh•I
1111 1111 1111 1
I II pll ••II 11111111g lr I 111 •' 11lr1 r,f llir t,,r,
I
U Oil (,I) 1'111 ( I, ) p 11fl,l'fl 1h,u11g h ,, Jl(fllll ,,, tlu r1ght ,,ru,
u l tn 11 H t. r n 1uh,· "' 11•11.:t h I 111 1!l I ti li>d ,. Ill" ,1,
1 1
1t 1111 11~ h 11 po111I lo Hu ldt ,,f th ,,
l \\\U}1111,n \If\ " " I I 1111 ,•• t,•d 111 11 1111'11'111 \ ( d ) hl'('ll 11 H' I II. I'll
..._ 1''111• lt•1 n• 1d1n1• ,,1111 I' 11t 1111 111 d in 1s 7l; 1·111. 17. ( '011is 1d«•1 I lw Rll ,wt 10llll()fhth1· rr1•v1r1:J J>rt1l,l1 rn r
x I Ill tuh1• 1 filh•d "1th 11 11 •1 n11, up 111 11; 1'111
l d t.t11 I\ \ I, SNi b , ll \ •'f" 1 JI JS Ill\ 1•1 ll•d Ill I I llll'I ('Ill"\
w11t 1•1 . p111-1 h I lw l<'ft wn y n ,r,r('(• f, and th, ril!Ji
up nd tlw , 1•1 k 1" 1, •nw, ,.,1 rlw lw1.:ht 111. nwn·un· hy lort·1• F,.
II

C'\ Iun1n an t lw t u !>,.• •,, 1'1• 1Ill' surlal·,• · · l>t' (t~l J<'.
F, ( h ) F 1 .> F 2 (cJ F 1 < p 2
111 thl• rup will
tdl <e. 76 rm. (d) th<' information 1s m s ufficient to know th e re}a
n 1 ·, (h) 7(, <-m t,•\ > 76 t'lll
12. \ '\' '\ ,1,•1 • ,I bl~-• . . d d l bctwct•n F, nnd F 2·
'"" 1" ,.,u,.,p,•n l' , , u :,;pnng balonrC'. A
l'M:':t "'' \''\lll t •\\lllll•' · lilt ,, ",l t Pr 1s p· , need on a weighing
.... ,... · · lS. Water enters through end A with a speed u 1 and 1 ,
m:idutll' "h1d1 n•·1d . -to N Tl1 . . through end B with a speed u2 of a cylindrical tu~d ta
1cm ,•n."\i , 0 h k ·
lt ~ sprmg b~lnnre 1s now
th,, "bl
1 oc · gets 1mmen,ed m the water.
• l' The tube is always comp_letely fille? ~th water. In At
Th,• ,-pnn~ halann• now reads 16 N. The reading of the I the tube is horizontal, m case II it 1s vertical . hca;,.
\\ l'lj:'11'1~ m.1ch1ne \\;u be end A upward and in case Ill it is vertical with~e ~
3 .36 ~ (bl 60 N tel 44 N t dl 56 N. B upward. We have u 1 = u2 for llC
13. A Pit><..'l' _o f \\Ood is floating in water kept in a bottle. The (a) case I (b) case II (c) case III (d) each case
bottle 18 ~~nnected to an air pump. Neglect the 19. Bernoulli theorem is based on conservation of ·
rompress1b1hty of water. When more air is pushed into (a) momentum (b) mass
t~e bottle from the pump, the piece of wood will float (c) energy (d) angular momentlllll.
,nth a long horizontal tube. Let
20. Water is flowing through
a) larger part in the water (b) lesser part in the water PA and P B be the pressures at two points A and B of the
(c) same part in the water (d) it will sink.
tube.
14. A metal cube is placed in an empty vessel. When water (a) PA must be equal to P 8 .
is filled i~ the vessel so that the cube is completely (b) PA must be greater than P B.
lfilIIlersed m the water, the force on the bottom of the (c) P,,, must be smaller than PB.
,·essel in contact with the cube
(d) P,,, = PB only if the cross-sectional area at A and B
(a \\-ill increase (b) will decrease
(c) will remain the same (d) will become zero. are equal.
21. Water and mercury are filled in two cylindrical vessel;
15. A wooden object floats in water kept in a beaker. The
up to same height. Both vessels have a hole in the wall
object is near a side of the beaker (figure 13-Q4). Let
near the bottom. The velocity of water and merCl.lr\·
P. P _. P 3 be the pressures at the three points A, B and
coming out of the holes are u 1 and u2 respectively. ·
C of the bottom as shown in the figure .
(a) u, = u2 • (b) u 1 = 13·6 u2•

t? .. 1
A B C
(c) u, = u 2 / 13·6.
22. A large cylindrical tank has a hole of area A at its
bottom. Water is poured in the tank by a tube of equal
(d) u 1 = ✓13·6 u2 •

cross-sectional area A ejecting water at the speed l'.

Figure 13-Q4 (a) The water level in the tank will keep on rising.
(b) No water can be stored in the tank
2
(c) The water level will rise to a height u /2g and then
(a) P =P, -P. (b) pl <P2 < P 3, stop.
(c) P > P > P. (d) P 2 = P 3 ~ P 1 • (d) The water level will oscillate.

OBJECTIVE Il

1. A solid floats in a liquid in a partially dipped position. Cc) The weight of the displaced liquid equals the weight
(a) The solid exerts a force equal to its weight on the of the solid. I
Cd) The weight of the dipped part of the solid is equs
liquid. to the weight of the displaced liquid.
(b) The liquid exerts a force of buoyancy on the solid
which is equal to the weight of the solid.
 273

Lh Lui
,pl. h ~\i "'11111 I ( ll 111 I 1111111,I 11 II f11t I I (I,, l•l11w 111 hoth t.h• Lil
\J, I1qm11 011 th, ,•ltd "'II fl
11) l•lo ,ly 111 A
I ll I \l\l Ill d ,i.~ \WI 111 11h• thf' ltq1111I
,f ll Ol'H'III II 1\\11 1 \'1111111,:pd
,.i, 1•111 J1ly 1n n 1,u A
7. Wnt•, 1 t1,,w111g
'( l ,r ~,
I t ,k,•11 p.11 t 111lh llllt 111 I lw ltq1111I
strenml1nc mat.ion thr
in
\\1th ,t 8 1,x,e hor1N1nt nl ( ' 1111 1tlcr two po nlA A
be 11tlrn H'riwnlh "P"nrd din•i·twn.
tlw 111'11• 111 th,. nrnc hr,n11int.nl lrvcl
d """"' ,s h.,lftilkd \\1th \\nt,•i· l'lwrp 1~ 11 hlllt• 111 1 'l'IH• fl" ' 11 11,cs ot. A nnd II ore fqunl for ony pc
th<' tnp ,,f tlw , 1•:-s,•l ,111d .ur ,s pumiwd out Imm
of tlw tuhe
, h1h (bl Thi· pn ,urcs nn• never cqunl
1 '111,• \\,tll•r 11•,d \\111 ris1• up m tlw "eHHt•I. (c) The pressures orr• cqu;tl 1f the tube has a uniform
Tlw pn•:--..uri• at tlw :-urfac1• of the wntcr will cross section.
(<'ft':t :--l' (d ) The pressures may be equal even ,f the tube has a
c The forcl.' h~ the "at('r on the bottom of the vessel nonuniform cross section.
\\'111 d1'Crl'a:-1•.
8. There is a small hole near the bottom of an open tank
d Thi• dm::-ity of the liquid will decrease.
filled with a liquid. The speed of the water ejected does
.i In 3 ::;treamline flow.
not depend on . .
11 the :,;p('ed of a particle always remains same (a) area of the hole (b l density of the hqmd
t, the velocity of a particle always remains same (c) height of the liquid from the hole
c the kinetic energies of all the particles arriving at a (d) acceleration due to gravity.

EXERCISES

t. The surface of water in a water tank on the top of a piston weighs 45 kg, find the difference in the level, of
I- use is 4 m above the tap level. Find the pressure of water in the two tubes.
water at the tap when the tap is closed. Is it necessary 4. A glass full of water has a bottom of area 20 cm %. top
..... to specify that the tap is closed ? Take g = 10 m s · •
2
of area 20 cm 2, height 20 cm and volume half a litre.
!, The heights of mercury surfaces in the two arms of the (a)_ Find the force exerted by the water on the bottom.
.., manometer shown in figure (13-El) are 2 cm and 8 cm. lb) Considering the equilibrium of the water. find the
Atmospheric pressure = 1·01 x 10 5 Nm 2 • Find (a) the resultant force exerted by the sides of the glass on the
pressure of the gas in the cylinder and (b) the pressure water. Atmospheric pressure= l·0 x 10 · Nm-·. Density
of mercury at the bottom of the U tube. of water= 1000 kg m ~ and g = 10 m s 2 • Take all num-
bers to be exact.

.LJ
" -r
gas '::: ?o -t- f~(r.) ?,0-1-
.... \.\ 'l
0 T
20cm

Figure 13-El
• ¼ ( e)
s.,_"" 20 cm2
1
~ Figure 13-E3
6, The area of cross section of the wider tube shown in
figure Cl3-E2) is 900 cm 2 . If the boy standing on the "'If. Suppose the glass of the previous problem 1s covered bv
a jar and the air inside the jar is completel) pumped
out. (a) What \vill be the answers to the problem'?
f~ PK
(b) Show that the answers do not change if a glas::; of
t. Po-t- !£. different shape is used pro,ided the height. the bottom
area and the volume are unchanged,
qoo m
~ - If water be US("d to construct n baronwtl'r. "hat "ould
- \o-t f ~IA be the height of ,..,· ntcr column nt ::;tnndard atmo,-.pheric
Figure 13-E2 pressure (76 rm of mercur~) '?
 m I Inn
1 1 ,,r , •• ,
I nl 1(!011 ol

ro<t n ul II t 111k 111 6 17 • hgur 18 E,

F nd tho tolnl fi,11 c t'Xl't I ,•ti h, t lu
m lll'li CC! or tJw I 111k ( h l ( '11111!1111•1 II
I'(' m 1 Ill r lk(l II l11111 7 llllt,,1 l(l'lfl
s lo tw conRlrucwd "':'1th 1rc,n
Ill lro Ill thu; t11 d,,, 11\1 lll'd 111 11 cl,•p(h or I u;. /\ rubiclll box IWhut can be th<· m1r11rnurn val
1 Lh1c urforc of \\lll<'r J.'111d tlw l111n• h't llw 111 thickness
cxl<'rnul edge so that the cube docs n1,t sink Ir
tnp ~) Find t lw 1,,n1u,• of t lw fon•t• 8
l)cmuty of iron 8000 kg m ~ and density
r n lhi IJf
cul too m p.'trt b} nt)(,ut tlw h,1ttom ,•d~,• of tlrn, side. ]

d :-'ind the totnl force h, tlw \\Ult'r on thii,; side. 1000 kg m ·

th i:d the tot • ''"\o/Ul' h~ thl.' ,, ntcr on the side about 16. A cubical block
of wood weighing 200 g has a ltad
ttom cdg~ '\, ~lcct the atmo8pheric pressure and fastened underneath . Find the mass ~f the lf:ad ~
take g 10 m.., which will just allow the block to float m water. S Pit(,.
gravity of wood is O·B and that of lead is ll·a ~
9. An omnment '' eighing 36 g in air, weighs only 34 g in
problem if the lead piece is fas
wat er. A;,,ummg that some copper is mLxed with gold 17. Solve the previous
· _"'-Pare the ornament. find the amount of copper in on the top surface of the block and the block 15 to~
• :,:'€-C1fic gra,ity of gold is 19·3 and that of copper is with its upper surface just dipping into water. ',ai
~ -l.
of edge 12 cm floats in m
18. A cubical metal block
10. Refer to the pre,ious problem. Suppose, the goldsmith with one fifth of the height inside the mercu.ry. ;CUry
argues th~t he has not mixed copper or any other is poured till the surface of the block is just inune ale!
matenal with gold, rather some cavities might have been in it. Find the height of the water column to be po~
left ms1de the ornament. Calculate the volume of the Specific gravity of mercury = 13·6. ed.
ca,itles left that will allow the weights given in that 19. A hollow spherical body of inner and outer
radii 6
problem. and 8 cm respectively floats half-submer ged in wa::
Find the density of the material of the sphere. r
1 I. A metal piece of mass 160 g lies in equilibrium inside
If
a glass of water figure 13-E4). The piece touches the 20. A solid sphere of radius 5 cm floats in water.
bottom of the glass at a small number of points. If the maximum load of 0· 1 k~ can be .put on it without wettin:
dem,ity of the metal is 8000 kg m-a, find the normal force the load, find the specific graVIty of the material of th;
exerted by the bottom of the glass on the metal piece. sphere.
21. Find the ratio of the weights, as measured by a spnn•0
balance, of a 1 kg block of iron and a 1 kg block of wooo

CJ
Figure 13-E4
Density of iron = 7800 kg m-3, density of wooo
= 800 kg m-3 and density of air= 1·293 kg m..:i.
22. A cylindrical object of outer diameter 20 cm and ma,,
2 kg floats in water with its axis vertical. If it is slight!)
depressed and then released, find the time period of the
resulting simple harmonic motion of the object.
23. A cylindrical object of outer diameter 10 cm, height
3
3 20 cm and density 8000 kg m is supported by a vertical
12. A ferry boat has internal volume 1 m and weight 50 kg. spring and is half dipped in water as shown Ill
(a Neglecting the thickness of the wood, find the figure(13-E6 ). (a) Find the elongation of the spring Ill
fract10n of the volume of the boat immersed in water. equilibrium condition. (b) If the object 1s slight)j
(b) If a leak develops in the bottom and water starts depressed and released, find the time period of resulting
coming in. \•,hat fraction of the boat's volume will be oscillations of the object. The spring constant
fiJJed with water before water starts coming in from the 1
500 Nm •
sides?
13. A cubical block of ice floating in water has to support a
metal piece weighing 0·5 kg. What can be the minimum
edge of the block so that it does not sink in water?
Specific gravity of ice = 0·9.
14. A cube of ice floats partly in water and partly in K.oil
(figure 13-E5). Find the ratio of the volume of ice
immersed in water to that in K.oil. Specific gra"ity of
K.oiJ ii': 0·8 and that of ice is 0·9. Figure 13-E6
 II I
I !.111 Ill n II I
\ II ,I I 1111 (I ) LI
:i.. ' I Ih lK l tI rl I l
I.) Imm, I 1
d I 1111 Ir 1111 lit Ir,
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hm
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,,. I'• 1111
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11
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' I I 1l I Ill \ I
nlt'lts \1mt.1n11h hom ,•ttrh 8 111 80 1 """' 1 11 ' 1
' •• t 1111 ,t 11lw11v11
"'' ,n Ill; ,·ul •11'll I "'\I\Jh•.
I H1•m1•mh1•111111 I I t
'-nn "\11'1' , find Ow l1•n~th of tlw l'1lu \ HI( tl(,'I' IM light,,,
l •• · ..,, n H ' 1cp cube I-,guro 1:1 E9
.,1 tht• 1 n:-t,ult 11 Just l1•n,,•~ contill't w,t1,
"tho 1 • 11ll' 1>ottom of
g n~,.
_ \ l tub,• cont,unmg a liquid is ucc"l"r11 t d h 33. Water flows through the tube shown m figure 13-El0
~ ' ' c onzontally
·
ith a con:.itant acc<'lcrutton O If ti The areas of cross section of the wide and the narrow
" . o· 1c separation 2
'-·•t\\l'l'l\ thl' ,crt1cal limbs is { find thn d "IY' . portions of the tube are 5 cm and 2 cm' respectr;eJ!,
'" h 1· · · ' " 111erence 1n the
ht.'t~ht:.; of t <' 1quid m the two arms. The rate of flow of water through the tube 1s 500 cm s
~ ,\t Oeoprayag {Garhwal , UP) river Alak d . Find the difference of mercury levels in the U-tube
;,. · · Bh · . nan a mixes
with the nver agirath1 .and becomes nver · G
----------=-
1u r
anga .
~
.::uppose
. Alaknanda has a width of 12 m , Bh agira · th·1 h as
1
8 " 1dth of 8 m and Ganga . has a width of 6 m. russume
A _

t hat t h e d ep th o f water 1s same in thn ,- th ree rivers. · L et

the ~yerage _spee~ of water i~ Alaknanda be 20 km h -'
1
and m B~agrrat~ be 16 km h • Find the average speed Figure 13-ElO
of water m the nver Ganga.
26• Water flows through a horizontal tube of variable cross 34. Water leaks out from an open tank through a hole of
section (figure 13-E7). The area of cross section at A and area 2 mm 2 in the bottom. Suppose water 1s filled up to
2 d 2
B are 4 mm an 2 mm respectively. If 1 cc of water a height of 80 cm and the area of cross section of the
enters per second through A , find (a) the speed of water 2
tank is 0·4 m • The pressure at the open surface and at
at A , (b) the speed of water at B and (c) the pressure the hole are equal to the atmospheric pressure. Neglect
difference P,.. - P 8 . the small velocity of the water near the open surface in
the tank. (a) Find the initial speed of water coming out
A
of the hole. (b) Find the speed of water coming out when
~ half of water has leaked out. (c) Find the volume of water
_.::-:.__--=:-- ti
- - 0.94 cm ~

Figure 13-E7
leaked out during a time interval dt after the height
remained is h. Thus find the decrease in height dh in
terms of h and dt.
•
(d) From the result of part (c) find the time required for
h alf of th e water to leak out.
29. Suppose the tube in the previous problem is kept vertical
with A upward but the other conditions remain the 35. Water level is maintained in a cylindrical vessel up to
same. The separation between the cross sections at A a fixed height H. The vessel is kept on a horizontal
and B is 15/16 cm. Repeat parts (a), (b) and (c) of the plane. At what height above the bottom should a hole
previous problem. Take g = 10 m s- •
2 be made in the vessel so that the water stream coming
out of the hole strikes the horizontal plane at the
30. Suppose the tube in the previous problem is kept vertical greatest distance from the vessel (figure 13-Ell)?
with B upward. Water enters through B at the rate of
3
1 cm s '. Repeat parts (a), (b) and (c). Note that the
speed decreases as the water falls down.
31 Water flows through a tube shown in figure (13-E8). The
2 2
areas of cross section at A and B are 1 cm and 0·5 cm
respectively. The height difference between A and B is
1
5 cm. If the speed of water at A is 10 cm s , find (a) the
speed at Band {b) the difference in pressures at A and B. Figure 13.11

A ~
~ B

Figure 13-EB
 , rt• , 1•11,,11 11

/\Nl-4 \\ I UH

01).11 ( I I\ I I I~ 11, l 'I.II

s
h
ft
(t')
•t (C')
◄ h i)
Ill r,·)
'
(cl) h '")
12, (!'l
I :J 17 ( ' 111

11 !ti l 14 I I
l4 ((') 15 (n) I c; kl 17, <h l I H. Cd l
15. 4 ·8 cm
cl) ~l (nl :.!:.!, k)
16. 54 ·8 g
17. 50g
OHJEC rt\ F II 18.10·4 cm
3
I a), (b), (c) 2. {a}. (c) 3. (cl, Cd) 19. 865 kg/m
4 bl. k) 5. (c). tdl 6. (a), (b), (c) 20. 0·8
7. (d. (d l 8 . (al, (bl 21. 1·0015
22.o·5 s
23.(a) 23·5 cm (b) 0·93 s
EXERCISES
24.(a) 2·5 cm (b) 1t 15 s
1
1. 40000 X m , Yes
25. 2·26 cm
2. (a) 1·09:x 10 5 N i m 2
( b) 1·12 x 10 Nim
5 2
26. a 0 l/g
3. 50 cm 27. 23 km/h
4. a 204 N (b) 1 N upward (b) 50 emfs (c) 94 N/m 2
28. (a) 25 emfs,
5. 4 X. 1 N upward 29. (a) 25 emfs, (b) 50 emfs, (d) zero
6. 1033·6 cm 30. (a) 25 emfs, (b) 50 emfs, (c) 188 N/m 2
2
7. 10 - N , No 31. (a) 20 emfs, (b) 485 Nim
8. ( a 60000 N , (b) 20000 x ox N 32. 146 cc/s,
(c ) 20000 x (l - x)ox N-m (d) 10000 N, 33. 1·97 cm
(e ) 1000013 N-m 34. (a) 4 mis, (b) ✓8 mis
✓2gh dt, ✓2gh x 5 x 10 - dt
6
9. 2·2g (c) (2 mm 2)
3
10. 0 · 112 cm (d) 6·5 hours
11. 1·4N 35. H/2.
□