Question:

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 (20oC) is 2.50×10^−2 Nm^−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:

1. Calculate the surface pressure of the bubble of soap solution: Pressure = 2.50 x 10^-2 Nm^-1 x (5.00 x 10^-3 m)^2 Pressure = 0.31375 x 10^5 Pa

2. Calculate the excess pressure inside the bubble of soap solution: Excess Pressure = 0.31375 x 10^5 Pa - 1.01 x 10^5 Pa Excess Pressure = 0.30375 x 10^5 Pa

3. Calculate the pressure inside the bubble of air at a depth of 40.0 cm: Pressure = 1.01 x 10^5 Pa + (1.20 x 9.81 m/s^2 x 0.400 m) Pressure = 1.637 x 10^5 Pa

4. Calculate the excess pressure inside the bubble of air: Excess Pressure = 1.637 x 10^5 Pa - 1.01 x 10^5 Pa Excess Pressure = 0.627 x 10^5 Pa

Question:

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) To keep a piece of paper horizontal, you should blow over, not under, it. This is because when you blow over the paper, the air pressure is greater on top of the paper and causes the paper to stay in place. When you blow under the paper, the air pressure is greater on the bottom of the paper, causing the paper to move in the direction of the air flow.

(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers. This is because when we close the tap, the pressure of the water increases and the jets of water become more powerful, forcing their way 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. This is because the size of the needle determines the size of the opening through which the fluid is passing, and a smaller opening will result in a slower flow rate. The thumb pressure of the doctor is not as effective in controlling the flow rate because it does not change the size of the opening.

(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel. This is because as the fluid flows out of the hole, it pushes against the walls of the vessel and causes a reaction force that pushes the vessel in the opposite direction.

(e) A spinning cricket ball in air does not follow a parabolic trajectory. This is because the spin of the cricket ball causes it to deviate from its normal trajectory and follow a curved path. The spin of the ball causes it to move in a different direction than it would if it were traveling in a straight line.

Question:

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:

1. Calculate the difference in levels between the two arms of the U-tube. This is 12.5 cm - 10.0 cm = 2.5 cm.

2. Calculate the density of water. This is 1000 kg/m^3.

3. Calculate the mass of the mercury. This is the difference in levels multiplied by the density of mercury, which is 13.6 g/cm^3. This gives us 33 g of mercury.

4. Calculate the volume of the spirit. This is the mass of mercury divided by the density of spirit. This gives us 0.25 m^3.

5. Calculate the specific gravity of spirit. This is the density of spirit divided by the density of water. This gives us 0.8 g/cm^3.

Question:

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^−3kgs^−1, what is the pressure difference between the two ends of the tube ? (Density of glycerine =1.3×10^3kgm^−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:

1. Calculate the flow rate of glycerine: Q = 4.0 × 10^−3 kgs^−1

2. Calculate the cross-sectional area of the tube: A = πr^2 = π × (1.0 × 10^−2 m)^2 = 3.14 × 10^−4 m^2

3. Calculate the velocity of the glycerine: v = Q/A = 4.0 × 10^−3 kgs^−1 / 3.14 × 10^−4 m^2 = 1.27 × 10^−2 m/s

4. Calculate the Reynold’s number: Re = ρvD/μ = (1.3 × 10^3 kgm^−3) × (1.27 × 10^−2 m/s) × (1.0 × 10^−2 m) / (0.83 Pa s) = 1560

5. Determine if the assumption of laminar flow is correct: Since the Reynold’s number is less than 2300, the assumption of laminar flow is correct.

6. Calculate the pressure difference between the two ends of the tube: ΔP = ρLv^2/2D = (1.3 × 10^3 kgm^−3) × (1.5 m) × (1.27 × 10^−2 m/s)^2 / (2 × (1.0 × 10^−2 m)) = 0.096 Pa

Question:

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:

1. Calculate the area of each hole: A = πr^2 = π (0.001 m)^2 = 7.85 x 10^-6 m^2

2. Calculate the total area of all 40 holes: A = 40 x 7.85 x 10^-6 m^2 = 3.14 x 10^-4 m^2

3. Calculate the velocity of the liquid inside the tube: v = Q/A = 1.5 m min^-1/ 8.0 cm^2 = 0.0001875 m s^-1

4. Calculate the speed of ejection of the liquid through the holes: v = Q/A = 0.0001875 m s^-1/ 3.14 x 10^-4 m^2 = 5.96 m s^-1

Question:

In deriving Bernoullis 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:

a) The largest average velocity of blood flow in an artery of diameter 2×10^−3 m if the flow must remain laminar can be calculated using the equation:

V = (2.0 × 10^-3 m)^2 × 8 × η / (π × μ)

where V is the velocity, η is the viscosity of the fluid, and μ is the dynamic viscosity.

b) Yes, the dissipative forces become more important as the fluid velocity increases. As the velocity increases, more energy is required to move the fluid, leading to an increase in the amount of energy dissipated in the form of heat. This is due to the increased friction between the molecules of the fluid as the velocity increases. This increased friction leads to a decrease in the efficiency of the fluid flow.

Question:

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

Answer:

1. Calculate the area of the heel: A = πr2 A = π (0.5 cm)2 A = 0.785 cm2

2. Calculate the force exerted by the girl on the heel: F = mg F = (50 kg) (9.8 m/s2) F = 490 N

3. Calculate the pressure exerted by the heel on the floor: P = F/A P = (490 N) / (0.785 cm2) P = 627.38 Pa

Question:

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:

1. Calculate the maximum stress due to the pressure of the ocean at a depth of 3 km.

2. Compare the maximum stress due to the pressure of the ocean to the maximum stress the structure is designed to withstand.

3. If the maximum stress due to the pressure of the ocean is less than the maximum stress the structure is designed to withstand, then the structure is suitable for putting up on top of an oil well in the ocean.

Question:

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:

1. Calculate the area of the soap film: The area of a U-shaped wire is equal to 2πrh, where r is the radius of the wire and h is the height of the U-shape. Since the length of the slider is 30 cm, the height of the U-shape is 30 cm.

2. Calculate the surface tension of the film: The surface tension of the film can be calculated using the formula F = 2πrT, where F is the force applied on the film, r is the radius of the wire and T is the surface tension of the film. Substituting the values, we get T = F/(2πr) = (1.5×10^−2 N)/(2πr) = (1.5×10^−2 N)/(2π0.15 m) = 0.2 N/m.

Therefore, the surface tension of the film is 0.2 N/m.

Question:

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 ? [Density of whole blood =1.06×10^3kgm^−3].

Answer:

1. Calculate the height of the blood container using the equation of hydrostatic pressure: P = ρgh

2. Substitute the given values into the equation: 2000 Pa = (1.06x10^3 kg/m^3)gh

3. Solve for h: h = 2000 Pa/(1.06x10^3 kg/m^3)g

4. Answer: The height of the blood container must be 18.87 m.

Question:

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-shaped 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^−2Nm^−1 Take the angle of contact to be zero and density of water to be 1.0×10^3kgm^3(g=9.8ms^−2) A 3 mm B 2 mm C 4 mm D 5.0 mm

Answer:

Answer: C 4 mm

Question:

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 blood pressure in humans is greater at the feet than at the brain because blood has to travel against gravity from the heart to the feet. This requires the heart to pump with greater pressure in order to ensure that the blood reaches the feet.

(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level because the atmosphere is composed of molecules that are attracted to each other and they settle down with gravity. As the height increases, the number of molecules decrease, thus resulting in lower atmospheric pressure.

(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area because hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. The force of gravity does not depend on the area, so the pressure is also a scalar quantity.

Question:

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

Answer:

Yes, Bernoulli’s equation can be used to describe the flow of water through a rapid in a river. The equation states that the sum of the pressure, kinetic energy, and potential energy of a fluid is constant at any point in a closed system. This means that, if the pressure of the water is increased due to the presence of a rapid, the kinetic energy of the water must also increase in order to keep the sum of the energies constant. Therefore, Bernoulli’s equation can be used to calculate the increased kinetic energy of the water due to the presence of a rapid in a river.

Question:

In a test experiment on a model airplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 ms^−1 and 63 ms^−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 kgm^−3.

Answer:

1. Calculate the difference in the flow speeds on the upper and lower surfaces of the wing: 70 ms^−1 - 63 ms^−1 = 7 ms^−1.

2. Calculate the lift on the wing using the formula: Lift = (1/2) x (density of air) x (area of wing) x (difference in flow speeds)^2.

3. Substitute the given values into the formula: Lift = (1/2) x (1.3 kgm^−3) x (2.5 m^2) x (7 ms^−1)^2

4. Calculate the lift on the wing: Lift = 0.735 N

Question:

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20o C) is 4.65×10^−1 Nm^−1. The atmospheric pressure is 1.01×10^5Pa. Also give the excess pressure inside the drop.

Answer:

Step 1: Calculate the surface area of the drop of mercury using the formula 4πr2, where r is the radius of the drop.

Step 2: Calculate the force due to the surface tension of mercury using the formula F = 2πrT, where T is the surface tension of mercury.

Step 3: Calculate the pressure inside the drop of mercury using the formula P = F/A, where F is the force due to the surface tension of mercury and A is the surface area of the drop.

Step 4: Calculate the excess pressure inside the drop of mercury by subtracting the atmospheric pressure from the pressure inside the drop.

Answer: The pressure inside the drop of mercury of radius 3.00 mm at room temperature (20o C) is 2.91×10^5Pa. The excess pressure inside the drop is 1.90×10^5Pa.

Question:

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

Answer:

1. Terminal speed of an uncharged drop: V_t = (2 * 9.8 * (2.0 x 10^-5)^2 * 1.2 x 10^3) / (18 x 10^-5) V_t = 0.2 m/s

2. Viscous force on the drop at that speed: F_v = 6 x 10^-5 * 1.8 x 10^-5 * 0.2 F_v = 0.00216 N

Question:

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 / Bernoullis 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 (b) increases, decreases (c) shear strain, rate of shear strain (d) Bernoulli’s principle (e) greater

Question:

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:

1. Calculate the volume of the tank by multiplying the area of the base (1.0 m^2) by the height of the tank (4.0 m), giving a total volume of 4.0 m^3.

2. Calculate the volume of the water compartment by multiplying the area of the base (1.0 m^2) by the height of the water (4.0 m), giving a total volume of 4.0 m^3.

3. Calculate the volume of the acid compartment by multiplying the area of the base (1.0 m^2) by the height of the acid (4.0 m), giving a total volume of 6.8 m^3.

4. Calculate the total mass of the water and acid by multiplying the volume of each compartment by the relative density of each liquid, giving a total mass of 6.8 kg for the water and 11.6 kg for the acid.

5. Calculate the force necessary to keep the door closed by multiplying the total mass of the liquids (18.4 kg) by the acceleration due to gravity (9.8 m/s^2), giving a total force of 180.52 N.

Question:

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:

1. No, the force exerted by the water on the base of the two vessels is not the same.

2. This is because the first vessel has a greater volume of water than the second vessel, and therefore exerts more pressure on the base.

3. This greater pressure exerted by the first vessel causes it to have a greater weight than the second vessel when both are filled to the same height. Therefore, when weighed on a scale, the first vessel will give a higher reading than the second.

Question:

(a) It is known that density of air decreases with height y as ρ=ρ0​e^−y/y0​ where ρ0​ = 1.25 kg m^−3 is the density at sea level, and y0​ is constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains a constant (isothermal conditions). Also, assume that the value of g remains 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 y0​ = 8000 m and ρHe​ = 0.18 kg m^−3].

Answer:

(a) The law of atmospheres under isothermal conditions can be obtained by combining the ideal gas law and the hydrostatic equation.

Ideal gas law: PV = nRT

Hydrostatic equation: dP/dy = -ρg

Substituting the ideal gas law in the hydrostatic equation, we get:

dP/dy = -nRT/V * g

Integrating both sides, we get:

P = -nRT/V * g * y + C

Where C is the integration constant.

Substituting the value of P at y = 0, we get:

P(y=0) = -nRT/V * g * 0 + C

=> C = P(y=0)

Therefore, the law of atmospheres under isothermal conditions can be written as:

P = -nRT/V * g * y + P(y=0)

Since density is inversely proportional to pressure, we can write the law of atmospheres as:

ρ = ρ0 * e^(-y/y0)

Where ρ0 = P(y=0)/RT

(b) To calculate the height to which the balloon rises, we need to calculate the buoyant force on the balloon. The buoyant force is equal to the weight of the displaced air.

Weight of displaced air = Volume of displaced air * density of air * gravitational acceleration

=> Weight of displaced air = V * ρ * g

Substituting the given values, we get:

Weight of displaced air = 1425 m^3 * 1.25 kg/m^3 * 9.8 m/s^2

=> Weight of displaced air = 17,812.5 N

Since the buoyant force is equal to the weight of the payload, we can write:

17,812.5 N = 400 kg * g

=> g = 44.53 m/s^2

Now, we can use the law of atmospheres to calculate the height to which the balloon rises.

ρ = ρ0 * e^(-y/y0)

Substituting the given values, we get:

0.18 kg/m^3 = 1.25 kg/m^3 * e^(-y/8000 m)

=> e^(-y/8000 m) = 0.144

Taking the natural logarithm of both sides, we get:

-y/8000 m = ln(0.144)

=> y = -8000 m * ln(0.144)

=> y = 10,904 m

Therefore, the balloon rises to a height of 10,904 m.

Question:

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:

1. Calculate the force (F) required to lift the car: F = m x g, where m is the mass of the car (3000 kg) and g is the acceleration due to gravity (9.81 m/s^2).

F = 3000 kg x 9.81 m/s^2 = 29423 N

2. Calculate the pressure (P) required to lift the car: P = F/A, where F is the force required to lift the car (29423 N) and A is the area of cross-section of the piston carrying the load (425 cm^2).

P = 29423 N / (425 cm^2) = 69.1 kPa

Question:

In the below given 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 ? (Specific gravity of mercury = 13.6) Problem:[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 ]

Answer:

Step 1: Calculate the weight of 10.0 cm of water and 12.5 cm of spirit. Weight of 10.0 cm of water = 10.0 cm x 1000 kg/m3 x 9.81 m/s2 = 98,100 kg Weight of 12.5 cm of spirit = 12.5 cm x 800 kg/m3 x 9.81 m/s2 = 78,800 kg

Step 2: Calculate the difference in weight of the two arms. Difference in weight = 98,100 kg – 78,800 kg = 19,300 kg

Step 3: Calculate the difference in the height of mercury in the two arms. Difference in height = 19,300 kg/13.6 kg/m3 = 1.42 m

Step 4: Calculate the difference in the levels of mercury in the two arms after 15.0 cm of water and spirit each are further poured into the respective arms. Difference in levels of mercury = 1.42 m + 15.0 cm = 1.57 m

Question:

(a) What is the largest average velocity of blood flow in an artery of radius 2×10^−3 m if the flow must remain lanimar? (b) What is the corresponding flow rate ? (Take viscosity of blood to be 2.084×10^−3 Pas).

Answer:

a) The largest average velocity of blood flow in an artery of radius 2×10^−3 m is given by the equation:

V = (2 × 10^−3)^2 × 8 × 2.084 × 10^−3/2

V = 1.66 × 10^−6 m/s

b) The corresponding flow rate is given by the equation:

Q = V × π × (2 × 10^−3)^2

Q = 3.25 × 10^−9 m^3/s

Question:

Mercury has an angle of contact equal to 140o with soda lime glass. A narrow tube of radius 1.00mm 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 0.465Nm^−1. Density of mercury =13.6×10^3kgm^−3.

Answer:

Step 1: Calculate the capillary rise, h, using the equation: h = (2Tcosθ)/(ρgr)

Step 2: Substitute the given values into the equation.

h = (2 × 0.465 × cos 140o) / (13.6 × 10^3 × 9.81 × 10^-3)

Step 3: Calculate the capillary rise, h.

h = 0.0015 m

Question:

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

Answer:

Answer:

Yes, it does matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation. Gauge pressure is the pressure measured relative to atmospheric pressure, whereas absolute pressure is measured relative to a perfect vacuum. This means that the gauge pressure value is always lower than the absolute pressure value. Therefore, when using Bernoulli’s equation, the gauge pressure value must be converted to absolute pressure before applying the equation. Otherwise, the result of the equation will be incorrect.

Question:

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 of contact of mercury with glass is obtuse because mercury is denser than water, and therefore has a higher surface tension. This causes the molecules of the mercury to be pulled together more strongly, resulting in a greater angle of contact. The angle of contact of water with glass is acute because water is less dense than mercury, and therefore has a lower surface tension. This causes the molecules of the water to be pulled apart more easily, resulting in a smaller angle of contact.

(b) Water on a clean glass surface tends to spread out because it has a lower surface tension than mercury. This causes the molecules of the water to be pulled apart more easily, resulting in a larger surface area of contact with the glass. On the other hand, mercury on the same surface tends to form drops because it has a higher surface tension than water. This causes the molecules of the mercury to be pulled together more strongly, resulting in a smaller surface area of contact with the glass.

(c) Surface tension of a liquid is independent of the area of the surface because it is a property of the liquid itself, and not of the surface. The surface tension of a liquid is determined by the intermolecular forces between the molecules of the liquid, and is independent of the size of the surface area.

(d) Water with detergent dissolved in it should have small angles of contact because detergents reduce the surface tension of the water. This causes the molecules of the water to be pulled apart more easily, resulting in a smaller angle of contact.

(e) A drop of liquid under no external forces is always spherical in shape because the surface tension of the liquid causes it to take the shape with the least surface area. A sphere has the smallest surface area for a given volume, and therefore the surface tension of the liquid causes it to take the spherical shape.

Question:

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:

1. Normal atmospheric pressure is 101.325 kPa.

2. Determine the density of mercury: mercury has a density of 13.6 g/cm^3, or 13600 kg/m^3.

3. Calculate the pressure difference between the mercury and wine columns:

Pressure difference = (13600 kg/m^3 - 984 kg/m^3) x g x height

4. Solve for the height of the wine column:

Height = Pressure difference / (984 kg/m^3 x g)

= 101.325 kPa / (984 kg/m^3 x 9.81 m/s^2)

= 10.37 m

Question:

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 km m^−3).

Answer:

1. Calculate the lift force on each wing: Lower wing lift force = 0.5 x 1 x (180)^2 x 25 = 162000 N Upper wing lift force = 0.5 x 1 x (234)^2 x 25 = 208800 N

2. Calculate the total lift force on the plane: Total lift force = Lower wing lift force + Upper wing lift force = 162000 + 208800 = 370800 N

3. Calculate the plane’s mass: Mass = Total lift force / 9.81 m/s^2 = 370800/9.81 = 37802.5 kg