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#### Page No 268:

#### Question 10.1:

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*
= *hρ*g

Where,

*P*
= Pressure

*h*
= Height of the liquid column

*ρ*
= 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 fixed direction and it is a scalar physical quantity.

#### Page No 268:

#### Question 10.2:

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 (*θ*), as shown in the given figure.

*S*_{la},
*S*_{sa}, 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.,

The
angle of contact *θ* ,
is obtuse if *S*_{sa} *< S*_{la} (as in
the case of mercury on glass). This angle is acute if *S*_{sl}*
< S*_{la} (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 (*θ*). This is because for a small *θ*,
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 (*θ*). 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 No 268:

#### Question 10.3:

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 fluids it is proportional to . ..
(shear strain / rate of shear strain)

**(d) **For a
fluid in a steady flow, the increase in flow 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 fluids 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 fluids, the shearing force is proportional to the rate of shear strain.

**(d) **Conservation
of mass/Bernoulli’s principle

For a steady-flowing fluid, an increase in its flow speed at a constriction follows the conservation of mass/Bernoulli’s principle.

**(e) **Greater

For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than it does for an actual plane. This follows from Bernoulli’s principle and different Reynolds’ numbers are associated with the motions of the two planes.

#### Page No 268:

#### Question 10.4:

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 fingers, fast jets of water gush
through the openings between our fingers

**(c)** The
size of the needle of a syringe controls flow rate better than the
thumb pressure exerted by a doctor while administering an injection

**(d)** A fluid
flowing 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 velocity 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 × Velocity = Constant

For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. This is because very small openings are left for the water to flow 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 flowing out. This is because of the equation of
continuity. At the constriction point of the syringe system, the flow
rate suddenly increases to a high value for a constant thumb pressure
applied.

**(d) **When a fluid flows out from a small hole in a vessel, the
vessel receives a backward thrust. A fluid flowing out from a small
hole has a large velocity according to the equation of continuity:

Area × 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 –
rotatory and linear. These two types of motion oppose the effect of
each other. This decreases the velocity of air flowing 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 No 268:

#### Question 10.5:

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 floor?

#### Answer:

Mass of the girl, *m*
= 50 kg

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

Radius of the heel, *r*

Area of the heel

= π (0.005)^{2}

= 7.85 × 10^{–5} m^{2}

Force exerted by the heel on the floor:

*F* = *m*g

= 50 × 9.8

= 490 N

Pressure exerted by the heel on the floor:

= 6.24 × 10^{6}
N m^{–2}

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

6.24 × 10^{6}
Nm^{–2}.

#### Page No 269:

#### Question 10.6:

Toricelli’s
barometer used mercury. Pascal duplicated it using French wine of
density 984 kg m^{–3}.
Determine the height of the wine column for normal atmospheric
pressure.

#### Answer:

10.5 m

Density of mercury, *ρ*_{1}
= 13.6 × 10^{3} kg/m^{3}

Height of the mercury
column, *h*_{1} = 0.76 m

Density of French wine,
*ρ*_{2} = 984 kg/m^{3}

Height of the French
wine column = *h*_{2}

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

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

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

= 10.5 m

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

#### Page No 269:

#### Question 10.7:

A
vertical off-shore structure is built to withstand a maximum stress
of 10^{9}
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.

#### Answer:

**Answer:** Yes

The maximum allowable
stress for the structure, *P* = 10^{9} Pa

Depth of the ocean, *d*
= 3 km = 3 × 10^{3} m

Density of water, *ρ*
= 10^{3} kg/m^{3}

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

The pressure exerted
because of the sea water at depth, *d *= *ρdg*

=
3 × 10^{3} × 10^{3} × 9.8

=
2.94 × 10^{7} Pa

The maximum allowable
stress for the structure (10^{9} Pa) is greater than the
pressure of the sea water (2.94 × 10^{7} 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 No 269:

#### Question 10.8:

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^{2}.
What maximum pressure would the smaller piston have to bear?

#### Answer:

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

Area of cross-section of the load-carrying piston, *A* = 425 cm^{2 }= 425 × 10^{–4} m^{2}

The maximum force exerted by the load, *F = m*g

= 3000 × 9.8

= 29400 N

The maximum pressure exerted on the load-carrying piston,

= 6.917 × 10^{5} Pa

Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 10^{5} Pa.

#### Page No 269:

#### Question 10.9:

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 specific gravity of spirit?

#### Answer:

The given system of water, mercury, and methylated spirit is shown as follows:

Height of the spirit
column, *h*_{1} = 12.5 cm = 0.125 m

Height of the water
column, *h*_{2} = 10 cm = 0.1 m

*P*_{0} =
Atmospheric pressure

*ρ*_{1}
= Density of spirit

*ρ*_{2}
= Density of water

Pressure at point B =

Pressure at point D =

Pressure at points B and D is the same.

Therefore, the specific gravity of spirit is 0.8.

#### Page No 269:

#### Question 10.10:

In problem 10.9, 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? (Specific gravity of mercury = 13.6)

#### Answer:

Height of the water column, *h*_{1} = 10 + 15 = 25 cm

Height of the spirit column, *h*_{2} = 12.5 + 15 = 27.5 cm

Density of water, *ρ*_{1} = 1 g cm^{–3}

Density of spirit, *ρ*_{2} = 0.8 g cm^{–3}

Density of mercury = 13.6 g cm^{–3}

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

Pressure exerted by height *h*, of the mercury column:

= *hρ*g

= *h* × 13.6g … (*i*)

Difference between the pressures exerted by water and spirit:

= g(25 × 1 – 27.5 × 0.8)

= 3g … (*ii*)

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

13.6 *h*g = 3g

*h* = 0.220588 ≈ 0.221 cm

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

#### Page No 269:

#### Question 10.11:

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

#### Answer:

**Answer: **No

Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river because of the turbulent flow of water. This principle can only be applied to a streamline flow.

#### Page No 269:

#### Question 10.12:

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

#### Answer:

**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 significantly different atmospheric pressures.

#### Page No 269:

#### Question 10.13:

Glycerine
flows 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^{–3}
kg s^{–1},
what is the pressure difference between the two ends of the tube?
(Density of glycerine = 1.3 × 10^{3}
kg m^{–3}
and viscosity of glycerine = 0.83 Pa s). [You may also like to check
if the assumption of laminar flow in the tube is correct].

#### Answer:

**Answer:** 9.8 ×
10^{2} Pa

Length of the
horizontal tube, *l* = 1.5 m

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

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

Glycerine is flowing at
a rate of 4.0 × 10^{–3} kg s^{–1}.

*M *= 4.0 ×
10^{–3} kg s^{–1}

Density of glycerine, *ρ*
= 1.3 × 10^{3} kg m^{–3}

Viscosity of glycerine,
*η* = 0.83 Pa s

Volume of glycerine flowing per sec:

= 3.08 ×
10^{–6} m^{3} s^{–1}

According to Poiseville’s formula, we have the relation for the rate of flow:

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

= 9.8 ×
10^{2} Pa

Reynolds’ number is given by the relation:

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

#### Page No 269:

#### Question 10.14:

In
a test experiment on a model aeroplane in a wind tunnel, the flow
speeds on the upper and lower surfaces of the wing are 70 m s^{–1}and
63 m s^{–1}
respectively. What is the lift on the wing if its area is 2.5 m^{2}?
Take the density of air to be 1.3 kg m^{–3}.

#### Answer:

Speed of wind on the
upper surface of the wing, *V*_{1} = 70 m/s

Speed of wind on the
lower surface of the wing, *V*_{2} = 63 m/s

Area of the wing, *A*
= 2.5 m^{2}

Density of air, *ρ*
= 1.3 kg m^{–3}

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

Where,

*P*_{1} =
Pressure on the upper surface of the wing

*P*_{2} =
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

= 1512.87

= 1.51 × 10^{3} N

Therefore, the lift on
the wing of the aeroplane is 1.51 × 10^{3} N.

#### Page No 269:

#### Question 10.15:

Figures 10.23 (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

#### Answer:

**Answer:** (a)

Take the case given in figure (b).

Where,

*A*_{1} =
Area of pipe1

*A*_{2} =
Area of pipe 2

*V*_{1} =
Speed of the fluid in pipe1

*V*_{2} =
Speed of the fluid in pipe 2

From the law of continuity, we have:

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 No 269:

#### Question 10.16:

The
cylindrical tube of a spray pump has a cross-section of 8.0 cm^{2}
one end of which has 40 fine holes each of diameter 1.0 mm. If the
liquid flow inside the tube is 1.5 m min^{–1},
what is the speed of ejection of the liquid through the holes?

#### Answer:

Area of cross-section
of the spray pump, *A*_{1} = 8 cm^{2 }= 8 ×
10^{–4} m^{2}

Number of holes, *n*
= 40

Diameter of each hole,
*d* = 1 mm = 1 × 10^{–3} m

Radius of each hole, *r*
= *d*/2 = 0.5 × 10^{–3} m

Area of cross-section
of each hole, *a *= π*r*^{2
}= π (0.5 ×
10^{–3})^{2} m^{2}

Total area of 40 holes,
*A*_{2 }= *n* × *a*

= 40 × π
(0.5 × 10^{–3})^{2} m^{2}

= 31.41 × 10^{–6} m^{2}

Speed of flow of liquid
inside the tube, *V*_{1 }= 1.5 m/min = 0.025 m/s

Speed of ejection of
liquid through the holes = *V*_{2}

According to the law of continuity, we have:

= 0.633 m/s

Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.

#### Page No 269:

#### Question 10.17:

A
U-shaped wire is dipped in a soap solution, and removed. The thin
soap film formed between the wire and the light slider supports a
weight of 1.5 × 10^{–2}
N (which includes the small weight of the slider). The length of the
slider is 30 cm. What is the surface tension of the film?

#### Answer:

The weight that the
soap film supports, *W* = 1.5 × 10^{–2} N

Length of the slider, *l*
= 30 cm = 0.3 m

A soap film has two free surfaces.

∴Total length =
2*l* = 2 × 0.3 = 0.6 m

Surface tension,

= **
**

Therefore, the surface
tension of the film is 2.5 × 10^{–2} N m^{–1}.

#### Page No 269:

#### Question 10.18:

Figure
10.24 (a) shows a thin liquid film supporting a small weight = 4.5 ×
10^{–2}
N. What is the weight supported by a film of the same liquid at the
same temperature in Fig. (b) and (c)? Explain your answer physically.

#### Answer:

Take case (a):

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

The weight supported by
the film, *W* = 4.5 × 10^{–2} N

A liquid film has two free surfaces.

∴Surface tension

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), i.e., 5.625 × 10^{–2}
N m^{–1}.

Since the length of the
film in all the cases is 40 cm, the weight supported in each case is
4.5 × 10^{–2} N.

#### Page No 270:

#### Question 10.19:

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^{–1} N m^{–1}. The atmospheric pressure is 1.01 × 10^{5} Pa. Also give the excess pressure inside the drop.

#### Answer:

**Answer:
**

Radius
of the mercury drop, *r*
= 3.00 mm = 3 × 10^{–3}
m

Surface
tension of mercury, *S*
= 4.65 × 10^{–1}
N m^{–1}

Atmospheric
pressure, *P*_{0}
= 1.01 × 10^{5}
Pa

Total pressure inside the mercury drop

= Excess pressure inside mercury + Atmospheric pressure

=
1.0131 × 10^{5}

=
1.01 ×10^{5}
Pa

Excess pressure

= 310 Pa

#### Page No 270:

#### Question 10.20:

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^{–2}
N m^{–1}?
If an air 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^{5}
Pa).

#### Answer:

Excess pressure inside the soap bubble is 20 Pa;

Pressure inside the air bubble is

Soap
bubble is of radius, *r*
= 5.00 mm = 5 ×
10^{–3}
m

Surface
tension of the soap solution, *S *=
2.50 ×
10^{–2}
Nm^{–1}

Relative density of the soap solution = 1.20

∴Density
of the soap solution, *ρ*
= 1.2 ×
10^{3}
kg/m^{3 }

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

Radius
of the air bubble, *r*
= 5 mm = 5 ×
10^{–3}
m

1
atmospheric pressure = 1.01 ×
10^{5}
Pa

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

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

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

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

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

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

=
Atmospheric pressure + *h**ρ*g
+ *P*’

Therefore, the pressure inside the air bubble is

#### Page No 270:

#### Question 10.21:

A
tank with a square base of area 1.0 m^{2}
is divided by a vertical partition in the middle. The bottom of the
partition has a small-hinged door of area 20 cm^{2}.
The tank is filled with water in one 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.

#### Answer:

Base area of the given tank, *A* = 1.0 m^{2}

Area of the hinged door, *a* = 20 cm^{2 }= 20 × 10^{–4} m^{2}

Density of water, *ρ*_{1} = 10^{3} kg/m^{3}

Density of acid, *ρ*_{2} = 1.7 × 10^{3} kg/m^{3}

Height of the water column, *h*_{1} = 4 m

Height of the acid column, *h*_{2} = 4 m

Acceleration due to gravity, g = 9.8

Pressure due to water is given as:

Pressure due to acid is given as:

Pressure difference between the water and acid columns:

Hence, the force exerted on the door = Δ*P* × *a*

= 2.744 × 10^{4} × 20 × 10^{–4}

= 54.88 N

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

#### Page No 270:

#### Question 10.22:

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:

**Answer: (a)** 96
cm of Hg & 20 cm of Hg; 58 cm of Hg & –18 cm of Hg

**(b)** 19
cm

**(a)** __For
figure____ (a)__

Atmospheric
pressure, *P*_{0 }= 76 cm of Hg

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

Hence, gauge pressure is 20 cm of Hg.

Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 + 20 = 96 cm of Hg

__For
figure____ (b)__

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

Hence, gauge pressure is –18 cm of Hg.

Absolute pressure = Atmospheric pressure + Gauge pressure

= 76 cm – 18 cm = 58 cm

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

Relative density of mercury = 13.6

Hence, a column of 13.6 cm of water is equivalent to 1 cm 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:

Atmospheric pressure + 1 cm of Hg

= 76 + 1 = 77 cm of Hg … (*i*)

The mercury column will rise in the left limb.

Hence, pressure in the left limb,

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

77
= 58 + *h*

∴*h*
= 19 cm

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

#### Page No 270:

#### Question 10.23:

Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill 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 filled with water to that same height give different readings on a weighing scale?

#### Answer:

**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 other vessel. Hence, when these vessels are filled with water to the same height, they give different readings on a weighing scale.

#### Page No 271:

#### Question 10.24:

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].

#### Answer:

Gauge
pressure, *P*
= 2000 Pa

Density
of whole blood,
*ρ*
= 1.06
×
10^{3
}kg
m^{–3}

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

Height
of the blood container
= *h*

Pressure
of the blood container,
*P*
= *h**ρ*g

The blood may enter the vein if the blood container is kept at a height greater than 0.1925 m, i.e., about 0.2 m.

#### Page No 271:

#### Question 10.25:

In
deriving Bernoulli’s equation, we equated the work done on the
fluid in the tube to its change in the potential and kinetic energy.
(a) What is the largest average velocity of blood flow in an artery
of diameter 2 × 10^{–3}
m if the flow must remain laminar? (b) Do the dissipative forces
become more important as the fluid velocity increases? Discuss
qualitatively.

#### Answer:

**Answer: (a)**
1.966 m/s **(b)** Yes

**(a) **Diameter
of the artery, *d*
= 2
×
10^{–3
}m

Viscosity of blood,

Density
of blood,
*ρ*
=
1.06
×
10^{3}
kg/m^{3}

Reynolds’
number for laminar flow,
*N*_{R}
=
2000

The largest average velocity of blood is given as:

Therefore, the largest average velocity of blood is 1.966 m/s.

**(b) **As the fluid velocity increases, the dissipative forces
become more important. This is because of the rise of turbulence.
Turbulent flow causes dissipative loss in a fluid.

#### Page No 271:

#### Question 10.26:

(a)
What is the largest average velocity of blood flow in an artery of
radius 2 ×
10^{–3}
m if the flow must remain laminar? (b) What is the corresponding flow
rate? (Take viscosity of blood to be 2.084 ×
10^{–3}
Pa s).

#### Answer:

**(a)**Radius
of the artery, *r*
= 2
×
10^{–3
}m

Diameter
of the artery, *d*
= 2
×
2
×
10^{–3
}m
= 4
×
10^{–
3 }m

Viscosity of blood,

Density
of blood,
*ρ*
= 1.06
×
10^{3}
kg/m^{3}

Reynolds’
number for laminar flow, *N*_{R}
= 2000

The largest average velocity of blood is given by the relation:

Therefore, the largest average velocity of blood is 0.983 m/s.

**(b) **Flow rate is
given by the relation:

*R*
= π *r*^{2}

Therefore, the corresponding flow rate is.

#### Page No 271:

#### Question 10.27:

A
plane is in level flight at constant speed and each of its two wings
has an area of 25 m^{2}.
If 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}).

#### Answer:

The
area of the wings of the plane, *A*
= 2 ×
25 = 50 m^{2}

Speed
of air over the lower wing, *V*_{1}
= 180 km/h = 50 m/s

Speed
of air over the upper wing, *V*_{2}
= 234 km/h = 65 m/s

Density
of air, *ρ*
= 1 kg m^{–3}

Pressure
of air over the lower wing = *P*_{1}

Pressure
of air over the upper wing=
*P*_{2}

The upward force on the plane can be obtained using Bernoulli’s equation as:

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

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

∼ 4400 kg

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

#### Page No 271:

#### Question 10.28:

In
Millikan’s oil drop experiment, what is the terminal speed of
an uncharged drop of radius 2.0 × 10^{–5}
m and density 1.2 × 10^{3}
kg m^{–3}?
Take the viscosity of air at the temperature of the experiment to be
1.8 × 10^{–5}
Pa s. How much is the viscous force on the drop at that speed?
Neglect buoyancy of the drop due to air.

#### Answer:

Terminal speed = 5.8
cm/s; Viscous force = 3.9 ×
10^{–10 }N

Radius of the given
uncharged drop, *r* = 2.0 ×
10^{–5} m

Density of the
uncharged drop, *ρ* =
1.2 × 10^{3} kg m^{–3}

Viscosity of air,

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

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

Terminal velocity (*v*)
is given by the relation:

Hence, the terminal
speed of the drop is 5.8 cm s^{–1}.

The viscous force on the drop is given by:

Hence, the viscous
force on the drop is 3.9 ×
10^{–10 }N.

#### Page No 271:

#### Question 10.29:

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^{–1}.
Density of mercury = 13.6 × 10^{3}
kg m^{–3}.

#### Answer:

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

Radius of the narrow
tube, *r* = 1 mm = 1 ×
10^{–3} m

Surface tension of
mercury at the given temperature, *s* = 0.465 N m^{–1 }

Density of mercury, *ρ*
=13.6 × 10^{3} kg/m^{3}

Dip in the height of
mercury = *h*

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

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

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

#### Page No 271:

#### Question 10.30:

**
**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

^{–2}N m

^{–1}. Take the angle of contact to be zero and density of water to be 1.0 × 10

^{3}kg m

^{–3}(g = 9.8 m s

^{–2}).

#### Answer:

Diameter of the first bore, *d*_{1} = 3.0 mm = 3 ×
10^{–3} m

Diameter of the second bore, = 6.0 mm

Surface tension of water, *s* = 7.3 × 10^{–2}
N m^{–1}

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

Density of water, *ρ* =1.0 × 10^{3} kg/m^{–3
}

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

Let *h*_{1} and *h*_{2 }be the heights
to which water rises in the first and second tubes respectively.
These heights are given by the relations:

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

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

#### Page No 271:

#### Question 10.31:

(a) It is known that density *ρ* of air decreases with height *y *as

Where *= *1.25 kg m^{–3} is the density at sea level, and *y*_{0} is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of gremains constant.

(b) A large He balloon of volume 1425 m^{3} is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?

[Take *y*_{0}= 8000 m and = 0.18 kg m^{–3}].

#### Answer:

**(a) **Volume of the balloon, *V* = 1425 m^{3}

Mass of the payload, *m* = 400 kg

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

Density of the balloon = *ρ*

Height to which the balloon rises = *y*

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

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.,

Where, *k* is the constant of proportionality

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

Integrating the sides between these limits, we get:

**(b) **

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

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