Chapter 19 Mechanical Properties of Fluids – Physics free study material by TEACHING CARE online tuition and coaching classes
Fluid is the name given to a substance which begins to flow when external force is applied on it. Liquids and gases are fluids. Fluids do not have their own shape but take the shape of the containing vessel. The branch of physics which deals with the study of fluids at rest is called hydrostatics and the branch which deals with the study of fluids in motion is called hydrodynamics.
Pressure.
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
 Units :
N / m^{2}
or Pascal (S.I.) and Dyne/cm^{2} (C.G.S.)
 Dimension :
[P] = [F] = [MLT 2 ] = [ML^{1}T ^{2} ]
 [L^{2} ]
 At a point pressure acts in all directions and a definite direction is not associated with So pressure is a tensor quantity.
 Atmospheric pressure : The gaseous envelope surrounding the earth is called the earth’s atmosphere and the pressure exerted by the atmosphere is called atmospheric Its value on the surface of the earth at sea level is
nearly 1.013 ´ 10^{5} N / m^{2} or Pascal in S.I. other practical units of pressure are atmosphere, bar and torr (mm of Hg)
1atm = 1.01 ´ 10^{5} Pa = 1.01bar = 760 torr
The atmospheric pressure is maximum at the surface of earth and goes on decreasing as we move up into the earth’s atmosphere.
 If P_{0} is the atmospheric pressure then for a point at depth h below the surface of a liquid of density r ,
hydrostatic pressure P is given by P = P0 + hr g
 Hydrostatic pressure depends on the depth of the point below the surface (h), nature of liquid ( r ) 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
PA = PB = PC
but
WA < WB < WC
 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
 Gauge pressure : The pressure difference between hydrostatic pressure P and atmospheric pressure P_{0} is
called gauge pressure.
P – P_{0} = hrg
Problem 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]
(a) 10 m (b) 20 m (c) 60 m (d) 30 m
Solution : (b) Pressure at bottom of the lake = P0 + hrg
and pressure ay half the depth of a lake = P0 + h rg
2
According to given condition
P + 1 hrg = 2 (P
+ hrg) Þ 1 P
= 1 hrg
Þ h = 2P0 =
2 ´ 10^{5}
= 20m .
0 2 3 0
3 ^{0} 6
rg 10^{3} ´ 10
Problem 2. 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 / cm^{3}. If the mass of the other is 48 g, its density in g / cm^{3} is [CBSE 1994]
(a) 4 3
(b)
3 (c) 3 (d) 5
2
Solution : (c) Apparent weight = V(r – s )g = m (r – s )g
r
where m =
mass of the body,
r = density of the body and s = density of water
If two bodies are in equilibrium then their apparent weight must be equal.
\ m1 (r_{1} – s )g = m2 (r _{2} – s )g Þ 36 (9 – 1) = 48 (r_{2} – 1)g . By solving we get
r_{2} = 3 .
r_{1} r _{2} 9 r_{2}
Problem 3. 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}) [CPMT 1989]
(a) 350 cm^{3} (b) 300 cm^{3} (c) 250 cm^{3} (d) 22 cm^{3}
Solution : (b) According to Boyle’s law, pressure and volume are inversely proportional to each other i.e. P µ 1
V
Þ P1V1 = P2V2
Þ (P0 + hr _{w} g)V1 = P0 V2
æ hr _{w} g ö
Þ V2 = ç1 +
è
÷V1
P0 ø
æ
Þ V2 = ç1 +
47.6 ´ 10^{2} ´ 1 ´ 1000 ö

V1
[As P_{2} = P_{0} = 70 cm of Hg = 70 ´ 13.6 ´ 1000 ]


ç 70 ´ 13.6 ´ 1000 ÷
Þ V2 = (1 + 5)50 cm^{3} = 300 cm^{3} .
Problem 4. A Utube in which the crosssectional area of the limb on the left is one quarter, the limb on the right contains mercury (density 13.6 g/cm^{3}). The level of mercury in the narrow limb is at a distance of 36 cm from the upper end of the tube. What will be the rise in the level of mercury in the right limb if the left limb is filled to the top with water
 2 cm
 35 cm
 56 cm
 8 cm
Solution : (c) If the rise of level in the right limb be x cm. the fall of level of mercury in left limb be 4x cm because the area of cross section of right limb 4 times as that of left limb.
\ Level of water in left limb is (36 + 4x) cm. Now equating pressure at interface of mercury and water (at A’ B’)
(36 + 4 x) ´ 1 ´ g = 5x ´ 13.6 ´ g
By solving we get x = 0.56 cm.
Problem 5. A uniformly tapering vessel is filled with a liquid of density 900 kg/m^{3}. The force that acts on the base of the vessel due to the liquid is (g = 10 ms^{–}^{2})
 6 N
 2 N
 0 N
(d) 14.4 N
Solution : (b) Force acting on the base
F = P ´ A = hdgA = 0.4 ´ 900 ´ 10 ´ 2 ´ 10^{–}^{3} = 7.2N
Problem 6. A tank 5 m high is half filled with water and then is filled to the top with oil of density 0.85 g/cm^{3}. The pressure at the bottom of the tank, due to these liquids is
(a) 1.85 g/cm^{2} (b) 89.25 g/cm^{2} (c) 462.5 g/cm^{2} (d) 500 g/cm^{2}
Solution : (c) Pressure at the bottom P = (h1d1 + h2d2) g = [250 ´ 1 + 250 ´ 0.85] = 250 [1.85] g = 462.5 g
cm^{2}
cm^{2}
cm^{2}
Problem 7. 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
(a) 10^{5} N/m
(b) 2 × 10^{5} N/m
 Zero
 Infinity
Solution : (c) 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.
Problem 8. The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ratio of density of mercury to that of air is 10^{4}. The height of the hill is
(a) 250 m (b) 2.5 km (c) 1.25 km (d) 750 m Solution : (b) Difference of pressure between sea level and the top of hill
DP = (h1 – h2) ´ r_{Hg} ´ g = (75 – 50) ´ 10^{–}^{2} ´ r_{Hg} ´ g
and pressure difference due to h meter of air DP = h ´ r _{air} ´ g
……(i)
……(ii)
By equating (i) and (ii) we get h ´ r_{air} ´ g = (75 – 50) ´ 10^{–}^{2} ´ r_{Hg} ´ g
2 æ rHg ö
2 4
Density.
\ h = 25 ´ 10
ç ÷ = 25 ´ 10
è rair ø
´ 10
= 2500 m \ Height of the hill = 2.5 km.
In a fluid, at a point, density r is defined as: r =
lim
Dm = dm
DV ®0 DV dV
 In case of homogenous isotropic substance, it has no directional properties, so is a
 It has dimensions [ML^{3}] and I. unit kg/m^{3} while C.G.S. unit g/cc with 1g / cc = 10^{3}kg / m^{3}
 Density of substance means the ratio of mass of substance to the volume occupied by the substance while density of a body means the ratio of mass of a body to the volume of the 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
 V ]
 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
 Sometimes instead of density we use the term relative density or specific gravity which is defined as :
RD = Density of body Density of water
 If m_{1} mass of liquid of density
r_{1} and m2 mass of density
r_{2} are mixed, then as
m = m1 + m2
and
V = (m1 / r_{1} ) + (m2 / r _{2} )
[As
V = m / r ]
r = m =
V
m1 + m2 =
(m1 / r_{1}) + (m2 / r_{2})
å mi
å(mi / pi )
If m = m
r = 2r_{1}r_{2}
= Harmonic mean

1 2
1
+ r_{2}
 If V_{1} volume of liquid of density
r_{1} and
V_{2} volume of liquid of density
r_{2} are mixed, then as:
m = r_{1}V1 + r_{2}V2 and
V = V1 + V2
[As
r = m / V ]
If V1 = V2 = V r = (r_{1} + r_{2}) / 2 = Arithmetic Mean

 With rise in temperature due to thermal expansion of a given body, volume will increase while mass will remain unchanged, so density will decrease, e.,
r = (m / V) = V0 = V0
[As
V = V (1 + gDq )]
r_{0} (m / V_{0}) V V_{0}(1 + gDq )
or r =
r_{0}
(1 + gDq )
~– r_{0}
(1 – gDq )
 With increase in pressure due to decrease in volume, density will increase, e.,
r = (m / V) = V0
[As r = m ]
r_{0} (m / V_{0}) V V
But as by definition of bulkmodulus
B = –V Dp i.e., V = V
é1 – Dp ù
^{0} DV
^{0} êë B úû
æ Dp ö^{–}^{1} ~
æ Dp ö

So r = r _{0} ç1 –
è
÷ – r _{0} ç1 + ÷

ø è ø
Problem 9. A homogeneous solid cylinder of length L(L < H / 2) . Crosssectional area
A / 5
is immersed such that it floats
with its axis vertical at the liquidliquid interface with length
L / 4
in the denser liquid as shown in the fig. The
lower density liquid is open to atmosphere having pressure P_{0} . Then density D of solid is given by [IITJEE1995]
 5 d
4
 4 d
5
 Ad
 d
5
Solution : (a) Weight of cylinder = upthrust due to both liquids
V ´ D ´ g = æ A . 3 Lö ´ d ´ g + æ A . L ö ´ 2d ´ g Þ æ A .Lö D.g = A L d g Þ D = d \ D = 5 d
ç ÷ ç ÷ ç ÷









è ø è ø è ø
Problem 10. Density of ice is r
and that of water is s . What will be the decrease in volume when a mass M of ice melts
(a)
M (b)
s – r
(c)
M é 1 – 1 ù
(d)
1 é 1 – 1 ù
s – r




M ë û
M ê r ú



M M M M æ 1 1 ö
Solution : (c) Volume of ice =
r , volume of water = s
\ Change in volume =
r – s
= M ç – ÷
r s
è ø
Problem 11. Equal masses of water and a liquid of density 2 are mixed together, then the mixture has a density of (a) 2/3 (b) 4/3 (c) 3/2 (d) 3
Solution : (b) If two liquid of equal masses and different densities are mixed together then density of mixture
r = 2r_{1}r_{2} = 2 ´ 1 ´ 2 = 4
r_{1} + r_{2} 1 + 2 3
Problem 12. Two substances of densities
r_{1} and
r_{2} are mixed in equal volume and the relative density of mixture is 4.
When they are mixed in equal masses, the relative density of the mixture is 3. The values of r_{l} and r_{2} are
(a)
r_{1} = 6
and r _{2} = 2
(b)
r_{1} = 3 and r _{2} = 5 (c)
r_{1} = 12 and r _{2} = 4
 None of these
Solution : (a) When substances are mixed in equal volume then density = r1 + r 2 = 4
2
Þ r_{1}
+ r _{2} = 8
…….(i)
When substances are mixed in equal masses then density = 2r1r2 = 3
r_{1} + r_{2}
Þ 2r_{1}r_{2} = 3(r_{1} + r_{2})
…….(ii)
By solving (i) and (ii) we get r_{1} = 6 and r_{2} = 2 .
Problem 13. A body of density d1 is counterpoised by M g
of the body is
of weights of density d_{2} in air of density d. Then the true mass
æ d ö
æ d ö
M(1 – d / d_{2})
 M (b)
M ç1 – ÷
d
(c)
M ç1 – ÷
d
(d)
(1 – d / d )
Solution : (d) Let
è 2 ø
M _{0} = mass of body in vacuum.
è 1 ø 1
Apparent weight of the body in air = Apparent weight of standard weights in air
Þ Actual weight – upthrust due to displaced air = Actual weight – upthrust due to displaced air
é d ù
æ M ö
æ M ö
M ê1 – d ú
Þ M _{0} g – ç^{ } ^{0} ÷ dg = Mg – ç
÷ dg Þ M0 = ë 2 û .
è d1 ø
è d2 ø
é d ù
ê1 – d ú
Pascal’s Law.
ë 1 û
It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same.
or
The increase in pressure at one point of the enclosed liquid in equilibrium of rest is transmitted equally to all other points of the liquid and also to the walls of the container, provided the effect of gravity is neglected.
Example : Hydraulic lift, hydraulic press and hydraulic brakes
Working of hydraulic lift : It is used to lift the heavy loads. If a small force f is applied on piston of C then the pressure exerted on the liquid
P = f / a [a = Area of cross section of the piston in C]
This pressure is transmitted equally to piston of cylinder D. Hence the upward force acting on piston of cylinder D.
F = P A = f
a
A = f æ A ö

a
è ø
As A >> a , therefore F >> f . So heavy load placed on the larger piston is easily lifted upwards by applying a
small force.
Archimedes Principle.
Accidentally Archimedes discovered that when a body is immersed partly or wholly in a fluid, in rest it is buoyed up with a force equal to the weight of the fluid displaced by the body. This principle is called Archimedes principle and is a necessary consequence of the laws of fluid statics.
When a body is partly or wholly dipped in a fluid, the fluid exerts force on the body due to hydrostatic pressure. At any small portion of the surface of the body, the force exerted by the fluid is perpendicular to the surface and is equal to the pressure at that point multiplied by the area. The resultant of all these constant forces is called upthrust or buoyancy.
To determine the magnitude and direction of this force consider a body immersed in a fluid of density s as shown in fig. The forces on the vertical sides of the body will cancel each other. The top surface of the body will experience a downward force.
F1 = AP1 = A(h1sg + P0 )
[As
P = hsg + P_{0} ]
While the lower face of the body will experience an upward force.
F = AP = A(h sg + P )
2 2 2 0
As h2 > h1, F2
will be greater than
F_{1} , so the body will experience a net upward force
F = F2 – F1 = Asg(h2 – h1 )
If L is the vertical height of the body
F = AsgL = Vsg
[As
V = AL = A(h_{2} – h_{1} )] ]
i.e., F = Weight of fluid displaced by the body.
This force is called upthrust or buoyancy and acts vertically upwards (opposite to the weight of the body) through the centre of gravity of displaced fluid (called centre of buoyancy). Though we have derived this result for a body fully submerged in a fluid, it can be shown to hold good for partly submerged bodies or a body in more than one fluid also.
 Upthrust is independent of all factors of the body such as its mass, size, density except the volume of the body inside the fluid.
 Upthrust depends upon the nature of displaced This is why upthrust on a fully submerged body is more in sea water than in fresh water because its density is more than fresh water.
 Apparent weight of the body of density ( r ) when immersed in a liquid of density (s ) .
æ s ö
Apparent weight = Actual weight – Upthrust = W – F_{up} = Vrg – Vsg = V(r – s )g = Vrgç1 – r ÷
è ø
æ s ö
\ WAPP
= Wç1 – r ÷
è ø
 If a body of volume V is immersed in a liquid of density s then its weight
W1 = Weight of the body in air, W2 = Weight of the body in water
Then apparent (loss of weight) W_{1}
 W2
= Vsg
\ V = W1 – W2
sg
 Relative density of a body (D.)= density of body
density of water
= Weight of body =
Weight of equal volume of water
Weight of body Water thrust
= Weight of body Loss of weight in water
= Weight of body in air Weight in air – weight in water
= W1 W1 – W2
 If the loss of weight of a body in water is ‘a ‘ while in liquid is ‘b‘
\ s L = Upthrust on body in liquid = Loss of weight in liquid = a = Wair – Wliquid
s _{W} Upthrust on body in water
Loss of weight in water
b Wair – Wwater
Floatation.
 Translatory equilibrium : When a body of density r and volume V is immersed in a liquid of density
s , the forces acting on the body are
Weight of body W = mg = Vrg, acting vertically downwards through centre of gravity of the body.
Upthrust force = Vsg
of buoyancy.
acting vertically upwards through the centre of gravity of the displaced liquid i.e., centre
If density of body is greater than that of  If density of body is equal to that of  If density of body is lesser than that of 
liquid r > s  liquid r = s  liquid r < s 
Weight will be more than upthrust so the body will sink 
Weight will be equal to upthrust so the body will float fully submerged in neutral equilibrium anywhere in the liquid. 
Weight will be less than upthrust so the body will move upwards and in equilibrium will float partially immersed in the liquid Such that, W = Vins g Þ Vr g = Vins g 
Vr = Vins Where Vin is the volume of body in the liquid 
 A body will float in liquid only and only if r £ s
 In case of floating as weight of body = upthrust So WApp= Actual weight – upthrust = 0
 In case of floating Vrg = Vins g
So the equilibrium of floating bodies is unaffected by variations in g though both thrust and weight depend on g.
 Rotatory Equilibrium : When a floating body is slightly tilted from equilibrium position, the centre of buoyancy B The vertical line passing through the new centre of buoyancy B¢ and initial vertical line meet at a point M called meta–centre. If the metacentre 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 to its original position. So for rotational equilibrium of floating body the metacentre must always be higher than the centre of gravity of the body.
However, if metacentre goes below 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 MC and so the body will topple if slightly tilted.
(3)Application of floatation
 When a body floats then the weight of body = Upthrust
Vrg = V sg Þ V
= æ r ö V
\ V = V – V
= æ1 – r ö V
in in
ç s ÷
out
_{in} ç s ÷
è ø è ø
i.e., Fraction of volume outside the liquid f
= Vout
= é1 – r ù
out V
 For floatation Vr = V s Þ r = Vin s = f s
ëê s úû
in V in
If two different bodies A and B are floating in the same liquid then
r A = ( fin )A
r B ( fin )B
 If the same body is made to float in different liquids of densities s _{A} and s _{B}
Vr = (V ) s
= (V ) s
\ s _{A} = (Vin )B
in A A
in B B
s _{B} (Vin )A
 If a platform of mass M and crosssection A is floating in a liquid of density s with its height h inside the liquid
Mg = hAsg…………………………………………………………………… (i)
Now if a body of mass m is placed on it and the platform sinks by y then
(M + m)g = (y + h) Asg
Subtracting equation (i) and (ii), mg = As y g , i.e., W µ y
……(ii)
……(iii)
So we can determine the weight of a body by placing it on a floating platform and noting the depression of the platform in the liquid by it.
Problem 14. A wooden block, with a coin placed on its top, floats in water as shown in fig. the distance l and h are shown there. After some time the coin falls into the water. Then [IITJEE (Screening) 2002]
 l decreases and h increases
 l increases and h decreases
 Both l and h increase
 Both l and h decrease
Solution : (d) As the block moves up with the fall of coin, l decreases, similarly h will also decrease because when the coin is in water, it displaces water equal to its own volume only.
Problem 15. A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 10^{3}kg/m^{3}. If outer diameter and the density of the bowl are 1 m and 2 × 10^{4} kg/m^{3} respectively, then the inner diameter of the bowl will be
(a) 0.94 m (b) 0.97 m (c) 0.98 m (d) 0.99 m
4 éæ D ö^{3} æ d ö^{3} ù
Solution : (c) Weight of the bowl = mg = Vrg = 3 p êç 2 ÷ – ç 2 ÷ ú rg
êëè ø è ø úû
where D is the outer diameter , d is the inner diameter and r is the density of bowl
4 æ D ö ^{3}
Weight of the liquid displaced by the bowl = Vsg = 3 p ç 2 ÷ s g
è ø

where s is the density of the liquid.
For the flotation
è ø êëè ø è ø úû è ø
êëè ø
è ø úû
By solving we get d = 0.98 m.
Problem 16. In making an alloy, a substance of specific gravity s_{1} and mass m_{1} is mixed with another substance of specific gravity s_{2} and mass m_{2} ; then the specific gravity of the alloy is [CPMT 1995]
æ m1 m2 ö
æ m + m ö
æ s s ö
m + m
ç + ÷
s s
(a)
ç 1 2 ÷
(b)
ç 1 2 ÷
(c)
1 2
(d)
è 1 2 ø
è s1 + s2 ø
è m1 + m2 ø
æ m1 m2 ö
m1 + m2
ç
è s1
+ ÷
s 2 ø
Solution : (c) Specific gravity of alloy =
Density of alloy Density of water
= Mass of alloy
Volume of alloy ´ density of water
= m1 + m2
= m1 + m2 = m1 + m2
éAs specific gravity of substance = density of substance ù
æ m m ö
m1
m2
m1 m2 ê
density of water ú
ç 1 + 2 ÷ ´ rw
r / r + r / r s + s ë û
è r1 p2 ø
1 w 2 w 1 2
Problem 17. 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 [AIIMS 1995]
(a) 8 (b) 4 (c) 3 (d) Zero
Solution : (b) Let specific gravities of concrete and saw dust are r_{1} and r _{2} respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere
4 p (R^{3} – r^{3})r_{1}g + 4 pr ^{3}r_{2}g = 4 pR^{3} ´ 1 ´ g Þ
R3 r1 – r 3 r1 + r 3 r2 = R3
3 3 3
3 3 R3
r_{1} – r_{2}
R^{3} – r ^{3}
r_{1} – r_{2} – r_{1} + 1
(R^{3} – r^{3})r_{1}
æ 1 – r_{2} ö r_{1}
Þ R (r_{1} – 1) = r
(r_{1} – r_{2}) Þ
r ^{3} = r – 1 Þ r ^{3} =
r – 1
Þ r^{3}r
= ç r
÷
– 1 r
Þ Mass of concrete
1
= æ 1 – 0.3 ö ´ 2.4 = 4
1 2 è 1 ø 2
Mass of saw dust
ç 2.4 – 1 ÷ 0.3
è ø
Problem 18. A vessel contains oil (density = 0.8 gm/cm^{3}) over mercury (density = 13.6 gm/cm^{3}). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in gm/cm^{3} is [IITJEE 1988]
(a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8
Solution : (c) As the sphere floats in the liquid. Therefore its weight will be equal to the upthrust force on it
Weight of sphere = 4 pR^{3}rg
3
2 _{3}
…… (i)
2 _{3}
Upthrust due to oil and mercury = pR
3
Equating (i) and (ii)
 s _{oil} g + 3 pR s _{Hg}g………….. (ii)
4 pR^{3}rg = 2 pR^{3}0.8g + 2 pR^{3} ´ 13.6g Þ 2r = 0.8 + 13.6 = 14.4 Þ r = 7.2
3 3 3
Problem 19. A body floats in a liquid contained in a beaker. The whole system as shown falls freely under gravity. The upthrust on the body due to the liquid is [IITJEE1982]
 Zero
 Equal to the weight of the liquid displaced
 Equal to the weight of the body in air
 Equal to the weight of the immersed position of the body
Solution : (a) Upthrust = Vr_{liquid}(g – a) ; where, a = downward acceleration, V = volume of liquid displaced But for free fall a = g \ Upthrust = 0
Problem 20. 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 will be
(a) 5 × 5 × 5 × 5 gf (b) 4 × 4 × 4 × 4 gf (c) 5 × 4 × 4 × 4 gf (d) 4 × 5 × 5 × 5 gf
Solution : (d) Apparent weight = V(r – s )g = l ´ b ´ h ´ (5 – 1) ´ g = 5 ´ 5 ´ 5 ´ 4 ´ g
Dyne or 4 ´ 5 ´ 5 ´ 5
gf.
Problem 21. A wooden block of volume 1000 cm^{3} is suspended from a spring balance. It weighs 12 N in air. It is suspended in water such that half of the block is below the surface of water. The reading of the spring balance is
(a) 10 N (b) 9 N (c) 8 N (d) 7 N
Solution : (d) Reading of the spring balance = Apparent weight of the block = Actual weight – upthrust
= 12 – Vinsg = 12 – 500 ´ 10 ^{–}^{6} ´ 10^{3} ´ 10 = 12 – 5 = 7 N.
Problem 22. An iceberg is floating in sea water. The density of ice is 0.92 gm/cm^{3} and that of sea water is 1.03g/cm^{3}. What percentage of the iceberg will be below the surface of water
(a) 3% (b) 11% (c) 89% (d) 92%
Solution : (c) For the floatation of iceberg, Weight of ice = upthrust due to displaced water
Vrg = V s g Þ V
= æ r öV = æ 0.92 ö V = 0.89V
\ Vin
= 0.89
or 89%.
in in
ç s ÷ ç 1.03 ÷ V
è ø è ø
Problem 23. 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
 1 2
 38
 23
 34
Solution : (a) Fraction of volume immersed in the liquid Vin
æ r ö

ç s ÷
i.e. it depends upon the densities of the block and liquid.
è ø
So there will be no change in it if system moves upward or downward with constant velocity or some acceleration.
Problem 24. 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 tension in the string in kg–wt is
(a) 1.6 (b) 1.94 (c) 3.1 (d) 5.25
M æ s ö æ 0.8 ö
Solution : (b) Apparent weight = V(r – s )g =
r (r – s )g = Mç1 – r ÷g = 2.1ç1 – 10.5 ÷g = 1.94 g
Newton = 1.94 Kgwt
è ø è ø
Problem 25. A sample of metal weighs 210 gm in air, 180 gm in water and 120 gm in liquid. Then relative density (RD) of
(a) Metal is 3 (b) Metal is 7 (c) Liquid is 3 (d) Liquid is 1
3
Solution : (b, c) Let the density of metal is r and density of liquid is s. If V is the volume of sample then according to problem
210 = Vrg
180 = V(r – 1)g
120 = V(r – s )g
……..(i)
……..(ii)
……..(iii)
By solving (i), (ii) and (iii) we get r = 7 and s = 3 .
Problem 26. 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
(a) 4 : 3 (b) 2 : 3 (c) 3 : 4 (d) 1 : 3
Solution : (c) If two different bodies A and B are floating in the same liquid then r A
= ( fin ) A
= 1 / 2 = 3
r _{B} ( fin )B
2 / 3 4
Problem 27. The fraction of a floating object of volume V_{0}
and density d _{0} above the surface of a liquid of density d will be
(a)
d0 (b)
d
dd0 d + d0
(c)
d – d0
d
(d)
dd0 d – d0
Solution : (c) For the floatation V d g = V d g Þ V
= V d0
0 0 in
in 0 d
\ V = V – V
= V – V
d_{0} = V é d – d_{0} ù Þ
Vout = d – d0 .
out
0 in 0 0 d
0 êë d úû
V0 d
Problem 28. A vessel with water is placed on a weighing pan and reads 600 g. Now a ball of 40 g and density 0.80 g/cc is sunk into the water with a pin as shown in fig. keeping it sunk. The weighing pan will show a reading
 600 g
 550 g
 650 g
 632 g
Solution : (c) Upthrust on ball = weight of displaced water
æ m ö 40
= V s g = ç r ÷ s g = 0.8 ´ 1 ´ g = 50g
Dyne or 50 gm
è ø
As the ball is sunk into the water with a pin by applying downward force equal to upthrust on it.
So reading of weighing pan = weight of water + downward force against upthrust = 600 + 50 = 650 gm.
Some Conceptual Questions.
Que.1 Why a small iron needle sinks in water while a large iron ship floats
Ans. For floatation, the density of body must be lesser or equal to that of liquid. In case of iron needle, the density of needle, i.e., iron is more than that of water, so it will sink. However, the density of a ship due to its large volume is lesser than that of water, so it will float.
Que.2 A man is sitting in a boat which is floating in a pond. If the man drinks some water from the pond, what will happen to the level of water in the pond
Ans. If the man drinks m g of water from the pond, the weight of (boat + man) system will increase by mg and so the system will displace mg more water for floating. So due to removal of water from pond, the water level in pond will fall but due to water displaced by the floating system the water level in the pond will rise and so the water removed from the pond is equal to the water displaced by the system; the level of water in the pond will remain unchanged.
Que.3 A boy is carrying a fish in one hand and a bucket full of water in the other hand. He then places the fish in the bucket thinking that in accordance with Archimedes’ principle he is now carrying less weight as the weight of the fish will reduce due to upthrust. Is he right
Ans. No, when he places the fish in water in the bucket, no doubt the weight of fish is reduced due to upthrust, but the weight of (water + bucket) system is increased by the same amount, so that the total weight carried by him remains unchanged.
Que. 4 A bucket of water is suspended from a spring balance. Does the reading of balance change (a) when a piece of stone suspended from a string is immersed in the water without touching the bucket? (b) when a piece of iron or cork is put in the water in the bucket?
Ans. (a) Yes, the reading of the balance will increase but the increase in weight will be equal to the loss in weight of the stone (Vsg) and not the weight of stone (Vrg)[> Vsg as r > s ] .
(b) Yes, the reading of the balance will increase but the increase in weight will be equal to the weight of iron or cork piece.
Que. 5 Why a soft plastic bag weighs the same when empty or when filled with air at atmospheric pressure? Would the weight be the same if measured in vacuum
Ans. If the weight of empty bag is W_{0} and the volume of bag is V, when the bag is filled with air of density r at NTP, its
weights will increase by Vrg . Now when the bag filled with air is weighed in air, the thrust of air Vrg
weight; so W = W0 + Vrg – Vrg = W0
will decrease its
i.e., the weight of the bag remains unchanged when it is filled with air at NTP and weighed in air. However if the bag is
weighed in vacuum will be W_{0}
when empty and
(W_{0} + Vrg)
when filled with air (as there is no upthrust), i.e., in
vacuum an airfilled bag will weigh more than an empty bag.
Que.6 Why does a uniform wooden stick or log float horizontally? If enough iron is added to one end, it will float vertically; explain this also.
Ans. When a wooden stick is made to float vertically, its rotational equilibrium will be unstable as its metacentre will be lower than its CG and with a slight tilt it will rotate under
the action of the couple formed by thrust and weight in the direction of tilt, till it becomes horizontal.
However, due to loading at the bottom, the CG of the stick (or log) will be lowered and so may be lower than the metacentre. In this situation the equilibrium will be stable and if the stick (or log) is tilted, it will come back to its initial vertical position.
Que.7 A boat containing some pieces of material is floating in a pond. What will happen to the level of water in the pond if on unloading the pieces in the pond, the piece (a) floats (b) sinks?
Ans. If M is the mass of boat and m of pieces in it, then initially as the system is floating M + m = VDs_{W}
i.e., the system displaces water VD = M + m
……(i)
s_{W} s_{W}
When the pieces are dropped in the pond, the boat will still float, so it displaces water M = V_{1}s_{W} ,
i.e,
V1 = (M / s_{W} )
 Now if the unloaded pieces floats in the pond, the water displaced by them m = V_{2}s _{W} ,
i.e,
V2 = (m / s_{W} )
So the total water displaced by the boat and the floating pieces
V1 + V2 = M + m
…..(ii)
s_{W} s_{W}
Which is same as the water displaced by the floating system initially (eqn. 1); so the level of water in the pond will remain unchanged.
 Now if the unloaded pieces sink the water displaced by them will be equal to their own volume, e,
V2¢ = m éas r = m ù
r êë V úû
and so in this situation the total volume of water displaced by boat and sinking pieces will be

æ
V1 + V2¢ = ç s
ö

+ r ÷
…..(iii)
è W ø
Now as the pieces are sinking
r > s_{W} , so this volume will be lesser than initial water displaced by the floating system (eq.
1); so the level of water in the pond will go down (or fall)
In this problem if the pieces (either sinking or floating) are unloaded on the ground, the water displaced after unloading,
V_{2} = M / s_{W} , will be lesser than before unloading. V = (M + m) / s_{W} ; so the level of water in the pond will fall.
Streamline, Laminar and Turbulent Flow.
 Stream line flow : Stream line flow of a liquid is that flow in which each element of the liquid passing through a point travels along the same path and with the same velocity as the preceeding element passes through that
A streamline may be defined as the path, straight or curved, the tangent to which at any point gives the direction of the flow of liquid at that point.
The two streamlines cannot cross each other and the greater is the crowding of streamlines at a place, the greater is the velocity of liquid particles at that place.
Path ABC is streamline as shown in the figure and v_{1} , v_{2}
and v3
are the
velocities of the liquid particles at A, B and C point respectively.
 Laminar flow : If a liquid is flowing over a horizontal surface with a steady flow and moves in the form of layers of different velocities which do not mix with each other, then the flow of liquid is called laminar flow.
In this flow the velocity of liquid flow is always less than the critical velocity of the liquid. The laminar flow is generally used synonymously with streamlined flow.
 Turbulent flow : When a liquid moves with a velocity greater than its critical velocity, the motion of the particles of liquid becomes disordered or Such a flow is called a
turbulent flow.
In a turbulent flow, the path and the velocity of the particles of the liquid change continuously and haphazardly with time from point to point. In a turbulent flow, most of the external energy maintaining the flow is spent in producing eddies in the liquid and only a small fraction of energy is available
for forward flow. For example, eddies are seen by the sides of the pillars of a river bridge.
Critical Velocity and Reynold’s Number.
The critical velocity is that velocity of liquid flow upto which its flow is streamlined and above which its flow becomes turbulent.
Reynold’s number is a pure number which determines the nature of flow of liquid through a pipe.
It is defined as the ratio of the inertial force per unit area to the viscous force per unit area for a flowing fluid.
N R =
Inertial force per unit area Viscous force per unit area
If a liquid of density r is flowing through a tube of radius r and cross section A then mass of liquid flowing
through the tube per second
dm =
dt
volume flowing per second × density =
Av ´ r
\ Inertial force per unit area =
dp / dt = v(dm / dt)
= v Av r = v2 r
Viscous force per unit area
A A A
F / A = hv
r
So by the definition of Reynolds number If the value of Reynold’s number
N = Inertial force per unit area
^{R} Viscous force per unit area
= v ^{2} r
hv / r
= vrr
h
 Lies between 0 to 2000, the flow of liquid is streamline or
 Lies between 2000 to 3000, the flow of liquid is unstable changing from streamline to turbulent
 Above 3000, the flow of liquid is definitely
Problem 29. In which one of the following cases will the liquid flow in a pipe be most streamlined
 Liquid of high viscosity and high density flowing through a pipe of small radius
 Liquid of high viscosity and low density flowing through a pipe of small radius
 Liquid of low viscosity and low density flowing through a pipe of large radius
 Liquid of low viscosity and high density flowing through a pipe of large radius
Solution : (b) For streamline flow Reynold’s number
N µ r r
R h
should be less.
For less value of
NR , radius and density should be small and viscosity should be high.
Problem 30. Two different liquids are flowing in two tubes of equal radius. The ratio of coefficients of viscosity of liquids is 52:49 and the ratio of their densities is 13:1, then the ratio of their critical velocities will be
(a) 4 : 49 (b) 49 : 4 (c) 2 : 7 (d) 7 : 2
Solution : (a) Critical velocity v = N
h Þ v_{1} = h_{1} ´ r _{2}
= 52 ´ 1 = 4 .
^{R} rr v2 h_{2} r_{1}
49 13 49
Equation of Continuity.
The equation of continuity is derived from the principle of conservation of mass.
A nonviscous liquid in streamline flow passes through a tube AB of varying cross section. Let the cross sectional area of the pipe at points A and B
be a1 and
a_{2} respectively. Let the liquid enter with normal velocity
v1 at A
and leave with velocity v_{2}
at B. Let
r_{1} and
r _{2} be the densities of the liquid
at point A and B respectively.
Mass of the liquid entering per second at A = Mass of the liquid leaving per second at B
a1v1 r1 = a2v2 r2
a1v1 = a2v2
or av = constant
or a µ 1
v
[If the liquid is incompressible (r_{1} = r_{2})]
This expression is called the equation of continuity for the steady flow of an incompressible and nonviscous liquid.
 The velocity of flow is independent of the liquid (assuming the liquid to be nonviscous)
 The velocity of flow will increase if crosssection decreases and viceversa. That is why :
 In hilly region, where the river is narrow and shallow (e., small crosssection) the water current will be faster, while in plains where the river is wide and deep (i.e., large crosssection)
the current will be slower, and so deep water will appear to be still.
 When water falls from a tap, the velocity of falling water under the
action of gravity will increase with distance from the tap (i.e,
v_{2} > v_{1} ). So in
accordance with continuity equation the cross section of the water stream will
decrease (i.e.,
A_{2} < A_{1}), i.e., the falling stream of water becomes narrower.
Problem 31. 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 [MNR 1980]
 4 times that in the other pipe (b)
1 times that in the other pipe
4
(c) 2 times that in the other pipe (d)
1 times that in the other pipe
2
Solution : (a)
d A = 2 cm and dB = 4 cm
\ rA = 1 cm and rB = 2 cm
v A a B
p (rB )^{2}
æ 2 ö ^{2}
From equation of continuity av = constant \ =
v a
=
p (r
)2 = ç 1 ÷
Þ v A = 4vB
B A A è ø
Problem 32. An incompressible liquid flows through a horizontal tube as shown in the following fig. Then the velocity v of the fluid is
(a) 3.0 m/s (b) 1.5 m/s (c) 1.0 m/s (d) 2.25 m/s
Solution : (c) If the liquid is incompressible then mass of liquid entering through left end, should be equal to mass of liquid coming out from the right end.
\ M = m1 + m2 Þ
Av_{1} = Av_{2} + 1.5 A . v Þ
A ´ 3 = A ´ 1.5 + 1.5 A . v Þ v = 1 m / s
Problem 33. Water enters through end A with speed v_{1} and leaves through end B with speed v_{2} 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 v_{1} = v_{2} for
 Case I (b) Case II (c) Case III (d) Each case
Solution : (d) 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.
Problem 34. Water is moving with a speed of 5.18 ms^{–1} through a pipe with a crosssectional area of 4.20 cm^{2}. The water gradually descends 9.66 m as the pipe increase in area to 7.60 cm^{2}. The speed of flow at the lower level is
(a) 3.0 ms^{–1} (b) 5.7 ms^{–1} (c) 3.82 ms^{–1} (d) 2.86 ms^{–1}
Solution : (d)
a 1 v 1
= a 2 v 2
Þ 4.20 ´ 5.18 = 7.60 ´ v2 Þ v2 = 2.86 m / s
Energy of a Flowing Fluid.
A flowing fluid in motion possesses the following three types of energy
Pressure Energy  Potential energy  Kinetic energy 
It is the energy possessed by a liquid by virtue of its pressure. It is the measure of work done in pushing the liquid against pressure without imparting any velocity to it.
Pressure energy of the liquid PV
Pressure energy per unit mass of the liquid P r Pressure energy per unit volume of the liquid P 
It is the energy possessed by liquid by virtue of its height or position above the surface of earth or any reference level taken as zero level.
Potential energy of the liquid mgh
Potential energy per unit mass of the liquid gh
Potential energy per unit volume of the liquid rgh 
It is the energy possessed by a liquid by virtue of its motion or velocity.
Kinetic energy of the liquid 1 mv ^{2} 2 Kinetic energy per unit mass of the liquid 1 v ^{2} 2 Kinetic energy per unit volume of the liquid 1 rv ^{2} 2 
Bernoulli’s Theorem.
According to this theorem the total energy (pressure energy, potential energy and kinetic energy) per unit volume or mass of an incompressible and nonviscous fluid in steady flow through a pipe remains constant throughout the flow, provided there is no source or sink of the fluid along the length of the pipe.
Mathematically for unit volume of liquid flowing through a pipe.
P + rgh + 1 rv^{2} = constant
2
To prove it consider a liquid flowing steadily through a tube of nonuniform area of crosssection as shown in
fig. If
P1 and
P_{2} are the pressures at the two ends of the tube respectively, work done in pushing the volume V of
incompressible fluid from point B to C through the tube will be W = P1 V – P2 V = (P1 – P2 )V
This work is used by the fluid in two ways.
 In changing the potential energy of mass m (in the volume V ) from mgh_{1} to mgh_{2},
……(i)
i.e.,
DU = mg(h_{2} – h_{1} )
……(ii)
 In changing the kinetic energy from
1 mv ^{2} to 1 mv^{2} , i.e., DK = 1 m(v ^{2} – v ^{2} )
…...(iii)
2 1 2 2 2 2 1
Now as the fluid is nonviscous, by conservation of mechanical energy
W = DU + DK
i.e.,
(P – P ) V = mg(h – h ) + 1 m(v ^{2} – v ^{2} )
1 2 2 1 2 2 1
or P – P = rg(h – h ) + 1 r(v ^{2} – v ^{2} )
[As
r = m / V ]
1 2 2 1 2 2 1
or P + rgh + 1 rv ^{2} = P + rgh + 1 rv ^{2}
1 1 2 1 2 2 2 2
or P + rgh + 1 rv ^{2} = constant
2
This equation is the so called Bernoulli’s equation and represents conservation of mechanical energy in case of moving fluids.
 Bernoulli’s theorem for unit mass of liquid flowing through a pipe can also be written as:
P + gh + 1 v ^{2} = constant
r 2
 Dividing above equation by g we get P+ h + v 2
= constant
rg 2g
Here
P is called pressure head, h is called gravitational head and
rg
v 2 is called velocity head. From this 2g
equation Bernoulli’s theorem can be stated as.
In stream line flow of an ideal liquid, the sum of pressure head, gravitational head and velocity head of every cross section of the liquid is constant.
Applications of Bernoulli’s Theorem.
(i) Attraction between two closely parallel moving boats (or buses)
: When two boats or buses move
side by side in the same direction, the water (or air) in the region between them moves faster than that on the remote sides. Consequently in accordance with Bernoulli’s principle the pressure between them is reduced and hence due to pressure difference they are pulled towards each other creating the so called attraction.
 Working of an aeroplane : This is also based on Bernoulli’s The wings of the aeroplane are of the shape as shown in fig. Due to this specific shape of wings when the
aeroplane runs, air passes at higher speed over it as compared to its lower surface. This difference of air speeds above and below the wings, in accordance with Bernoulli’s principle, creates a pressure difference, due to which an upward force called ‘dynamic lift’ (= pressure difference × area of wing) acts on the plane. If this force becomes greater than the weight of the plane, the plane will rise up.
 Action of atomiser: The action of carburetor, paintgun, scentspray or insectsprayer is based on Bernoulli’s principle. In all these, by means of motion of a piston P in a cylinder
C, high speed air is passed over a tube T dipped in liquid L to be sprayed. High speed air creates low pressure over the tube due to which liquid (paint, scent, insecticide or petrol) rises in it and is then blown off in very small droplets with expelled air.
 Blowing off roofs by wind storms : During a tornado or hurricane, when a high speed wind blows over a straw or tin roof, it creates a low pressure (P) in accordance with
Bernoulli’s principle.
However, the pressure below the roof (i.e., inside the room) is still atmospheric ( = P_{0} ). So due to this difference of pressure the roof is lifted up and is then blown off by the wind.
 Magnus effect : When a spinning ball is thrown, it deviates from its usual path in flight. This effect is called Magnus effect and plays as important role in tennis, cricket and soccer, as by applying appropriate spin the moving ball can be made to curve in any desired direction.
If a ball is moving from left to right and also spinning about a horizontal axis perpendicular to the direction of motion as shown in fig. then relative to the ball, air will be moving from right to left.
The resultant velocity of air above the ball will be
(v + rw)
while below it
(v – rw). So in accordance with
Bernoulli’s principle pressure above the ball will be less than below it. Due to this difference of pressure an upward
force will act on the ball and hence the ball will deviate from its usual path
OA0
and will hit the ground at A_{1}
following the path OA_{1}
i.e., if a ball is thrown with backspin, the pitch will curve less sharply prolonging the flight.
Similarly if the spin is clockwise i.e., the ball is thrown with topspin, the force due to pressure difference will act in the direction of gravity and so the pitch will curve more sharply shortening the flight.
Furthermore, if the ball is spinning about a vertical axis, the curving will be sideways as shown in producing the so called out swing or in swing.
 Venturimeter : It is a device based on Bernoulli’s theorem used for measuring the rate of flow of liquid through
It consists of two identical coaxial tubes A and C connected by a narrow coaxial tube B. Two vertical tubes D and E are mounted on the tubes A and B to measure the pressure of the following liquid.
When the liquid flows in the tube ABC, the velocity of flow in part B will be larger than in the tube A or C. So the pressure in part B will be less than that in tube A or C. By measuring the pressure difference between A and B, the rate of flow of the liquid in the tube can be calculated.
Let a1 , a2 = area of cross section of tube A and B respectively
v_{1}, v_{2} = Velocity of flow of liquid through A and B respectively
P1 , P2 = Liquid pressure at A and B respectively
\ P_{1} – P_{2} = hrg
……(i) [ r = density of flowing liquid]
From Bernoulli’s theorem for horizontal flow of liquid
P + 1 rv ^{2} = P + 1 rv ^{2}
1 2 1 2 2 2
P – P = 1 r(v ^{2} – v ^{2} )
……(ii)
1 2 2 2 1
From (i) and (ii) hrg = 1 r(v ^{2} – v ^{2} ) = 1 r é V 2 – V 2 ù
[As
V = a v
= a v ]
2 2 1
2 ê a ^{2}
a 2 ú
1 1 2 2


2a ^{2}a ^{2}hg
V a ^{2} – a ^{2}
ë 2 1 û
or V = a1a2
1 2
Problem 35. The velocity of kerosene oil in a horizontal pipe is 5 m/s. If g = 10m / s^{2}
then the velocity head of oil will be
(a) 1.25 m (b) 12.5 m (c) 0.125 m (d) 125 m
Solution : (a) Velocity head h = v2
2g
= (5)^{2}
2 ´ 10
= 1.25 m
Problem 36. In the following fig. is shown the flow of liquid through a horizontal pipe. Three tubes A, B and C are connected to the pipe. The radii of the tubes A, B and C at the junction are respectively 2 cm, 1 cm and 2 cm.
It can be said that the
 Height of the liquid in the tube A is maximum
 Height of the liquid in the tubes A and B is the same
 Height of the liquid in all the three tubes is the same
 Height of the liquid in the tubes A and C is the same
Solution : (d) As crosssection areas of both the tubes A and C are same and tube is horizontal. Hence according to equation
of continuity
v A = vC
and therefore according to Bernoulli’s theorem
PA = PC i.e. height of liquid is same in
both the tubes A and C.
Problem 37. A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is r and angular velocity of rotation is w , then the difference in the heights of the liquid at the centre of the vessel and the edge is
(a)
rw (b)
2g
r 2w 2
2g
(c)
(d)
w 2
2gr ^{2}
Solution : (b) From Bernoulli’s theorem, PA + 1 dv^{2} + dgh
= P + 1 dv^{2} + dgh
2 A A
1 _{2} 1 _{2}
B 2 B B
1 2 2
Here,
hA = hB
\ PA + 2 dvA = PB + 2 dvB
Þ PA – PB = 2 d[vB – vA]
Now,
vA = 0, vB
= rw
and PA
 PB
= hdg
\ hdg = 1 dr ^{2}w ^{2} or h = r 2w 2
2 2g
Problem 38. A manometer connected to a closed tap reads 3.5 × 10^{5} N/m^{2}. When the valve is opened, the reading of manometer falls to 3.0 × 10^{5} N/m^{2}, then velocity of flow of water is
(a) 100 m/s (b) 10 m/s (c) 1 m/s (d) 10
Solution : (b) Bernoulli’s theorem for unit mass of liquid P + 1 v ^{2} = constant
m/s
r 2
As the liquid starts flowing, it pressure energy decreases
1 v 2 = P_{1} – P_{2}
2 r
Þ 1 v 2 = 3.5 ´ 10^{5} – 3 ´ 10^{5} Þ v 2 = 2 ´ 0.5 ´ 10^{5} Þ v 2 = 100 Þ v = 10 m / s
2 10^{3} 10^{3}
Problem 39. Water flows through a frictionless duct with a crosssection varying as shown in fig. Pressure p at points along the axis is represented by
(a)
(b) P
(c)
(d) P
Solution : (a) When cross section of duct decreases the velocity of water increases and in accordance with Bernoulli’s theorem the pressure decreases at that place.
Problem 40. 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 metre^{3} and the wing is 10 m long and has an average width of 2 m, then the difference of the pressure on the two sides of the wing of
(a) 4095.0 Pascal (b) 409.50 Pascal (c) 40.950 Pascal (d) 4.0950 Pascal
Solution : (a) From the Bernoulli’s theorem P1 – P2 = 1 r(v^{2} – v^{2} ) = 1 ´ 1.3 ´ [(120)^{2} – (90)^{2} ] = 4095 N / m^{2} or Pascal
2 2 1 2
Velocity of Efflux.
If a liquid is filled in a vessel up to height H and a hole is made at a depth h below the free surface of the liquid as shown in fig. then taking the level of hole as reference level (i.e., zero
point of potential energy) and applying Bernoulli’s principle to the liquid just inside and outside the hole (assuming the liquid to be at rest inside) we get
\ (P0
+ hrg) + 0 = P_{0}
+ 1 rv ^{2}
2
or v =
Which is same as the speed that an object would acquire in falling from rest through a distance h and is called velocity of efflux or velocity of flow.
This result was first given by Torricelli so this is known as Torricelli’s theorem.
 The velocity of efflux is independent of the nature of liquid, quantity of liquid in the vessel and the area of
 Greater is the distance of the hole from the free surface of liquid greater will be the velocity of efflux
[i.e., v µ h]
 As the vertical velocity of liquid at the orifice is zero and it is at a height taken by the liquid to reach the baselevel t =
(H – h)
from the base, the time
 Now during time t liquid is moving horizontally with constant velocity v, so it will hit the base level at a horizontal distance x (called range) as shown in
Such that x = vt = ´ = 2
For maximum range
dx = 0
dh
\ h = H
2
H
i.e., range x will be maximum when h = 2 .
\ Maximum range
xmax = 2 = H
 If the level of free surface in a container is at height H from the base and there are two holes at depth h and
y below the free surface, then
x = 2
and
x¢ = 2
Now if
i.e.,
or
i.e.,
x = x¢ , i.e., h(H – h) = y(H – y)
y ^{2} – Hy + h(H – h) = 0
y = 1 [H ± (H – 2h)] ,
2
y = h or (H – h)
i.e., the range will be same if the orifice is at a depth h or (H – h) below the free surface. Now as the distance
(H – h)
from top means
H – (H – h) = h
from the bottom, so the range is same for liquid coming out of holes at
same distance below the top and above the bottom.
 If
A_{0} is the area of orifice at a depth y below the free surface and A that of container, the volume of liquid
coming out of the orifice per second will be (dV / dt) = vA_{0} = A_{0}
[As v = ]
Due to this, the level of liquid in the container will decrease and so if the level of liquid in the container above
the hole changes from y to
y – dy
in time t to
t + dt
then – dV = Ady
So substituting this value of dV in the above equation
 A dy = A
dt ^{0}
i.e., ò
dt = – A
1 ò y 1 / 2 dy
So the time taken for the level to fall from H to H ¢


A 1 H + 1 / 2 A 2
A0 H
A0 g
If the hole is at the bottom of the tank, time t taken to make the tank empty :
t = [As here H ¢ = 0 ]
Problem 41. A large tank filled with water to a height ‘h’ is to be emptied through a small hole at the bottom. The ratio of
times taken for the level of water to fall from h to
h and from
2
h to zero is [EAMCET (Engg.) 2003]
2
(a)
(b)
 – 1
(d)
Solution : (c) Time taken for the level to fall from H to H‘
t = A
A0
[ H –
H‘ ]
According to problem the time taken for the level to fall from h to h
2
t = A
1 A0
é h ù


êê – 2 ú
and similarly time taken for the level to fall from
1 – 1
h to zero
2
t = A
2 A0
é ù
ê – 0ú
ëê úû
\ t1
t 2
= 2 =
1 – 0
– 1.
Problem 42. A cylinder of height 20 m is completely filled with water. The velocity of efflux of water (in m/s) through a small hole on the side wall of the cylinder near its bottom is [AIEEE 2002]
(a) 10 (b) 20 (c) 25.5 (d) 5
Solution : (b) v = = = 20 m / s
Problem 43. There is a hole in the bottom of tank having water. If total pressure at bottom is 3 atm (1 atm = 10^{5}N/m^{2}) then the velocity of water flowing from hole is [CPMT 2002]
(a)
400 m / s
(b)
m / s
(c)
60 m / s
 None of these
Solution : (b) Pressure at the bottom of tank P = hrg = 3 ´ 10^{5} N
m2
and velocity of water v =
\ v = = = 600 m / s
Problem 44. 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 the holes are the same. Then R is equal to
[IITJEE (Screening) 2000]
(a)
2p L
(b)
L (c) L (d) L
2p
Solution : (b) Velocity of efflux when the hole is at depth h, v =
Rate of flow of water from square hole Q1 = a1v1 = L^{2}
Rate of flow of water from circular hole Q2 = a2v2 = pR ^{2}
and according to problem Q1 = Q2 Þ L^{2}
= pR ^{2}
Þ R = L
Problem 45. There is a hole of area A at the bottom of cylindrical vessel. Water is filled up to a height h and water flows out in t second. If water is filled to a height 4h, it will flow out in time equal to [MP PMT 1997]
(a) t (b) 4t (c) 2 t (d) t/4
Solution : (c) Time required to emptied the tank
t = A
A0
Þ t 2 =
t1
= = 2
\ t 2
= 2t
Problem 46. A cylinder containing water up to a height of 25 cm has a hole of crosssection
 cm^{2}
4
in its bottom. It is
counterpoised in a balance. What is the initial change in the balancing weight when water begins to flow out
 Increase of 5 gmwt
 Increase of 25 gmwt
 Decrease of 5 gmwt
 Decrease of 25 gmwt
Solution : (c) Let A = The area of cross section of the hole, v = Initial velocity of efflux, d = Density of water, Initial volume of water flowing out per second = Av
Initial mass of water flowing out per second = Avd
Rate of change of momentum = Adv^{2} \ Initial downward force on the out flowing water = Adv^{2}
So equal amount of reaction acts upwards on the cylinder.
\ Initial upward reaction = Adv^{2}
[As
v = ]

\ Initial decrease in weight = Ad (2gh) = 2Adgh = 2 ´ æ 1 ö ´ 1 ´ 980 ´ 25 = 12.5 gmwt.
4
è ø
Problem 47. A cylindrical tank has a hole of 1 cm^{2} in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of 70 cm^{3}/sec. then the maximum height up to which water can rise in the tank is
(a) 2.5 cm (b) 5 cm (c) 10 cm (d) 0.25 cm
Solution : (a) The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second.
Volume of water flowing out per second = Av = A
and volume of water flowing in per second = 70 cm^{3} / sec .
\ A = 70 Þ 1 ´
= 70 Þ 1´
= 70
\ h = 4900 = 2.5 cm.
1960
Viscosity and Newton’s law of Viscous Force.
In case of steady flow of a fluid when a layer of fluid slips or tends to slip on adjacent layers in contact, the two layer exert tangential force on each other which tries to destroy the relative
motion between them. The property of a fluid due to which it opposes the relative motion between its different layers is called viscosity (or fluid friction or internal friction) and the force between the layers opposing the relative motion is called viscous force.
Consider the two layers CD and MN of the liquid at distances x and x + dx from the fixed surface AB, having
the velocities v and v + dv respectively. Then as velocity gradient.
dv denotes the rate of change of velocity with distance and is known
dx
According to Newton’s hypothesis, the tangential force F acting on a plane
parallel layer is proportional to the area of the plane A and the velocity gradient
dv in a direction normal to the layer, i.e.,
dx
F µ A
and
F µ dv \
dx
F µ A dv
dx
or F = –hA dv dx
Where h is a constant called the coefficient of viscosity. Negative sign is employed because viscous force acts in a direction opposite to the flow of liquid.
If A = 1, dv = 1 then h = F.
dx
Hence the coefficient of viscosity is defined as the viscous force acting per unit area between two layers moving with unit velocity gradient.
 Units : dynescm^{–2} or Poise (C.G.S. system); Newtonsm^{–2} or Poiseuille or decapoise (S.I. system) 1 Poiseuille = 1 decapoise = 10 Poise
 Dimension : [ML^{–1} T^{–1}]
 Viscosity of liquid is much greater (about 100 times more) than that of gases e. h _{L} > h_{G}
Example : Viscosity of water = 0.01 Poise while of air = 200 m Poise
 With increase in pressure, the viscosity of liquids (except water) increases while that of gases is practically independent of The viscosity of water decreases with increase in pressure.
 Difference between viscosity and solid friction : Viscosity differs from the solid friction in the respect that the viscous force acting between two layers of the liquid depends upon the area of the layers, the relative velocity of two layers and distance between two layers, but the friction between two solid surfaces is independent of the area of surfaces in contact and the relative velocity between
 From kinetic theory point of view viscosity represents transport of momentum, while diffusion and conduction represents transport of mass and energy
 The viscosity of thick liquids like honey, glycerin, coaltar etc. is more than that of thin liquids like
 The cause of viscosity in liquids is cohesive forces among molecules where as in gases it is due to diffusion.
 The viscosity of gases increases with increase of temperature, because on increasing temperature the rate of diffusion
 The viscosity of liquid decreases with increase of temperature, because the cohesive force between the liquid molecules decreases with increase of temperature
Relation between coefficient of viscosity and temperature; Andrade formula h =
AeCr / T
r 1 / 3
Where T= Absolute temperature of liquid, r = density of liquid, A and C are constants.
Problem 48. A square plate of 0.1 m side moves parallel to a second plate with a velocity of 0.1 m/s, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise, distance between the plates in m is [EAMCET (Med.) 2003]
(a) 0.1 (b) 0.05 (c) 0.005 (d) 0.0005
Solution : (d)
A = (0.1)^{2} = 0.01m^{2} ,
h = 0.01 Poise = 0.001
decapoise
(M.K.S. unit), dv = 0.1 m/s and F = 0.002 N
F = hA dv
\ dx = hAdv = 0.001´ 0.01´ 0.1 = 0.0005m .
dx F 0.002
Problem 49. The diagram shows a cup of tea seen from above. The tea has been stirred and is now rotating without turbulence. A graph showing the speed v with which the liquid is crossing points at a distance X from O along a radius XO would look like
 V
 V
 V
 V
X X X X
Solution : (d) When we move from the centre to the circumference the velocity of liquid goes on decreasing and finally becomes zero
Stoke’s Law and Terminal Velocity.
When a body moves through a fluid, the fluid in contact with the body is dragged with it. This establishes relative motion in fluid layers near the body, due to which viscous force starts operating. The fluid exerts viscous force on the body to oppose its motion. The magnitude of the viscous force depends on the shape and size of the body, its speed and the viscosity of the fluid. Stokes established that if a sphere of radius r moves with velocity v through a fluid of viscosity h, the viscous force opposing the motion of the sphere is
F = 6phrv
This law is called Stokes law.
If a spherical body of radius r is dropped in a viscous fluid, it is first accelerated and then it’s acceleration becomes zero and it attains a constant velocity called terminal velocity.
Force on the body
 Weight of the body (W) = mg = (volume × density) × g = 4 pr ^{3} rg
3
 Upward thrust (T) = weight of the fluid displaced
= (volume × density) of the fluid × g =
 Viscous force (F) = 6phrv
4 pr ^{3}sg
3
When the body attains terminal velocity the net force acting on the body is zero. \W–T–F =0 or F= W – T
Þ 6phrv = 4 p r ^{3} r g – 4 p r ^{3}s g = 4 pr ^{3}(r – s ) g
3 3 3
\ Terminal velocity
2 r ^{2} (r – s ) g
v =
9 h
 Terminal velocity depend on the radius of the sphere so if radius is made n – fold, terminal velocity will become n^{2}
 Greater the density of solid greater will be the terminal velocity
 Greater the density and viscosity of the fluid lesser will be the terminal
 If r > s then terminal velocity will be positive and hence the spherical body will attain constant velocity in downward
 If r < s then terminal velocity will be negative and hence the spherical body will attain constant velocity in upward Example : Air bubble in a liquid and clouds in sky.
 Terminal velocity graph :
Problem 50. Spherical balls of radius ‘r‘ are falling in a viscous fluid of viscosity ‘h‘ with a velocity ‘v‘. The retarding viscous force acting on the spherical ball is [AIEEE 2004]
 Inversely proportional to ‘r‘ but directly proportional to velocity ‘v‘
 Directly proportional to both radius ‘r‘ and velocity ‘v‘
 Inversely proportional to both radius ‘r‘ and velocity ‘v‘
 Directly proportional to ‘r‘ but inversely proportional to ‘v‘
Solution : (b) F = 6 ph rv
Problem 51. A small sphere of mass m is dropped from a great height. After it has fallen 100 m, it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100 m of fall is [MP PMT 1990]
 Greater than the work done by air friction in the second 100 m
 Less than the work done by air friction in the second 100 m
 Equal to 100 mg
 Greater than 100 mg
Solution : (b) In the first 100 m body starts from rest and its velocity goes on increasing and after 100 m it acquire maximum velocity (terminal velocity). Further, air friction i.e. viscous force which is proportional to velocity is low in the beginning and maximum at v = vT .
Hence work done against air friction in the first 100 m is less than the work done in next 100 m.
Problem 52. Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be [MP PMT 1990]
(a) 10 cm per sec (b) 2.5 cm per sec (c)
5 ´ (4)^{1/} ^{3} cm per sec (d) 5 ´
 cm per sec
Solution : (c) If two drops of same radius r coalesce then radius of new drop is given by R
4 pR^{3} = 4 pr ^{3} + 4 pr ^{3}
Þ R ^{3} = 2r ^{3} Þ R = 2^{1/} ^{3} r
3 3 3
If drop of radius r is falling in viscous medium then it acquire a critical velocity v and v µ r ^{2}
v æ R ö ^{2}
æ 21 / 3 r ö 2
2 = ç
÷ = ç ÷
Þ v2 = 2^{2} ^{/} ^{3} ´ v1 = 2^{2} ^{/} ^{3} ´ (5) = 5 ´ (4)^{1/} ^{3} m / s


v1 è r ø ç ÷
Problem 53. A ball of radius r and density r falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is h, the value of h is given by
2 _{2} æ 1 – r ö
 r ç ÷ g
9 è h ø
2 r 2 æ r – 1 ö g
ç ÷
81 è h ø
2 æ r – 1 ö 2
r ^{4} ç ÷ g
81 è h ø
2 æ r – 1 ö 2
r ^{4} ç ÷ g
9 è h ø
Solution : (c) Velocity of ball when it strikes the water surface v =…………………………………. (i)
Terminal velocity of ball inside the water v = 2 r ^{2} g (r – 1)
……….(ii)
9
2 r ^{2} g
h
2 æ r – 1 ö 2
Equating (i) and (ii) we get
Poiseuille’s Formula.
= 9 h
(r – 1) Þ
h = 81
r ^{4} ç ÷ g
è h ø
Poiseuille studied the streamline flow of liquid in capillary tubes. He found that if a pressure difference (P) is maintained across the two ends of a capillary tube of length ‘l ‘ and radius r, then the volume of liquid coming out of the tube per second is
 Directly proportional to the pressure difference (P).
 Directly proportional to the fourth power of radius (r) of the capillary tube
 Inversely proportional to the coefficient of viscosity (h) of the
 Inversely proportional to the length (l) of the capillary
i.e.
V µ P r ^{4}
hl
or V =
KP r ^{4}
hl
\ V = pP r ^{4}
8hl
[Where
K = p
8
is the constant of proportionality]
This is known as Poiseulle’s equation. This equation also can be written as,
R is called as liquid resistance.
(1) Series combination of tubes
V = P
R
where
R = 8hl
p r ^{4}
 When two tubes of length l_{1} , l_{2} and radii r_{1} r_{2} are connected in series across a pressure difference P,
Then P = P_{1}+P_{2} ……(i)
Where P_{1} and P_{2} are the pressure difference across the first and second tube respectively
 The volume of liquid flowing through both the tubes e. rate of flow of liquid is same. Therefore V = V_{1} = V_{2}
pP r ^{4} pP r ^{4}
i.e.,
V = 1 1 = 2 2
…..(ii)
8hl1 8hl2
Substituting the value of P_{1} and P_{2} from equation (ii) to equation (i) we get
P = P + P
= V é 8hl1
+ 8hl 2 ù
1 2 ê pr 4
pr ^{4} ú
ëê 1 2 úû
\ V =
P
é 8hl1 + 8hl2 ù
= P =
R1 + R2
P
Reff
Where R_{1} and R_{2} are the liquid resistance in tubes
ê pr ^{4} pr ^{4} ú
ë 1 2 û
 Effective liquid resistance in series combination
(2) Parallel combination of tubes
 P = P_{1} = P_{2}
Reff
= R1 + R2
Ppr ^{4} Ppr ^{4} é pr ^{4} pr ^{4} ù
(ii) V= V_{1}+V_{2} = ^{1} + ^{2} = P ê 1 + ^{ } ^{2} ú
8hl1 8hl2 ë 8hl1 8hl2 û
\ V = P é 1 + 1 ù = P


ê R1
R2 ú
Reff
 Effective liquid resistance in parallel combination
1 = 1 + 1
or R
= R1 R2
Reff
R1 R2
eff
R1 + R2
Problem 54. The rate of steady volume flow of water through a capillary tube of length ‘l‘ and radius ‘r‘ under a pressure difference of P is V. This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P)
[EAMCET (Engg.) 2003]
 V
16
V (c)
17
16V (d)
17
17V
16
Solution : (b) Rate of flow of liquid V = P
R
where liquid resistance R = 8hl
pr ^{4}
For another tube liquid resistance R‘ = 8hl = 8hl .16 = 16R
æ r ö^{4}
p ç 2 ÷
pr ^{4}
è ø
For the series combination VNew = P = P = P
= V .
R + R‘
R + 16R
17R 17
Problem 55. A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is
[EAMCET 2001]
 P (b) 3P

4
(c)
P (d) P
2 4
Solution : (d) From V =
2 ´ 2 ´ ç ÷ =
Þ P2 = ^{1} = .
Problem 56. Two capillary tubes of same radius r but of lengths l_{1} and l_{2} are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the length of a single tube that can replace the two tubes so that the rate of flow is same as before
(a)
l + l
1 + 1
l1l 2
 1
1 2 l1 l2
l1 + l 2
l1 + l 2
Solution : (c) For parallel combination 1 = 1 + 1
Þ pr ^{4}
= pr ^{4}
+ pr ^{4}
Þ 1 = 1 + 1
\ l =
l1l 2
Reff
R1 R2
8hl
8hl1
8hl2
l l1 l 2
l1 + l 2
Problem 57. We have two (narrow) capillary tubes T_{1} and T_{2}. Their lengths are l_{1} and l_{2} and radii of crosssection are r_{1} and r_{2} respectively. The rate of flow of water under a pressure difference P through tube T_{1} is 8cm^{3}/sec. If l_{1} = 2l_{2} and r_{1} =r_{2}, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)
 4 cm^{3}/sec (b) (16/3) cm^{3}/sec (c) (8/17) cm^{3}/sec (d) None of these
Solution : (b)
V = p Pr ^{4}
8hl
= 8cm^{3}
sec

For composite tube V =
Ppr ^{4}
= 2 pPr ^{4}
= 2 ´ 8 = 16 cm^{3}
é l = l = 2l
or l
= l ù

8hæl +
è
l ö

÷
ø
3 8hl 3
Q 1

 sec ë
2 2 2 úû
Problem 58. A capillary tube is attached horizontally to a constant head arrangement. If the radius of the capillary tube is increased by 10% then the rate of flow of liquid will change nearly by
(a) + 10% (b) + 46% (c) – 10% (d) – 40%
Ppr ^{4}
V æ r ö4
æ 110 ö^{4}
Solution : (b) V =
Þ 2 = ç 2 ÷
Þ V_{2} = V_{1}ç ÷ = V_{1}(1.1)^{4}
= 1.4641V
8hl
V1 ç r1 ÷
è 100 ø
è ø
DV = V_{2} – V_{1} = 1.4641V – V = 0.46 or
46% .
V V V