Fluid Mechanics E18-1 Measurement and calculation of pressure Q.1 An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm 3 of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm 3 ) (1) 350 cm 3 (2) 300 cm 3 (3) 250 cm 3 (4) 22 cm 3 Ans. (2) Sol. According to Boyle’s law, pressure and volume are inversely proportional to each other i.e. V P 1 P2V2 (P1 V1) h 2 2 1 1 V P V P 2 0 1 0 ) ( V P V g h P w 1 0 2 1 V P g h V w 1 2 2 1000 6 . 13 70 1000 1 10 6 . 47 1 V V . 300 50 ) 5 1 ( 3 3 2 cm cm V [As cm P P 70 0 2 of Hg 1000 6 . 13 70 ] Q.2 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 R Q 10 cm P S 20 cm (1) 10 5 N/m (2) 2 × 10 5 N/m (3) Zero (4) Infinity Ans. (3) Sol. As the both points are at the surface of liquid and these points are in the open atmosphere. So both point possess similar pressure and equal to 1 atm. Hence the pressure difference will be zero. Q.3 Density of ice is and that of water is . What will be the decrease in volume when a mass M of ice melts (1) M (2) M (3) 1 1 M (4) 1 1 1 M Ans. (3) Fluid Mechanics EXERCISE-I
Transcript
Fluid Mechanics
E18-1
Measurement and calculation of pressure
Q.1 An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The bell is brought to the surface
of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm3)
(1) 350 cm3 (2) 300 cm3 (3) 250 cm3 (4) 22 cm3
Ans. (2)
Sol. According to Boyle’s law, pressure and volume are inversely proportional to each other i.e.V
P1
P2V2
(P1 V1)
h
2211 VPVP
2010 )( VPVghP w
10
2 1 VP
ghV w
1
2
210006.1370
10001106.471 VV
.30050)51( 332 cmcmV
[As cmPP 7002 of Hg 10006.1370 ]
Q.2 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
RQ
10
cm
P
S
20 cm
(1) 105 N/m (2) 2 × 105 N/m (3) Zero (4) Infinity
Ans. (3)
Sol. As the both points are at the surface of liquid and these points are in the open atmosphere. So both point possess similar
pressure and equal to 1 atm. Hence the pressure difference will be zero.
Q.3 Density of ice is and that of water is . What will be the decrease in volume when a mass M of ice melts
(1)
M(2)
M
(3)
11M (4)
111
M
Ans. (3)
Fluid Mechanics
EXERCISE-I
Fluid Mechanics
E18-2
Sol. Volume of ice
M , volume of water
M .
Change in volume
11M
MM
Q.4 The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface.Assuming atmospheric
pressure to be 75 cm of Hg and the density of water to be 1/10 of the density of mercury, the depth of the lake is
(1) 5 m (2) 10 m (3) 15 m (4) 20 m
Ans. (3)
Sol. 2211 VPVP VghP )( 0 = VP 30
02Pgh g
gh
10
6.13
6.13752
= 15 m
Q.5 A closed rectangular tank is completely filled with water and is accelerated horizontally with an acceleration a towards
right. Pressure is (i) maximum at, and (ii) minimum at
aA D
B C
(1) (i) B (ii) D (2) (i) C (ii) D (3) (i) B (ii) C (4) (i) B (ii) A
Ans. (1)
Sol.a
A D
B C
Due to acceleration towards right, there will be a pseudo force in a left direction. So the pressure will be more on rear side
(Points A and B) in comparison with front side (Point D and C).
Q.6 If two liquids of same volume but different densities 1 and 2 are mixed, then density of mixture is given by
(1)2
21
(2)
21
21
2
(3)
21
212
(4)
21
21
Ans. (1)
Sol.
V
V
V
mm
22meTotal volu
massTotal 2121
221
Fluid Mechanics
E18-3
Q.7 From the adjacent figure, the correct observation is
Water Water
(a) (b)
(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 (a) 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
Ans. (4)
Sol. Pressure = hg i.e. pressure at the bottom is independent of the area of the bottom of the tank. It depends on the height
of water upto which the tank is filled with water. As in both the tanks, the levels of water are the same, pressure at the
bottom is also the same.
Q.8 Why the dam of water reservoir is thick at the bottom
(1) Quantity of water increases with depth
(2) Density of water increases with depth
(3) Pressure of water increases with depth
(4) Temperature of water increases with depth
Ans. (3)
Sol. A torque is acting on the wall of the dam trying to make it topple. The bottom is made very broad so that the dam will be
stable.
Pascal law
Q.9 The value of g at a place decreases by 2%. The barometric height of mercury
(1) Increases by 2%
(2) Decreases by 2%
(3) Remains unchanged
(4) Sometimes increases and sometimes decreases
Ans. (1)
Sol. g
Ph
gh
1 (P and are constant)
If value of g decreased by 2% then h will increase by 2%.
Q.10 Avertical U-tube of uniform inner cross section contains mercury in both sides of its arms. Aglycerin (density = 1.3 g/cm3)
column of length 10 cm is introduced into one of its arms. Oil of density 0.8 gm/cm3 is poured into the other arm until the
upper surfaces of the oil and glycerin are in the same horizontal level. Find the length of the oil column, Density of mercury
= 13.6 g/cm3
Mercury
Gly
ceri
ne
10 cm
Oil h
(1) 10.4 cm (2) 8.2 cm (3) 7.2 cm (4) 9.7 cm
Ans. (4)
Fluid Mechanics
E18-4
Sol.
Mercury
Gly
ceri
ne
10 cm
Oil h
BA
10–h
At the condition of equilibrium
Pressure at point A = Pressure at point B
BA PP ghghg 6.13)10(8.03.110
By solving we get h = 9.7 cm
Archemedies Principle and force of buoyancy
Q.11 A body of density 1d is counterpoised by Mg of weights of density 2d in air of density d. Then the true mass of the
body is
(1) M (2)
2
1d
dM (3)
1
1d
dM (4) )/1(
)/1(
1
2
dd
ddM
Ans. (4)
Sol. Let 0M mass of body in vacuum.
Apparent weight of the body in air = Apparent weight of standard weights in air
Actual weight – upthrust due to displaced air
= Actual weight – upthrust due to displaced air
dgd
MMgdg
d
MgM
21
00
1
20
1
1
d
d
d
dM
M
Q.12 A concrete sphere of radius R has a cavity of radius r which is packed 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. Ratio of
mass of concrete to mass of sawdust will be
(1) 8 (2)4 (3)3 (4) Zero
Ans. (2)
Sol. Let specific gravities of concrete and saw dust are 1 and 2 respectively..
According to principle of floatation weight of whole sphere = upthrust on the sphere
gRgrgrR 13
4
3
4)(
3
4 32
31
33
32
31
31
3 RrrR
)()1( 213
13 rR
11
213
3
r
R
1
1
1
1213
33
r
rR
2
1
1
2
23
133
1
1)(
r
rR
43.0
4.2
14.2
3.01
dustsawofMass
concreteofMass
Fluid Mechanics
E18-5
Q.13 A metallic block of density 5 gm cm–3 and having dimensions 5 cm × 5 cm × 5 cm is weighed in water. Its apparent weight
Q.14 A cubical 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
3
g
(1)2
1(2)
8
3(3)
3
2(4)
4
3
Ans. (1)
Sol. Fraction of volume immersed in the liquid VVin
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.
Q.15 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
Ans. (2)
Sol. Effective weight )(' agmW which is less than actual weight mg, so the length of spring decreases.
Q.16 A solid sphere of density ( > 1) times lighter than water is suspended in a water tank by a string tied to its base as shown
in fig. If the mass of the sphere is m then the tension in the string is given by
(1) mg
1(2) mg (3) 1
mg(4) mg)1(
Ans. (4)
Sol. Tension in spring T = upthrust – weight of sphere
gVgV gVgV )As(
gV )1( = .)1( mg
Q.17 A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one,
how will it affect the level of water
(1) It will remain unchanged (2) It will rise
(3) It will fall (4) First it will first rise and then fall
Ans. (3)
Fluid Mechanics
E18-6
Q.18 A wooden cylinder floats vertically in water with half of its length immersed. The density of wood is(1) Equal of that of water(2) Half the density of water(3) Double the density of water(4) The question is incomplete
Ans. (2)
Sol. gV
gV 2
2
( = density of water)
Continuity equation & bernoulli theorem and their applicationQ.19 In which one of the following cases will the liquid flow in a pipe be most streamlined
(1) Liquid of high viscosity and high density flowing through a pipe of small radius(2) Liquid of high viscosity and low density flowing through a pipe of small radius(3) Liquid of low viscosity and low density flowing through a pipe of large radius(4) Liquid of low viscosity and high density flowing through a pipe of large radius
Ans. (2)
Sol. For streamline flow, Reynold’s number
rN R should be less. For less value of RN , radius and density should be
small and viscosity should be high.
Q.20 Two water pipes of diameters 2 cm and 4 cm are connected with the main supply line. The velocity of flow of water in thepipe of 2 cm diameter is(1) 4 times that in the other pipe
(2)4
1times that in the other pipe
(3) 2 times that in the other pipe
(4)2
1times that in the other pipe
Ans. (1)
Sol. cmd A 2 and cmdB 4 cmrA 1 and cmrB 2
From equation of continuity, av = constant
BAA
B
A
B
B
A vvr
r
a
a
v
v4
1
2
)(
)(2
2
2
Q.21 An incompressible liquid flows through a horizontal tube as shown in the following fig. Then the velocity v of the fluidis
v1 = 3 m/s
v2 = 1.5 m/s
A
A
1.5 Av
(1) 3.0 m/s (2) 1.5 m/s (3) 1.0 m/s (4) 2.25 m/sAns. (3)
Sol. If the liquid is incompressible then mass of liquid entering through left end, should be equal to mass of liquid coming outfrom the right end.
21 mmM vAAvAv .5.121
vAAA .5.15.13 smv /1
Fluid Mechanics
E18-7
Q.22 Water enters through end A with speed 1v and leaves through end B with speed 2v of a cylindrical tube AB. The tube
is always completely filled with water. In case I tube is horizontal and in case II it is vertical with end A upwards and in
case III it is vertical with end B upwards. We have 21 vv for
(1) Case I (2) Case II (3) Case III (4) Each case
Ans. (4)
Sol. 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.
Q.23 Water is moving with a speed of 5.18 ms–1 through a pipe with a cross-sectional area of 4.20 cm2. The water gradually
descends 9.66 m as the pipe increase in area to 7.60 cm2. The speed of flow at the lower level is
Q.2 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 1000kg 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
Ans. (1)Sol. F = [rgh] [A]
= (1000) (10) (6) (10) (8).
Q.3 Two vessels A and B of different shapes have the same base area and are filled with water up to the same height h (seefigure). The force exerted by water on the base is F
Afor vessel A and F
Bfor vessel B. The respective weights of the water
filled in vessels are WA
and WB. Then
(1) FA
> FB; W
A> W
B(2*) F
A= F
B; W
A> W
B
(3) FA
= FB; W
A< W
B(4) F
A> F
B; W
A= W
B
Ans. (2)Sol. W
A> W
Bas mass of water in A is more than in B
PA
= PB
Area of A = Area of Bor P
AArea
A= P
BArea
B
or FA= F
B.
Q.4 A bucket contains water filled upto a height = 15 cm. The bucket is tied to a rope which is passed over a frictionless lightpulley and the other end of the rope is tied to a weight of mass which is half of that of the (bucket + water). The waterpressure above atmosphere pressure at the bottom is :(1) 0.5 kPa (2) 1 kPa (3) 5 kPa (4) None
Ans. (2)
Sol. Pressure = h geff.
a = g/3g
eff= g – g/3 = 2g/3
P =0.15 1000 2 10
3
P = 1KPa
Measurement and calculation of pressureQ.1 Figure here shows the vertical cross-section of a vessel filled with a liquid of density r. The normal thrust per unit area on
Fluid Mechanics
E18-11
Q.5 A uniformly tapering vessel shown in Fig. is filled with liquid of density 900 kg/m3. The force that acts on the base of thevessel due to liquid is (Take g = 10 m/s2) -
AREA = 10–3
m2
AREA = 2 × 10–3m2
0.4 m
(1) 3.6 N (2) 7.2 N (3) 9.0 N (4) 12.6 NAns. (2)Sol. GivenA= 2 × 10–3, h = 0.4 m, r = 900 Kg/m3
F = mg = Vrg = (pr2h)rg= 2 × 10–3 × 0.4 × 900 × 10= 7.2 N
Q.6 A liquid of mass 1 kg is filled in a flask as shown in figure. The force exerted by the flask on the liquid is (g = 10 m/s2)[Neglect atmospheric pressure]
(1) 10 N (2) greater than 10 N (3) less than 10 N (4) zeroAns (1)Sol. F = mg
F = 10 N
Q.7 A light semi cylindrical gate of radius R is piovted at its mid point O, of the diameter as shown in the figure holding liquidof density r. The force F required to prevent the rotation of the gate is equal to
F
O
R
(1) 2pR3rg (2) 2rgR3l (3)2
3
2R gl(4) none of these
Ans. (4)Sol. O(zero) all the forces passes through O
no torque.
Q.8 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)
Ans. (4)
Sol. Dv = vf– v
i= x
m
y
m .
Q.9 The pressure at the bottom of a tank of water is 3P where P is the atmospheric pressure. If the water is drawn out till thelevel of water is lowered by one fifth., the pressure at the bottom of the tank will now be(1)2P (2) (13/5) P (3) (8/5) P (4) (4/5) P
Ans. (2)
Sol. h g = 2P
Fluid Mechanics
E18-12
5
h4 g =
5
P8 After loweringP dueto liquid.
PT
=5
P8+ P (Atmospheric pressure)
=5
P13
Q.10 Two stretched membranes of areas 2 and 3 m2 are placed in a liquid at the same depth. The ratio of the pressure on themis -
(1) 1 : 1 (2) 2 : 3 (3) 2 : 3 (4) 22 : 32
Ans. (1)Sol. At same depth pressure is same
So ratio P1: P
2= 1 : 1.
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.
Ans. (1)
Sol.1
1
A
gm=
2
2
A
gm
Solving, m2= 3.75 kg.
Q.12 An open-ended U-tube of uniform cross-sectional area contains water (density 1.0 gram/centimeter3) standing initially 20centimeters from the bottom in each arm. An immiscible liquid of density 4.0 grams/centimeter3 is added to one arm untila layer 5 centimeters high forms, as shown in the figure above. What is the ratio h
2/h
1of the heights of the liquid in the
two arms ?
h2
5cm
h1
(1) 3/1 (2) 5/2 (3) 2/1 (4) 3/2Ans. (3)
Sol.
PA
= PB
5 × 4 × g + x × 1 × g= (40 - x) × 1 × g
x =10Now, h
1= x + 5 = 15 cm
h2= 40 – x = 30 cm
h2/h
1= 2
Pascal lawQ.11 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
Fluid Mechanics
E18-13
Q.13 The area of cross-section of the wider tube shown in figure is 800 cm2. If a mass of 12 kg is placed on the massless piston,the difference in heights h in the level of water in the two tubes is :
12kgh
(1) 10 cm (2) 6 cm (3) 15 cm (4) 2 cmAns. (3)Sol. Given m = 12 kg,A= 800 cm2, r = 1000 kg/m3
P = rgh
mg
A= rgh
4
12 10
800 10
= 1000 × 10 × h
12
80= h
h =1200
80= 15 cm
(1) start oscillating (2) sink to the bottom(3) come back to the same position immediately (4) come back to the same position slowly
Ans. (2)Sol. F
b= rVg – rvg = 0
Q.15 The reading of a spring balance when a block is suspended from it in air is 60 newton. This reading is changed to 40newton 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
Ans. (1)Sol. mg = 60 .................(i)
mg – rivg = 40 .................(ii)
mg
gvmg ρ=
3
2or
ρ
ρ0= 3
where r0
= density of the block and rl= density of the liquid.
Q.16 A metallic sphere floats in an immiscible mixture of water (rw
= 103 kg/m3) and a liquid(r
L= 13.5 × 103) with (1/5)th portion by volume in the liquid. The density of the metal is :
Q.17 An ice block floats in a liquid whose density is less than water. A part of block is outside the liquid. When whole of icehas melted, the liquid level will -(1) rise (2) go down (3) remain same (4) first rise then go down
Ans. (2)
Archemedies Principle and force of buoyancyQ.14 A body is just floating in a liquid (their densities are equal) If the body is slightly pressed down and released it will -
Fluid Mechanics
E18-14
Q.18 Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and itsdensity is 9 g/cc. If the mass of the other is 48 g, its density in g/cc is :(1) 4/3 (2) 3/2 (3*) 3 (4)5
Ans. (3)Sol. [36 – r
lv
l]g = [48 – r
lv
2]g
g9
3636 i
= g
4848
0i
ρρ
Solving, r0= 3.
Q.19 In order that a floating object be in a stable 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
Ans. (1)Sol. In stable equilibrium the object comes to its original state if disturbed.
Q.20 A cubical block of wood 10 cm on a side, floats at the interface of oil and water as shown in figure. The density of oil is0.6 g cm–3 and density of water is 1 g cm–3. The mass of the block is
(1) 706 g (2) 607 g (3*) 760 g (4) 670 gAns. (3)Sol. As, weight = Buoyant force
Q.21 Two bodies having volumes V and 2V are suspended from the two arms of a common balance and they are found tobalance each other. If larger body is immersed in oil (density d
1= 0.9 gm/cm3) and the smaller body is immersed in an
unknown liquid, then the balance remain in equilibrium. The density of unknown liquid is given by :(1) 2.4 gm/cm3 (2) 1.8 gm/cm3 (3) 0.45 gm/cm3 (4) 2.7 gm/cm3
Ans. (2)Sol. r
1V=r
22V
m1= m
2
m1g = 0.92 Vg = m
2g – xVg
x = 1.8 gm/cm3
Q.22 A boy carries a fish in one hand and a bucket (not full) of water in the other hand. If he places the fish in the bucket, theweight now carried by him (assume that water does not spill) :(1) is less than before (2) is more than before(3) is the same as before (4) depends upon his speed
Ans. (3)
Q.23 A piece of steel has a weight W in air, W1
when completely immersed in water and W2
when completely immersed in anunknown liquid. The relative density (specific gravity) of liquid is :
(1)1
2
W W
W W
(2)2
1
W W
W W
(3)1 2
1
W W
W W
(4)1 2
2
W W
W W
Ans. (2)Sol. W – v × 1 × g = W
1
W – v × x × g = W2
W – (W – W1) × x= W
2
x =2
1
W – W
W – W
Fluid Mechanics
E18-15
Q.24 A metal ball of density 7800 kg/m3 is suspected to have a large number of cavities. It weighs 9.8 kg when weighed directlyon a balance and 1.5 kg less when immersed in water. The fraction by volume of the cavities in the metal ball isapproximately :(1) 20% (2) 30% (3) 16% (4) 11%
Ans. (3)Sol. Volume where metal is present
=7800
8.9= 1.256 × 10–3
Buoyancy = vrg = 1.5 g v × 1000 = 1.5v = 1.5 × 10–3
fraction of volume =
3–
3–3–
105.1
10256.1–105.1
× 100
= 16%
Q.25 A sphere of radius R and made of material of relative density has a concentric cavity of radius r. It just floats whenplaced in a tank full of water. The value of the ratio R/r will be
(1)
1 /3
1
(2)
11 3/
(3)
11 3/
(4)
1
1
1 3/
Ans. (1)
Sol. ×3
4 (R3 – r3)g = 1 ×
3
4R3g
r
R=
3/1
1
Q.26 A beaker containing water is placed on the platform of a spring balance. The balance reads 1.5 kg. Astone of mass 0.5 kgand density 500 kg/m3 is immersed in water without touching the walls of beaker. What will be the balance reading now?(1) 2 kg (2) 2.5 kg (3) 1 kg (4) 3 kg
Q.27 A cube of iron whose sides are of length L, is put into mercury. The weight of iron cube is W. The density of iron is rI, thatof mercury is rM. The depth to which the cube sinks is given by the expression –
(1)WL2rI (2) WL2rM (3)I
2L
W
(4)
gL
W
M2ρ
Ans. (4)Sol. Equilibrium Position W = F
B
W = L2h rM
g
h = 2
M
W
L g
Fluid Mechanics
E18-16
Q.28 A cylindrical block of area of cross-section A and of material of density is placed in a liquid of density one-third ofdensity of block. The block compresses a spring and compression in the spring is one-third of the length of the block. Ifacceleration due to gravity is g, the spring constant of the spring is
(1) Ag (2) 2Ag (3) 2Ag/3 (4) Ag/3Ans. (2)Sol.
___3
Kx =
g3
V
V =A..
NowA g
3
+
3
K= AA
K = 2Ag
Q.29 Ram and Shyam are having the same weight when measured outside the water. When measured under water, it is foundthat weight of Ram is more than that of Shyam, then we can say that(1) Ram is having more fat content than Shyam.(2*) Shyam is having more fat content that Ram.(3) Ram and Shyam both are having the same fat content.(4) None of these.
Ans. (2)Sol. Apparent weight (W
app.) = W – V
g
Since, Wapp. (Ram)
> Wapp. (Shyam)
W(Ram)
> W(Shyam)
Therefore, from given passage shyam has more fat than Ram.
Q.30 Ram is being weighed by the spring balance in two different situations. First when he was fully immersed in water and thesecond time when he was partially immersed in water, then(1) Reading will be more in the first case.(2*) Reading will be more in the second case.(3) Reading would be same in both the cases.(4) Reading will depend upon experimental setup.
Ans. (2)Sol. V
1> V
2 W
app. (1)< W
app. (2)
(Since Wapp.
= W – V g)
Hence (2)
Q.31 Salt water is denser than fresh water. If you were immersed fully first in salt water and then in fresh water and weighed,then(1*) Reading would be less in salt water.(2) Reading would be more in salt water.(3) Reading would be the same in both the cases.(4) reading could be less or more.
Ans. (1)Sol.
Salt waver>
Fresh waver W
app. (s)< W
app. (F)
Hence (1)
Fluid Mechanics
E18-17
Q.32 A person of mass 165 Kg having one fourth of his volume consisting of fat (relative density 0.4) and rest of the volume
consisting of everything else (average relative density3
4) is weighed under water by the spring balance. The reading
shown by the spring balance is -(1) 15 N (2) 65 N (3*) 150 N (4) 165 N
Ans. (3)Sol. Let 'V' be the total volume of the person
Then ;
4
V(0.4 × 103) +
3103
4V
4
3= 165 V =
1100
165
Reading on spring balance under water is :
Wapp
= [165 × 10] –
1100
165[103] [10]
= 150 N
Q.33 In the above question if the spring is cut, the acceleration of the person just after cutting the spring is(1) zero (2) 1 m/s2 (3) 9.8 m/s2 (4*) 0.91 m/s2
Ans. (4)Sol. Just after the string is cut :
a =165
150= 0.91 m/s2 Ans.
the surface of water as shown in the figure. 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
Ans. (3)Sol. R = vt
= gD2 g
)DH(2
= )DH(D2 .
Q.35 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 surfaceof water. For maximum horizontal range h is equal to :(1) H (2) 3H/4 (3*) H/2 (4) H/4
Ans. (3)
Sol. x = 2 )hH(h
for xmax
,dh
dx= 0 or h =
2
H
Continuity equation & bernoulli theorem and their applicationQ.34 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
Fluid Mechanics
E18-18
Q.36 For a fluid which is flowing steadily in the figure shown, the level in the vertical tubes is best represented by
(1*) (2)
(3) (4)
Ans. (1)Sol. From continuity equation, velocity at cross-section (1) is more than that at cross-section (2).
Hence ; P1< P
2
Q.37 There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. Thedifference in height of the two holes is h as shown in the figure. 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
Ans. (2)Sol. F
thrust= av2
Fnet
= F1– F
2= a[2g(h
1– h
2)]
= a(2gh)or F h
Q.38 Acylindrical 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) 0.46 sec.
Ans. (1)
Sol.
A1
v1
= A2v
2
Fluid Mechanics
E18-19
R2 dh/dt = r2 v ....(i)
v = gh2 ....(ii)
from equation (ii) put the value of v in equation (i)
R2 dh/dt = r2 gh2
gh2r
dhR2
2
= dt
0
h
2
2
h
dh
g2r
R=
t
0
dt
on solvingt = 46.26 second.
Q.39 Water is flowing in a horizontal pipe of non-uniform cross - section. At the most contracted place of the pipe –(1) Velocity of water will be maximum and pressure minimum(2) Pressure of water will be maximum and velocity minimum(3) Both pressure and velocity of water will be maximum(4) Both pressure and velocity of water will be minimum
Ans. (1)Sol. AV = constant
AV
P + gh + 21v
2 = constant
VP
Q.40 The cross sectional area of a horizontal tube increases along its length linearly, as we move in the direction of flow. Thevariation of pressure, as we move along its length in the direction of flow (x-direction), is best depicted by which of thefollowing graphs
(1)
P
x
(2)
P
x
(3)
P
x
(4)
P
x
Ans. (1)
Sol.
P
xV
A = ax + bContinuty equation bV = (ax + b) V
2
By bernaulies equation = P2
+1
2v
22 = cosntant
P2
= Costant -1
2v
22
P2
= Costant -1
2
2 2
2
b V
ax b
P2
= Cosntant – 1
2
C
ax b
Where C1
= Constant
Fluid Mechanics
E18-20
Q.41 Water is flowing steadily through a horizontal tube of non uniform cross-section. If the pressure of water is 4 × 104 N/m2
at a point where cross-section is 0.02 m2 and velocity of flow is 2 m/s, what is pressure at a point where cross-sectionreduces to 0.01 m2
Q.42 A tube is attached as shown in closed vessel containing water. The velocity of water coming out from a small hole is :
20cm
(1) 2 m s/ (2) 2 m/s
(3) depends on pressure of air inside vessel (4) None of theseAns. (2)
Sol. 1010202 2– = 2 m/sec.
Q.43 In the case of a fluid, Bernoulli’s theorem expresses the application of the principle of conservation of(1) linear momentum (2) energy (3) mass (4) angular momentum
Ans. (2)
Q.44 Acylindrical vessel open at the open at the top is 20 cm high and 10 cm in diameter. A circular hole whose cross-sectionalarea 1 cm2 is cut at the centre of the bottom of the vessel. Water flows from a tube above it into the vessel at the rate 100cm3s–1. The height of water in the vessel under state is (Take g = 1000 cms–2)(1) 20 cm (2) 15 cm (3) 10 cm (4) 5 cm
Ans. (4)Sol. Inlet = outlet
dt = a 2gh dt
h =
2
22ga
=
2100
2 1000 (1)= 5 cm
Fluid Mechanics
E18-21
Q.45 A fire hydrant delivers water of density at a volume rate L. The water travels vertically upward through the hydrant andthen does 90° turn to emerge horizontally at speed V. The pipe and nozzle have uniform cross-section throughout. Theforce exerted by the water on the corner of the hydrant is :
v
v
(1)VL (2) zero (3)2VL (4) 2 VL
Ans. (4)Sol. Force exerted by the water on the corner
= change in momentum in 1 sec
= 2 mv
mv
mv
= 2 vL
Q.46 A vertical tank, open at the top, is filled with a liquid and rests on a smooth horizontal surface. A small hole is opened atthe centre of one side of the tank. The area of cross-section of the tank is N times the area of the hole, where N is a largenumber. Neglect mass of the tank itself. The initial acceleration of the tank is
(1)g
N2(2)
g
N2(3)
g
N(4)
g
N2
Ans. (3)
Sol.
H
Na
a
H/2
Force = a 22/gh2
acceleration =agh
Na.H
= g/N
Q.47 Water flows into a cylindrical vessel of large cross-sectional area at a rate of 10–4m3/s. It flows out from a hole of area10–4 m2, which has been punched through the base. How high does the water rise in the vessel ?(1) 0.075 m (2) 0.051 m (3) 0.031 m (4) 0.025 m
Ans. Sol. dt = Av dt
10–4 = 10–4 gh2 h =g2
1
h = 0.051 m
Q.48 A cyclindrical vessel of cross-sectional area 1000 cm2, is fitted with a frictionless piston of mass 10 kg, and filled with
water completely. A small hole of cross-sectional area 10 mm2 is opened at a point 50 cm deep from the lower surface of
the piston. The velocity of efflux from the hole will be
(1) 10.5 m/s (2) 3.4 m/s (3) 0.8 m/s (4) 0.2 m/s
Ans. (2)
Fluid Mechanics
E18-22
Sol. By Bernaulie’s Theorem
P0+ 4
10 10
1000 10
+1000 × 10 ×
50
100
= P0+
1
2× 1000 × v2
6000 =1
2× 1000 × v2 v=?
v = 12 = 3.4 m/s
Q.49 A horizontal right angle pipe bend has cross-sectional area = 10 cm2 and water flows through it at speed = 20 m/s. The
force on the pipe bend due to the turning of water is :
(1) 565.7 N (2) 400 N (3) 20 N (4) 282.8 N
Ans. (1)
Sol. Change in momentum is/sec.
2 Avv2 = 565.7 N.
Q.50 A jet of water having velocity = 10 m/s and stream cross-section = 2 cm2 hits a flat plate perpendicularly, with the water
splashing out parallel to plate. The plate experiences a force of
(1) 40 N (2) 20 N (3) 8 N (4) 10 N
Ans. (2)
Sol. AV2=1000 × 2 × 10–4× (10)2
= 20 N
Q.51 Water is pumped from a depth of 10 m and delivered through a pipe of cross section 10–2m2. If it is needed to deliver avolume of 10–1m3 per second the power required will be :(1) 10 kW (2) 9.8 kW (3) 15 kW (4) 4.9 kW
Ans. (3)Sol. Energy required in one second is the power
10–1=A.V. 10–1= 10–2 × VV = 10 m/sec.
mgh 2
1mV2 = P
Here m = mass in one second
P = AVgh +2
1AVV3
P = AV[10 × 10 + 50]= 15 Kwatt
Q.52 Which of the following is not an assumption for an ideal fluid flow for which Bernoulli’s principle is valid(1) Steady flow (2) Incompressible (3) Viscous (4) Irrotational
Ans. (3)
Q.53 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 (2) 1.5 m/s (3*) 1.0 m/s (4) 2.25 m/s
Fluid Mechanics
E18-23
Ans. (3)Sol. from equation of continuity,
(A× 3) = (A× 1.5) + (1.5 A× V) V = 1 m/s2
Q.54 Water is flowing in a tube of non-uniform radius. The ratio of the radii at entrance and exit ends of tube is 3 : 2. The ratioof the velocities of water entering in and exiting from the tube will be –(1) 8 : 27 (2) 4 : 9 (3) 1 : 1 (4) 9 : 4
Ans. (2)
Sol. A1V
1=A
2V
2(Given
1
2
r
r =3
2)
1
2
v
v =2
1
A
A =
22
21
r
r
=
22
3
=4
9
Q.55 Fountains usually seen in gardens are generated by a wide pipe with an enclosure at one end having many small holes.Consider one such fountain which is produced by a pipe of internal diameter 2 cm in which water flows at a rate 3ms–1. Theenclosure has 100 holes each of diameter 0.05 cm. The velocity of water coming out of the holes is (in ms–1) :(1)0.48 (2)96 (3)24 (4)48
Ans. (4)Sol. A
1V
1=A
2V
2
(1 × 10–2)2 × 3 = 100 ×
4
1005.022–
× V2V
2= 48 m/sec.
Q.56 Two water pipes P and Q having diameters 2 × 10–2 m and 4×10–2 m, respectively, are joined in series with the main supplyline of water. The velocity of water flowing in pipe P is(1) 4 times that of Q (2) 2 times that of Q(3) 1/2 times of that of Q (4) 1/4 times that of Q
Ans. (1)Sol. FromA
PV
P=A
QV
Q
Q
P
V
V=
P
Q
A
A= 22–
22–
)101(
)102(
VP= 4V
Q
Q.57 A tank has an orifice near its bottom. The volume of the liquid flowing per second out of the orifice does not dependupon–(1) Area of the orifice (2) Height of the liquid level above the orifice(3) Density of liquid (4) Acceleration due to gravity
Ans. (3)
Sol.dV
A 2ghdt
Q.58 The rate of flowing of water from the orifice in a wall of a tank will be more if the orific is –(1) Near the bottom (2) Near the upper end(3) Exactly in the middle (4) Does not depend upon the position of orific
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