Fluid Mechanics for JEE Main

Target : JEE(NITs)

Fluid Mechanics

Fluid Mechanics

Contents Topic

Page No.

Theory

1 – 20

Exercise - 1

21 – 22

JEE (main) problems Exercise - 2

22 – 24

JEE (advanced) problems Exercise - 3

24 – 25

Miscellaneous problems Exercise - 4

25 – 26

Level - 1 : AIEEE Problems - JEE (main) Level - 2 : JEE Problems - JEE (Advance) Answer Key

26

Advanced Level Problems

27 – 30

JEE (advanced) problems Answer Key

30

Solutions of ALP

31 – 35

JEE(NITs) Syllabus – 2012 Pressure due to a fluid column; Pascalís law and its applications. Streamline and turbulent flow, Reynolds number. Bernoulliís principle and its applications.

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

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

P =

F A

F = normal force on a surface area A.

The pressure is a scalar quantity. This is because hydrostatic pressure is transmitted equally in all directions when force is applied, which shows that a definite direction is not associated with pressure. 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 that surface. The normal force (or thrust) exerted by liquid at rest per unit area of the surface in contact with it, is called pressure 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 on this surface is P = F/A (1) Units : N / m2 or Pascal (S.I) and Dyne/cm2 (C.G.S) (2) Dimension : (P) =

[F] [MLT –2 ]   [ML–1T – 2 ] 2 [A] [L ]

(3) At a point pressure acts in all directions and a definite direction is not associated with it. So pressure is a tensor quantity. (4) Atmospheric pressure : The gaseous envelope surrounding the earth is called the earth's atmosphere and the pressure exerted by the atmosphere is called atmospheric pressure its value on the surface of the earth at sea level is nearly 1.013 × 105 N/m2 or Pascal in S.I. other practical units of pressure are atmosphere, bar and torr (mm of Hg) 1 atm = 1.01 × 105 Pa = 1.01 bar = 760 torr The atmospheric pressure is maximum at the surface of earth and goes on decreasing as we move up into the 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 + hg.

RESONANCE

AIEEE-FLUID MECHANICS - 1

(6) Hydrostatic pressure depends on the depth of the point below the surface (h). nature of liquid () and acceleration due to gravity (g) while it is independent of the amount of liquid, shape of the container or crosssectional area considered. So if a given liquid is filled in vessels of different shapes to same height, the pressure at the base in each vessel's will be the same, though the volume or weight of the liquid in different vessels 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 liquid cannot be at rest. This is why the height of liquid is the same in vessels of different shapes containing different 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 P0 is called gauge pressure. P – P0 = hg

C ON SE QUE N C E S OF P RE SSURE (i)

(ii)

(iii)

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 the ground and hence prevents the yielding of ground under the weight of the train. 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 compared to the blunt knife. Hence the pressure exerted in case of sharp knife is more than in case of blunt knife. A camel walks easily on sand but a man cannot inspite of the fact that a camel is much heavier than man. This is because the area of camel’s feet is large as compared to man’s feet. So the pressure exerted by camel on the sand is very small as compared to the pressure exerted by man. Due to large pressure, sand under the feet of man yields and hence he cannot walk easily on sand.

VARIATI ON OF PRESSURE WITH HEI GHT Assumptions : (i) unaccelerated liquid (ii) uniform density of liquid (iii) uniform gravity W eight of the small element dh is balanced by the excess pressure. It means P



h

dp  g

Pa



dp = g. dh

 dh 0

P = P a + gh

P A SC AL’ S L AW if the pressure in a liquid is changed at a particular, point the change is transmitted to the entire liquid 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 – P a) = hg. This is Pascal’s Law which states that Hydraulic lift is common application of Pascal’s Law.

RESONANCE


1.

Hydraulic press. p=

f W W  or f  a a A A

as A >> a then f << W.. This can be used to lift a heavy load placed on the platform of larger piston or to press the things placed between the piston and the heavy platform. The work done by applied force is equal to change in potential energy of the weight in hydraulic press.

Density In a fluid, at a point, density  is defined as : m dm  V  0  V dV (1) In case of homogenous isotropic substance, it has no directional properties, so is a scalar. (2) It has dimensions (ML–3) and S.I. unit kg/m3 while C.G.S. unit g/cc with 1g /cc = 103 kg/m3 (3) Density of substance means the rotio of mass of substance to the means the ratio of mass of a body to the volume of the body. So for a solid body. Density of body = Density of substance While for a hollow body, density of body is lesser than that of substance [As Vbody > Vsub.] (4) When immiscible liquids of different densities are poured in a container, the liquid of highest density will be 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 :

= lim

Density of body RD = Density of water (6) If m1 mass of liquid of density 1 and m2 mass of density 2 are mixed. then as m = 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 as m = 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 =

Example 2. Solution :

F mg  A A

(i)

P1 =

(ii)

P2 =

1 10N 10  10  2 m 2 1 10N 2  10  4

 10 4 N / m 2

= 5 × 10 4 (= 5p 1)

For a hydraulic system A car of mass 2000 kg standing on the platform of Area 10m 2 while the area other side platform 10 cm 2 find the mass required to balance the car According to the Pascal Law P 1 = P2



 A  m =  A  car

RESONANCE

m car g mg  A car A 2    m car = 10cm  2000 kg = 0.2 kg = 200 gm  10m 2 


Example 3.

If two liquids of same masses but densities P1 and P2 respectively are mixed, then density of mixture is given by (1)  =

1   2 2

Total mass 2m   = Total volume = V1  V2

Solution :





Example 4.

21 2 (3)  =    1 2

1   2 (2)  = 2  1 2 2m  1 m  1   2

1 2 (4)  =    1

2

  

21 2 1   2

If two liquids of same masses but different densities 1 and 2 are mixed, then density of mixture is given by (1)  =

1   2 2

21 2 (3)  =    1 2

1   2 (2)  = 2  1 2

1 2 (4)  =    1

2

m1  m 2 V(1   2 ) 1   2 Total mass   = Total volume  2V 2V 2

Solution :

Example 5.

Solution :

If pressure at the half depth of a lake equal to 2/3 pressure at the bottom of the lake, then the depth of the lake [water = 103 kgm–3, P0 = 105 N/m2 ] (1) 10 m (2) 20 m (3) 60 m (4) 30 m Pressure at bottom of the lake = Po + hg Pressure at half the depth of a lake = Po +

h g 2

According to given condition Po +

Example 6.

1 2 hg  (Po  hg) 2 3



1 1 Po  hg 3 6



h=

2Po 2  105  3 = 20m. g 10  10

A uniformly tapering vessel is filled with a liquid of uniform density 900 kg/m3. The force that acts on the base of the vessel due to the liquid is (g = 10 ms–2) (1) 3.6 N (2) 7.2 N (3) 9.0 N (4) 14.4 N Force acting on the base F = P × A = hdgA = 0.4 × 900 × 10 × 2 × 10–3 = 7.2N

Solution :

Example 7.

The area of cross-section of the two arms of a hydraulic press are 1 cm 2 and 10 cm 2 respectively (figure). A force of 5 N is applied on the water in the thinner arm. W hat force should be applied on the water in the thicker arms so that the water may remain in equilibrium?

RESONANCE


Solution :

In equilibrium, the pressures at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is P and a force F is applied to maintain the equilibrium, the pressures are 5N

P0 +

2

and P 0 +

1 cm This givens F = 50 N.

F 10 cm 2

respectively..

 2.

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

HY DRO STATI C P A RAD OX Pressure is directly proportional to depth and by applying pascal’s law it can be seen that pressure is independent of the size and shape of the containing vessel. (In all the three cases the heights are same).

A

B

C

PA = PB = PC

A T M OS P H E R I C P R E SS U R E Definition. The atmospheric pressure at any point is numerically equal to the weight of a column of air of unit cross-sectional area extending from that point to the top of the atmosphere. At 0ºC, density of mercury = 13.595 g cm –3 , and at sea level, g = 980.66 cm s –2 Now P = hg. Atmospheric pressure = 76 × 13.595 × 980.66 dyne cm –2 = 1.013 × 10 –5 N-m 2 (p a) Height of Atmosphere The standard atmospheric pressure is 1.013 × 10 5 Pa (N m –2 ). If the atmosphere of earth has a uniform density  = 1.30 kg m –3 , then the height h of the air column which exerts the standard atmospheric pressure is given by  hg = 1.013 × 10 5

1.013  105 1.013  10 5 h= = m = 7.95 × 10 3 m  8 km. g 1.13  9.8 In fact, density of air is not constant but decreases with height. The density becomes half at

1 th at about 12 km and so on. Therefore, we can not draw a clear cut line above 4 which there is no atmosphere. Anyhow the atmosphere extends upto 1200 km. This limit is considered for all practical purposes. about 6 km high,

M E A SURE M E N T OF ATMOS P HE RI C P RE SS URE 1.

Mercury Barometer. To measure the atmospheric pressure experimentally, torricelli invented a mercury barometer in 1643. p a =hg The pressure exerted by a mercury column of 1mm high is called 1 Torr. 1 Torr = 1 mm of mercury column

RESONANCE


2.

Open tube Manometer Open-tube manometer is used to measure the pressure gauge. W hen equilibrium is reached, the pressure at the bottom of left limb is equal to the pressure at the bottom of right limb.

i.e. p + y1 g = p a + y2 g p – p a = g (y2 – y1) = gy p – p a = g (y2 – y1) = gy p = absolute pressure, p – p a = 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 column will be (1) 152 cm

Solution :

cos 60º =

  Example 9.

(2) 76 cm

(3) 38 cm

(4) 38 3cm

h 

h 76  cos 60 º 1/ 2  = 152 cm

=

When a large bubble rises from the bottom of a lake to the surface. Its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is [AIIMS 1995 ; AFMC 1997]

Solution :

(1) H (3) 7H P1V1 = P2V2 

(Po + hg) ×

(2) 2H (4) 8H 4 2 4 r  Po  ( 2r )3 3 3

Example 10. A beaker containing liquid is kept inside a big closed jar If the air inside the jar is continuously pumped 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 liquid P = P0 + hg As air is continuously pumped out from jar (container), P0 decreases and hence P decreases.

RESONANCE


Example 11. W rite the pressure inside the tube

Solution :

P A = P 0 + g

8 = PB 100

P tube = P B – g

6 100

...(i)

...(ii)

(i) & (ii) 8  6    g P tube =  P0  g 100 100   g    P tube =  P0  50  

Ans.

P0 

g P 50

Example 12. Find the pressure inside the tube

RESONANCE


Solution :

P A = (P 0 + gh) + 3g (2h) = P B  2h h    P tube = P B – 8g   3 4

 5  = P 0 + gh + 6gh – 8g  h   12 

= P 0 + gh + 6gh – = P0 

Ans.

10 gh 3

( 21  10 )gh 11 gh = P0 + 3 3

P0 +

11 gh 3

Example 13. The m anom eter shown below is us ed to m easur e the dif fer ence in water level between the two tanks. Calculate this difference for the conditions indicated.

Solution :

p a + h 1 g – 40 1g + 40g = p a + h 2 g h 2 g – h 1 g = 40 g – 40  1g

as  1 = 0.9 (h 2 – h 1) g = 40g – 36g h 2 – h 1 = 4 cm

 3.

Water Barometer. Let us suppose water is used in the barometer instead of mercury. hg = 1.013 × 10 5

or

h=

1.013  10 5 g

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

RESONANCE


Example 14. I n a g i v e n U - t u b e ( o p e n a t o n e - e n d ) f i n d o u t r e l a t i o n b e t w e e n p a n d p a . Given d 2 = 2 × 13.6 gm/cm 3 d 1 = 13.6 gm/cm 3

Solution :

Pressure in a liquid at same level is same i.e. at A – A–,

p a  d2 yg  xd1g  p

In C.G.S. p a + 13.6 × 2 × 25 × g + 13.6 × 26 × g = p p a + 13.6 × g [50 + 26] = p 2p a = p [p a = 13.6 × g × 76] Example 15. Truck start from rest with acceleration 2.5 ms–2 then the angle (acute) between vertical and surface at the liquid, In equilibrium (assume that liquid is at with respect to truck)

4 (1) sin =

1 (2) cos =

17

17 (4) None of these

(3) tan = 4

Solution :

Consider a particle on the liquid surface mg cos  = ma cos  gcos = a sin tan = Ans.

g a ABC

RESONANCE



tan =

10 =4 2.5


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 surface in 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 the X-axis (1) –4

(2) –0.25

(3) –2.5

(4) 

1 4

Solution : (16 to 19) In the frame of truck liquid is in equilibrium, liquid surface is  to the geff and pressure increase along the line of g(eff) as P = P0 + geff(h), where h is depth along the geff or  to the surface, all the line which is parallel to the liquid surface have same magnitude of pressure and magnitude increase as one move along the geff inside the liquid. So PA > PB > PC & PA = PB = PC = P0 (pressure is highest at A) and slope tan(90 + ) = – cot = –

1 1 =– tan  4

 ARC HI M E DE S’ P RI N C I P L E According to this principle, when a body is immersed wholly or partially in a fluid, it loses its weight 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 liquid weight of the floating solid = weight of the liquid displaced V1 1 g = V2 2 g or



1 V2  2 V1

Density of solid Volume of the immeresed portion of the solid  Density of liquid Total Volume of the solid

RESONANCE


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 W eight 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 immersed in the liquid is called its centre of buoyancy. This will coincide with the centre of gravity if the solid body is homogeneous. On the other hand if the body is not homogeneous, then the centre of gravity may not lie on the line of the upward thrust and hence there may be a torque that causes rotation in the body. If the centre of gravity of the body and the centre of buoyancy lie on the same straight line, the body is in equilibrium. If the centre of gravity of the body does not coincide with the centre of buoyancy (i.e., the line of upthrust), then torque acts on the body. This torque causes the rotational motion of the body.

Floatation 1.

Translatory equilibrium : When an body of density p and volume V is immersed in a liquid of density , the forces acting on the body are Weight of body W = mg = Vg, acting vertically downwards through centre of gravity of the body. Upthrust force = Vg acting vertical upwards through the centre of gravity of the displaced liquid i.e., centre of buoyancy.

(i) A body will float in liquid only and only if  (ii) In case of floating as weight of body = upthrust So W App = Actual weight – upthrust = 0 (iii) In case of floating Vg = Ving So the equilibrium of floating bodies is unaffected by variations in g though both thrust and weight depend on g.

RESONANCE


(2) Rotatory Equilibrium : When a floating body is slightly tied from equilibrium position, the centre of buoyancy B shift. The vertical line passing through the new ecntre of buoyancy B' and initial vertical line meet at a point M called meta - centre. If the meta-centre M is above the centre of gravity the couple due to forces at G (weight of body W) and at B' (upthrust) tends to bring the body back body the meta-centre must always be 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 sitting passengers stand on it. In these situations CG becomes higher than MG and so the body will topple if slightly tilted.

Example 20. A concrete sphere of redius R has cavity of radius r which is paked with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume 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 4 4 4 (R 3 – r 3 )1g  r 3  2 g  R 3  1 g 5 3 3  R31 – r3 1 + r3 2 = R3



R3 (1 – 1) = r3 (1 – 2 ) =

R3 – r 3 

r

3





3

r 2

r

3



1 –  2 1 – 1

1 –  2 – 1  1 1 – 1

(R 3 – r 3 )1 

R3

 1 –  2  1     1 – 1   2

Mass of concrete  1 – 0.3  2.4   =4 Mass of saw dust  2.4 – 1  0.3

Example 21. A metallic block of density 5 gm cm–3 and having dimensions 5 cm × 5 cm × 5cm is weighed in water. 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×g Dyne = 4 × 5 × 5 × 5 gf

RESONANCE


Example 22. A cobical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with acceleration of g/3, the fraction of volume immersed in the liquid will be (1)

Solution :

1 2

(2)

3 8

(3)

2 3

(4)

3 4

 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 if system 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 of relative 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 

 0.8     g = 1.94 gN T = M  1 – g  2.1 1 –  10 .5    

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 Solution :

(2) Metal is 7

(3) Liquid is 3

(4) Liquid is

1 3

Density of metal = . Density of liquid =  If V is the volume of sample then according to problem 210 = Vg ......(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 lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting it. Density of wood = 800 kg/m 3 and spring constant of the spring = 50 N/m. Take g = 10 m/s 2. Solution : The specific gravity of the block = 0.8. Hence the height inside water = 3 cm × 0.8 = 2.4 cm. The height outside ater = 3 cm – 2.4 = 0.6 cm. Suppose the maximum weight that can be 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 × 10 –6 m 3. Hence, the force of buoyancy = (27 × 10 –6 m 3) × 1(1000 kg/m3) × (10 m/s 2) = 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 block plus the weight W put on the block. The weight of the block is W = (27 × 10 –6 m) × (800 kg/m 3) × (10 m/s 2) = 0.22 N. Thus, W = 0.27 N + 0.3 N – 0.22 N = 0.35 N.

RESONANCE


 PRE SSURE IN CASE OF ACCE LERATI NG FLUID (i)

Liquid Placed in elevator : When elevator accelerates upward with acceleration a 0 then pressure in the fluid, at depth ‘h’ may be given by, p = h [g + a 0 ] and force of buoyancy, B = m (g + a 0 )

(ii)

Free surface of liquid in horizontal acceleration :

tan  =

a0 g

p 1 – p 2 =  a 0 where p 1 and p 2 are pressures at point 1 & 2. Then h 1 – h 2 =

a0 g

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

a 1 tan  = g = 3 Depth at corner ‘A’ = 1 – 1.5 tan = 0.5 m Depth at corner ‘B’ = 1 + 1.5 tan  = 1.5 m

Ans. Ans.

 (iii)

(constant)

Free surface of liquid in case of rotating cylinder.

 2r 2 v2 h = = 2g 2g

r

A

B h

C

STR E AML I N E F L OW The path taken by a particle in flowing fluid is called its line of flow. 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 which is also called streamline.

RESONANCE


C HARAC TE RI STI C S OF STRE AML I N E 1. 2.

A tangent at any point on the stream line gives the direction of the velocity of the fluid particle at that point. Two steamlines never intersect each other. Laminar flow : If the liquid flows over a horizontal surface in the form of layers of different velocities, then the flow of liquid is called Laminar flow. The particle of one layer do not go to another layer. 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 not same and the motion of the fluid becomes disorderly or irregular is called turbulent flow.

RE YN OL D ’ S N U MB E R According to Reynold, the critical velocity (v c ) 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 v c  D

or

R vc =  D

or =

v c D 

...............(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 lies between 2000 and 3000, the flow is unstable and may change from streamline flow to turbulent flow.

E QUATI ON OF C ONTI N UI T Y The equation of continuity expresses the law of conservation of mass in fluid dynamics.

a 1 v 1 = a 2 v2 In general av = constant . This is called equation of continuity and states that as the area of cross section of the tube of flow becomes larger, the liquid’s (fluid) speed becomes smaller and vice-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. W here water is deep the area of cross section increases hence velocity decreases.

ENERGY OF A LIQUID A 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



1 mv 2. 2

1 mv 2 1 2 2 K.E. per unit mass = = v. 2 m

RESONANCE


(ii)

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

(iii)

P.E. per unit mass =

mgh = gh m

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 a liquid. Let ‘A’ be the area of cross section of the piston and P be the pressure experienced by the liquid. The force acting on the piston = PA If dx be the distance moved by the piston, then work done by the force = PA dx = PdV where 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 =

PdV P = . dV 

B E RN OU L L I ’ S TH E O RE M It states that the sum of pressure energy, kinetic energy and potential energy per unit mass or per unit volume or per unit weight is always constant for an ideal (i.e. incom pressible and non-viscous) fluid having stream-line flow. P 1 2 i.e. + v + gh = constant.  2

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

v 1v 2 v1  v 2

(4) v =

v 12  v 22

m = m1 + m2 V = V1 + V2 Av = 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 cylindrical tube AB. The tude is always completely filled with water. In case I tude is horizontal and is case II it is 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 the principle of conservation of mass and it is true in every case, either tube remain horizontal or vertical.

RESONANCE


Example 29. Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600 N/m 2 between A and B where the areas of cross-section are 30cm 2 and 15cm 2 respectively. Find the rate of flow of water through the tube.

Solution :

Let the velocity at A = v A and that at B = v B.

vB 30 cm 2 By the equation of continuity, v = = 2. 15 cm 2 A By Bernoulli’s equation, PA +

1 1  v A2 = P B + v B2 2 2

or,

PA – PB =

or,

600

N m

2

1 1 3 (2v A) 2 – v A2 = v 2 2 2 2 A =

3 1000 kg    v 2 m3  A 2 

0.4 m 2 / s 2 = 0.63 m/s. The rate of flow = (30 cm 2) (0.63 m/s) = 1800 cm 3/s.

or,

vA =

 AP PL I C ATION OF B E RN OUL L I ’S THE ORE M (i) (ii) (iii) (iv) (v) (vi)

Bunsen burner Lift of an airfoil. Spinning of a ball (Magnus effect) The sprayer. A ping-pong ball in an air jet Torricelli’s theorem (speed of efflux) At point A, P 1 = P, v 1 = 0 and h 1 = h

At point B, P 2 = P, v 2 = v (speed of efflux) and h = 0

Using Bernoulli’s theorem

P1 P2 1 2 1 2 + gh 1 + + gh 2 = v1 = v , we have   2 2 2

P P 1 2 + gh + 0 =   +0 + 2 v



1 2 v = gh 2

or

v=

2gh

 (VII)

Venturimeter. It is a gauge put on a flow pipe to measure the flow of speed of a liquid (Fig). Let the liquid of density  be flowing through a pipe of area of cross section A 1. Let A 2 be the area of cross section at the throat and a manometer is attached as shown in the figure. Let v 1 and P 1 be the velocity of the flow and pressure at point A, v 2 and P 2 be the corresponding quantities at point B.

RESONANCE


Using Bernoulli’s theorem :

P1 P2 1 2 1 2 + gh 1 + = + gh 2 + v v , we get 1   2 2 2

P1 P2 1 2 1 2 + gh + v1 = + gh + v (Since h 1 = h 2 = h)   2 2 2

1 ( v 22 – v 12) ....(1) 2 According to continuity equation, A 1 v 1 = A 2v 2 or

(P 1 – P 2) =

or

A  v 2   1  v  A2  1

Substituting the value of v 2 in equation (1) we have 2   1  A 1  v 2  v 2  1 2 1  1 (P 1 – P 2) =    2  v1 2  A 2  

 A  2   1   1  A 2    

Since A 1 > A 2, therefore, P 1 > P 2 or

v 12

=

2 A 22 (P1  P2 ) 2(P1  P2 ) =  A  2   ( A 12  A 22 )   1   1  A 2    

where (P 1 – P 2) =  m gh and h is the difference in heights of the liquid levels in the two tubes. 2 m gh v1 =

 A  2    1   1  A 2    

The flow rate (R) i.e., the volume of the liquid flowing per second is given by R = v 1 A 1. (viii)

During wind storm, The velocity of air just above the roof is large so according to Bernoulli’s theorem, the pressure just above the roof is less than pressure below the roof. Due to this pressure difference an upward force 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 tendency to fall towards the train. This is because a fast moving train produces large velocity in air between person and the train and hence pressure decreases according to Bernoulli’s theorem. Thus the excess pressure on the other side pushes the person towards the train.

RESONANCE


Example 30. Water flows through a horizontal tube of variable cross-section (figure). The area of cross-section at A 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 + Ans.

1 1 v 2 + 0 = r2 + v22 + gh. 2 1 2 (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 different liquids of densities 2and as shown in figureis

(1) 6gh Solution :

(2) 2 gh

(3) 2 2gh

(4) gh

(2) Pressure at (1) : P1 = Patm +  g (2h) Applying Bernoulli's theorum between points (1) and (2) [Patm + 2  g h] + g(2h) +

1 (2 ) v2 2

= Patm + (2 ) g (0) + 

1 (2 ) (0)2 2

v = 2 gh

Ans.

Example 32. A horizontal pipe line carries water in a streamline flow. At a point along the pipe where the cross-sectional area is 10 cm², the water velocity is 1 ms 1 and the pressure is 2000 Pa. The pressure of water at another point where the cross-sectional area is 5 cm² will be :. Solution :

[Density of water = 103 kg. m 3 ] From continuity equation A1 v1 = A2 v2 

[JEE - 94, 2] 1

 A1   10  v2 =  A  v1 =   (1) = 2 m/s  5   2

2

v1

Applying Bernoulli’s theorem at 1 and 2 P2 +

1 1 v22 = P 1 +  v12 2 2 P2 = P1 +

 

1

1 (v12 – v22 ) 2 

3 =  2000  2  10 (1  4)  Ans.



RESONANCE



500 Pa AIEEE-FLUID MECHANICS - 19

v2

Example 33.

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

1 (1)

(2) 0.5

2 2

1 (3) 0.25 Solution :

(4)

2

for hole (i)

h V12 = P0 + g 2 2 for hole (ii) P0 + 

P0 + 

V22 = P0 + gh + 2h 2

V22 = 2gh. 2



V1 =



gh

h   2

V2 = 2 gh

V1 1  V2 2 = 0.5 Example 34.

A tank is filled upto a height h with a liquid and is placed on a platform of height h from the ground. To get maximum range xm a small hole is punched at a distance of y from the free surface of the liquid. Then: (1) xm = 2h (3) y = h

Solution :

(2) xm = 1.5 h (4) y = 0.75 h

2(2h – y ) g

R = Vx t =

2gy

R2 = 2gy ×

2(2h – y ) = 4 (2hy – y2) g

dR for Rmax = dy = 0 dR 2R dy = 4 (2h – 2y) = 0



y=h

x = 2h m

~~~~~~

RESONANCE


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 filled with a liquid of density . The normal thrust per unit area on the walls 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 s –2 and density of water is 1000 kg m –3, 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 and are filled with water up to the same height h (see figure). The force exerted by water on the base is FA for vessel A and FB for vessel B. The respective weights of the water filled in vessels are W A and W B. Then (1) FA > FB ; W A > W B (2) FA = FB ; W A > W B (3) FA = FB ; W A < W B (4) FA > FB ; W A = W B

SECTION (B) : ARCHEMEDIES PRINCIPLE AND FORCE OF BUOYANCY B-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)

d0 d

dd0 (2) d  d 0

(3)

d  d0 d

dd 0 (4) d  d 0

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, when m 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 elevator moving 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 the minimum 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

RESONANCE


B-6.

A body floats in liquid contained in a beaker. If the whole system (shown in fig.) falls under gravity then the up-thrust on the body is(1) 2 mg (3) mg

(2) zero (4) less than mg

SECTION (C) : CONTINUIT Y EQUATION AND BERNOULLI THEOREM & THEIR APPLICATION C-1

Bernoulli’s principle is based on the law of conservation of: (1) mass (2) momentum (3) energy

C-2

Bernoulli’s 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

Bernoulli’s equation is based upon: (1) isochoric process (2) isobaric process

C-4

(4) adiabatic process

Two water pipes of diameters 2 cm and 4 cm are connected with the main supply line. The velocity of flow of water in the pipe of 2 cm diameter is (1) 4 time that in the other pipe

C-5

(3) isothermal process

(4) none of these

(2)

1 times that in the other pipe 4

1 times that in the other pipe 2 A tank is filled with water up to height H. Water is allowed to come out of 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 : (3) 2 times that in the other pipe

(4)

(1) x =

(2) x =

D(H  D )

(3) x = 2 D(H  D)

D(H  D) 2

(4) x = 4 D(H  D )

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 from the 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 tube as shown in the figure. Then the velocity ' v ' of the fluid is : (1) 3.0 m/s (3) 1.0 m/s

(2) 1.5 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 moved upward upto a height h and it is still in liquid. The increase in gravitational potential energy of the system 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/m 3) 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/m 3 (2) 4.0 × 103 kg/m 3 (3) 3.5 × 103 kg/m 3 (4) 1.9 × 103 kg/m 3

RESONANCE


3.

A fire hydrant delivers water of density  at a volume rate L. The water travels vertically upward through the hydrant and then does 900 turn to emerge horizontally at speed V. The pipe and nozzle have uniform crosssection throughout. The force exerted by the water on the corner of the hydrant is (1) VL (2) zero (3) 2VL

(4) 2  VL

4.

A tube in vertical plane is shown in figure. It is filled with a liquid of density and its end B is closedThen the force exerted by the fluid on the tube at end B will be : [Neglect atmospheric pressure and assume the radius of the tube to be negligible in comparison to ] (1) 0 (2) g A0 (3) 2g A0 (4) Cannot be determined

5.

A block of iron is kept at the bottom of a bucket full of water at 2°C. The water exerts buoyant force on the block. If the temperature of water is increased by 1°C the temperature of iron block also increases by 1°C. 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 and incompressible) fluid, moves in a gravity free space with a acceleration of a = a0 ( ˆi  ˆj  kˆ ) where a0 is a positive constant. Then the only point in the container where pressure is maximum, is (1) B (2) C (3) E (4) F

7.

8.

In previous question pressure will be minimum at point – (1) A (2) B (3) H

(4) F

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

1 1 (3) M    

(4)

1 1 1    M  

9.

The reading of a spring balance when a block is suspended from it in air is 60 newton. This reading is changed to 40 newton when the block is submerged in water. The specific gravity of the block must be therefore : (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. Its apparent 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 one body is 36 g and its density is 9 g/cc. If the m ass of the other is 48 g, its density in g/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

RESONANCE


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 containing a liquid. The tank is open at the top. The difference in height between the two holes is h. As the liquid comes out of the two holes, the tank will experience 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 of cylinders as shown in figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by (1)

gH 2L

(2)

(3)

2gH L

(4)

gH L gH 2L

MISCELLANEOUS PROBLEMS ASSERTION / REASON 1.

STATEMENT-1 : Any pressure increase at one point of a static connected fluid passed to each point undiminished. 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 the other contains two ice cubes. Now both the containers are filled with water to same level as shown. Then both the containers shall weigh the same. STATEMENT-2 : The weight of volume of water displaced by ice cube floating in water is equal to the weight 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

RESONANCE


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, the object moves up. STATEMENT-2 :As the oil is poured in the situation of statement-1, pressure inside the water will increase 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 ms–1) 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 of oscillation 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 the jar. 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 of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position? [AIEEE 2010, 4/144]

(1)

5.

(2)

(3)

(4)

Water is flowing continuously from a tap having an internal diameter 8 × 10–3 m. The water velocity as it leaves the tap is 0.4 ms–1. The diameter of the water stream at a distance 2 × 10–1 m below the tap is close to : [AIEEE - 2011, 4/120, –1] (1) 5.0 × 10–3 m (2) 7.5 × 10–3 m (3) 9.6 × 10–3 m (4) 3.6 × 10–3 m

LEVEL - 2 : PREVIOUS YEARS JEE PROBLEMS (JEE-ADVANCE) * Marked Questions are More than one correct 1.

An application of Bernoulli’s equation for fluid flow is found in (A) Dynamic lift of an aeroplane (B) Viscosity meter (C) Capillary rise (D) Hydraulic press

RESONANCE

[IIT–JEE (Screening) 1994]


2.

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

L (A)

(B) 2 L

2

(C) L

(D)

L 2

3.

A wooden block with a coin placed on its top, floats in water as shown in figure. The distance and h are shown here. After some time the coin falls 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 -1 The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down. [JEE-2008' 3/162] and STATEMENT -2 In 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 # 1

EXERCISE # 4 LEVEL -I

SECTION (A) : A-1

(3)

A-2

(3)

A-4

(1)

A-5

(2)

A-3

(1)

1.

(2)

2.

(3)

4.

(2)

5.

(4)

3.

(3)

3.

(D)

LEVEL -II

SECTION (B) : B-1

(3)

B-2

(3)

B-3

(4)

1.

(A)

B-4

(2)

B-5

(1)

B-6.

(2)

4.

(A)

2.

(A)

SECTION (C) : C-1

(3)

C-2

(4)

C-3

(3)

C-4

(1)

C-5

(3)

C-6

(3)

C-7

(3)

EXERCISE # 2 1.

(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 # 3 1.

(1)

2.

(1)

RESONANCE

3.

(1)


OBJECTIVE PROBLEMS (JEE-ADVANCE) * Marked Questions are More than one correct 1. 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 the system 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/m 3) 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/m 3 (2) 4.0 × 103 kg/m 3 3 3 (3) 3.5 × 10 kg/m (4) 1.9 × 103 kg/m 3

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 the mixture is (1)

11d 7

(2)

18d 11

(3)

13d 9

5.

Figure shows a weigh-bridge, with a beaker P with water on one pan and a balancing weight R on the other. A solid ball Q is hanging with a thread outside water. It has volume 40 cm3 and weighs 80 g. If this solid is lowered to sink fully in water, but not touching the beaker anywhere, 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 of densities  and 2 is moving with an acceleration “a” on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by:

a (1) 2g  7.

3a (2) 2g 

According to Bernoulli’s equation

a (3) g 

(4)

23d 18

2a (4) 3g 

P 1 h = Constant pg 2g

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 of manometer falls to 3.0 × 105 N/m2, then velocity of flow of water is (1) 100 m/s

9.

(2) 10 m/s

(3) 1 m/s

(4) 10 10 m/s

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 up to a height of 0.16 m. how long it will take to empty the tank through a hole of radius 5×10–3 m in its bottom. (1) 46.26 sec. (2) 4.6 sec. (3) 462.6 sec. (4) .46 sec.

RESONANCE


10.

A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 rev/s, The difference in the height of the liquid 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 90 m/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 an average width of 2 m and negligible height difference, then the difference of the presure on the two sides of the wing 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 unknown liquid then it flows with

(1) 4.9 gm/c.c

(2)

1 rd of it outside the water, When it is inserted in an 3

3 th of it outside, then density of unknown liquid is : 4

9 gm/c.c 4

(3)

8 gm/c.c 3

(4)

[RPMT 2001]

3 gm/c.c 8

14.

Water is flowing inside a tube of an uniform radius ratio of radius of entry and exit terminals of the tube is 3 : 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 exposed height will be : [RPMT 2005] 2 5 1 1 (1) (2) (3) (4) 3 6 6 3

16.*

An ice-cube of density 900 kg/m3 is floating in water of density 1000 kg/m3. The percentage of volume of icecube 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, what will be the horizontal range of water jet ? [RPMT 2006] (1) 2 hH – h

(2) 4 hH  h

(3) 4 hH – h

(4) 2 hH  h

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 : (draft = 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 ? (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 Nm–2), then velocity of water flowing from hole is : [RPMT 2007]

21.

(1)

400 ms–1

(2)

(3)

60 ms–1

(4) none of these

[RPMT 2007]

600 ms–1

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 depth of the lake : [RPET 2000] (1) 10 m (2) 20 m (3) 60 m (4) 30 m

RESONANCE


22.

The pressure at the bottom of a tank containing a liquid does not depend on (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 (1) Archimedes principle (3) Pascal's law

24.

25.

[Kerala (Engg.) 2002]

[Kerals PMT 2005]

(2) Bernoulli's theorem (4) Newton's laws

From the adjacent figure, the correct observation is [KCET 2005] (1) The pressure on the bottom of tank (a) is greater than at 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

mA 2 In the arrangement shown in figure m  3 and the ratio of density of B 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-section varying as shown in fig. Pressure p at points along the 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 the mixture is (1)

29.

11d 7

(2)

18d 11

(3)

13d 9

(4)

23d 18

A cylindrical container of radius ' R ' and height ' h ' is completely filled with a liquid. Two horizontal L shaped pipes of small cross-section area ' a ' are connected to the cylinder as shown in the figure. Now the two pipes are opened and fluid starts coming out of the pipes horizontally in opposite directions. Then the torque due to ejected liquid on the system is: (1) 4 a g h  R (3) 2 a g h  R

RESONANCE

(2) 8 a g h  R (4) none of these AIEEE-FLUID MECHANICS - 29

30.

According to Bernoulli’s equation

P 1 h = Constant The terms A, B, and C are generally called pg 2g

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 water and is mounted on a rotatable shaft as shown in figure. If the tube is rotated with 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 of diameter 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 a cylinder of radius R. The volume of the remaining cylinder is V and its mass M. It is suspended by a string in a liquid of density  where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is : [I.I.T. 2001, 3/105 Screening] (1) Mg (2) Mg – Vg (3) Mg + R2hg (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 the container 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/s 2) [I.I.T. 2005 Screening , 3/60]

(1) 50

1. 8. 15. 22. 29.

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

(2) 51

2. 9. 16. 23. 30.

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

RESONANCE

3. 10. 17. 24. 31.

(3) 48

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

4. 11. 18. 25. 32.

(2) (1) (1) (1) (2)

(4) 51.5

5. 12. 19. 26. 33.

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

6. 13. 20. 27. 34.

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

7. 14. 21. 28.


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

1.

U = mgh

U = (b – ) Vgh

2.

 4v   4v   g   g =  (V)g m 5    5 

on solving m = 3.5 × 103 . m1  m 2  m 3 V( d  2d  3d)  = 2d. 3V 3V

3.

mix =

4.

3m 3m 36 18   d d m m m 11 11 mix = V1  V2  V3   d 2d 3 d

5.

Since not touching, So

6.

R = Fb = l(vg) = 40g.

For the given situation, liquid of density 2 r should be behind that of r. From right limb PA = Patm +  gh PB = PA +  a

  = Patm +  gh +  a 2 2

PC = PB + (2) a

 3 = Patm +  gh +  a  2 2

But from left limb : PC = Patm + (2) gh From (1) and (2) : Patm +  gh +

.... (2)

3  a  = Patm + 2  gh 2

3a h = 2g  8.

.... (1)

Ans.

Bernoulli's theorem for unit mass of liquid

P 1 2  u = constant  2 As the liquid starts flowing, it pressure energy decreases

1 2 P1 – P2 u   2



= 

1 2 3.5  10 5 – 3  10 5 u  2 10 3 2  0.5  10 5 10 3

u2 = 100

RESONANCE



u = 10 m/s AIEEE-FLUID MECHANICS - 31

9.

v=

....(ii)

2gh

from equation (ii) put the value of v in equation (i) R2 dh/dt = r2



–

–

R2dh

r

2

2gh

R2

0

r 2 2g



on solving

10.

h

2gh

 dt

=

t

dh

=

h

 dt 0

t = 46.26 second.

 2r 2 y= 2y Put values and get y = 2cm.

11.

From the Benoulli's theorem P1 – P2 =

1 2 1  u2 – u12   1.3  [(120 )2 – (90 )2 ] 2 2





= 4095 N/m2 or Pascal

12.

mass 120 3 Volume of log of wood V = density  600  0.2m Let x weight that can be put on the log of weed. So weight of the body = (120 + x) × 10 N Weight of displaced liquid = Vg = 0.2 × 103 × 10 N The 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.

Re lative density of body volume of inserted part of body = Re lative density of water total volume of body d 2V  1 3V

d=



2 3

Again using the same formula 1 2  4 3d

d= 14.

8 gm/c.c. 3

According to rule for flowing of liquid product of area and velocity is same A1v1 = A2v2 2

v1 A 2 r22  r2  4      = =4:9 2 v2 A1 9 r2  r1 

RESONANCE


15.

Here area is uniform, so portion of immersed height =

1 3

1 ) 3 1 2 1– = 3 3

(Here : the ratio of density of body and density of water = Therefore, fraction of exposed height will 16.

17.

The percentage of volume of ice-cube outside the water is =

 water –  ice  100  water

=

1000 – 900 ×100 = 10% 1000

Applying Bernoulli’s theorem, the velocity of water at point A v=

2gh

Time taken to reach point C is t So,

H–h=

t=

1 2 gt 2

2H – h g

Now, horizontal range R = vt =

2gh ×

=2 18.

2H – h g

H – h h

Volume of raft is given by V=

m 120 = = 0.2 m3 d 600

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.

Bernoulli’s equation for flowing liquid be written as p+

1 v2 + gh = constant 2

… (i)

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

 v2 + + h = constant g 2g

RESONANCE


It this expression

p v2 is velocity head and g is pressure head. 2g

It is given that, velocity head = pressure head

20.

ie.

p v2 = g 2g

or

v2 =

or

v2 =



v = 10.32 m/s

2p  2  13.6  10 3  40  10 –2  9.8 10 3

Let height of water column in the tank be h. Total pressure (p) = atmospheric pressure (p0) + pressure due to water column in tank (p’)  p’ = p – p0 = 3 – 1 = 2 atm or hg = 2 × 105 or h × 103 × 10 = 2 × 105 or h = 20 m Hence, velocity of water coming from hole ie, velocity of efflux is v=

2gh

=

2  10  20

=

400 ms–1

26.

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

27.

mix =

28.

3m 3m 36 18   d d m m m V  V  V 11 11 mix = 1 2 3   d 2d 3d

m1  m 2  m 3 V( d  2d  3d)  = 2d. 3V 3V

29.

Velocity of efflux of water (v) =

h 2g   = 2

gh

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

RESONANCE


Torque of these forces about central line = (av2) 2R . 2 = 4av2 R = 4 agh R 33.

[Flower – Fupper] by liquid = Upthrust F2 – F1 = upthrust  F2 = F1 + upthrust F2 = gh (R2) + Vg or F2 = g(V + R2h)

F1 Upthrust

F2 In this problem, we did not take the force due to air pressure on the cylinder. This is because force due to air pressure is cancelled. At top and bottom of the cylinder the force due to air pressure is equal and opposite. 34.

Apply continuity equation A1V1 = A2V2 and Bernauli theorum

p0 v2 + + gh = constant at the top and  2

at the hole.

RESONANCE


Fluid Mechanics for JEE Main

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