GSEB Class 11 Physics Solutions Chapter 10 Mechanical Properties of Fluids

Question 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 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 height of the blood column in the human body is more for the feet compared to that for the brain. As a result, the blood pressure in humans is greater at the feet than at the brain, meaning the blood exerts more pressure at the feet than at the brain.
(b) We understand that the density of air is highest near the Earth's surface and quickly drops with height. At a height of about \(6\) km, it falls to nearly half its sea-level value. Above \(6\) km, the air density drops very slowly with more height. For this reason, the atmospheric pressure at about \(6\) km height reduces to nearly half its value at sea level.
(c) Because of the force applied to the liquid, pressure is sent out equally in all directions within the liquid, according to Pascal's law. Therefore, there is no set direction for the liquid's pressure. This makes hydrostatic pressure a scalar quantity.
In simple words: Blood pressure is higher in the feet because there's a taller column of blood pressing down. Air pressure drops quickly at lower altitudes because air density decreases fast. Hydrostatic pressure is a scalar because it pushes equally in all directions, not just one.

Exam Tip: Remember to link pressure differences to the height of the fluid column and density changes. For scalar/vector, hydrostatic pressure acts in all directions, making it scalar despite force being a vector.

Question 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 detergents 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) When a small amount of liquid is poured onto glass, three interfaces are formed: liquid-air, solid-air, and solid-liquid. The surface tensions corresponding to these three interfaces, namely \(T_{LA}\), \(T_{SA}\), and \(T_{SL}\), are connected to the angle of contact \( \theta \) of a liquid with a solid by:
\( \cos \theta = \frac{T_{SA}-T_{SL}}{T_{LA}} \)
For mercury-glass, \(T_{SA} < T_{SL}\). Therefore, from the equation, \( \cos \theta \) is negative, meaning \( \theta > 90^\circ \) (obtuse). Conversely, for water-glass, \(T_{SA} > T_{SL}\). So from the equation, \( \cos \theta \) is positive, meaning \( \theta < 90^\circ \) (acute).
(b) For a liquid drop to be in equilibrium on a solid surface, the equation \( T_{SA} = T_{SL} + T_{LA} \cos \theta \) must be met. For mercury-glass, the angle of contact is obtuse. To achieve this obtuse value of angle of contact, mercury often forms a drop. However, for water-glass, the angle of contact is acute, so the equation is not met. To achieve this acute value, water tends to spread out.
(c) Surface tension is defined as the force acting per unit length along an imaginary line drawn tangentially on the liquid's surface. Since this force does not depend on the liquid surface's area, surface tension also does not depend on the liquid surface's area.
(d) We know that cloth has tiny spaces that act as fine capillaries. The rise of liquid in a capillary tube is described by the formula \( h = \frac{2T \cos \theta}{r \rho g} \), which shows that \( h \propto \cos \theta \). This implies that if \( \theta \) is small, \( \cos \theta \) will be large, and the detergent will rise more in the narrow spaces of the cloth. Since detergents have small angles of contact, they can penetrate deeper, thus helping to remove dirt from the cloth.
(e) Without any outside forces, the only force acting on a liquid drop is due to surface tension. A liquid drop naturally tries to achieve the smallest possible surface area because of surface tension. Since a sphere has the smallest surface area for a given liquid volume, the liquid drop will always take a spherical shape.
In simple words:
(a) Mercury's contact angle with glass is big (obtuse) because its surface tension favors sticking to itself more than to glass, while water's is small (acute) because it prefers glass.
(b) Water spreads on glass because it "wets" it, meaning it likes glass more than itself. Mercury forms drops because it "doesn't wet" glass and prefers to stick together.
(c) Surface tension is about the force per edge, not the total size of the liquid surface.
(d) Detergents make water spread out more (small contact angle) so it can get into small spaces and clean better.
(e) Liquids want the smallest possible surface, and a sphere is the shape with the least surface for its volume, so drops become round.

Exam Tip: When explaining phenomena related to surface tension and contact angles, always refer to the balance of adhesive and cohesive forces and how they determine the shape and behavior of liquid on a surface. Bernoulli's principle is key for understanding fluid flow dynamics.

Question 3. Fill in the blanks using the word(s) from the list appended with each statement.
1. Surface tension of liquids generally .................... with temperature, (increases/decreases)
2. Viscosity of gases .................... with temperature, whereas viscosity of liquids .................... with temperature. (increases/decreases)
3. 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)
4. For a fluid in a steady flow, the increase in flow speed at a constriction follows from .................... while the decrease of pressure there follows from .................... (conservation of mass/Bernoulli's principle)
5. For the model of a plane in a wind tunnel, turbulence occurs at a .................... speed than the critical speed for turbulence for an actual plane, (greater/smaller)
Answer:
1. decreases.
2. increases, decreases.
3. shear strain, rate of shear strain.
4. conservation of mass, Bernoulli's principle.
5. greater.
In simple words:
1. Liquids get less "sticky" on the surface when they get hotter.
2. Gases get thicker when hot, but liquids get thinner when hot.
3. For solids, the push needed to twist them depends on how much they twist; for liquids, it's about how fast they twist.
4. When water speeds up in a narrow pipe, it's because mass is saved, and when pressure drops there, it's explained by Bernoulli's rule.
5. Small model planes in wind tunnels will start to get turbulent air currents at a higher speed compared to what a real plane needs.

Exam Tip: When filling in blanks, always consider the direct relationship between the quantities. For example, temperature usually reduces intermolecular forces, affecting surface tension and liquid viscosity, but increases molecular collisions, affecting gas viscosity.

Question 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) If you blow over a piece of paper, the air velocity above the paper becomes greater than the air velocity below it. As the kinetic energy of the air above the paper rises, its pressure energy and therefore its pressure decrease, according to Bernoulli's theorem \( (P + \frac{1}{2} \rho v^2 = \text{constant}) \). Because the pressure below the paper (atmospheric pressure) is higher, the paper stays horizontal and does not fall. Conversely, if you blow under the paper, the pressure on the lower side drops. The atmospheric pressure above the paper would then bend the paper downwards, so the paper would not stay horizontal.
(b) This can be explained using the equation of continuity, \( a_1 v_1 = a_2 v_2 \). When we try to close a water tap with our fingers, the cross-sectional area of the water jet's outlet is significantly reduced because the openings between our fingers create constrictions (smaller area regions). As a result, the water's velocity greatly increases, causing fast jets of water to gush through the finger openings.
(c) According to Bernoulli's Theorem, we know that \( P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \). Here, the needle's size influences the flow velocity, while the thumb's pressure affects the pressure. Since \( P \) has a power of one and velocity \( (v) \) has a power of two in the equation, velocity has a stronger effect. That is why the syringe needle provides better control over the flow rate.
(d) When a fluid flows out of a small hole in a vessel, it gains a high velocity and therefore has significant momentum. Since no external force acts on the system, the vessel must acquire a backward velocity (following the law of conservation of momentum). This results in an impulse (backward thrust) being experienced by the vessel.
(e) This phenomenon is due to the Magnus effect. If a ball moving to the right is also spinning at the top, the air velocity above the ball is higher than the air velocity below it. According to Bernoulli's Theorem, the pressure above the ball is less than the pressure below the ball. Therefore, there is a net upward force on the spinning ball, causing it to follow a curved path. This dynamic lift created by spinning is called the Magnus effect.
In simple words:
(a) Blowing over paper lowers air pressure above it, so higher pressure below pushes it up, keeping it flat. Blowing under it raises pressure, pushing it down.
(b) Squeezing a tap makes the water exit faster because the area is smaller, forcing the same amount of water through a smaller gap.
(c) The needle's size affects how fast the liquid moves, which has a bigger impact on flow rate than the pressure applied by the thumb.
(d) When fluid squirts out of a hole, the vessel gets a push backward, similar to a rocket, due to momentum conservation.
(e) A spinning ball creates different air speeds above and below it, leading to a pressure difference that lifts or pushes it sideways, changing its path from a simple arc.

Exam Tip: For "Explain why" questions in fluid dynamics, always connect the observation to a core principle such as Bernoulli's theorem, the equation of continuity, or the Magnus effect. Clearly define the principle and then apply it to the specific scenario.

Question 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 on the horizontal floor?
Answer:
Here, force \( F = mg = 50 \times 9.8 N = 490 N \).
Diameter \( d = 1.0 \text{ cm} \)
So, radius \( r = \frac{d}{2} = 0.5 \text{ cm} = 0.5 \times 10^{-2} \text{ m} \)
\( r = 5 \times 10^{-3} \text{ m} \)
Area of the horizontal floor where the heel exerts pressure is
\( A = \pi r^2 = 3.142 \times (5 \times 10^{-3})^2 \)
\( = 3.142 \times 25 \times 10^{-6} m^2 \)
If \( P \) is the pressure exerted by the heel on the horizontal floor,
Then \( P = \frac{F}{A} = \frac{490}{3.142 \times 25 \times 10^{-6}} \)
\( = 6.24 \times 10^6 \text{ Pascal} \)
Therefore, \( P = 6.24 \times 10^6 \text{ Pa} \).
In simple words: First, calculate the force the girl applies (her weight). Then, find the area of the heel. Finally, divide the force by the area to get the pressure on the floor.

Exam Tip: Remember that pressure is force per unit area. Pay close attention to unit conversions (cm to m, mass to force using gravity) and correctly apply the formula for the area of a circle.

Question 6. Torricelli'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:
Normal atmospheric pressure \( P = 1.013 \times 10^5 \text{ Pa} \)
Let \( h \) be the height of the French wine column that Earth's atmosphere can support.
If \( P' \) is the pressure corresponding to height \( h \) of wine column,
Then, \( P' = h \rho_w g \)
where \( \rho_w = \text{density of wine} = 984 \text{ kg m}^{-3} \)
Now according to the given statement,
\( h \rho_w g = P \)
or \( h = \frac{P}{\rho_w g} \)
\( = \frac{1.013 \times 10^{5}}{984 \times 9.8} \)
\( = 10.5 \text{ m} \).
In simple words: To find the height of the wine column, divide the normal atmospheric pressure by the product of the wine's density and the acceleration due to gravity.

Exam Tip: This question directly tests your understanding of fluid pressure, \( P = h \rho g \). Ensure you use standard values for atmospheric pressure and acceleration due to gravity, and perform accurate unit conversions.

Question 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:
Here,
Depth of water column, \( h = 3 \text{ km} = 3 \times 10^3 \text{ m} \).
Density of water, \( \rho = 10^3 \text{ kg m}^{-3} \).
If \( P \) is the pressure exerted by this water column at this depth,
Then \( P = h \rho g = 3 \times 10^3 \times 10 \times 9.8 \)
\( = 29.4 \times 10^6 \text{ Pa} = 30 \times 10^7 \text{ Pa} \).
\( = 3 \times 10^7 \text{ Pa} \).
When the structure is placed in the sea, the seawater will exert an upward thrust of \( 3 \times 10^7 \text{ Pa} \).
The maximum stress that the vertical off-shore structure can withstand is \( 10^9 \text{ Pa} \).
Then \( 3 \times 10^7 \text{ Pa} < 10^9 \text{ Pa} \)
Therefore, we conclude that the structure is appropriate as the stress applied by the water is much lower than the maximum stress it can endure.
In simple words: Calculate the pressure the ocean exerts at the given depth. Compare this pressure to the maximum stress the structure can handle. If the ocean's pressure is less, the structure is safe.

Exam Tip: This problem combines the concept of fluid pressure (\( P = h \rho g \)) with material strength (maximum stress). Ensure you convert units correctly and compare the calculated pressure directly with the given withstandable stress.

Question 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:
Here, the maximum force that the bigger piston can bear is,
\( F = 3000 \text{ kgf} = 3000 \times 9.8N \)
Area of piston, \( A = 425 \text{ cm}^2 = 425 \times 10^{-4} \text{ m}^2 \).
If \( P \) is the maximum pressure on the bigger piston,
Then \( P = \frac{F}{A} = \frac{3000 \times 9.8}{425 \times 10^{-4}} = 6.92 \times 10^5 \text{ Pa} \)
Since the liquid transfers pressure equally in all directions, the smaller piston must bear the maximum pressure of \( 6.92 \times 10^5 \text{ Pa} \).
In simple words: First, calculate the maximum force the lift needs to handle (the weight of the car). Then, divide this force by the area of the large piston to find the pressure. Because hydraulic systems transmit pressure equally, this is the same pressure the smaller piston will experience.

Exam Tip: Remember Pascal's principle: pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. Convert all units to SI (kg, m, N, Pa) before calculations.

Question 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:
Here, for the water column in one arm of the U-tube:
\( h_1 = 10.0 \text{ cm} \)
\( \rho_1 = 1 \text{ gm cm}^{-3} \)
For the spirit column in the other arm of the U-tube:
\( h_2 = 12.5 \text{ cm} \)
\( \rho_2 = ? \)
If \( P_1 \) and \( P_2 \) are the pressures exerted by the water column and spirit column, respectively,
Then \( P_1 = h_1 \rho_1 g \) (i)
and \( P_2 = h_2 \rho_2 g \) (ii)
Also, the mercury columns in the two arms of the U-tube are at the same level, meaning they are in equilibrium,
\( P_1 = P_2 \)
or \( h_1 \rho_1 g = h_2 \rho_2 g \)
or \( \rho_2 = \frac{h_1 \rho_1}{h_2} = \frac{10 \times 1}{12.5} = \frac{4}{5} = 0.8 \text{ g cm}^{-3} \)
Now, specific gravity of spirit \( = \frac{\text{density of spirit}}{\text{density of water}} \)
\( = \frac{0.8 \text{ g cm}^{-3}}{1 \text{ g cm}^{-3}} = 0.800 \).

In simple words: Since the mercury levels are equal, the pressure from the water column in one arm is the same as the pressure from the spirit column in the other. Using the heights and density of water, we can calculate the density of the spirit, and then its specific gravity.

Exam Tip: For U-tube problems with different liquids, remember that the pressure at the same horizontal level in a continuous fluid must be equal. Use \( P = h \rho g \) and ensure consistent units for height and density.

Question 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? (Specific gravity of mercury = \(13.6\)).
Answer:
When \(15\) cm of water and spirit are each poured into the U-tube's respective arms, the mercury level will rise in the arm containing the spirit.
Let \( \rho_m \) be the density of mercury.
Let's choose two points \( A \) and \( B \) on the same horizontal plane. Thus, according to Pascal's law,
Pressure at \( A \) = Pressure at \( B \)
or \( P_0 + h_w \rho_w g = P_0 + h_s \rho_s g + h_m \rho_m g \)
where \( P_0 \) is atmospheric pressure
or \( h_w \rho_w = h_s \rho_s + h_m \rho_m \)
or \( h_m \rho_m = h_w \rho_w - h_s \rho_s \) (i)
Specific gravity of mercury \( = 13.6 \)
Here, \( h_w = \text{height of water column} = 10 + 15 = 25 \text{ cm} \)
\( h_s = \text{height of spirit column} = 12.5 + 15 = 27.5 \text{ cm} \)
\( \rho_w = 1 \text{ g cm}^{-3} \)
\( \rho_s = 0.8 \text{ g cm}^{-3} \)
\( \rho_m = \text{density of mercury} = 13.6 \text{ g cm}^{-3} \)
From equation (i), we get
\( h_m \times 13.6 = 25 \times 1 - 27.5 \times 0.8 \)
or \( h_m = \frac{25 - 22.00}{13.6} = 0.2206 \text{ cm} \).
So, \( h_m = 0.221 \text{ cm} \).
This means mercury will rise in the arm containing the spirit.
In simple words: We add more water and spirit to the U-tube. Then, we use the principle that pressure is equal at the same horizontal level. By calculating the new total pressures from the water and spirit columns, we can find the height difference in the mercury levels.

Exam Tip: This is an extension of the U-tube problem. Carefully track the new heights of the liquid columns after adding more fluid. The key is still balancing the pressures at the lowest interface (mercury in this case).

Question 11. Can Bernoulli's equation be used to describe the flow of water through a rapid in a river? Explain?
Answer:
No, Bernoulli's equation cannot be used to describe the flow of water through a rapid river. This is because Bernoulli's equation applies to streamline flow, and in a rapid river, the water flow is not streamline.
In simple words: No, Bernoulli's equation is for smooth, orderly flow. A rapid river has chaotic, turbulent flow, so the equation doesn't apply there.

Exam Tip: Always remember the assumptions for Bernoulli's equation: incompressible fluid, non-viscous fluid, and streamline (laminar) flow. Turbulent flow violates the streamline assumption.

Question 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 instead of absolute pressure when applying Bernoulli's equation, unless the atmospheric pressure at the two points where Bernoulli's equation is applied is significantly different.
In simple words: It usually doesn't matter if you use gauge or absolute pressure in Bernoulli's equation because the reference pressure (like atmospheric pressure) often cancels out. It only matters if the atmospheric pressure changes a lot between the two points you're looking at.

Exam Tip: Gauge pressure is the pressure relative to atmospheric pressure, while absolute pressure is relative to a perfect vacuum. In Bernoulli's equation, if the atmospheric pressure is constant at both points, it will subtract out, so the choice of gauge or absolute pressure doesn't change the difference. If it's not constant, then absolute pressure must be used.

Question 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 \(4.0 \times 10^{-3}\) kgs\(^{-1}\). What is the pressure difference between the two ends of the tube? (Density of glycerine = \(1.3 \times 10^3\) kgm\(^{-3}\) and viscosity of glycerine = \(0.83\) Pas). (You may also like to check if the assumption of laminar flow in the tube is correct)
Answer:
Here, radius, \( r = 1.0 \text{ cm} = 10^{-2} \text{ m} \)
Length of tube, \( l = 1.5 \text{ m} \)
Density of glycerine, \( \rho = 1.3 \times 10^3 \text{ kgm}^{-3} \)
Mass of glycerine flowing per second, \( M = 4 \times 10^{-3} \text{ kg s}^{-1} \)
Viscosity of glycerine, \( \eta = 0.83 \text{ Pas} = 0.83 \text{ Nm}^{-2} \text{ s} \).
Let \( P \) be the pressure difference between the two ends of the tube.
We need to find the pressure difference \( P \). This problem requires Poiseuille's formula, which is \( P = \frac{8 \eta l Q}{\pi r^4} \), where \( Q \) is the volume flow rate.
First, calculate the volume flow rate \( Q \).
\( Q = \frac{M}{\rho} = \frac{4.0 \times 10^{-3} \text{ kg s}^{-1}}{1.3 \times 10^3 \text{ kg m}^{-3}} \approx 3.077 \times 10^{-6} \text{ m}^3 \text{ s}^{-1} \)
Now, substitute the values into Poiseuille's formula:
\( P = \frac{8 \times 0.83 \text{ Nm}^{-2} \text{ s} \times 1.5 \text{ m} \times 3.077 \times 10^{-6} \text{ m}^3 \text{ s}^{-1}}{\pi \times (10^{-2} \text{ m})^4} \)
\( P = \frac{8 \times 0.83 \times 1.5 \times 3.077 \times 10^{-6}}{3.14159 \times 10^{-8}} \)
\( P \approx 961.4 \text{ Pa} \)
To check if the assumption of laminar flow in the tube is correct, we calculate the Reynolds number, \( N_R \).
\( N_R = \frac{2 \rho Q}{\pi r \eta} \)
\( N_R = \frac{2 \times 1.3 \times 10^3 \text{ kg m}^{-3} \times 3.077 \times 10^{-6} \text{ m}^3 \text{ s}^{-1}}{\pi \times 10^{-2} \text{ m} \times 0.83 \text{ Nm}^{-2} \text{ s}} \)
\( N_R \approx 0.308 \)
For laminar flow, \( N_R \) should be less than \( 2000 \). Since \( 0.308 < 2000 \), the flow is laminar.
In simple words: We used Poiseuille's formula to find the pressure difference needed to push the glycerine through the tube at the given rate. We first calculated the volume of glycerine flowing per second, then plugged all the values into the formula. We also checked that the flow was smooth and not turbulent by calculating the Reynolds number.

Exam Tip: For problems involving viscous fluid flow through a pipe, recall Poiseuille's equation for pressure difference \( (\Delta P) \) and the Reynolds number \( (N_R) \) for checking flow type. Ensure all units are in SI before calculations.

Question 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 \(70\) 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 \(1.3\) kg m\(^{-3}\)?
Answer:
Let \( v_1 \) and \( v_2 \) be the speeds on the upper and lower surfaces of the wings, and \( P_1 \) and \( P_2 \) be the pressures on the upper and lower surfaces, respectively.
Here, \( v_1 = 70 \text{ ms}^{-1} \)
\( v_2 = 63 \text{ ms}^{-1} \)
\( \rho = 1.3 \text{ kg m}^{-3} \)
The height levels of the upper and lower surfaces of the wings from the ground can be considered the same, so \( h_1 = h_2 \).
Area of wing, \( A = 2.5 \text{ m}^2 \)
Thus, from Bernoulli's Theorem,
\( P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2 \)
Since \( h_1 = h_2 \), the terms \( \rho g h_1 \) and \( \rho g h_2 \) cancel out.
\( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \)
Rearranging for pressure difference:
\( P_2 - P_1 = \frac{1}{2} \rho (v_1^2 - v_2^2) \) (i)
This pressure difference provides the lift to the aeroplane wing.
If \( F \) is the lift on the wing, then
\( F = (P_2 - P_1) \times A \)
\( F = \frac{1}{2} \rho (v_1^2 - v_2^2) \times A \) (using (i))
\( F = \frac{1}{2} \times 1.3 \times (70^2 - 63^2) \times 2.5 \)
\( F = \frac{1}{2} \times 1.3 \times (4900 - 3969) \times 2.5 \)
\( F = \frac{1}{2} \times 1.3 \times 931 \times 2.5 = 1512.9 \text{ N} \)
\( F = 1.5129 \times 10^3 \text{ N} \approx 1.513 \times 10^3 \text{ N} \)
\( F = 1.5 \times 10^3 \text{ N} \).
In simple words: We used Bernoulli's principle to find the difference in pressure between the top and bottom of the wing, which is caused by the different air speeds. Then, we multiplied this pressure difference by the wing's area to calculate the total lift force.

Exam Tip: This is a classic application of Bernoulli's principle for aerodynamic lift. Remember that higher speed leads to lower pressure. The lift force is simply the pressure difference multiplied by the wing area.

Question 15. Fig (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?