NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Physics
Chapter : 10
Chapter Name : Mechanical Properties of Fluids

Q10.1 Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea
level, though the height of the atmosphere is more than 100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Answer. (a) The pressure Of a liquid is given by the relation:
p = hpg
where,
p = Pressure
h = Height of the liquid column
p = Density of the liquid
g = Acceleration due to the gravity
It can be inferred that pressure is directly proportional to height. Hence, the blood
pressure in human vessels depends on the height Of the blood column in the body. The
height Of the blood column is more at the feet than it is at the brain. Hence, the blood
pressure at the feet is more than it is at the brain.
(b) Density of air is the maximum near the sea level. Density of air decreases with
increase in height from the surface. At a height of about 6 km, density decreases to
nearly half of its value at the sea level. Atmospheric pressure is proportional to density.
Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level.
(c) When force is applied on a liquid, the pressure In the liquid is transmitted in all
directions. Hence, hydrostatic pressure does not have a xed direction and it is a scalar
physical quantity.

Page : 272 , Block Name : Exercise

Q10.2 Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to
form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape

Answer. (a) The angle between the tangent to the liquid surface at the point of contact and the
surface inside the liquid is called the angle of contact (B), as shown in the given gure.

S lat S sat and S sl are the respective interfacial tensions between the liquid-air, solid-air,

and solid-liquid interfaces. At the line of contact, the surface forces between the three

media must be in equilibrium, i.e.,
Sa −Ssl
cos θ =
Sla

The angle of contact θ, is obtuse if Ssa < S1a ( as in the case of mercury on glass). This

angle is acute if Ssl < Sla ( as in the case of water on glass).

(b) Mercury molecules (which make an obtuse angle with glass) have a strong force Of
attraction between themselves and a weak force of attraction toward solids. Hence, they
tend to form drops.
On the other hand, water molecules make acute angles with glass. They have a weak
force of attraction between themselves and a strong force of attraction toward solids.
Hence, they tend to spread out.
(c) Surface tension is the force acting per unit length at the interface between the plane
Of a liquid and any other surface. This force is independent Of the area Of the liquid
surface. Hence, surface tension is also independent Of the area Of the liquid surface.
(d) Water with detergent dissolved in it has small angles of contact (9). This is because
for a small 9, there is a fast capillary rise of the detergent in the cloth. The capillary rise
of a liquid is directly proportional to the cosine of the angle of contact (B). If θ is small,
then cos θ will be large and the rise of the detergent water in the cloth will be fast.
(e) A liquid tends to acquire the minimum surface area because of the presence of
surface tension. The surface area Of a sphere is the minimum for a given volume. Hence,
under no external forces, liquid drops always take spherical shape.

Page : 272 , Block Name : Exercise

Q10.3 Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally ... with temperatures (increases / decreases)
(b) Viscosity of gases ... with temperature, whereas viscosity of liquids ... with temperature
(increases / decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to ... , while for
uids it is proportional to ... (shear strain / rate of shear strain)
(d) For a uid in a steady ow, the increase in ow speed at a constriction follows (conservation of

mass / Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an
actual plane (greater / smaller)

Answer. (a) decreases
The surface tension of a liquid is inversely proportional to temperature.
(b) increases; decreases
Most uids Offer resistance to their motion. This is like internal mechanical friction,
known as viscosity. Viscosity Of gases increases with temperature, while viscosity Of
liquids decreases with temperature.
(c) Shear strain; Rate of shear strain
With reference to the elastic modulus of rigidity for solids, the shearing force is
proportional to the shear strain. With reference to the elastic modulus of rigidity for
uids, the shearing force is proportional to the rate of shear strain.

Page : 272 , Block Name : Exercise

Q10.4 Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our ngers, fast jets of water gush through the openings
between our ngers
(c) The size of the needle of a syringe controls ow rate better than the thumb pressure exerted by
a doctor while administering an injection
(d) A uid owing out of a small hole in a vessel results in a backward thrust on the vessel
(e) A spinning cricket ball in air does not follow a parabolic trajectory

Answer. (a) When air is blown under a paper, the veloc ty Of air is greater under the paper than it is
above it. As per Bernoulli's principle, atmospheric pressure reduces under the paper. This makes
the paper fall. To keep a piece of paper horizontal, one should blow over it, This increases the
velocity of air above the paper. As per Bernoulli's principle, atmospheric pressure reduces above the
paper and the paper remains horizontal.
(b) According to the equation of continuity: Area x Velocity = Constant
For a smaller opening, the velocity Of ow Of a uid is greater than it is when the
opening is bigger. When we try to close a tap Of water with our ngers, fast jets Of water gush
through the openings between our ngers, This is because very small openings are left for the
water to ow out of the pipe. Hence, area and velocity are inversely proportional to each other.
(c) The small opening Of a syringe needle controls the velocity Of the blood owing Out.
This is because Of the equation Of continuity. At the constriction point Of the syringe
system, the ow rate suddenly increases to a high value for a constant thumb pressure
applied.
(d) When a uid ows out from a small hole in a vessel, the vessel receives a backward
thrust. A uid owing out from a small hole has a large velocity according to the
equation of continuity : Area x Velocity Constant According to the law of conservation of
momentum, the vessel attains a backward velocity because there are no external forces acting on
the system.
(e) A spinning cricket ball has two simultaneous motions — rotary and linear. These
two types of motion oppose the effect of each other. This decreases the velocity of air

owing below the ball. Hence, the pressure on the upper side of the ball becomes lesser
than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes
a curved path. It does not follow a parabolic path.

Page : 272 , Block Name : Exercise

Q10.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a
diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal oor ?

Mass of the girl, m = 50kg

Diameter of the heel, d = 1cm = 0.01m

d
Radius of the heel, r = = 0.005m
2
2
Area of the heel = πr

2
= π(0.005)

−5 2
= 7.85 × 10 m

Force exerted by the heel on the floor:

F = mg

= 50 × 9.8

= 490N

Pressure exerted by the heel on the floor:

Force
P =
Area
490
=
−5
7.85×10

6 −2
= 6.24 × 10 Nm

Therefore, the pressure exerted by the heel on the horizontal floor is

6 −2
6.24 × 10 Nm

Page : 272 , Block Name : Exercise

Q10.6 Torricelli's barometer used mercury. Pascal duplicated it using French wine of density 984 kg
m . Determine the height of the wine column for normal atmospheric pressure.
–3

10.5m

3 3
Density of mercury, ρ1 = 13.6 × 10 kg/m

3
Height of the mercury column, h1 = 0.76m

3
Density of French wine, ρ2 = 984kg/m

Height of the French wine column = h2

2
Acceleration due to gravity, g = 9.8m/s

The pressure in both the columns is equal, i.e.,

Pressure in the mercury column = Pressure in the French wine column

ρ 1 h1 g = ρ 2 h2 g

ρ1 h1
h2 =
ρ2

3
13.6×10 ×0.76
=
984

= 10.5m

Hence, the height of the French wine column for normal atmospheric pressure is 10.5m.

Page : 273 , Block Name : Exercise

Q10.7 A vertical offshore structure is built to withstand a maximum stress of 109 Pa. Is the
structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to
be roughly 3 km, and ignore ocean currents.

Yes

9
The maximum allowable stress for the structure, P = 10 Pa

3
Depth of the ocean, d = 3km = 3 × 10 m

3 3
Density of water, ρ = 10 kg/m

2
Acceleration due to gravity, g = 9.8m/s

The pressure exerted because of the sea water at depth, d = ρdg
3 3
= 3 × 10 × 10 × 9.8

7
= 2.94 × 10 Pa

9
The maximum allowable stress for the structure (10 Pa) is greater than the pressure of

7
the sea water (2.94 × 10 Pa). The pressure exerted by the ocean is less than the

pressure that the structure can withstand. Hence, the structure is suitable for putting up

on top of an oil well in the ocean.

Page : 273 , Block Name : Exercise

Q10.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area
of cross-section of the piston carrying the load is 425 cm . What maximum pressure would the
2

smaller piston have to bear ?

The maximum mass of a car that can be lifted, m = 3000kg

2 −4 2
Area of cross-section of the load-carrying piston, A = 425cm = 425 × 10 m

The maximum force exerted by the load, F = mg

= 3000 × 9.8

= 29400N

F
The maximum pressure exerted on the load-carrying piston, P =
A

29400
=
−4
425×10

5
= 6.917 × 10 Pa

Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum

5
pressure that the smaller piston would have to bear is 6.917 × 10 Pa.

Page : 273 , Block Name : Exercise

Q10.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns
in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other.
What is the speci c gravity of spirit ?

Answer. The given system of water, mercury, and methylated spirit is +10wn as follows:

Height of the spirit column, h1 = 12.5cm = 0.125m

Height of the water column, h2 = 10cm = 0.1m

P0 = Atmospheric pressure

ρ1 = Density of spirit

ρ2 = Density of water

Pressure at point B = P0 + h1 ρ1 g

Pressure at point D = P0 + h2 ρ2 g

Pressure at points B and D is the same.

P0 + h 1 ρ 1 g = h 2 ρ 2 g

ρ1 h2
=
ρ2 h1

ρ1 h2
=
ρ2 h1

10
= = 0.8
12.5

Therefore, the speci c gravity of spirit is 0.8.
Page : 273 , Block Name : Exercise

Q10.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into the
respective arms of the tube, what is the difference in the levels of mercury in the two arms ?
(Speci c gravity of mercury = 13.6)

Height of the water column, h1 = 10 + 15 = 25cm

Height of the spirit column, h2 = 12.5 + 15 = 27.5cm

−3
Density of water, ρ1 = 1gcm

−3
Density of spirit, ρ2 = 0.8gcm

−3
Density of mercury = 13.6gcm

Let h be the difference between the levels of mercury in the two arms

Pressure exerted by height h, of the mercury column:

= hρg

= h × 13.6g … (i)

Difference between the pressures exerted by water and spirit:

= h1 ρ 1 g − h1 ρ 1 g

= g(25 × 1 − 27.5 × 0.8)

= 3g … (ii)

Equating equations (i) and (ii), we get:

13.6hg = 3g

h = 0.220588 ≈ 0.221cm

Hence, the difference between the levels of mercury in the two arms is 0.221cm

Page : 273 , Block Name : Exercise

Q10.11 Can Bernoulli’s equation be used to describe the ow of water through a rapid in a river ?
Explain.

Answer. NO
Bernoulli's equation cannot be used to describe the ow Of water through a rapid in a
river because Of the turbulent ow Of water. This principle can only be applied to a
streamline ow.

Page : 273 , Block Name : Exercise

Q10.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s
equation ? Explain

Answer. No
It does not matter if one uses gauge pressure instead of absolute pressure while
applying Bernoulli's equation. The two points where Bernoulli's equation is applied should have
signi cantly different atmospheric pressures.

Page : 273 , Block Name : Exercise

Q10.13 Glycerine ows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the
amount of glycerine collected per second at one end is 4.0 × 10 kg s , what is the pressure
–3 –1

difference between the two ends of the tube ? (Density of glycerine = 1.3 × 103 kg m and viscosity
–3

of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar ow in the tube is
correct].

2
9.8 × 10 Pa

Length of the horizontal tube, l = 1.5m

Radius of the tube, r = 1cm = 0.01m

Diameter of the tube, d = 2r = 0.02m

−3 −1
Glycerine is flowing at a rate of 4.0 × 10 kgs

−3 −1
M = 4.0 × 10 kgs
3 −3
Density of glycerine, ρ = 1.3 × 10 kgm

viscosity of glycerine, η = 0.83Pa s

Volume of glycerine flowing per sec:

M
V =
ρ

−3
4.0×10
= 3
1.3×10

−6 3 −1
= 3.08 × 10 m s

According to Poiseville's formula, we have the relation for the rate of flow:
4
πpr
V =
8nl

Where, p is the pressure difference between the two ends of the tube

V 8nl
∴ p =
4
πr
−6
3.08 × 10 × 8 × 0.83 × 1.5
=
4
π × (0.01)

2
= 9.8 × 10 Pa

Reynolds' number is given by the relation:
4ρV
R =
πdη

3 −6
4×1.3×10 ×3.08×10
= = 0.3
π×(0.02)×0.83

Reynolds' number is about 0.3. Hence, the flow is laminar.

Page : 273 , Block Name : Exercise

Q10.14 In a test experiment on a model aeroplane in a wind tunnel, the ow speeds on the upper
and lower surfaces of the wing are 70 m s and 63 m s respectively. What is the lift on the wing if
–1 –1

its area is 2.5 m ? Take the density of air to be 1.3 kg m .
2 –3

Speed of wind on the upper surface of the wing, V1 = 70m/s

Speed of wind on the lower surface of the wing, V2 = 63m/s

2
Area of the wing, A = 2.5m

−3
Density of air, ρ = 1.3kgm

According to Bernoulli's theorem, we have the relation:

1 2 1 2
P1 + ρV = P2 + ρV
2 1 2 2

1 2 2
P2 − P1 = ρ (V − V )
2 1 2

Where,

P1 = Pressure on the upper surface of the wing

P2 = Pressure on the lower surface of the wing

The pressure difference between the upper and lower surfaces of the wing provides lift to

the aeroplane.

Lift on the wing = (P2 − P1 ) A

1 2 2
= ρ (V − V )A
2 1 2

1 2 2
= 1.3 ((70) − (63) ) × 2.5
2

= 1512.87

3
= 1.51 × 10 N

3
Therefore, the lift on the wing of the aeroplane is 1.51 × 10 N

Page : 273 , Block Name : Exercise

Q10.15 Figures 10.23(a) and (b) refer to the steady ow of a (non-viscous) liquid. Which of the two
gures is incorrect ? Why ?

(a)

Take the case given in figure (b).

Where,

A1 = Area of pipel

A2 = Area of pipe 2

V2 = Speed of the fluid in pipe 2

V2 = Speed of the fluid in pipe 2

From the law of continuity, we have:

A1 V 1 = A2 V 2

When the area of cross-section in the middle of the venturimeter is small, the speed of

the flow of liquid through this part is more. According to Bernoulli's principle, if speed is

more, then pressure is less.

Pressure is directly proportional to height. Hence, the level of water in pipe 2 is less.

Therefore, figure (a) is not possible.

Page : 273 , Block Name : Exercise

Q10.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm one end of which has 40 2

ne holes each of diameter 1.0 mm. If the liquid ow inside the tube is 1.5 m min , what is the –1

speed of ejection of the liquid through the holes ?

2 −4 2
Area of cross-section of the spray pump, A1 = 8cm = 8 × 10 m

Number of holes, n = 40

−3
Diameter of each hole, d = 1mm = 1 × 10 m

−3
Radius of each hole, r = d/2 = 0.5 × 10 m

2
2 −3 2
Area of cross-section of each hole, a = πr = π(0.5 × 10 ) m

Total area of 40 holes, A2 = n × a

2
−3 2
= 40 × π(0.5 × 10 ) m

−6 2
= 31.41 × 10 m

Speed of flow of liquid inside the tube, V1 = 1.5m/min = 0.025m/s

Speed of ejection of liquid through the holes = V2

According to the law of continuity, we have:

A1 V 1 = A2 V 2

A1 V1
V2 =
A2

−
8×10 ×0.025
=
−6
31.61×10

= 0.633m/s

Therefore, the speed Of ejection Of the liquid through the holes is O. 633 m/s.

Page : 273 , Block Name : Exercise

Q10.17 A U-shaped wire is dipped in a soap solution, and removed. The thin soap lm formed
between the wire and the light slider supports a weight of 1.5 × 10 N (which includes the small
–2

weight of the slider). The length of the slider is 30 cm. What is the surface tension of the lm ?

−2
The weight that the soap film supports, W = 1.5 × 10 N

Length of the slider, I = 30cm = 0.3m

A soap film has two free surfaces.

∴ Total length = 2l = 2 × 0.3 = 0.6m
Force or Weight
S =
2l

Surface tension,

−2
1.5×10 −2
= = 2.5 × 10 N/m
0.6

−2 −1
Therefore, the surface tension of the film is 2.5 × 10 Nm

Page : 273 , Block Name : Exercise

Q10.18 Figure 10.24 (a) shows a thin liquid lm supporting a small weight = 4.5 × 10 N. What is
–2

the weight supported by a lm of the same liquid at the same temperature in Fig. (b) and (c) ?
Explain your answer physically.

Take case ( a ):

The length of the liquid film supported by the weight, I = 40cm = 0.4cm

−2
The weight supported by the film, W = 4.5 × 10 N

A liquid film has two free surfaces.
W
∴ Surface tension =
2l

−2
4.5×10 −2 −1
= = 5.625 × 10 Nm
2×0.4

In all the three figures, the liquid is the same. Temperature is also the same for each

case. Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a),

−2 −1
i.e., 5.625 × 10 Nm .

since the length of the film in all the cases is 40cm, the weight supported in each case

−2
is 4.5 × 10 N.

Page : 274 , Block Name : Exercise

Q10.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ?
Surface tension of mercury at that temperature (20 °C) is 4.65 × 10 N m . The atmospheric
–1 –1

pressure is 1.01 × 10 Pa. Also give the excess pressure inside the drop.
5

5
1.01 × 10 Pa; 310Pa

−3
Radius of the mercury drop, r = 3.00mm = 3 × 10 m

−1 −1
Surface tension of mercury, S = 4.65 × 10 Nm

5
Atmospheric pressure, P0 = 1.01 × 10 Pa

Total pressure inside the mercury drop

= Excess pressure inside mercury + Atmospheric pressure
2S
= + P0
r

−1
2×4.65×10 5
= −3
+ 1.01 × 10
3×10

5
= 1.0131 × 10

5
= 1.01 × 10 Pa

2S
Excess pressure =
r

Page : 272 , Block Name : Exercise

Q10.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that
the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10 N m ? If an air –2 –1

bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the
soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1
atmospheric pressure is 1.01 × 10 Pa) 5

Excess pressure inside the soap bubble is 20Pa ;

5
Pressure inside the air bubble is 1.06 × 10 Pa

−3
Soap bubble is of radius, r = 5.00mm = 5 × 10 m

−2 −1
Surface tension of the soap solution, S = 2.50 × 10 Nm

−2 −1
Relative density of the soap solution = 1.20 × 10 Nm
3 3
∴ Density of the soap solution, ρ = 1.2 × 10 kg/m

Air bubble formed at a depth, h = 40cm = 0.4m

−3
Radius of the air bubble, r = 5mm = 5 × 10 m

5
1 atmospheric pressure = 1.01 × 10 Pa

2
Acceleration due to gravity, g = 9.8m/s

Hence, the excess pressure inside the soap bubble is given by the relation:

4S
P =
r
−2
4 × 2.5 × 10
=
−3
5 × 10

= 20Pa

Therefore, the excess pressure inside the soap bubble is 20 Pa.

The excess pressure inside the air bubble is given by the relation:

2S
′
P =
r
−2
2 × 2.5 × 10
=
−3
5 × 10

= 10Pa

Therefore, the excess pressure inside the air bubble is 10 Pa.

At a depth of 0.4m, the total pressure inside the air bubble

′
= Atmospheric pressure + hρg + P

5 3
= 1.01 × 10 + 0.4 × 1.2 × 10 × 9.8 + 10

5
= 1.057 × 10 Pa

5
= 1.06 × 10 Pa

Therefore, the pressure inside the air bubble is 1.06 × 10 Pa 5

Page : 274 , Block Name : Exercise

Q10.21 A tank with a square base of area 1.0 m is divided by a vertical partition in the middle. The
2

bottom of the partition has a small-hinged door of area 20 cm . The tank is lled with water in one
2

compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute
the force necessary to keep the door close.

2
Base area of the given tank, A = 1.0m

2 −4 2
Area of the hinged door, a = 20cm = 20 × 10 m

3 3
Density of water, ρ1 = 10 kg/m

3 3
Density of acid, ρ2 = 1.7 × 10 kg/m

Height of the water column, h1 = 4m

Height of the acid column, h2 = 4m

Acceleration due to gravity, g = 9.8

Pressure due to water is given as:

P1 = h 1 ρ 1 g
3
= 4 × 10 × 9.8

4
= 3.92 × 10 Pa

Pressure due to acid is given as:

P2 = h 2 ρ 2 g

3
= 4 × 1.7 × 10 × 9.8

4
= 6.664 × 10 Pa

Pressure difference between the water and acid columns:

ΔP = P2 − P1
4 4
= 6.664 × 10 − 3.92 × 10
4
= 2.744 × 10 Pa

Hence, the force exerted on the door = ΔP × a

4 −4
=2.744 × 10 × 20 × 10

=54.88N

=54.88N

Therefore, the force necessary to keep the door closed is 54.88N

Page : 274 , Block Name : Additional Exercises

Q10.22 A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a
pump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in the
manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of
cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are
poured into the right limb of the manometer ? (Ignore the small change in the volume of the gas).

Answer. (a) 96 cm of Hg & 20 cm of Hg; 58 cm of Hg & —18 cm of Hg
(b) 19 cm
(a) For fiaure (a)

Atmospheric pressure, P0 = 76cm of Hg

Difference between the levels of mercury in the two limbs gives gauge pressure

Hence, gauge pressure is 20cm of Hg.

Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 + 20 = 96cm of Hg

For figure(b)

Difference between the levels of mercury in the two limbs = −18cm

Hence, gauge pressure is − 18cm of Hg.

Absolute pressure = Atmospheric pressure + Gauge pressure

= 76cm − 18cm = 58cm

(b) 13.6cm of water is poured into the right limb of figure (b)

Relative density of mercury = 13.6

Hence, a column of 13.6cmcm of water is equivalent to 1cm of mercury.

Let h be the difference between the levels of mercury in the two limbs.

The pressure in the right limb is given as:

PR = Atmospheric pressure + 1cm of Hg

= 76 + 1 = 77cm of Hg … (i)

The mercury column will rise in the left limb.

Hence, pressure in the left limb, PL = 58 + h

Equating equations (i) and (ii), we get:

77 = 58 + h

∴ h = 19cm

Hence, the difference between the levels of mercury in the two limbs will be 19cm.

Page : 274 , Block Name : Additional Exercises

Q10.23 Two vessels have the same base area but different shapes. The rst vessel takes twice the
volume of water that the second vessel requires to ll upto a particular common height. Is the force
exerted by the water on the base of the vessel the same in the two cases ? If so, why do the vessels
lled with water to that same height give different readings on a weighing scale ?

Answer. yes
Two vessels having the same base area have identical force and equal pressure acting
on their common base area. Since the shapes of the two vessels are different, the force
exerted on the sides of the vessels has non-zero vertical components. When these
vertical components are added, the total force on one vessel comes out to be greater
than that on the vessel. Hence, when these vessels are lled with to the
same height, they give different readings

Page : 274 , Block Name : Additional Exercises

Q10.24 During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000
Pa. At what height must the blood container be placed so that blood may just enter the vein ? [Use
the density of whole blood from Table 10.1]

Gauge pressure, P = 2000Pa

3 −3
Density of whole blood, ρ = 1.06 × 10 kgm

2
Acceleration due to gravity, g = 9.8m/s

Height of the blood container = h

Pressure of the blood container, P = hpg

P
∴ h =
ρg

2000
=
3
1.06 × 10 × 9.8

= 0.1925m

The blood may enter the vein if the blood container is kept at a height greater than

0.1925m, i.e., about 0.2m.

Page : 275 , Block Name : Additional Exercises

Q10.25 In deriving Bernoulli’s equation, we equated the work done on the uid in the tube to its
change in the potential and kinetic energy. (a) What is the largest average velocity of blood ow in
an artery of diameter 2 × 10 m if the ow must remain laminar ? (b) Do the dissipative forces
–3

become more important as the uid velocity increases ? Discuss qualitatively.

(a) 1.966m/s (b) Yes

−3
(a) Diameter of the artery, d = 2 × 10 m

−3
Viscosity of blood, η = 2.084 × 10 Pas

3 3
Density of blood, ρ = 1.06 × 10 kg/m

Reynolds' number for laminar flow, NR = 2000

The largest average velocity of blood is given as:
NR η
Varg =
ρd

−3
2000×2.084×10
=
3 −3
1.06×10 ×2×10

= 1.966m/s

Therefore, the largest average velocity of blood is 1.966 m/s.
(b) As the uid velocity increases, the dissipative forces become more important. This is
because of the rise of turbulence. Turbulent ow causes dissipative loss in a uid.

Page : 275 , Block Name : Additional Exercises

Q10.26 (a) What is the largest average velocity of blood ow in an artery of radius 2× 10 m if the
–3

ow must remain lanimar? (b) What is the corresponding ow rate ? (Take viscosity of blood to be
2.084 ×10 Pa s)
–3

−3
(a)Radius of the artery, r = 2 × 10 m

−3 −3
Diameter of the artery, d = 2 × 2 × 10 m = 4 × 10 m

−3
Viscosity of blood, η = 2.084 × 10 Pa s

3 3
Density of blood, ρ = 1.06 × 10 kg/m

Reynolds' number for laminar flow, NR = 2000

The largest average velocity of blood is given by the relation

NR η
V arg =
ρd

−3
2000×2.084×10
= 3 −3
1.06×10 ×4×10

= 0.983m/s

Therefore, the largest average velocity Of blood is 0.983 m/s.
(b) Flow rate is given by the relation:

2
R = πr V arg

2
−3
= 3.14 × (2 × 10 ) × 0.983

−5 3 −1
= 1.235 × 10 m s

−5 3 −1
Therefore, the corresponding flow rate is 1.235 × 10 m s

Page : 275 , Block Name : Additional Exercises

Q10.27 A plane is in level ight at constant speed and each of its two wings has an area of 25 m . If
2

the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface,
determine the plane’s mass. (Take air density to be 1 kg m ). –3

2
The area of the wings of the plane, A = 2 × 25 = 50m

Speed of air over the lower wing, V1 = 180km/h = 50m/s

Speed of air over the upper wing, V2 = 234km/h = 65m/s

−3
Density of air, ρ = 1kgm

Pressure of air over the lower wing = P1

Pressure of air over the lower wing = P2

The upward force on the plane can be obtained using Bernoull's equation as:

1 2 1 2
P1 + ρV = P2 + ρV
2 1 2 2

1 2 2
P1 − P2 = ρ (V − V )
2 2 1

The upward force (F) on the plane can be calculated as:

(P1 − P2 ) A

1 2 2
= ρ (V − V )A Using equation (i)
2 2 1

1 2 2
= × 1 × ((65) − (50) ) × 50
2

= 43125N

Using Newton's force equation, we can obtain the mass (m) of the plane as:

F = mg

43125
∴ m =
9.8

= 4400.51kg

∼ 4400kg

Hence, the mass of the plane is about 4400kg.

Page : 275 , Block Name : Additional Exercises

Q10.28 In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius
2.0 × 10 m and density 1.2 × 10 kg m . Take the viscosity of air at the temperature of the
–5 3 –3

experiment to be 1.8 × 10 Pa s. How much is the viscous force on the drop at that speed ? Neglect
5

buoyancy of the drop due to air

−10
Terminal speed = 5.8cm/s; Viscous force = 3.9 × 10 N

−5
Radius of the given uncharged drop, r = 2.0 × 10 m

3 −3
Density of the uncharged drop, ρ = 1.2 × 10 kgm

−5
viscosity of air, η = 1.8 × 10 Pa s

Density of air (ρo ) can be taken as zero in order to neglect buoyancy of air.

2
Acceleration due to gravity, g = 9.8m/s

Terminal velocity (v) is given by the relation:
2
2r × (ρ − ρ0 ) g
v =
9η

2
−5 3
2 × (2.0 × 10 ) (1.2 × 10 − 0) × 9.8
=
−3
9 × 1.8 × 10
−2 −1
= 5.807 × 10 ms
−1
= 5.8cms
−1
Hence, the terminal speed of the drop is 5.8cms .

The viscous force on the drop is given by:

F = 6πηrv

−5 −5 −2
∴ F = 6 × 3.14 × 1.8 × 10 × 2.0 × 10 × 5.8 × 10

−10
= 3.9 × 10 N

−10
Hence, the viscous force on the drop is 3.9 × 10 N .

Page : 275 , Block Name : Additional Exercises

Q10.29 Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius
1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the
mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at
the temperature of the experiment is 0.465 N m . Density of mercury = 13.6 × 10 kg m .
–1 3 –3

∘
Angle of contact between mercury and soda lime glass, θ = 140

−3
Radius of the narrow tube, r = 1mm = 1 × 10 m

−1
Surface tension of mercury at the given temperature, s = 0.465Nm

3 3
Density of mercury, ρ = 13.6 × 10 kg/m

Dip in the height of mercury = ℏ

2
Acceleration due to gravity, g = 9.8m/s

Surface tension is related with the angle of contact and the dip in the height as:

hρgr
s =
2 cos θ

2s cos θ
∴ h =
rρg

2s cos θ
∴ h =
rρg

2 × 0.465 × cos 140
=
−3 3
1 × 10 × 13.6 × 10 × 9.8

= −0.00534m

= −5.34mm

Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by
5.34 mm.

Page : 275 , Block Name : Additional Exercises

Q10.30 Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube
open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs
of the tube ? Surface tension of water at the temperature of the experiment is 7.3 × 10 N m . –2 –1

Take the angle of contact to be zero and density of water to be 1.0 × 10 kg m (g = 9.8 m s ) .
3 –3 –2

−3
Diameter of the first bore, d1 = 3.0mm = 3 × 10 m

d1 −3
Hence, the radius of the first bore, r1 = = 1.5 × 10 m
2

Diameter of the second bore, d2 = 6.0mm

d2 −3
^
Hence, the radius of the second bore, r2 = = 3 × 10 m
2
−2 −1
Surface tension of water, s = 7.3 × 10 Nm

Angle of contact between the bore surface and water, θ = 0

3 −3
Density of water, ρ = 1.0 × 10 kg/m

2
Acceleration due to gravity, g = 9.8m/s

Let h1 and h2 be the heights to which water rises in the first and second tubes

respectively. These heights are given by the relations:

2s cos θ
h1 =
r1 ρg

2x cos θ
h2 =
r2 ρg

The difference between the levels of water in the two limbs of the tube can be calculated

as:

2s cos θ 2s cos θ
= −
r1 ρg r2 ρg

2s cos θ 1 1
= [ − ]
ρg r1 r2

2s cos θ 1 1
= [ − ]
ρg ri r2
−2
2 × 7.3 × 10 × 1 1 1
= [ − ]
3 −3 −3
1 × 10 × 9.8 1.5 × 10 3 × 10
−3
= 4.966 × 10 m

= 4.97mm

Hence, the difference between levels of water in the two bores is 4.97 mm.

Page : 275 , Block Name : Additional Exercises

Q10.31 (a) It is known that density ρ of air decreases with height y as 0 y/yo e − ρ = ρ where ρ0 =
1.25 kg m is the density at sea level, and y0 is a constant. This density variation is called the law
–3

of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant
(isothermal conditions). Also assume that the value of g remains constant. (b) A large He balloon of
volume 1425 m is used to lift a payload of 400 kg. Assume that the balloon maintains constant
3

radius as it rises. How high does it rise ? [Take y0 = 8000 m and ρHe = 0.18 kg m ]. –3

3
(a) Volume of the balloon, V = 1425m

Mass of the payload, m = 400kg

2
Acceleration due to gravity, g = 9.8m/s

y0 = 8000m

−3
ρlk = 0.18kgm

3
ρb = 1.25kg/m

Density of the balloon = ρ

Helght to which the balloon rises = y

Density (ρ) of air decreases with height (y) as:

−y/γ0
ρ = ρ0 e
ρ
−yf0
= e
ρ0

This density variation is called the law of atmospherics.

It can be inferred from equation (i) that the rate of decrease of density with height is

directly proportional to ρ, i.e.,
dρ
− ∝ ρ
dy

dρ
= −kρ
dy

dρ
= −kdy
ρ

Where, k is the constant of proportionality

Height changes from 0 to y , while density changes from ρ oto ρ.

Integrating the sides between these limits, we get:
ρ dρ y
∫ = −∫ kdy
μ ρ 0

p
[log ρ] = −ky
c nn

ρ
loge = −ky
ρ0

ρ
−ky
= e
ρ0

Comparing equations (i) and (ii), we get:

1
y0 =
k

1
k =
y0

From equations (i) and (iii), we get:

−y/y0
ρ = ρ0 e

(b)
Mass
Density ρ =
Volume

Mass of the payload + Mass of helium
=
Volume

m+V ρ he
=
V

400+1425×0.18
=
1425

2
= 0.46kg/m

From equations (ij) and (iii), we can obtain y as

−y/yn
ρ = ρ0 e
ρ y
logd = −
p0 y0

0.46
∴ y = −8000 × log
e 1.25

= −8000 × −1

= 8000m = 8km

Hence, the balloon will rise to a height of 8 km.

Page : 275 , Block Name : Additional Exercises