Mechanical Properties Of Fluids

Human body is filled with blood which constitutes for the hydrostatic pressure. Hydrostatic pressure depends on the height of the liquid column.

P = ρgh

Where, P = Pressure

h = height of the liquid column

g = Acceleration due to gravity = 9.8 m/s

Height of the blood column at feet is greater than the height of blood column at the brain.

Therefore blood pressure in humans is greater at the feet than at the brain.

The air at sea level is compressed by the atmospheric air column above that level. At higher altitudes the air column above decreases. So the air will expand and its density decreases. Also at higher altitudes gravity is less because of which air molecules can spread freely. Hence density of air decreases with increasing altitude. So the density of air is less at a height of about 6 km from the ground than the density of air at sea level.

We know that hydrostatic pressure is directly proportional to density of the fluid and the height of the liquid column above that level.

P = ρgh

As both density of air and height of the air column 6 km above sea level are less compared to that at the sea level , Atmospheric pressure decreases to nearly half of its value at 6 lm above the sea level.

Pascal's law states that pressure exerted anywhere on a confined incompressible fluid is transmitted equally and undiminished throughout the entire fluid. So when force is applied on a liquid, pressure is transmitted in all directions. A vector has only one direction but hydrostatic pressure does not have a specific direction . So it is considered a scalar quantity.

Figure showing the pictorial representation of Pascal's law

The angle between the glass surface and tangent to the liquid surface at the point of contact with the glass surface is known as angle of contact (θ)

Figure showing various tensions at the interfaces

Let S_la = Interfacial tension between liquid-air

S_sa = Interfacial tension between solid-air

S_sl = Interfacial tension between solid-liquid

At the line of contact, the surface forces between the three media should be in equilibrium.

S_laCosθ = S_sa - S_sl

cosθ = (S_sa - S_sl)/S_la

We also know that surface force is directly proportional to the density of the substance.

In case of mercury, the density of mercury is greater than that of density of glass, therefore cohesive forces exerted by mercury molecules are greater than adhesive forces between the glass and the mercury molecules .But density of air is less that of density of glass; so adhesive forces are greater than the cohesive forces. The surface tension of a liquid results from the imbalance of the cohesive forces between molecules. Therefore interfacial tension between the glass and mercury is greater than the tension between solid-air surface.

So, S_la>S_sa.

cosθ = (S_sa - S_sl)/S_la is negative. Hence θ is an obtuse.

In case of water, the density of water is lesser than that of density of glass molecules, therefore cohesive forces exerted by water are lesser than adhesive forces between the glass and water molecules. Therefore the interfacial tension between glass-water is less than tension between air-glass surface.

So, S_la<S_sa

cosθ = (S_sa - S_sl)/S_la is positive. Hence θ is acute.

Mercury molecules have strong forces of attraction between themselves (cohesive forces) and the forces of attraction between molecules of mercury and glass are weak (Adhesive forces).So they form a droplets on the glass surface.

Water molecules have weak forces of attraction between themselves (cohesive forces) and the forces of attraction between the water molecules and glass molecules is strong (adhesive forces). So water spreads on glass.

The force acting per unit length at the interface between the plane of a liquid and any other surface is known as surface tension. The force acting does not depend on the area of the surface. Therefore surface tension is independent of area of the surface.

If angle of contact (θ) is small, cosθ will be large. We know that the capillary rise is directly proportional to cosθ

As water enters into the pores of our clothes by capillary action, the capillary rise of a detergent dissolved in a liquid will be more if the angle of contact is acute.

The most stable state of any system is that which possesses the least amount of energy. A liquid tends to have minimum surface area because it has less surface energy. Surface area of a sphere is minimum for a given volume. Hence under no external pressure drop of a liquid always forms a spherical surface.

(a) decreases

Explanation: Surface tension decreases with increase in temperature because cohesive forces in the liquid decrease with increasing molecular thermal activity. As surface tension depends on cohesive forces of the liquid, it's value decreases at higher temperatures.

(b) increases, decreases

Explanation: Usually gases are free flowing in the atmosphere. When temperature increases collisions between the molecules increase and as a result the their free flow is resisted. Resistance to free flow is known as viscosity. Therefore viscosity increases in gases due to increase in temperature.

(c) shear strain, rate of shear strain

Explanation: In solids

Therefore in solids, shear stress is directly proportional to shear strain.

In fluids according to continuum mechanics, in a Newtonian fluid shear stress is directly proportional to rate of shear strain.

(d) Conservation of mass, Bernoulli's principle

Explanation: A steady flowing fluid's increase in flow speed at a constriction depends on both conservation of mass and Bernoulli's principle.

(e) greater

Explanation: For the model of plane in a wind tunnel turbulence occurs at a greater speed than for an actual plane. This is due to the difference in Reynolds's number associated with the planes.

When we blow over a paper velocity of air above the plane of the paper increases. As a result Kinetic energy above the plane of paper increases.

According to Bernoulli's principle

As K.E increases, Pressure energy above the plane of paper decreases below the atmospheric pressure. The pressure below the paper is atmospheric pressure As a result pressure above the paper is less than pressure below the paper. So the paper remains horizontal and does not fall.

We know that according to equation of continuity

Area × velocity = constant

By closing a water tap with our fingers, the area of outlet of water jet is reduced. As a result velocity of the outlet jet increases according to the above equation of continuity.

From Bernoulli's theorem, we know that

We know that diameter of the needle is very less. So flow velocity through the needle will be very high according to equation of continuity; Area × velocity = constant

So needle controls the velocity of the fluid flow

By using our thumb we apply more pressure to the flow of fluid . So thumb controls pressure.

The power of pressure is 1 and the power of velocity is 2. Thus size of the needle has more control over the flow rate than the thumb pressure.

The area through which water is flowing out of the hole in the vessel is small. According to equation of continuity ,

Area × velocity = constant

So the velocity of water flowing through the hole is large.

According to law of conservation of momentum,

m₁v₁ = m₂v₂

Due to this the vessel attains a backward velocity as there are no external forces acting on the system. So we can that water exerts a backward thrust on the vessel.

A spinning cricket ball has velocity both due to linear(v) and rotational motion(u). when a ball is moved forward, the air occupies the space left by the ball with a velocity v (say). When the ball spins, the layer of air around it also moves with the ball, say with the velocity 'u'. So the resultant velocity of air above the ball becomes (v-u) and below the ball becomes (v + u). Hence, the Due to this the velocity of air above the ball decreases and below the ball increases. According to Bernoulli's principle decrease in velocity increases pressure on the upper side of the ball.

So there is a downward force on the ball and because of this and it fails to follow parabolic trajectory.

Given:

Mass of the girl (m) = 50kg

Diameter of the heel (d) = 1.0cm

Radius of the heel (r) = d/2 = 0.5cm

Area of the heel(A) = π × r²

= π × (0.5 × 10^-2 m)²

= 7.8539 × 10^-5 m²

We know that Pressure (P) = F/A

F = Force exerted by heel on the floor

A = Area of the heel

F = m × g where m = mass of the girl

= 50 kg × 9.8 m/s² g = acceleration due to gravity

= 490 N

P = F/A

= 490 N/(7.8539 × 10^-5) m²

= 6.24 × 10⁶ N/m²

Pressure exerted by heel on the horizontal floor is 6.24 × 10⁶N/m²

In Toricelli's barometer mercury is used

Density of mercury (ρ₁) = 13.6 × 10³ kg/m³

Height of mercury column at atmospheric pressure (h₁) = 0.76 m

Density of French wine (ρ₂) = 984 kg/m³

Height of the French wine column at atmospheric pressure = h₂

We know that pressure in a column of height h and density ρ = ρgh

where, g = acceleration due to gravity = 9.8 m/s²

Pressure in both the columns is the same i.e. atmospheric pressure

ρ₁gh₁ = ρ₂gh₂

ρ₁h₁ = ρ₂h₂

(13.6 × 10³) × 0.76 = (984) × h₂

h₂ = 10.504 m

Hence the height of French white column at normal atmospheric pressure is 10.504 m.

Given:

Maximum stress the structure can withstand is (P) = 109 Pa

Depth of the ocean (d) = 3 km = 3 × 10³ m

Density of water (ρ) = 1000 kg/m³

We know that pressure due to depth d = ρgd

Pressure due to depth d of water = ρgd

= 1000 kg/m³ × 9.8 m/s² × (3 × 10³m)

= 2.94 × 10⁷ Pa

The maximum allowable stress for the structure(10⁹Pa ) is greater than the pressure of the sea (2.94 × 10⁷Pa). Hence the structure can be put on top of an oil well in the ocean.

Maximum mass of a car that can be lifted (m) = 3000 kg

Area of cross section of piston carrying the load (A₁) = 452 cm²

= 0.0452 m²

Maximum force exerted by the car due to its weight (F) = mg

= 3000kg × 9.8m/s²

= 29400 kgm/s²

Pressure on the piston lifting the car P = F/A

=

= 6.504 × 10⁵ Pa

Pressure is transmitted equally in all the directions of a fluid. Therefore the maximum pressure on the smaller piston is 6.504 × 10⁵ Pa.

Figure showing heights of liquids in an U-Tube

Given:

Height of spirit column (h₁) = 12.5 cm

Height of water column (h₂) = 10.0 cm

Let P₀ = Atmospheric pressure

ρ₁ = density of the spirit

ρ₂ = density of the water column

Pressure due to spirit column at B = P₀ + ρ₁gh₁

Pressure due to water column at D = P₀ + ρ₂gh₂

Pressure is equal at the points having the same height from the ground i.e. at points B and D.

P₀ + ρ₁gh₁ = P₀+ ρ₂gh₂

ρ₁h_{1 =} ρ₂h₂

We know that

Therefore specific gravity of spirit is 0.8.

Given:

Height of the spirit column (h₁) = 12.5 + 15 = 27.5 cm

Height of the water column (h₂) = 10 + 15 = 25 cm

Density of spirit (ρ₁) = 1 gm/cc

Density of spirit (ρ₂) = 0.8 gm/cc

Density of mercury (ρ) = 13.6 gm/cc

Mercury rises because of the differences in pressure exerted by the columns of water and spirit.

Pressure due to height h of mercury column = ρgh

= (13.6gm/cc) × (980 cm/s) × (h cm)

Differences in pressures due to water and spirit = ρ₁gh₁-ρ₂gh₂

= (1 × 980 × 25)-(0.8 × 980 × 27.5)

= 2940 gm/cm²s

Pressure due to height h of mercury = Differences in pressures due to water and spirit

13.6 × 980 × h = 2940

h = 0.2205 cm

Difference in the levels of mercury in two columns is 0.2205.

The flow of water through a rapid in a river consist of turbulent flow of water. The Bernoulli’s equation cannot be applied to the turbulent flow. It can only be applied to the streamline flow.

Thus, The Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river.

The excess of pressure, P − Pa, at depth h is called a gauge pressure at that point.

The bernoulli’s equation works when the atmospheric pressure at the two points at which it is applied is significantly different. Using gauge pressure instead of absolute pressure in bernoulli’s equation not matter as we are using the difference only.

The Poiseuille’s law states the relation:

Where,

‘p’ is the difference in the pressure between the two ends.

‘r’ is the radius of the tube.

‘ɳ’ is the viscosity of the fluid.

‘l’ is the horizontal length of the tube.

Given,

Length of the tube, l = 1.5 m

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

Mass of glycerine flowing per second, M = 4.0 × 10^-3 kg/s

Density of the glycerine, ρ = 1.3 × 10³ kg/m³

Viscosity of the glycerine, ɳ = 0.83 Pa s

Volume of glycerine flowing per second is given as,

⇒V=

⇒ V=3.08 × 10^-6 m³/sec

Applying Poiseuille’s law we get the relation,

⇒

⇒ p=9.8 × 10² Pa

To check whether the assumption of laminar flow in the tube is correct we use the Reynolds’s number relation given as,

Where,

and if the Reynolds number is less than 2000, the flow is laminar.

Where,

‘ρ’ is the density of the fluid.

‘V’ is the volume of fluid flowing per sec

‘d’ is the diameter of the pipe

‘ɳ’ is the viscosity of the fluid

Thus,

⇒

⇒ R =0.3

Since, the Reynolds’s number is less than 2000. Hence, the flow is laminar.

The pressure difference between the two ends of the tube is9.8 × 10²Pa

The Bernoulli’s theorem states the relation,

P_a + V_a²= P_b + V_b²

Thus we get,

P_a – P_b =(v_a² – v_b²) …..(i)

Where,

‘P_a’ is the pressure on the upper surface of the wing

‘P_b’ is the pressure on the lower surface of the wing.

‘v₁’ is the speed on the upper surface of the wing.

‘v₂’ is the speed on the lower surface of the wing.

‘ρ’ is the density of the air.

The reason for the lift of the airplane is the difference between the pressure differences between the upper and lower surface,

Thus,

The force which is responsible for the airplane lift, F = (P_b-P_a) A

Replacing the value of Pressure difference from equation (i),

⇒F = (v_a²– v_b²) A

Given,

Speed of the wind on the upper surface, v₁ = 70 m/s

Speed of the wind on the lower surface , v₂ = 63 m/s

Area of the wing, A = 2.5 m²

Density of the air, ρ = 1.3 kg / m³

⇒

⇒ F = 1512.87 N = 1.512 × 10³ N

the lift on the wing of the airplane is 1.512 × 10³ N.

The law of continuity states that,

A v= A’ v’

As taken in the figure,

Where,

Area of ‘Pipe a’ is A

Area of ‘Pipe b’ is A’

Speed of the fluid in the ‘pipe a’ = v

Speed of the fluid in the ‘pipe b’ = v_’

The law thus states that if the area of cross section in the Venturimeter is small, the speed of the flowing liquid through the part will be more.

And the Bernoulli’s principle states that if the speed is more around a part, the pressure will be less.

The pressure is directly proportional to the height.

In the figure, the area A’ is less than area A, Thus, the speed in the is higher and thus the height of the in ‘pipe b’ must be higher.

The figure ‘a’ is the incorrect option.

The law of continuity states that,

A₁v₁ = A₂v₂

⇒ v₂ = A₁v₁ / A₂ ……….(i)

Where,

‘A₁’ is the area of cross section of the spray pump

‘A₂’ is the area of cross section of the end with the 40 holes

‘v₁’is the speed of flow of fluid inside the tube

‘v₂’ is the speed of flow of the fluid through the holes

Given,

Area of cross section of spray pump, A₁ = 8 cm² = 8 × 10^-4 m²

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

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

Velocity of the liquid flow inside the tube, v₁ = 1.5 m min^-1

⇒ v₁ = 1.5/60 m sec^-1 = 0.025 m/s

Thus,

Area of cross section of each hole, a = πr²

⇒ a = 3.14 × (0.5 × 10^-3 m)²

⇒ a = 0.785 × 10^-6 m²

Total area of cross section of 40 holes, A₂ = 40 × a

⇒ A₂ = 40 × 0.785 × 10^-6 m²

⇒ A₂ = 31.41 × 10^-6 m²

Therefore, from equation (i),

⇒ v_{1 =}

⇒ 0.633 m/s

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

Surface tension, S is written as

S=

Where,

‘F’ is the Force exerted or weight supported between the wire and the slider

‘l’ is the length of the slider.

Given,

The force exerted, F =1.5 × 10^-2 N

The length of the slider, l =30 cm = 0.30 m

The soap film has two free surfaces thus,

Total length = 2l = 2 × 0.30 m = 0.6 m

Therefore, the surface tension, S =

⇒ S = 2.5 × 10^-2 N / m

The surface tension of the file is 2.5 × 10^-2 N /m

The surface tension =

Where,

‘F’ is the weight supported by the films,

‘l’ is the length of the liquid film.

From the equation, we can see that the surface tension only depends on the weight and the length of the film. The liquid is same and at the same temperature in all the three cases (a), (b), (c).

Thus, the surface tension will be same in all the three cases and thus, the weight supported by will be same in all the three cases i.e 4.5 × 10^-2 N

Total pressure inside a mercury drop is given as ‘P’,

P = Excess pressure inside the mercury drop + Atmospheric pressure

Excess pressure inside the mercury drop, P₁ =

Where,

‘S’ is the surface tension of the mercury

‘r’ is the radius of the mercury drop.

Thus,

P = P₁ + P₀

Given,

Surface tension of the mercury, S = 4.65 × 10^-1 N

Radius of the mercury drop, r = 3.0 mm = 3.0 × 10^-3 m

Atmospheric pressure, P₀ = 1.01 × 10⁵ Pa

⇒ P₁ =

⇒ P₁ = 3.1 × 10² Pa = 0.0031 × 10⁵ Pa

Therefore,

P = 0.0031 × 10⁵ Pa + 1.01 × 10⁵ Pa

⇒ P = 1.0131 × 10⁵ Pa

The pressure inside the drop of mercury is 1.0131 × 10⁵ Pa and the excess pressure inside the drop is 310 Pa.

Excess pressure inside the soap bubble:

Excess pressure inside the air bubble:

Where,

S is the surface tension of the soap solution.

r is the radius of the soap bubble.

r’ is the radius of the air bubble.

Total Pressure inside the air bubble at a depth of 0.4 m= P₁

P₁ = P₀ + hρg + P’

Where,

‘P₀’ is the atmospheric pressure

Given,

Radius of the soap bubble, r = 5.0 mm = 5 × 10^-3 m

Surface tension of the soap solution, S = 2.50 × 10^-2 N / m

Relative density to the air of the soap solution, ρ = 1.2 × 10³ kg /m³

The height at which the air bubble is formed, h = 40 cm = 0.4 m

Radius of the air bubble, r’ = r = 5 × 10^-3 m

Acceleration due to gravity = 9.8 m/s²

The atmospheric pressure, P₀ = 1.01 × 10⁵ Pa

Therefore,

Excess pressure inside the soap bubble

⇒

⇒ P = 20 Pa

Excess pressure inside the air bubble,

⇒

⇒ P’ = 10 Pa

The total pressure inside an air bubble at the depth of 0.4 m, P₁

P₁ = 1.01 × 10⁵ Pa + (0.4 m × 1.2 × 10³ kg/m³ × 9.8 m/s² ) + 10 Pa

P₁ = 1.057 × 10⁵ Pa

The excess pressure inside the soap bubble is 20 Pa, the excess pressure inside the air bubble 10 Pa and the total pressure inside the air bubble is 1.057 × 10⁵ Pa

Given,

Base area of the tank = 1 m²

Area of the hinged door = 20 cm²

Height of both water and acid columns, h = 4 m

Density of the water, ρ_w = 1000 kg/m³

Density of the acid, ρ_a = 1.7 × ρ_w = 1700 kg/m³

Pressure, P = ρgh

Where,

ρ = density

g = acceleration due to gravity

h = height of the column

∴ Pressure exerted by water, P_w = 1000 × 9.81 × 4

= 3.92 × 10⁴ pa

Pressure exerted by the acid, P_a = 1700 × 9.81 × 4

= 6.664 × 10⁴ pa

∴ Pressure difference, ΔP = P_a – P_w

= (6.664-3.92) × 10⁴ pa

= 2.744 × 10⁴ pa

Thus, Force exerted on the hinged plate,

⇒ = 54.88 N

(a) Local atmospheric pressure, P_atm = 76 cm of Hg

Gauge pressure = mercury height in (Right limb – Left limb)

Absolute pressure = Atmospheric pressure + Gauge pressure

For figure a:

Gauge pressure = 20 cm of Hg

Absolute pressure = 76 + 20 = 96 cm of Hg

For figure b:

Gauge pressure = -18 cm of Hg

Absolute pressure = 76 - 18 = 56 cm of Hg

(b) Relative density of the mercury = 13.6

∴ 13.6 cm of water exerts same pressure as 1cm of mercury exerts.

Thus,

Pressure in the right limb, P_r = Atmospheric + 1 cm of Hg

= 76 + 1 = 77 cm of Hg

Pressure in the left limb, P_l = 58 + h cm of Hg

∴ Height difference between limbs, h = 77 - 58 = 19 cm

Given,

Two vessels are having same base area.

Two vessels are filled with water.

We have,

Pressure, P = ρgh

Where,

ρ = density

g = acceleration due to gravity

h = height of the column

Thus, for given two vessels pressure only depends on height of the water column only.

Force at the base,

Hence, Force exerted on the base of two vessels is same.

Weight = mg

Where,

m = mass

g = acceleration due to gravity

Different volumes of two vessels have different mass. So, on weighing scale they show different readings.

From the table 10.1,

Density of the blood, ρ = 1.06 × 10³ kg/m³

Gauge pressure, P = 2000 pa

We have,

Pressure, P = ρgh

Where,

ρ = density

g = acceleration due to gravity

h = height of the column

It is clear that, pressure exerted by the column of the blood should at least equal to inside vein pressure to enter blood into the vein from bottle.

Thus,

Required height,

=

= 0.1925 m

So, h≈0.2m required to make blood enter into the vein.

(a) Reynolds’s number,

Where,

ρ = density of the fluid

v = largest average velocity of the fluid

D = diameter of the duct

μ = viscosity of the fluid

We have,

Density of the blood, ρ = 1.06 × 10³ kg/m³

Reynolds’s number for laminar flow in duct, Re = 2000

Viscosity of the blood, μ = 2.084 × 10^-3 Nm/s

Thus,

Largest average velocity,

= 1.962 m/s

(b) Dissipative forces are more important when flow velocity is increasing because increase in flow leads to turbulence.

Reynolds’s number,

Where,

ρ = density of the fluid

v = largest average velocity of the fluid

D = diameter of the duct

μ = viscosity of the fluid

We have,

Density of the blood, ρ = 1.06 × 10³ kg/m³

Reynolds’s number for laminar flow in duct, Re = 2000

Viscosity of the blood, μ = 2.084 × 10^-3 Nm/s

Thus,

Largest average velocity,

= 0.982 m/s

Flow rate, Q = Av

Where,

A = Area of the duct

v = velocity of the fluid

Thus, flow rate Q = πr²v

= π × (2 × 10^-3)² m² × 0.982 m/s

= 1.235 × 10^-5 m³/s

Given,

Area of the two wings, A = 2 × 25 = 50 m²

Air Velocity over wings, v₁ = 234 km/h = 65 m/s

Air Velocity below wings, v₂ = 180 km/h = 50 m/s

Density of the air, ρ = 1 kg/m³

Bernoulli’s equation,

(∵ neglecting potential energy)

Let, P₁ = pressure over wings

P₂ = pressure below wings

⇒

⇒

= 862.5 pa

We have,

Where,

ΔP = change in pressure = P₂ – P₁

A = area

∴ F = 862.5 pa × 50 m²

= 43125 N

As per Newton’s second law,

F = mg

Where,

F = force

g = acceleration due to gravity

⇒ mass of the plane,

Thus,

Mass of the plane, m ≈ 4400 kg

Radius of the drop = 2 × 10^-5 m

Density of the drop = 1.2 × 10³ kg/m³

Viscosity of the air, μ = 1.8 × 10^-5 Pa.s

*Consider density of air to be zero in order to neglect buoyancy force create by air on drop.

Terminal velocity is given by the relation,

Where,

r = radius

ρ = highest density = density of drop

ρ₀ = Lowest density = density of air

g = acceleration due to gravity = 9.81 m/s

μ = viscosity

∴

= 5.8 × 10^-2 m/s

= 5.8 cm/s

Hence, the terminal speed of the drop = 5.8 cm/s

The viscous force on the drop is given by,

Where,

μ = viscosity

r = radius

v = velocity

∴

= 3.9 × 10^-10 N

Hence, the viscous force on drop = 3.9 × 10^-10 N

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

Where,

h = dip height

ρ = density = 13.6 × 10³ kg/m³

g = acceleration due to gravity = 9.81 m/s

r = radius of the dipped tube = 1 × 10^-3 m

θ = contact angle = 140⁰

and, Surface Tension S = 0.465 N/m

Thus, height of dip down,

= -0.00534 m

= -5.34 mm

Negative sign indicates decrease in level of mercury in tube. Hence the mercury level is dipped by 5.34 mm.

Diameter of the first tube = 3 mm

Diameter of the second tube = 6 mm

Thus, Radius of the first tube, r₁ = 1.5 mm = 1.5 × 10^-3 m

Radius of the first tube, r₂ = 3 mm = 3 × 10^-3 m

Surface tension of water, S = 7.3 × 10^-2 N/m

Density of water, ρ = 1000 kg/m³

Angle of contact between bore surface and water, θ = 0⁰

Acceleration due to gravity, g = 9.8 m/s

Height of water rise,

Where,

S = Surface tension

θ = Contact angle

ρ = Density

g = acceleration due to gravity

r = radius

Let h₁ and h₂ be the height rise of water in tube₁ and tube₂ respectively. Then the height are given by,

;

Thus, the difference between heights is given by,

Δh = h₁ – h₂

= 4.996 × 10^-3 m

≈ 4.97 mm

∴ The difference between the water levels in two bores = 4.97 mm