fluid mechanics for jee main

Target : JEE(NITs) Fluid Mechanics

ContentsTopic Page No.

Theory 1 20

Exercise - 1 21 22JEE (main) problems

Exercise - 2 22 24JEE (advanced) problems

Exercise - 3 24 25Miscellaneous problems

Exercise - 4 25 26Level - 1 : AIEEE Problems - JEE (main)Level - 2 : JEE Problems - JEE (Advance)

Answer Key 26

Advanced Level Problems 27 30JEE (advanced) problems

Answer Key 30

Solutions of ALP 31 35

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Fluid Mechanics

JEE(NITs) Syllabus 2012Pressure due to a fluid column; Pascals law and its applications. Streamline and turbulent flow, Reynoldsnumber. Bernoullis principle and its applications.

RESONANCE AIEEE-FLUID MECHANICS - 1

FLUID MECHANICS

DEFINITION OF FLUIDThe term fluid refers to a substance that can flow and does not have a shape of its own. Forexample liquid and gases.Fluid includes property (A) Density (B) Viscosity (C) Bulk modulus of elasticity (D) pressure(E) specific gravity

PRESSURE IN A FLUIDThe pressure p is defined as the magnitude of the normal force acting on a unit surface area.

P = AF

F = normal force on a surface area A.

The pressure is a scalar quantity. This is because hydrostatic pressure is transmitted equally inall directions when force is applied, which shows that a definite direction is not associated withpressure.Thrust. The total force exerted by a liquid on any surface in contact with it is called thrust of the liquid.

Note :The normal force exerted by liquid at rest on a given surface in contact with it is called thrust of liquid on thatsurface.The normal force (or thrust) exerted by liquid at rest per unit area of the surface in contact with it, is calledpressure of liquid or hydrostatic pressure.If F be the normal force acting on a surface of area A in contact with liquid, then pressure exerted by liquid onthis surface is P = F/A(1) Units : N / m2 or Pascal (S.I) and Dyne/cm2 (C.G.S)

(2) Dimension : (P) = ]TML[]L[]MLT[

]A[]F[ 21

2

2

(3) At a point pressure acts in all directions and a definite direction is not associated with it. So pressure isa tensor quantity.(4) Atmospheric pressure : The gaseous envelope surrounding the earth is called the earth's atmosphere andthe pressure exerted by the atmosphere is called atmospheric pressure its value on the surface of the earthat sea level is nearly 1.013 105 N/m2 or Pascal in S.I. other practical units of pressure are atmosphere, barand torr (mm of Hg)1 atm = 1.01 105 Pa = 1.01 bar = 760 torrThe atmospheric pressure is maximum at the surface of earth and goes on decreasing as we move up intothe earth's(5) If P0 is the atmospheric pressure then for a point at depth h below the surface of a liquid of density .hydrostatic pressure P is given by P = P0 + hg.


(6) Hydrostatic pressure depends on the depth of the point below the surface (h). nature of liquid () andacceleration due to gravity (g) while it is independent of the amount of liquid, shape of the container or cross-sectional area considered. So if a given liquid is filled in vessels of different shapes to same height, thepressure at the base in each vessel's will be the same, though the volume or weight of the liquid in differentvessels will be different.

(7) In a liquid at same level, the pressure will be same at all points, if not, due to pressure difference the liquidcannot be at rest. This is why the height of liquid is the same in vessels of different shapes containingdifferent amounts of the same liquid at rest when they are in communication with each other.

(8) Gauge pressure : The pressure difference between hydrostatic pressure P and atmospheric pressure P0is called gauge pressure.

P P0 = hg

CONSEQUENCES OF PRESSURE(i) Railway tracks are laid on large sized wooden or iron sleepers. This is because the weight (force)

of the train is spread over a large area of the sleeper. This reduces the pressure acting on theground and hence prevents the yielding of ground under the weight of the train.

(ii) A sharp knife is more effective in cutting the objects than a blunt knife.The pressure exerted = Force/area. The sharp knife transmits force over a small area as com-pared to the blunt knife. Hence the pressure exerted in case of sharp knife is more than in case ofblunt knife.

(iii) A camel walks easily on sand but a man cannot inspite of the fact that a camel is much heavierthan man.This is because the area of camels feet is large as compared to mans feet. So the pressureexerted by camel on the sand is very small as compared to the pressure exerted by man. Due tolarge pressure, sand under the feet of man yields and hence he cannot walk easily on sand.

VARIATION OF PRESSURE WITH HEIGHTAssumptions : (i) unaccelerated liquid (ii) uniform density of liquid (iii) uniform gravity

Weight of the small element dh is balanced by the excess pressure. It means dhdp

= g.

P

Pa

gdp h

0

dh

P = Pa + gh

PASCALS LAWif the pressure in a liquid is changed at a particular, point the change is transmitted to the entireliquid without being diminished in magnitude. In the above case if P

a is increased by some amount

than P must increase to maintained the difference (P Pa) = hg. This is Pascals Law which

states that Hydraulic lift is common application of Pascals Law.


1. Hydraulic press.

p = aAWfor

AW

a

f

as A >> a then f Vsub.](4) When immiscible liquids of different densities are poured in a container, the liquid of highest density willbe at the bottom while that of lowest density at the top and interfaces will be plane.(5) Sometimes instead of density we use the term relative density or specific gravity which is defined as :

RD = waterofDensitybodyofDensity

(6) If m1 mass of liquid of density 1 and m2 mass of density 2 are mixed. then asm = m1 + m2 and V = (m1 / 1) + (m2 / 2)[As V = m/]

(7) If V1 volume of liquid of density 1 and V2 volume of liquid of density 2 are mixed, then asm = 1V1 + 2V2 and V = V1 + V2 [As = m/V]If V1 = V2 = V 1(1 + 2 )/2 = Arithmetic Mean

Example 1. A body of one kg placed on two object of negligible mass.Calculate pressure due to force on its bottom.

Solution : P = Amg

AF

(i) P1 = 24

22 m/N10m1010N101

(ii) P2 = 4102N101

= 5 104 (= 5p1)

Example 2. For a hydraulic system A car of mass 2000 kg standing on the platform of Area 10m2 whilethe area other side platform 10 cm2 find the mass required to balance the car

Solution : According to the Pascal Law

P1 = P2 Amg

Agm

car

car

m = carcar

mA

A

= kg2000

m10cm10

2

2

= 0.2 kg = 200 gm


Example 3. If two liquids of same masses but densities P1 and P2 respectively are mixed, then density ofmixture is given by

(1) = 221 (2) =

21

212

(3) =

21

212

(4) =21

21

Solution : = volumeTotalmassTotal

=

21

21 1m

m2VV

m2

21

212

Example 4. If two liquids of same masses but different densities 1 and 2 are mixed, then density of mixture isgiven by

(1) = 221 (2) =

21

212

(3) =

21

212

(4) =21

21

Solution : = 2V2)(V

V2mm

volumeTotalmassTotal 212121

Example 5. If pressure at the half depth of a lake equal to 2/3 pressure at the bottom of the lake, then the depthof the lake [

water = 103 kgm3, P0 = 105 N/m2 ](1) 10 m (2) 20 m (3) 60 m (4) 30 m

Solution : Pressure at bottom of the lake = Po + hg

Pressure at half the depth of a lake = Po + 2

h g

According to given condition

Po + )ghP(3

2gh21

o

gh61P

31

o

h = 1010102

gP2

3

5o

= 20m.

Example 6. A uniformly tapering vessel is filled with a liquid of uniform density900 kg/m3. The force that acts on the base of the vessel due to the liquid is(g = 10 ms2)

(1) 3.6 N (2) 7.2 N(3) 9.0 N (4) 14.4 N

Solution : Force acting on the baseF = P A = hdgA = 0.4 900 10 2 103 = 7.2N

Example 7. The area of cross-section of the two arms of a hydraulicpress are 1 cm2 and 10 cm2 respectively (figure). A force of5 N is applied on the water in the thinner arm. What forceshould be applied on the water in the thicker arms so thatthe water may remain in equilibrium?


Solution : In equilibrium, the pressures at the two surfaces should be equal as they lie in the samehorizontal level. If the atmospheric pressure is P and a force F is applied to maintain theequilibrium, the pressures are

P0 + 2cm1N5

and P0 + 2cm10F

respectively..

This givens F = 50 N.

2. Hydraulic Brake.

Hydraulic brake system is used in automobiles to retard the motion.

HYDROSTATIC PARADOXPressure is directly proportional to depth and by applying pascals law it can be seen that pres-sure is independent of the size and shape of the containing vessel. (In all the three cases theheights are same).

A B CPA = PB = PC

ATMOSPHERIC PRESSUREDefinition.The atmospheric pressure at any point is numerically equal to the weight of a column of air of unitcross-sectional area extending from that point to the top of the atmosphere.At 0C, density of mercury = 13.595 g cm3, and at sea level, g = 980.66 cm s2Now P = hg.Atmospheric pressure = 76 13.595 980.66 dyne cm2 = 1.013 105 N-m2 (p

a)

Height of AtmosphereThe standard atmospheric pressure is 1.013 105 Pa (N m2). If the atmosphere of earth has auniform density = 1.30 kg m3, then the height h of the air column which exerts the standardatmospheric pressure is given by hg = 1.013 105

h = g10013.1 5

= 8.913.110013.1 5

m = 7.95 103 m 8 km.

In fact, density of air is not constant but decreases with height. The density becomes half at

about 6 km high, th41

at about 12 km and so on. Therefore, we can not draw a clear cut line above

which there is no atmosphere. Anyhow the atmosphere extends upto 1200 km. This limit is con-sidered for all practical purposes.

MEASUREMENT OF ATMOSPHERIC PRESSURE1. Mercury Barometer.

To measure the atmospheric pressure experimentally,torricelli invented a mercury barometer in 1643.

pa =hg

The pressure exerted by a mercury column of 1mmhigh is called 1 Torr.

1 Torr = 1 mm of mercury column


2. Open tube ManometerOpen-tube manometer is used to measure the pressure gauge. When equilibrium is reached, thepressure at the bottom of left limb is equal to the pressure at the bottom of right limb.

i.e. p + y1 g = pa + y2 gp p

a = g (y2 y1) = gy

p pa = g (y2 y1) = gy

p = absolute pressure, p pa = gauge pressure.

Thus, knowing y and (density of liquid), we can measure the gauge pressure.Example 8. A barometer tube reads 76 cm of mercury. If the tube is gradually inclined at an angle of 60 with

vertical, keeping the open end immersed in the mercury reservoir, the length of the mercury columnwill be(1) 152 cm (2) 76 cm (3) 38 cm (4) cm338

Solution : cos 60 =

h

= 2/176

60cosh

= 152 cm

Example 9. When a large bubble rises from the bottom of a lake to the surface. Its radius doubles. If atmosphericpressure is equal to that of column of water height H, then the depth of lake is

[AIIMS 1995 ; AFMC 1997](1) H (2) 2H(3) 7H (4) 8H

Solution : P1V1 = P2V2

(Po + hg) 3o2 )r2(3

4Pr34

Example 10. A beaker containing liquid is kept inside a big closed jar If the air inside the jar is continuouslypumped out, the pressure in the liquid near the bottom of the liquid will(1) Increase(2) Decreases(3) Remain constant(4) First decrease and then increase

Solution : Total pressure at (near) bottom of the liquidP = P0 + hgAs air is continuously pumped out from jar (container), P0 decreases and hence P decreases.


Example 11. Write the pressure inside the tube

Solution : PA = P0 + g 1008

= PB ...(i)

Ptube = PB 1006g

...(ii)(i) & (ii)

Ptube = 1006g

1008gP0

Ptube =

50

gP0

Ans. P50gP0

Example 12. Find the pressure inside the tube


Solution : PA = (P0 + gh) + 3g (2h) = PB

Ptube = PB 8g

4h

3h2

= P0 + gh + 6gh 8g

h125

= P0 + gh + 6gh gh310

= 3gh)1021(P0 = P0 + gh3

11

Ans. P0 + gh311

Example 13. The manometer shown below is used to measure the dif ference in water levelbetween the two tanks. Calculate this difference for the conditions indicated.

Solution : pa + h1 g 401g + 40g = pa + h2 g

h2 g h1 g = 40 g 40 1g

as 1 = 0.9(h2 h1) g = 40g 36gh2 h1 = 4 cm

3. Water Barometer.

Let us suppose water is used in the barometer instead of mercury.

hg = 1.013 105 or h = g

10013.1 5

The height of the water column in the tube will be 10.3 m. Such a long tube cannot be managedeasily, thus water barometer is not feasible.


Example 14. In a g iven U- tube (open a t one-end) f ind ou t re la t ion be tween p and pa.

Given d2 = 2 13.6 gm/cm3 d1 = 13.6 gm/cm3

Solution : Pressure in a liquid at same level is same i.e. at A A,pgxdygdp 12a

In C.G.S.p

a + 13.6 2 25 g + 13.6 26 g = p

pa + 13.6 g [50 + 26] = p

2pa = p [p

a = 13.6 g 76]

Example 15. Truck start from rest with acceleration 2.5 ms2 then the angle (acute) between vertical and surfaceat the liquid, In equilibrium (assume that liquid is at with respect to truck)

(1) sin = 174

(2) cos = 171

(3) tan = 4 (4) None of these

Solution :

Consider a particle on the liquid surfacemg cos = ma cos gcos = a sin

tan = a

g tan = 5.2

10 = 4

Ans. ABC


Example 16. In previous question pressure at the point A, B and C(1) PA = PB = PC (2) PA > PB > PC(3) PA < PB < PC (4) Non of these

Example 17. In previous question, three different point, above the point A,B and C of an accelerated liquid surfacein equilibrium called A' B', C' then pressure at the point A' B' and C'(1) PA' = PB' = PC' (2) PA' > PB' > PC'(3) PA' < PB' < PC' (4) Po = atmosphere pressure

Example 18. Highest pressure at the point inside the liquid :(1) A (2) C(3) Pressure at A, B and C are equal and highest (4) None of these

Example 19. Slope of the line on which pressure is same consider the direction of acceleration of truck as theX-axis

(1) 4 (2) 0.25 (3) 2.5 (4) 41

Solution : (16 to 19)In the frame of truck liquid is in equilibrium, liquid surface is to the g

eff and pressure increase alongthe line of g(eff) as P = P0 + geff(h), where h is depth along the geff or to the surface, all the line whichis parallel to the liquid surface have same magnitude of pressure and magnitude increase as onemove along the g

eff inside the liquid.So PA > PB > PC & PA = PB = PC = P0 (pressure is highest at A)

and slope tan(90 + ) = cot = tan

1 = 4

1

ARCHIMEDES PRINCIPLE

According to this principle, when a body is immersed wholly or partially in a fluid, it loses itsweight which is equal to the weight of the fluid displaced by the body.Up thrust = buoyancy = V

g

V = volume submerged

= density of liquid.

Relation between density of solid and liquidweight of the floating solid = weight of the liquid displaced

V1 1 g = V2 2 g 12

2

1VV

orsolidtheofVolumeTotal

solidtheofportionimmeresedtheofVolumeliquidofDensitysolidofDensity


This relationship is valid in accelerating fluid also. Thus, the force acting on the body are :(i) its weight Mg which acts downward and(ii) net upward thrust on the body or the buoyant force (mg)Hence the apparent weight of the body = Mg mg = weight of the body weight of the displaced liquid.Or Actual Weight of body Apparent weight of body = weight of the liquid displaced.The point through which the upward thrust or the buoyant force acts when the body is immersedin the liquid is called its centre of buoyancy. This will coincide with the centre of gravity if the solidbody is homogeneous. On the other hand if the body is not homogeneous, then the centre ofgravity may not lie on the line of the upward thrust and hence there may be a torque that causesrotation in the body.If the centre of gravity of the body and the centre of buoyancy lie on the same straight line, thebody is in equilibrium.If the centre of gravity of the body does not coincide with the centre of buoyancy (i.e., the line ofupthrust), then torque acts on the body. This torque causes the rotational motion of the body.

Floatation1. Translatory equilibrium : When an body of density p and volume V is immersed in a liquid of density , the

forces acting on the body areWeight of body W = mg = Vg, acting vertically downwards through centre of gravity of the body.Upthrust force = Vg acting vertical upwards through the centre of gravity of the displaced liquid i.e., centreof buoyancy.

(i) A body will float in liquid only and only if (ii) In case of floating as weight of body = upthrustSo WApp = Actual weight upthrust = 0(iii) In case of floating Vg = VingSo the equilibrium of floating bodies is unaffected by variations in g though both thrust and weight depend on g.


(2) Rotatory Equilibrium : When a floating body is slightly tied from equilibrium position, the centre ofbuoyancy B shift. The vertical line passing through the new ecntre of buoyancy B' and initial vertical line meetat a point M called meta - centre. If the meta-centre M is above the centre of gravity the couple due to forcesat G (weight of body W) and at B' (upthrust) tends to bring the body back body the meta-centre must alwaysbe higher then the centre of gravity of the body.

However, if meta-centre goes CG, the couple due to forces at G and B' tends to topple the floating body.That is why a wooden log cannot be made to float vertical in water or a boat is likely to capsize if the sittingpassengers stand on it. In these situations CG becomes higher than MG and so the body will topple ifslightly tilted.

Example 20. A concrete sphere of redius R has cavity of radius r which is paked with sawdust. The specificgravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entirevolume submerged under water. Ration of mass of concrete to mass of sawdust will be

[AIIMS 1995](1) 8 (2) 4 (3) 3 (4) Zero

Solution : Let specific gravities of consrete and saw dust are 1 and 2 respectiviely.According to principle of floatation weight of whole sphere = upthrust on the sphere

g1R34gr

34g)rR(

54 3

23

133

R31 r3 1 + r3 2 = R3

R3 (1 1) = r3 (1 2 ) = 1

r

R1

213

3

11

r

rR1

1213

33

2

1

1

2

23

133

11

r

)rR(

3.04.2

14.23.01

dustsawofMassconcreteofMass

= 4

Example 21. A metallic block of density 5 gm cm3 and having dimensions 5 cm 5 cm 5cm is weighed inwater. Its apparent weight will be(1) 5 5 5 5 gf (2) 4 4 4 4 gf (3) 5 4 4 4 gf (4) 4 5 5 5 gf

Solution : Apparent weight= V ( ) g = 1 b h (5 1) g= 5 5 5 4 gDyne = 4 5 5 5 gf


Example 22. A cobical block is floating in a liquid with half of its volume immersed in the liquid. When the wholesystem accelerates upwards with acceleration of g/3, the fraction of volume immersed in the liquidwill be

(1) 21 (2) 8

3 (3) 32 (4) 4

3

Solution : Fraction of volume immersed in the liquid Vin =

V

i.e. it depends upon the densities of the block and liquid. So there will be no change in it ifsystem moves upward or downward with constant velocity or some acceleration.

Example 23. A silver ingot weighing 2.1 kg is held by a string so as to be completely immersed in a liquid ofrelative density 0.8. The relative density of silver is 10.5. The rension in the string in kg-wt is(1) 1.6 (2) 1.94 (3) 3.1 (4) 5.25

Solution : T = Apparent weight = V( ) g = M

( ) g

T = M

5.108.011.2g1

g = 1.94 gN

T = 1.94 Kg-wt

Example 24. A sample of metal weighs 210 gm in air, 180 gm in water and 120 gm in liquid. Then density (RD) of

(1) Metal is 3 (2) Metal is 7 (3) Liquid is 3 (4) Liquid is 31

Solution : Density of metal = . Density of liquid = If V is the volume of sample then according to problem210 = Vg ......(i)180 = V ( 1)g ......(ii)120 = V ( )g ......(iii)By solving (i), (ii) and (iii) we get = 7 and = 3.

Example 25. A cubical block of wood of edge 3 cm floats in water. The lowersurface of the cube just touches the free end of a vertical springfixed at the bottom of the pot. Find the maximum weight that can beput on the block without wetting it. Density of wood = 800 kg/m3 andspring constant of the spring = 50 N/m. Take g = 10 m/s2.

Solution : The specific gravity of the block = 0.8. Hence the height inside water = 3 cm 0.8 = 2.4cm. The height outside ater = 3 cm 2.4 = 0.6 cm. Suppose the maximum weight that canbe put without wetting it is W. The block in this case is completely immersed in the water.The volume of the displaced water

= volume of the block = 27 106 m3.Hence, the force of buoyancy

= (27 106 m3) 1(1000 kg/m3) (10 m/s2) = 0.27 N.The spring is compressed by 0.6 cm and hence the upward force exerted by the spring

= 50 N/m 0.6 cm = 0.3 N.The force of buoyancy and the spring force taken together balance the weight of the blockplus the weight W put on the block. The weight of the block is

W = (27 106 m) (800 kg/m3) (10 m/s2) = 0.22 N.Thus, W = 0.27 N + 0.3 N 0.22 N = 0.35 N.


PRESSURE IN CASE OF ACCELERATING FLUID(i) Liquid Placed in elevator :

When elevator accelerates upward with acceleration a0 then pressure in the fluid, at depth h maybe given by,

p = h [g + a0]and force of buoyancy, B = m (g + a0)

(ii) Free surface of liquid in horizontal acceleration :

tan = ga0

p1 p2 = a0 where p1 and p2 are pressures at point 1 & 2. Then h1 h2 = ga0

Example 26. An open rectangular tank 1.5 m wide 2m deep and 2m long is half filled with water. It isaccelerated horizontally at 3.27 m/sec2 in the direction of its length. Determine the depthof water at each end of tank. [g = 9.81 m/sec2]

Solution : tan = ga

= 31

Depth at corner A= 1 1.5 tan= 0.5 m Ans.

Depth at corner B= 1 + 1.5 tan = 1.5 m Ans.

(iii) Free surface of liquid in case of rotating cylinder.

h = g2v2

= g2r22

(constant)r

hA

C

B

STREAMLINE FLOWThe path taken by a particle in flowing fluid is called its line of flow. In the case of steady flow allthe particles passing through a given point follow the same path and hence we have a unique lineof flow passing through a given point which is also called streamline.


CHARACTERISTICS OF STREAMLINE1. A tangent at any point on the stream line gives the direction of the velocity of the fluid particle at

that point.2. Two steamlines never intersect each other.

Laminar flow : If the liquid flows over a horizontal surface in the form of layers of different veloci-ties, then the flow of liquid is called Laminar flow. The particle of one layer do not go to anotherlayer. In general, Laminar flow is a streamline flow.Turbulent Flow : The flow of fluid in which velocity of all particles crossing a given point is notsame and the motion of the fluid becomes disorderly or irregular is called turbulent flow.

REYNOLD S NUMBERAccording to Reynold, the critical velocity (vc) of a liquid flowing through a long narrow tube is(i) directly proportional to the coefficient of viscosity () of the liquid.(ii) inversely proportional to the density of the liquid and(iii) inversely proportional to the diameter (D) of the tube.

That is vc D

or vc = DR

or = Dvc

...............(1)where R is the Reynold number.If R < 2000, the flow of liquid is streamline or laminar. If R > 3000, the flow is turbulent. If R liesbetween 2000 and 3000, the flow is unstable and may change from streamline flow to turbulent flow.

EQUATION OF CONTINUIT YThe equation of continuity expresses the law of conservation of mass in fluid dynamics.

a1v1 = a2v2In general av = constant . This is called equation of continuity and states that as the area ofcross section of the tube of flow becomes larger, the liquids (fluid) speed becomes smaller andvice-versa.

Illustrations -(i) Velocity of liquid is greater in the narrow tube as compared to the velocity of the liquid in

a broader tube.(ii) Deep waters run slow can be explained from the equation of continuity i.e., av = constant.

Where water is deep the area of cross section increases hence velocity decreases.

ENERGY OF A LIQUIDA liquid can posses three types of energies :

(i) Kinetic energy :The energy possessed by a liquid due to its motion is called kinetic energy. The kinetic

energy of a liquid of mass m moving with speed v is 21

mv2.

K.E. per unit mass = m

mv21 2

= 21

v2.


(ii) Potential energy :The potential energy of a liquid of mass m at a height h is m g h.

P.E. per unit mass = m

mgh = gh

(iii) Pressure energy :The energy possessed by a liquid by virtue of its pressure is called pressure energy.Consider a vessel fitted with piston at one side (figure). Let this vessel is filled with aliquid. Let A be the area of cross section of the piston and P be the pressure experiencedby the liquid.The force acting on the piston = PAIf dx be the distance moved by the piston, then work done by the force = PA dx = PdVwhere dV = Adx, volume of the liquid swept.This work done is equal to the pressure energy of the liquid. Pressure energy of liquid in volume dV = PdV.The mass of the liquid having volume dV = dV, is the density of the liquid.

Pressure energy per unit mass of the liquid = dVPdV =

P.

BERNOULLI S THEOREMIt states that the sum of pressure energy, kinetic energy and potential energy per unit mass or perunit volume or per unit weight is always constant for an ideal ( i.e. incompressible andnon-viscous) fluid having stream-line flow.i.e.

P + 2

1 v2 + gh = constant.

Example 27. An incompressible liquid flows through a horizontaltube as shown in the following fig. Then the velocity of the fluid is (1) v = 2v1 + v2 (2) v = v1 + v2(3) v =

21

21vv

vv

(4) v = 2221 vv

Solution : m = m1 + m2V = V1 + V2Av = Av1 + Av2 v = 2v1 + v2

Example 28. Water enters through end A with speed 1 and leaves through end B with speed 2 of a cylindricaltube AB. The tude is always completely filled with water. In case I tude is horizontal and is case II itis vertical with end A upwards and in case III it is vertical with end B upwards. We have 1 = 2(1) Case I (2) Case II (3) Case III (4) Each case

Solution : This happens in accordance with equation of continuity and this equation was derived on theprinciple of conservation of mass and it is true in every case, either tube remain horizontal or vertical.


Example 29. Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600N/m2 between A and B where the areas of cross-section are 30cm2 and 15cm2 respectively.Find the rate of flow of water through the tube.

Solution : Let the velocity at A = vA and that at B = vB.

By the equation of continuity, A

Bv

v = 2

2

cm15cm30

= 2.

By Bernoullis equation,

PA + 21 vA

2 = PB + 2

1 vB

2

or, PA PB = 21

(2vA)2 21

vA2 = 2

3vA

2

or, 600 2mN

= 23

3m

kg1000 vA

2

or, vA = 22 s/m4.0 = 0.63 m/s.

The rate of flow = (30 cm2) (0.63 m/s) = 1800 cm3/s.

APPLICATION OF BERNOULLI S THEOREM

(i) Bunsen burner(ii) Lift of an airfoil.(iii) Spinning of a ball (Magnus effect)(iv) The sprayer.(v) A ping-pong ball in an air jet(vi) Torricellis theorem (speed of efflux)

At point A, P1 = P, v1 = 0 and h1 = h

At point B, P2 = P, v2 = v (speed of efflux) and h = 0

Using Bernoullis theorem 1P

+ gh1 + 21 2

1v = 2P

+ gh2 = 21 2

2v , we have

P

+ gh + 0 = P

+ 0 + 21

v2 21

v2 = gh or v = gh2

(VII) Venturimeter.

It is a gauge put on a flow pipe to measure the flow of speed of a liquid (Fig). Let the liquid ofdensity be flowing through a pipe of area of cross section A1. Let A2 be the area of cross sectionat the throat and a manometer is attached as shown in the figure. Let v1 and P1 be the velocity ofthe flow and pressure at point A, v2 and P2 be the corresponding quantities at point B.


Using Bernoullis theorem :

1P

+ gh1 + 21

21v =

2P + gh2 + 2

1 22v , we get

1P

+ gh + 21

v12 =

2P + gh + 2

1 22v (Since h1 = h2 = h)

or (P1 P2) = 21

( 22v v12) ....(1)According to continuity equation, A1 v1 = A2v2

or

2

12 A

Av

v1

Substituting the value of v2 in equation (1) we have

(P1 P2) = 21

21

21

2

2

1 vvAA

21 v1

2

1

AA 2

2

1

Since A1 > A2, therefore, P1 > P2

or 21v =

1AA

)PP(22

2

1

21 = )AA(

)PP(A222

21

2122

where (P1 P2) = m gh and h is the difference in heights of the liquid levels in the two tubes.

1v =

1AA

gh22

2

1

m

The flow rate (R) i.e., the volume of the liquid flowing per second is given by R = v1 A1.(viii) During wind storm,

The velocity of air just above the roof is large so according to Bernoullis theorem, the pressurejust above the roof is less than pressure below the roof. Due to this pressure difference an upwardforce acts on the roof which is blown of without damaging other parts of the house.

(ix) When a fast moving train cross a person standing near a railway track, the person has a tendencyto fall towards the train. This is because a fast moving train produces large velocity in air betweenperson and the train and hence pressure decreases according to Bernoullis theorem. Thus theexcess pressure on the other side pushes the person towards the train.


Example 30. Water flows through a horizontal tube of variable cross-section (figure). The area of cross-section atA and B are 4 mm2 and 2 mm2 respectively. If 1 cc of water enters per second through A, find(i) the speed of water at A,(ii) the speed of water at B and(iii) the pressure difference PA PB.

Solution : A1v1 = A2v2

1 + 21v1

2 + 0 = r2 + 2

1v2

2 + gh.

Ans. (i) 25 cm/s, (ii) 50 cm/s (iii) 94 N/m2

Example 31. The velocity of the liquid coming out of a small hole of a large vessel containing two differentliquids of densities 2and as shown in figureis

(1) gh6 (2) gh2 (3) gh22 (4) ghSolution : (2) Pressure at (1) :

P1 = Patm + g (2h)Applying Bernoulli's theorum between points (1) and (2)

[Patm + 2 g h] + g(2h) + 2

1 (2

) (0)2

= Patm + (2 ) g (0) + 2

1 (2

) v2

v = 2 gh Ans.

Example 32. A horizontal pipe line carries water in a streamline flow. At a point along the pipe where thecross-sectional area is 10 cm, the water velocity is 1 ms1 and the pressure is 2000 Pa. Thepressure of water at another point where the cross-sectional area is 5 cm will be :.[Density of water = 103 kg. m3 ] [JEE - 94, 2]

Solution : From continuity equationA1 v1 = A2 v2

v2 =

2

1AA

v1 =

5

10 (1) = 2 m/s

Applying Bernoullis theorem at 1 and 2 v2v1

1 2

P 2 + 2

1v2

2 = P 1 + 2

1 v1

2

P2 = P1 + 21(v12 v22 )

=

)41(1021

2000 3 Ans. 500 Pa


Example 33. Equal volumes of two immiscible liquids of densities and 2 arefilled in a vessel as shown in figure. Two small holes are punchedat depth h/2 and 3h/2 from the surface of lighter liquid. If v1 and v2are the velocities of efflux at these two holes, then v1/v2 is :

(1) 221

(2) 0.5

(3) 0.25 (4) 21

Solution : for hole (i)

P0 + 2V21

= P0 + 2hg V1 = gh

for hole (ii)

P0 + 2V22

= P0 + gh + 2h

2h

2V22

= 2gh. V2 = gh2

21

VV

2

1 = 0.5

Example 34. A tank is filled upto a height h with a liquid and is placed on aplatform of height h from the ground. To get maximum rangex

m a small hole is punched at a distance of y from the free

surface of the liquid. Then:

(1) xm = 2h (2) x

m = 1.5 h

(3) y = h (4) y = 0.75 h

Solution : R = Vx t = gy2 g

)yh2(2

R2 = 2gy g)yh2(2

= 4 (2hy y2)

for Rmax

= dydR

= 0

2R dydR

= 4 (2h 2y) = 0 y = hx

m = 2h

~~~~~~


OBJECTIVE PROBLEMS (JEE-MAIN)SECTION (A) : MEASUREMENT AND CALCULATION OF PRESSURE

A-1 A siphon in use is demonstrated in the following figure.The density of the liquid flowing in siphon is 1.5 gm/cc.The pressure difference between the point P and S will be (1) 105 N/m (2) 2 105 N/m(3) Zero (4) Infinity

A-2 Figure here shown the vertical cross-section of a vessel filledwith a liquid of density . The normal thrust per unit area on thewalls of the vessel at point. P, as shown, will be (1) h

g (2) H g(3) (H h)

g (4) (H h) g cos

A-3 A tank with length 10 m, breadth 8 m and depth 6m is filled with water to the top. If g = 10 m s2 anddensity of water is 1000 kg m3, then the thrust on the bottom is(1) 6 1000 10 80 N (2) 3 1000 10 48 N(3) 3 1000 10 60 N (4) 3 1000 10 80 N

A-4 In a hydraulic lift, used at a service station the radius of the large and small piston are in the ratio of 20: 1. What weight placed on the small piston will be sufficient to lift a car of mass 1500 kg ?(1) 3.75 kg (2) 37.5 kg (3) 7.5 kg (4) 75 kg.

A-5 Two vessels A and B of different shapes have the same base area andare filled with water up to the same height h (see figure). The forceexerted by water on the base is FA for vessel A and FB for vessel B. Therespective weights of the water filled in vessels are WA and WB. Then (1) FA > FB ; WA > WB (2) FA = FB ; WA > WB(3) FA = FB ; WA < WB (4) FA > FB ; WA = WB

SECTION (B) : ARCHEMEDIES PRINCIPLE AND FORCE OF BUOYANCYB-1 Two solids A and B float in water. It is observed that A floats with half its volume immersed and B floats with

2/3 of its volume immersed. Compare the densities of A and B(1) 4 : 3 (2) 2 : 3 (3) 3 : 4 (4) 1 : 3

B-2 The fraction of a floating object of volume V0 and density d0 above the surface of a liquid of density d will be

(1) dd0 (2)

0

0dd

dd (3) d

dd 0 (4) 0

0dd

dd

B-3 The density of ice is x gm/cc and that of water is y gm/cc. What is the change in volume in cc, whenm gm of ice melts ?(1) M (y x) (2) (y x)/m (3) mxy (x y) (4) m (1/y 1/x)

B-4 A cork is suberged in water by a spring attached to the bottom of a bowl. When the bowl is kept in an elevatormoving with acceleration downwards, the length of spring(1) Increases (2) Decreases (3) Remains unchanged (4) None of these

B-5 A hollow sphere of volume V is floating on water surface with half immersed in it. What should be theminimum volume of water poured inside the sphere so that the sphere now sinks into the water(1) V / 2 (2) V / 3 (3) V / 4 (4) V


B-6. A body floats in liquid contained in a beaker. If the whole system (shownin fig.) falls under gravity then the up-thrust on the body is-

(1) 2 mg (2) zero(3) mg (4) less than mg

SECTION (C) : CONTINUITY EQUATION AND BERNOULLI THEOREM & THEIR APPLICATIONC-1 Bernoullis principle is based on the law of conservation of:

(1) mass (2) momentum (3) energy (4) none of these

C-2 Bernoullis equation is applicable to points:(1) in a steadily flowing liquid(2) in a stream line(3) in a straight line perpendicular to a stream line(4) for ideal lequid stream line flow on a stream line

C-3 Bernoullis equation is based upon:(1) isochoric process (2) isobaric process (3) isothermal process (4) adiabatic process

C-4 Two water pipes of diameters 2 cm and 4 cm are connected with the main supply line. The velocity of flow ofwater in the pipe of 2 cm diameter is

(1) 4 time that in the other pipe (2) 41

times that in the other pipe

(3) 2 times that in the other pipe (4) 21

times that in the other pipe

C-5 A tank is filled with water up to height H. Water is allowed to come outof a hole P in one of the walls at a depth D below the surface of water.Express the horizontal distance x in terms of H and D :

(1) x = )DH(D (2) x = 2)DH(D

(3) x = )DH(D2 (4) x = )DH(D4

C-6 A fixed cylindrical vessel is filled with water up to height H. A hole is bored in the wall at a depth h fromthe free surface of water. For maximum horizontal range h is equal to :(1) H (2) 3H/4 (3) H/2 (4) H/4

C-7 An incompressible liquid flows through a horizontal tubeas shown in the figure. Then the velocity '

v

' of the fluid is :

(1) 3.0 m/s (2) 1.5 m/s(3) 1.0 m/s (4) 2.25 m/s

1. A block of volume V and of density b is placed in liquid of density l(l > b), then block is movedupward upto a height h and it is still in liquid. The increase in gravitational potential energy of thesystem is :(1) bVgh (2) (b + l)Vgh (3) (b l)Vgh (4) none of these

2. A metallic sphere floats (just sink) in an immiscible mixture of water (w = 103 kg/m3) and a liquid

(L = 13.5 103) with (1/5)th portion by volume in the liquid. The density of the metal is :(1) 4.5 103 kg/m3 (2) 4.0 103 kg/m3 (3) 3.5 103 kg/m3 (4) 1.9 103 kg/m3


3. A fire hydrant delivers water of density at a volume rate L. The watertravels vertically upward through the hydrant and then does 900 turn toemerge horizontally at speed V. The pipe and nozzle have uniform cross-section throughout. The force exerted by the water on the corner of thehydrant is(1) VL (2) zero(3) 2VL (4) VL2

4. A tube in vertical plane is shown in figure. It is filled with a liquid ofdensity and its end B is closedThen the force exerted by thefluid on the tube at end B will be : [Neglect atmospheric pressureand assume the radius of the tube to be negligible in comparison to ] (1) 0 (2) g A0(3) 2g A0 (4) Cannot be determined

5. A block of iron is kept at the bottom of a bucket full of water at 2C. The water exerts buoyant force onthe block. If the temperature of water is increased by 1C the temperature of iron block also increasesby 1C. The buoyant force on the block by water(1) will increase(2) will decrease(3) will not change(4) may decrease or increase depending on the values of their coefficient of expansion

6. The cubical container ABCDEFGH which is completely filled with an ideal (nonviscous andincompressible) fluid, moves in a gravity free space with a acceleration ofa = a0 )kji( where a0 is a positive constant. Then the only point inthe container where pressure is maximum, is (1) B (2) C(3) E (4) F

7. In previous question pressure will be minimum at point (1) A (2) B (3) H (4) F

8. Density of the ice is and that of water is . What will be the decreasein volume when a mass M of ice melts.

(1) M (2) M

(3)

11M (4)

11M1

9. The reading of a spring balance when a block is suspended from it in air is 60 newton. This reading ischanged to 40 newton when the block is submerged in water. The specific gravity of the block must betherefore :(1) 3 (2) 2 (3) 6 (4) 3/2

10. A block of steel of size 5 cm 5 cm 5 cm is weighed in water. If the relative density of steel is 7. Itsapparent weight is :(1) 6 5 5 5 gf (2) 4 4 4 7 gf (3) 5 5 5 7 gf (4) 4 4 4 6 gf

11. Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of onebody is 36 g and its density is 9 g/cc. If the mass of the other is 48 g, its density ing/cc is :(1) 4/3 (2) 3/2 (3) 3 (4) 5

12. In order that a floating object be in a stable rotation at equilibrium, its centre of buoyancy should be(1) vertically above its centre of gravity (2) vertically below its centre of gravity(3) horizontally in line with its centre of gravity (4) may be anywhere


13. For a fluid which is flowing steadily, the level in the vertical tubes is best represented by

(1) (2)

(3) (4)

14. There are two identical small holes on the opposite sides of a tank containinga liquid. The tank is open at the top. The difference in height between thetwo holes is h. As the liquid comes out of the two holes, the tank willexperience a net horizontal force proportional to:(1) h1/2 (2) h (3) h3/2 (4) h2

15. A narrow tube completely filled with a liquid is lying on a series ofcylinders as shown in figure. Assuming no sliding between any surfaces,the value of acceleration of the cylinders for which liquid will not comeout of the tube from anywhere is given by

(1) L2gH (2) L

gH

(3) LgH2 (4) L2

gH

MISCELLANEOUS PROBLEMSASSERTION / REASON

1. STATEMENT-1 : Any pressure increase at one point of a static connected fluid passed to each pointundiminished.STATEMENT-2 : Fluid is assumed to be incompressible.(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(3) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is True

2. STATEMENT-1 : One of the two identical container is empty and theother contains two ice cubes. Now both the containers are filled withwater to same level as shown. Then both the containers shall weighthe same.STATEMENT-2 : The weight of volume of water displaced by ice cube floating in water is equal to theweight of ice cube. Hence both the container in above situation shall weigh the same.(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(3) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is True


3. STATEMENT-1 : Consider an object that floats in water but sinks in oil. When the object floats in water,half of it is submerged. If we slowly pour oil on top of water till it completely covers the object, theobject moves up.STATEMENT-2 :As the oil is poured in the situation of statement-1, pressure inside the water willincrease everywhere resulting in an increase in upward force on the object.(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(3) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is True

LEVEL -1 : PREVIOUS YEARS AIEEE PROBLEMS (JEE-MAIN)1. A cylinder of height 20m is completely filled with water. The velocity of efflux of water (in ms1) through a small

hole on the side wall of the cylinder near its bottom, is : [AIEEE 2002](1) 10 (2) 20 (3) 25.5 (4) 5

2. The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period ofoscillation of the bob is t0 in air. Neglecting frictional force of water and given that the density of the bob is(4/3) 1000 kg/m3. What relationship between t and t0 is true? [AIEEE 2004](1) t = t0 (2) t = t0/2 (3) t = 2t0 (4) t = 4t0

3. A jar is filled with two non-mixing liquids 1 and 2 having densities 1 and 2,respectively. A solid ball, made of a material of density 3, is dropped in thejar. It comes to equilibrium in the position shown in the figure.

[AIEEE 2008, 4/300] Which of the following is true for 1, 2 and 3 ?(1) 1 > 3 > 2 (2) 1 < 2 < 3(3) 1 < 3 < 2 (4) 3 < 1 < 2

4. A ball is made of a material of density where oil < < water with oil and water representing the densities ofoil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture ofthis oil and water, which of the following pictures represents its equilibrium position? [AIEEE 2010, 4/144]

(1) (2) (3) (4)

5. Water is flowing continuously from a tap having an internal diameter 8 103 m. The water velocity as itleaves the tap is 0.4 ms1. The diameter of the water stream at a distance 2 101 m below the tap is closeto : [AIEEE - 2011, 4/120, 1](1) 5.0 103 m (2) 7.5 103 m (3) 9.6 103 m (4) 3.6 103 m

LEVEL - 2 : PREVIOUS YEARS JEE PROBLEMS (JEE-ADVANCE)* Marked Questions are More than one correct1. An application of Bernoullis equation for fluid flow is found in [IITJEE (Screening) 1994]

(A) Dynamic lift of an aeroplane (B) Viscosity meter(C) Capillary rise (D) Hydraulic press


2. A large open tank has two holes in the wall. One is a square hole ofside L at a depth y from the top and the other is a circular hole ofradius R at a depth 4y from the top. When the tank is completely filledwith water, the quantities of water flowing out per second from bothholes are the same. Then radius R, is equal to :[JEE-2000,2/105]

(A) 2

L(B) 2 L (C) L (D) 2

L

3. A wooden block with a coin placed on its top, floats in water as shownin figure. The distance and h are shown here. After some time the coinfalls into the water. Then : [I.I.T. 2002, 3/105 Screening](A) decreases and h increase (B) increases and h decreases(C) both and h increases (D) both and h decrease

4. STATEMENT -1The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when heldvertically up, but tends to narrow down when held vertically down. [JEE-2008' 3/162]andSTATEMENT -2In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.(volume flow rate)(A) STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT -2 is a correct explanation

for STATEMENT -1(B) STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT -2 is NOT a correct explanation for

STATEMENT -1(C) STATEMENT -1 is True, STATEMENT -2 is False(D) STATEMENT -1 is False, STATEMENT -2 is True.

EXERCISE # 1SECTION (A) :A-1 (3) A-2 (3) A-3 (1)A-4 (1) A-5 (2)SECTION (B) :B-1 (3) B-2 (3) B-3 (4)B-4 (2) B-5 (1) B-6. (2)SECTION (C) :C-1 (3) C-2 (4) C-3 (3)C-4 (1) C-5 (3) C-6 (3)C-7 (3)

EXERCISE # 21. (3) 2. (3) 3. (4)4. (2) 5. (1) 6. (1)7. (3) 8. (3) 9. (1)10. (1) 11. (3) 12. (1)13. (1) 14. (2) 15. (1)

EXERCISE # 31. (1) 2. (1) 3. (1)

EXERCISE # 4LEVEL -I

1. (2) 2. (3) 3. (3)4. (2) 5. (4)

LEVEL -II1. (A) 2. (A) 3. (D)4. (A)


OBJECTIVE PROBLEMS (JEE-ADVANCE)* Marked Questions are More than one correct1. A block of volume V and of density b is placed in liquid of density l(l > b), then block is moved

upward upto a height h and it is still in liquid. The increase in gravitational potential energy of thesystem is :(1) bVgh (2) (b + l)Vgh (3) (b l)Vgh (4) none of these

2. A metallic sphere floats (just sink) in an immiscible mixture of water (w = 103 kg/m3) and a liquid

(L = 13.5 103) with (1/5)th portion by volume in the liquid. The density of the metal is :(1) 4.5 103 kg/m3 (2) 4.0 103 kg/m3(3) 3.5 103 kg/m3 (4) 1.9 103 kg/m3

3. Three liquids of densities d, 2d and 3d are mixed in equal volumes. Then the density of the mixture is(1) d (2) 2d (3) 3d (4) 5d

4. Three liquids of densities d, 2d and 3d are mixed in equal proportions of weights. The relative density of themixture is

(1) 7d11 (2) 11

d18 (3) 9d13 (4) 18

d23

5. Figure shows a weigh-bridge, with a beaker P with water on one panand a balancing weight R on the other. A solid ball Q is hanging witha thread outside water. It has volume 40 cm3 and weighs 80 g. If thissolid is lowered to sink fully in water, but not touching the beakeranywhere, the balancing weight R' will be (1) same as R (2) 40 g less than R(3) 40 g more than R (4) 80 g more than R

6. A U-tube of base length l filled with same volume of two liquids ofdensities and 2 is moving with an acceleration a on the horizontalplane. If the height difference between the two surfaces (open toatmosphere) becomes zero, then the height h is given by:

(1) g2a (2) g2

a3 (3) ga (4) g3

a2

7. According to Bernoullis equation g21h

pgP

= Constant

The terms A, B, and C are generally called respectively :(1) Gravitational head, pressure head and velocity head(2) Gravity, gravitational head and velocity head(3) Pressure head, gravitational head and velocity head(4) Gravity, Pressure and velocity head

8. A manometer connected to a closed tap reads 3.5 105 N/m2, When the value is opened, the reading ofmanometer falls to 3.0 105 N/m2, then velocity of flow of water is(1) 100 m/s (2) 10 m/s (3) 1 m/s (4) 10 10 m/s

9. A cylindrical tank of height 0.4 m is open at the top and has a diameter 0.16 m. Water is filled in it upto a height of 0.16 m. how long it will take to empty the tank through a hole of radius 5103 m in itsbottom.(1) 46.26 sec. (2) 4.6 sec. (3) 462.6 sec. (4) .46 sec.


10. A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If theradius of the vessel is 0.05 m and the speed of rotation is 2 rev/s, The difference in the height of theliquid at the centre of the vessel and its sides will be (2 = 10) : [REE 1987](1) 3 cm (2) 2 cm (3) 3/2 cm (4) 2/3 cm

11. Air is streaming past a horizontal air plane wing such that its speed in 120 m/s over the upper surface and 90m/s at the lower surface. If the density of air is 1.3 kg per metre3 and the wing is 10 m long and has anaverage width of 2 m and negligible height difference, then the difference of the presure on the two sides of thewing of(1) 4095.0 Pascal (2) 409.50 Pascal (3) 40.950 Pascal (4) 4.0950 Pascal

12. A log of wood of mass 120 Kg floats in water. The weight that can be put on the raft to make it just sink,should be (density of wood = 600 Kg/m3) [CPMT 2004](1) 80 Kg (2) 50 Kg (3) 60 Kg (4) 30 Kg

13. If a sphere is inserted in water, then it flows with 31

rd of it outside the water, When it is inserted in an

unknown liquid then it flows with 43

th of it outside, then density of unknown liquid is : [RPMT 2001]

(1) 4.9 gm/c.c (2) 49

gm/c.c (3) 38

gm/c.c (4) 83

gm/c.c

14. Water is flowing inside a tube of an uniform radius ratio of radius of entry and exit terminals of the tube is3 : 2. Then the ratio of velocities at entry and exit terminals will be : [RPMT 2001](1) 4 : 9 (2) 9 : 4 (3) 8 : 27 (4) 1 : 1

15. A body of uniform cross-sectional area floats in a liquid of density thrice its value. The fraction of exposedheight will be : [RPMT 2005]

(1) 32 (2) 6

5 (3) 61 (4) 3

1

16.* An ice-cube of density 900 kg/m3 is floating in water of density 1000 kg/m3. The percentage of volume of ice-cube outside the water is : [RPMT 2006](1) 20% (2) 35% (3) 10% (4) 25%

17. A tank is filled with water upto height H. When a hole is made at a distance h below the level of water, whatwill be the horizontal range of water jet ? [RPMT 2006](1) hHh2 (2) hHh4 (3) hHh4 (4) hHh2

18. A raft of wood of mass 120 kg floats in water. The weight that can be put on the raft to make it just sing,should be : (d

raft = 600 kg/m3) [RPMT 2006](1) 80 kg (2) 50 kg (3) 60 kg (4) 30 kg

19. At what speed, the velocity head of water is equal to pressure head of 40 cm of hg ? [RPMT 2007](1) 10.3 m/s (2) 2.8 m/s (3) 5.6 m/s (4) 8.4 m/s

20. A hole is in the bottom of the tank having water. If total pressure at the bottom is 3 atm (1 atm = 105 Nm2),then velocity of water flowing from hole is : [RPMT 2007](1) 400 ms1 (2) 600 ms1

(3) 60 ms1 (4) none of these

21. If pressure at half the depth of a lake is equal to 2/3 pressure at the bottom of the lake then what is the depthof the lake : [RPET 2000](1) 10 m (2) 20 m (3) 60 m (4) 30 m


22. The pressure at the bottom of a tank containing a liquid does not depend on [Kerala (Engg.) 2002](1) Acceleration due to gravity (2) Height of the liquid column(3) Area of the bottom surface (4) Nature of the liquid

23. Construction of submarines is based on [Kerals PMT 2005](1) Archimedes principle (2) Bernoulli's theorem(3) Pascal's law (4) Newton's laws

24. From the adjacent figure, the correct observation is [KCET 2005](1) The pressure on the bottom of tank (a) is greater thanat the bottom of (b)(2) The pressure on the bottom of the tank (1) is smaller than at the bottom of (b)(3) The pressure depend on the shape of the container(4) The pressure on the bottom of (a) and (b) is the same

25. In the arrangement shown in figure 32

m

m

B

A and the ratio of density of

block B and of liquid is 2 : 1. The system is released from rest. Then:(1) block B will oscillate but not simple harmonically(2) block B will oscillate simple harmonically(3) the system will remain in equilibrium(4) none of the above

26. Water flows through a frictionless duct with a cross-sec-tion varying as shown in fig. Pressure p at points alongthe axis is represented by

(1) (2) (3) (4)

27. Three liquids of densities d, 2d and 3d are mixed in equal volumes. Then the density of the mixture is(1) d (2) 2d (3) 3d (4) 5d

28. Three liquids of densities d, 2d and 3d are mixed in equal proportions of weights. The relative density of themixture is

(1) 7d11 (2) 11

d18 (3) 9d13 (4) 18

d23

29. A cylindrical container of radius '

R

' and height '

h

' is completely filled witha liquid. Two horizontal L shaped pipes of small cross-section area '

a

' areconnected to the cylinder as shown in the figure. Now the two pipes areopened and fluid starts coming out of the pipes horizontally in oppositedirections. Then the torque due to ejected liquid on the system is:

(1) 4 a

g

h

R (2) 8 a

g

h

R(3) 2 a

g

h

R (4) none of these


30. According to Bernoullis equation g21h

pgP

= Constant The terms A, B, and C are generally called

respectively :(1) Gravitational head, pressure head and velocity head(2) Gravity, gravitational head and velocity head(3) Pressure head, gravitational head and velocity head(4) Gravity, Pressure and velocity head

31. A given shaped glass tube having uniform cross section is filled with waterand is mounted on a rotatable shaft as shown in figure. If the tube is rotatedwith a constant angular velocity then : [AIIMS 2005](1) Water levels in both sections A and B go up(2) Water level in Section A goes up and that in B comes down(3) Water level in Section A comes down and that in B it goes up(4) Water levels remains same in both sections

32. A candle of diameter d is floating on a liquid in a cylindrical container ofdiameter D (D > > d) as shown in figure. If it is burning at the rate of 2cm/hour then the top of the candle will [AIIMS 2005](1) Remain at the same height (2) Fall at the rate of 1 cm/hour(3) Fall at the rate of 2 cm/hour (4) Go up the rate of 1 cm/hour

33. A hemispherical portion of radius R is removed from the bottom of acylinder of radius R. The volume of the remaining cylinder is V and itsmass M. It is suspended by a string in a liquid of density where itstays vertical. The upper surface of the cylinder is at a depth h belowthe liquid surface. The force on the bottom of the cylinder by the liquidis : [I.I.T. 2001, 3/105 Screening] (1) Mg (2) Mg Vg(3) Mg + R2hg (4) g(V + R2 h)

34. Water is filled in a container upto height 3m. A small hole of area 'a' is punched in the wall of thecontainer at a height 52.5 cm from the bottom. The cross sectional area of the container is A. If a/A = 0.1 then v2 is : (where v is the velocity of water coming out of the hole) (g = 10 m/s2)

[I.I.T. 2005 Screening , 3/60](1) 50 (2) 51 (3) 48 (4) 51.5

1. (3) 2. (3) 3. (2) 4. (2) 5. (3) 6. (2) 7. (3)8. (2) 9. (1) 10. (2) 11. (1) 12. (1) 13. (3) 14. (1)15. (1) 16. (3)(4) 17. (1) 18. (1) 19. (1) 20. (1) 21. (2)22. (3) 23. (1) 24. (4) 25. (1) 26. (1) 27. (2) 28. (2)29. (1) 30. (3) 31. (1) 32. (2) 33. (4) 34. (1)


1. U = mgh U = (b ) Vgh

2. g5v4g

5v4

=

m(V)g

on solving m = 3.5 103 .

3. mix = V3

)d3d2d(VV3

mmm 321

= 2d.

4. mix =

d1118d

1163

d3m

d2m

dm

m3VVV

m3321

5. Since not touching,So R = Fb = l(vg) = 40g.

6. For the given situation, liquid of density 2

r should be behind that of r.From right limbPA = Patm + gh

PB = PA + a 2

= Patm + gh + a 2

PC = PB + (2) a 2

= Patm + gh + 2

3

a .... (1)But from left limb :

PC = Patm + (2) gh .... (2)From (1) and (2) :

Patm + gh + 2

3

a = Patm + 2 gh

h = g2a3 Ans.

8. Bernoulli's theorem for unit mass of liquid

2u21P

= constant

As the liquid starts flowing, it pressure energy decreases

212

PPu

21

3

552

10103105.3

u21

= 3

5

10105.02

u2 = 100 u = 10 m/s


9. v = gh2 ....(ii)from equation (ii) put the value of v in equation (i)

R2 dh/dt = r2 gh2

gh2rdhR

2

2= dt

0

h2

2

hdh

g2rR

= t

0

dt

on solving t = 46.26 second.

10. y = y2r22

Put values and get y = 2cm.

11. From the Benoulli's theorem

P1 P2 = ])90()120[(3.121

uu21 222

122

= 4095 N/m2 or Pascal

12. Volume of log of wood V = 3m2.0

600120

densitymass

Let x weight that can be put on the log of weed. So

weight of the body = (120 + x) 10 NWeight of displaced liquid = Vg = 0.2 103 10 NThe body will just sink in liquid if the weight of the body will be equal to the weight of displaced liquid. (120 x) 10 = 0.2 103 10 120 + x = 200 x = 80 kg

13. waterofdensitylativeRebodyofdensitylativeRe

= bodyofvolumetotalbodyofpartinsertedofvolume

V3V2

1d

d = 32

Again using the same formula

d32

41

d = 38

gm/c.c.

14. According to rule for flowing of liquid product of area and velocity is sameA1v1 = A2v2

1

2

2

1AA

v

v

= 94

r

r

r

r2

1

222

22

= 4 : 9


15. Here area is uniform, so portion of immersed height = 31

(Here : the ratio of density of body and density of water = 31 )

Therefore, fraction of exposed height will 1 31

= 32

16. The percentage of volume of ice-cube outside the water is

=100

water

icewater

= 10009001000

100 = 10%

17. Applying Bernoullis theorem, the velocity of water at point A

v = gh2Time taken to reach point C is t

So, H h = 21 gt2

t =

ghH2

Now, horizontal range R = vt

= gh2

ghH2

= 2 hhH

18. Volume of raft is given by

V = dm

= 600120

= 0.2 m3

If the raft is immersed fully in water. Then weight which can be put on the raft= 0.2 103 m= 200 kg

so extra weight put on the raft= 200 kg 120 kg= 80 kg

19. Bernoullis equation for flowing liquid be written as

p + 21

v2 + gh = constant (i)

Here, p = pressure energy per unit volume of liquid = density of liquid (water)h = height of liquid columnv = velocity of liquidand g = acceleration due to gravity Dividing Eq. (i) by g, we have

g

+ g2v2

+ h = constant


It this expression g2v2

is velocity head and gp is pressure head.

It is given that,velocity head = pressure head

ie. g2v2

= gp

or v2 = p2

or v2 = 3

23

108.91040106.132

v = 10.32 m/s

20. Let height of water column in the tank be h. Total pressure (p) = atmospheric pressure (p0) + pressure due towater column in tank (p) p = p p0 = 3 1 = 2 atmor hg = 2 105or h 103 10 = 2 105or h = 20 mHence, velocity of water coming from hole ie, velocity of efflux is

v = gh2

= 20102

= 400 ms1

26. When cross-section of duct is decreased, the velocity of water increased and in accordance with Bernoulli'stheorem, the pressure P decreased at that place.

27. mix = V3

)d3d2d(VV3

mmm 321

= 2d.

28. mix =

d1118d

1163

d3m

d2m

dm

m3VVV

m3321

29.

Velocity of efflux of water (v) =

2hg2

= gh

force on ejected water = Rate of change of momentum of ejected water.= (av) (v)= av2


Torque of these forces about central line= (av2) 2R . 2= 4av2 R = 4 agh R

33. [Flower Fupper] by liquid = Upthrust F2 F1 = upthrust F2 = F1 + upthrust

F2 = gh (R2) + Vgor F2 = g(V + R2h)

F2

F1

Upthrust

In this problem, we did not take the force due to air pressure on thecylinder. This is because force due to air pressure is cancelled. At topand bottom of the cylinder the force due to air pressure is equal andopposite.

34. Apply continuity equation A1V1 = A2V2 and Bernauli theorum 0p

+ 2v2

+ gh = constant at the top and

at the hole.

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