Why does a large ship float on water while a small stone sinks?
A ship floats because the water exerts an upward buoyant force on it. Although the ship is made of materials that are denser than water, its hollow structure makes its average density less than the density of water. The ship also displaces a large volume of water, producing enough buoyant force to balance its weight. A small stone, however, has a density greater than water and displaces only a small volume, so its weight is greater than the buoyant force and it sinks.
Why can a small needle float on water even though it is denser than water?
A small needle can float when it is placed gently on the water surface because of surface tension. The surface of water behaves like a stretched elastic film and can provide an upward force along the surface of the needle. If the surface film is disturbed or the needle is pushed below the surface, this support is lost and the needle sinks.
Why does honey flow more slowly than water?
Honey flows more slowly than water because honey has a much higher viscosity. Viscosity is the property of a fluid that opposes the relative motion between its adjacent layers. The greater internal resistance in honey makes its layers move less easily past one another, so honey flows slowly compared with water.
Why does the viscosity of a liquid decrease with temperature?
When the temperature of a liquid increases, its molecules gain more kinetic energy and move more vigorously. This reduces the effective influence of intermolecular cohesive forces that resist the relative motion of adjacent layers. Therefore, the layers can move more easily and the viscosity of the liquid decreases.
Why does the viscosity of a gas increase with temperature?
When the temperature of a gas increases, its molecules move faster and transfer momentum more rapidly between adjacent layers moving at different velocities. This increases the resistance to relative motion between the layers. Therefore, unlike liquids, the viscosity of a gas increases with temperature.
Why does a small sphere falling through a viscous liquid eventually move with constant velocity?
When a small sphere falls through a viscous liquid, its weight acts downward while buoyant force and viscous drag act upward. As the sphere speeds up, the viscous drag increases. Eventually, the upward forces become equal to the weight of the sphere. The net force then becomes zero, so the sphere continues to fall with a constant velocity called terminal velocity.
Why does a narrower pipe offer much greater resistance to flow?
According to Poiseuille’s law, the resistance to flow through a cylindrical tube is inversely proportional to the fourth power of its radius. Therefore, even a small decrease in radius produces a very large increase in resistance. This is why a narrow pipe offers much greater resistance to the flow of a liquid than a wider pipe.
\[ R \propto \frac{1}{r^4} \]
Why does water flow faster through a narrow section of a pipe?
For steady flow of an incompressible liquid, the same volume of liquid must pass through every cross-section of the pipe in a given time. Therefore, when the cross-sectional area decreases, the velocity must increase. This follows from the equation of continuity.
\[ A_1v_1=A_2v_2 \]
Why does the pressure of a flowing fluid decrease when its speed increases?
According to Bernoulli’s principle, for steady flow along a streamline, the total mechanical energy per unit volume remains constant. Therefore, when the speed of a fluid increases, some of the pressure energy is converted into kinetic energy. As a result, the static pressure decreases.
\[ P+\frac{1}{2}\rho v^2+\rho gh=\text{constant} \]
How does Bernoulli’s principle help lift an aircraft wing?
The shape and angle of an aircraft wing cause air to flow differently around its upper and lower surfaces. The resulting pressure distribution produces a net upward aerodynamic force called lift. Bernoulli’s principle helps explain the relationship between the air velocity and pressure in the flow around the wing.
Why does a man standing close to a railway track get pushed toward a fast-moving train?
When a fast-moving train passes close to a person, the air in the narrow region between the person and the train is disturbed and moves rapidly. The pressure in this region can become lower than the pressure on the outer side of the person. The higher pressure outside then produces a net force toward the train. This is why a person should maintain a safe distance from a moving train.
How does a paint gun work using Bernoulli’s principle?
In a paint gun, rapidly moving air passes over the opening of a tube containing paint. The high-speed air produces a region of lower pressure near the opening. The higher pressure on the paint surface pushes the paint upward, where the fast-moving air breaks it into fine droplets and sprays it onto the surface.
Why does a flag flutter when wind blows?
When wind flows past a flag, the flexible cloth experiences continuously changing aerodynamic forces and pressure differences. Small disturbances in the cloth are amplified by the airflow, producing waves and oscillations along the fabric. Therefore, the flag keeps changing its shape and appears to flutter.
Why are race cars aerodynamically designed to produce downforce?
Race cars are designed so that their aerodynamic surfaces produce a downward force called downforce. This increases the normal force between the tyres and the road without significantly increasing the car’s mass. The increased tyre grip helps the car maintain better traction and stability, especially while turning at high speed.
How does Pascal’s law help transmit pressure through an enclosed liquid?
Pascal’s law states that pressure applied to a confined liquid is transmitted equally and undiminished in all directions throughout the liquid. Thus, when pressure is applied at one point of an enclosed liquid, the pressure is transmitted to every part of the liquid and to the walls of its container.
\[ P=\frac{F}{A} \]
How does a hydraulic lift lift a heavy vehicle using a small force?
A hydraulic lift uses Pascal’s law. A small force applied to a small piston produces pressure in the enclosed liquid. The same pressure acts on a larger piston, producing a much larger force. Therefore, a relatively small input force can lift a heavy vehicle.
\[ \frac{F_1}{A_1}=\frac{F_2}{A_2} \]
How does Archimedes’ principle explain buoyant force?
When an object is partially or completely immersed in a fluid, the fluid exerts pressure on all surfaces of the object. The pressure is greater at greater depths, so the upward pressure force on the lower surface is greater than the downward pressure force on the upper surface. The difference produces an upward buoyant force equal to the weight of the fluid displaced by the object.
\[ F_B=\rho Vg \]
Why does an object float, sink, or remain suspended in a liquid?
An object floats when the buoyant force can balance its weight, sinks when its weight is greater than the maximum buoyant force available, and remains suspended when its weight equals the buoyant force while fully immersed. In terms of average density, an object generally floats when its average density is less than the liquid’s density and sinks when it is greater.
Why does an object appear to lose weight when immersed in water?
When an object is immersed in water, water exerts an upward buoyant force on it. This upward force opposes the object’s weight, so the tension in the supporting string or the reading of a spring balance decreases. Therefore, the object appears to weigh less in water.
\[ W_{\text{apparent}}=W-F_B \]
Why does a liquid rise in a narrow capillary tube?
A liquid rises in a capillary tube when the liquid wets the tube wall and the angle of contact is less than \(90^\circ\). Surface tension acts along the contact line and has an upward component that pulls the liquid upward. The liquid rises until the upward force is balanced by the weight of the liquid column.
\[ h=\frac{2T\cos\theta}{r\rho g} \]
Why does mercury get depressed in a glass capillary tube?
Mercury does not wet clean glass effectively because the cohesive forces between mercury molecules are stronger than the adhesive forces between mercury and glass. Therefore, the mercury surface becomes convex and the angle of contact is greater than \(90^\circ\). The surface-tension force has a downward component, causing the mercury level to fall below the outside level.
Why does water form a concave meniscus in glass?
Water forms a concave meniscus in a clean glass tube because the adhesive force between water molecules and glass is stronger than the cohesive force between water molecules. Water therefore wets the glass surface and rises near the walls, producing a concave curved surface.
Why is a small water drop nearly spherical?
Surface tension tends to reduce the free surface area of a liquid. For a given volume, a sphere has the minimum possible surface area. Therefore, surface tension pulls a small water drop toward a nearly spherical shape, which minimizes its surface energy.
Why does a large raindrop become flattened instead of remaining perfectly spherical?
Surface tension tends to make a raindrop spherical, but as the drop becomes larger and falls faster, air resistance becomes significant. The airflow around the drop produces aerodynamic forces that deform it. Large raindrops therefore become flattened or oblate rather than remaining perfectly spherical.
Why don’t swan feathers get wet easily in water?
Swan feathers are coated with natural oils during preening, making their surfaces hydrophobic. Water therefore does not easily spread or penetrate the feather structure. Surface tension also helps water remain in droplets on the feather surface, allowing the water to roll off more easily.
Why does soap help water spread more easily?
Soap reduces the surface tension of water. With lower surface tension, water can spread more readily over surfaces instead of remaining as nearly spherical droplets. This improves the wetting and cleaning action of water.
Why does a liquid tend to minimize its free surface area?
Molecules at the free surface of a liquid experience unbalanced intermolecular forces, giving the surface excess energy. The liquid tends to reduce this surface energy by reducing its surface area. Therefore, surface tension makes a free liquid surface tend to occupy the minimum possible area.
Why is there excess pressure inside a liquid drop?
Surface tension acts along the curved surface of a liquid drop and tends to contract the surface. To balance this inward tendency, the pressure inside the drop must be greater than the pressure outside. For a spherical liquid drop of radius \(R\), the excess pressure is:
\[ \Delta P=P_{\text{inside}}-P_{\text{outside}}=\frac{2T}{R} \]
Why is excess pressure inside a soap bubble greater than inside an air bubble in a liquid?
A soap bubble has two liquid-air surfaces, an inner surface and an outer surface. Both surfaces contribute to the excess pressure inside the bubble. An air bubble inside a liquid has only one liquid-air surface. Therefore, for the same radius and surface tension, a soap bubble has twice the excess pressure.
Air bubble in liquid: \[ \Delta P=\frac{2T}{R} \]
Soap bubble: \[ \Delta P=\frac{4T}{R} \]
Why does liquid pressure increase with depth?
As depth increases, the amount of liquid above a point increases. The weight of this liquid produces greater pressure on the layers below. Therefore, pressure in a liquid increases with depth. For a liquid of density \(\rho\), the pressure at depth \(h\) due to the liquid column is:
\[ P=\rho gh \]
If atmospheric pressure is also considered, the total pressure is:
\[ P_{\text{total}}=P_0+\rho gh \]
10 Solved Numerical Problems – Fluid Mechanics
The following solved numerical problems cover the important numerical applications of Fluid Mechanics. Each problem is solved step by step using the appropriate formula.
1. Calculate the Coefficient of Viscosity of a Liquid
A liquid is flowing between two parallel layers. A tangential force of \(0.12\,\text{N}\) is required to maintain a velocity gradient of \(3\,\text{s}^{-1}\) between layers of area \(0.04\,\text{m}^2\). Calculate the coefficient of viscosity of the liquid.
Given:
\[ F=0.12\,\text{N} \]
\[ A=0.04\,\text{m}^2 \]
\[ \frac{dv}{dy}=3\,\text{s}^{-1} \]
Formula:
\[ \eta=\frac{F}{A\left(\frac{dv}{dy}\right)} \]
Substitution:
\[ \eta=\frac{0.12}{0.04\times3} \]
\[ \eta=\frac{0.12}{0.12} \]
\[ \boxed{\eta=1\,\text{Pa}\cdot\text{s}} \]
Answer: The coefficient of viscosity of the liquid is \(1\,\text{Pa}\cdot\text{s}\).
2. Calculate the Viscous Force Using Stokes’ Law
A small spherical particle of radius \(0.5\,\text{mm}\) moves through a liquid of viscosity \(0.9\,\text{Pa}\cdot\text{s}\) with a speed of \(0.02\,\text{m/s}\). Calculate the viscous force acting on the particle.
Given:
\[ r=0.5\,\text{mm}=5\times10^{-4}\,\text{m} \]
\[ \eta=0.9\,\text{Pa}\cdot\text{s} \]
\[ v=0.02\,\text{m/s} \]
Formula:
\[ F_v=6\pi\eta rv \]
Substitution:
\[ F_v=6\pi(0.9)(5\times10^{-4})(0.02) \]
\[ F_v\approx1.70\times10^{-4}\,\text{N} \]
\[ \boxed{F_v\approx1.7\times10^{-4}\,\text{N}} \]
Answer: The viscous force acting on the particle is approximately \(1.7\times10^{-4}\,\text{N}\).
3. Calculate the Terminal Velocity of a Small Sphere
A small spherical particle of radius \(1\,\text{mm}\) and density \(7800\,\text{kg/m}^3\) falls through a liquid of density \(1000\,\text{kg/m}^3\) and viscosity \(1.2\,\text{Pa}\cdot\text{s}\). Calculate its terminal velocity. Take \(g=9.8\,\text{m/s}^2\).
Given:
\[ r=1\times10^{-3}\,\text{m} \]
\[ \rho_s=7800\,\text{kg/m}^3 \]
\[ \rho_l=1000\,\text{kg/m}^3 \]
\[ \eta=1.2\,\text{Pa}\cdot\text{s} \]
\[ g=9.8\,\text{m/s}^2 \]
Formula:
\[ v_t=\frac{2r^2g(\rho_s-\rho_l)}{9\eta} \]
Substitution:
\[ v_t= \frac{2(1\times10^{-3})^2(9.8)(7800-1000)} {9(1.2)} \]
\[ v_t\approx1.24\times10^{-2}\,\text{m/s} \]
\[ \boxed{v_t\approx0.0124\,\text{m/s}} \]
Answer: The terminal velocity of the sphere is approximately \(0.0124\,\text{m/s}\).
4. Calculate the Reynolds Number of a Flowing Liquid
Water flows through a pipe of diameter \(1\,\text{cm}\) with an average velocity of \(0.5\,\text{m/s}\). If the density of water is \(1000\,\text{kg/m}^3\) and its coefficient of viscosity is \(1\times10^{-3}\,\text{Pa}\cdot\text{s}\), calculate the Reynolds number and identify the nature of the flow.
Given:
\[ \rho=1000\,\text{kg/m}^3 \]
\[ v=0.5\,\text{m/s} \]
\[ d=1\,\text{cm}=0.01\,\text{m} \]
\[ \eta=1\times10^{-3}\,\text{Pa}\cdot\text{s} \]
Formula:
\[ Re=\frac{\rho vd}{\eta} \]
Substitution:
\[ Re= \frac{(1000)(0.5)(0.01)} {1\times10^{-3}} \]
\[ \boxed{Re=5000} \]
Since the Reynolds number is greater than approximately \(3000\) for this standard pipe-flow classification, the flow is turbulent.
Answer: The Reynolds number is \(5000\), indicating turbulent flow.
5. Calculate the Volume of Liquid Flowing Through a Capillary Tube
Water flows through a capillary tube of radius \(0.5\,\text{mm}\) and length \(20\,\text{cm}\). The pressure difference between its ends is \(4000\,\text{Pa}\). If the viscosity of water is \(1\times10^{-3}\,\text{Pa}\cdot\text{s}\), calculate the volume of water flowing through the tube per second.
Given:
\[ r=0.5\,\text{mm}=5\times10^{-4}\,\text{m} \]
\[ l=20\,\text{cm}=0.20\,\text{m} \]
\[ \Delta P=4000\,\text{Pa} \]
\[ \eta=1\times10^{-3}\,\text{Pa}\cdot\text{s} \]
Formula:
\[ Q=\frac{\pi\Delta P r^4}{8\eta l} \]
Substitution:
\[ Q= \frac{\pi(4000)(5\times10^{-4})^4} {8(1\times10^{-3})(0.20)} \]
\[ Q\approx1.23\times10^{-4}\,\text{m}^3/\text{s} \]
\[ \boxed{Q\approx1.23\times10^{-4}\,\text{m}^3/\text{s}} \]
Answer: Approximately \(1.23\times10^{-4}\,\text{m}^3\) of water flows through the tube every second.
6. Calculate the Capillary Rise of Water
Water rises in a capillary tube of radius \(0.5\,\text{mm}\). The surface tension of water is \(0.072\,\text{N/m}\), and the angle of contact is \(0^\circ\). Calculate the height to which water rises. Take the density of water as \(1000\,\text{kg/m}^3\) and \(g=9.8\,\text{m/s}^2\).
Given:
\[ T=0.072\,\text{N/m} \]
\[ r=0.5\,\text{mm}=5\times10^{-4}\,\text{m} \]
\[ \theta=0^\circ \]
\[ \rho=1000\,\text{kg/m}^3 \]
\[ g=9.8\,\text{m/s}^2 \]
Formula:
\[ h=\frac{2T\cos\theta}{r\rho g} \]
Since \(\theta=0^\circ\),
\[ \cos0^\circ=1 \]
Therefore,
\[ h=\frac{2T}{r\rho g} \]
Substitution:
\[ h= \frac{2(0.072)} {(5\times10^{-4})(1000)(9.8)} \]
\[ h\approx0.0294\,\text{m} \]
\[ \boxed{h\approx2.94\,\text{cm}} \]
Answer: Water rises approximately \(2.94\,\text{cm}\) in the capillary tube.
7. Determine Surface Tension Using the Capillary Rise Method
A liquid rises to a height of \(3\,\text{mm}\) in a capillary tube of radius \(0.4\,\text{mm}\). The density of the liquid is \(1000\,\text{kg/m}^3\), and the angle of contact is \(0^\circ\). Calculate the surface tension of the liquid. Take \(g=9.8\,\text{m/s}^2\).
Given:
\[ h=3\,\text{mm}=3\times10^{-3}\,\text{m} \]
\[ r=0.4\,\text{mm}=4\times10^{-4}\,\text{m} \]
\[ \rho=1000\,\text{kg/m}^3 \]
\[ \theta=0^\circ \]
\[ g=9.8\,\text{m/s}^2 \]
Formula:
\[ T=\frac{hr\rho g}{2\cos\theta} \]
Since \(\theta=0^\circ\),
\[ \cos0^\circ=1 \]
Therefore,
\[ T=\frac{hr\rho g}{2} \]
Substitution:
\[ T= \frac{(3\times10^{-3})(4\times10^{-4})(1000)(9.8)} {2} \]
\[ T=5.88\times10^{-3}\,\text{N/m} \]
\[ \boxed{T=5.88\times10^{-3}\,\text{N/m}} \]
Answer: The surface tension of the liquid is \(5.88\times10^{-3}\,\text{N/m}\).
8. Calculate the Work Done in Increasing the Surface Area of a Liquid
The surface tension of a liquid is \(0.05\,\text{N/m}\). Calculate the work required to increase its surface area by \(0.02\,\text{m}^2\).
Given:
\[ T=0.05\,\text{N/m} \]
\[ \Delta A=0.02\,\text{m}^2 \]
Formula:
\[ W=T\Delta A \]
Substitution:
\[ W=(0.05)(0.02) \]
\[ W=1\times10^{-3}\,\text{J} \]
\[ \boxed{W=1\times10^{-3}\,\text{J}} \]
Answer: The work required is \(1\times10^{-3}\,\text{J}\).
9. Calculate Excess Pressure in a Liquid Drop and a Soap Bubble
The surface tension of a liquid is \(0.072\,\text{N/m}\). Calculate the excess pressure inside a spherical liquid drop of radius \(2\,\text{mm}\). Also calculate the excess pressure inside a soap bubble of the same radius.
Given:
\[ T=0.072\,\text{N/m} \]
\[ R=2\,\text{mm}=2\times10^{-3}\,\text{m} \]
(a) Liquid drop
Formula:
\[ \Delta P=\frac{2T}{R} \]
Substitution:
\[ \Delta P= \frac{2(0.072)} {2\times10^{-3}} \]
\[ \boxed{\Delta P=72\,\text{Pa}} \]
(b) Soap bubble
A soap bubble has two surfaces, so its excess pressure is:
\[ \Delta P=\frac{4T}{R} \]
Substitution:
\[ \Delta P= \frac{4(0.072)} {2\times10^{-3}} \]
\[ \boxed{\Delta P=144\,\text{Pa}} \]
Answer: The excess pressure is \(72\,\text{Pa}\) inside the liquid drop and \(144\,\text{Pa}\) inside the soap bubble.
10. Use Continuity and Bernoulli’s Principle to Find Pressure
Water flows through a horizontal pipe. The cross-sectional area at point 1 is four times the cross-sectional area at point 2. If the speed of water at point 1 is \(2\,\text{m/s}\) and the pressure there is \(2.0\times10^5\,\text{Pa}\), calculate the speed and pressure of water at point 2. Take the density of water as \(1000\,\text{kg/m}^3\).
Given:
\[ A_1=4A_2 \]
\[ v_1=2\,\text{m/s} \]
\[ P_1=2.0\times10^5\,\text{Pa} \]
\[ \rho=1000\,\text{kg/m}^3 \]
Since the pipe is horizontal,
\[ h_1=h_2 \]
Step 1: Find the velocity at point 2
Using the equation of continuity,
\[ A_1v_1=A_2v_2 \]
Since \(A_1=4A_2\),
\[ 4A_2v_1=A_2v_2 \]
\[ v_2=4v_1 \]
\[ v_2=4(2) \]
\[ \boxed{v_2=8\,\text{m/s}} \]
Step 2: Find the pressure at point 2
Bernoulli’s equation is:
\[ P_1+\frac{1}{2}\rho v_1^2+\rho gh_1 = P_2+\frac{1}{2}\rho v_2^2+\rho gh_2 \]
Since \(h_1=h_2\),
\[ P_1+\frac{1}{2}\rho v_1^2 = P_2+\frac{1}{2}\rho v_2^2 \]
Therefore,
\[ P_2=P_1+\frac{1}{2}\rho(v_1^2-v_2^2) \]
Substitution:
\[ P_2= 2.0\times10^5 +\frac{1}{2}(1000)(2^2-8^2) \]
\[ P_2=2.0\times10^5-30000 \]
\[ \boxed{P_2=1.70\times10^5\,\text{Pa}} \]
Answer: The speed at point 2 is \(8\,\text{m/s}\), and the pressure at point 2 is \(1.70\times10^5\,\text{Pa}\).