Maharashtra Board Class 12 Physics Chapter 2 Mechanical Properties of Fluids Exercise Solutions

1. Multiple Choice Questions

(i) A hydraulic lift is designed to lift heavy objects of a maximum mass of 2000 kg. The area of cross-section of piston carrying the load is \( 2.25 \times 10^{-2} \text{ m}^2 \). What is the maximum pressure the piston would have to bear?
(a) \( 0.8711 \times 10^6 \text{ N/m}^2 \)
(b) \( 0.5862 \times 10^7 \text{ N/m}^2 \)
(c) \( 0.4869 \times 10^5 \text{ N/m}^2 \)
(d) \( 0.3271 \times 10^4 \text{ N/m}^2 \)
Answer: (a) \( 0.8711 \times 10^6 \text{ N/m}^2 \)
In simple words: Pressure is force divided by area. Here, the force is the weight of the 2000 kg mass, and when you divide that weight by the piston's area, you get the pressure it must withstand.

📝 Teacher's Note: Remind students that force is \( F = mg \). Using \( g = 9.8 \text{ m/s}^2 \) is standard unless the problem specifies \( 10 \text{ m/s}^2 \).

🎯 Exam Tip: Always double-check your power-of-ten calculations, as these are the most common places to lose marks in pressure problems.

(ii) Two capillary tubes of radii 0.3 cm and 0.6 cm are dipped in the same liquid. The ratio of heights through which the liquid will rise in the tubes is
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer: (b) 2:1
In simple words: In narrow tubes, liquid rises higher. Since one tube is half as wide as the other, the liquid will rise twice as high in the thinner tube.

📝 Teacher's Note: This is a direct application of Jurin's Law, where height \( h \) is inversely proportional to radius \( r \) (\( h \propto 1/r \)).

🎯 Exam Tip: When the question asks for a ratio of \( h_1:h_2 \), ensure you use the inverse ratio of the radii \( r_2:r_1 \).

(iii) The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly
(a) \( 0.9 \times 10^{-3} \text{ J} \)
(b) \( 0.4 \times 10^{-3} \text{ J} \)
(c) \( 0.7 \times 10^{-3} \text{ J} \)
(d) \( 0.5 \times 10^{-3} \text{ J} \)
Answer: (a) \( 0.9 \times 10^{-3} \text{ J} \)
In simple words: A soap bubble has two surfaces (inside and outside), so the total surface area is doubled. The energy is the surface tension multiplied by this total area.

📝 Teacher's Note: Students often forget the factor of 2 for soap bubbles. Emphasize that bubbles have two liquid-air interfaces, unlike drops which have only one.

🎯 Exam Tip: Use the radius (3 cm = 0.03 m) in calculations, not the diameter (6 cm). Converting units to SI (meters) is essential before starting.

(iv) Two hail stones with radii in the ratio of 1:4 fall from a great height through the atmosphere. Then the ratio of their terminal velocities is
(a) 1:2
(b) 1:12
(c) 1:16
(d) 1:8
Answer: (c) 1:16
In simple words: Terminal velocity depends on the square of the radius. If the radius increases by 4 times, the velocity increases by \( 4 \times 4 \), which is 16 times.

📝 Teacher's Note: Use Stokes' Law to show that terminal velocity \( v \propto r^2 \). This helps students visualize why larger raindrops fall faster than smaller mist droplets.

🎯 Exam Tip: Identify the relationship between variables first (\( v \propto r^2 \)) to quickly solve ratio-based MCQs without full derivation.

(v) In Bernoulli’s theorem, which of the following is conserved?
(a) linear momentum
(b) angular momentum
(c) mass
(d) energy
Answer: (d) energy
In simple words: Bernoulli's principle states that for a flowing fluid, the total mechanical energy (pressure + kinetic + potential) stays the same along a streamline.

📝 Teacher's Note: Contrast this with the Equation of Continuity, which is based on the conservation of mass, to avoid confusion.

🎯 Exam Tip: "Bernoulli = Energy" is a high-frequency exam concept. Memorize this fundamental link.

2. Answer in brief.

(i) Why is the surface tension of paints and lubricating oils kept low?
Answer: For better wettability (surface coverage), the surface tension and angle of contact of paints and lubricating oils must be low. Low surface tension allows these liquids to spread easily over a surface rather than forming droplets.
In simple words: Low surface tension helps paint spread smoothly on a wall and helps oil get into tiny cracks in an engine to lubricate them better.

📝 Teacher's Note: Explain "wetting" using the analogy of water on a waxed car (high surface tension, poor wetting) vs. soapy water (low surface tension, good wetting).

🎯 Exam Tip: The keyword here is "wettability" or "surface coverage." Mentioning "angle of contact" also adds value to your answer.

(ii) How much amount of work is done in forming a soap bubble of radius r?
Answer: Let T be the surface tension of a soap solution. The initial surface area of soap bubble = 0. The final surface area of soap bubble = \( 2 \times 4\pi r^2 \) (as a bubble has two surfaces).
\( \implies \) The increase in surface area \( \Delta A = 8\pi r^2 \).
The work done in blowing the soap bubble is \( W = \text{surface tension} \times \text{increase in surface area} \)
\( \implies W = T \times 8\pi r^2 = 8\pi r^2 T \)
In simple words: To make a bubble, you have to stretch the soap film. The energy required to do this stretching is the surface tension multiplied by the total area of the two film surfaces created.

📝 Teacher's Note: Derive this step-by-step on the board, emphasizing why the area is \( 2 \times 4\pi r^2 \) and not just \( 4\pi r^2 \).

🎯 Exam Tip: In derivations, clearly define your symbols (e.g., "Let T be surface tension"). Always end with the formula clearly boxed or highlighted.

(iii) What is the basis of the Bernoulli’s principle?
Answer: Conservation of energy.
In simple words: It is simply the Law of Conservation of Energy applied to moving fluids.

📝 Teacher's Note: Bernoulli's principle assumes the fluid is non-viscous and incompressible for the conservation of mechanical energy to be perfect.

🎯 Exam Tip: A one-line answer is sufficient for this specific question type. Don't over-explain unless marks allow.

(iv) Why is a low density liquid used as a manometric liquid in a physics laboratory?
Answer: An open tube manometer measures the gauge pressure, \( p - p_0 = h\rho g \), where \( p \) is the pressure being measured, \( p_0 \) is the atmospheric pressure, \( h \) is the difference in height between the manometric liquid of density \( \rho \) in the two arms. For a given pressure difference, the height \( h \) is inversely proportional to density \( \rho \). That is, \( \rho \) should be small for \( h \) to be large. Therefore, for a noticeably large and easily readable \( h \), laboratory manometers use a low density liquid.
In simple words: If the liquid is heavy (dense), it won't move much when pressure changes. A lighter liquid moves more, making it easier to see and measure small changes in pressure.

📝 Teacher's Note: Use the analogy of trying to move a heavy rock vs. a light ball with the same puff of air. The light ball (low density) shows more reaction (height).

🎯 Exam Tip: Use the formula \( h = \frac{p-p_0}{\rho g} \) to justify that \( h \propto 1/\rho \). Mathematical proof always strengthens a "Why" answer.

(v) What is an incompressible fluid?
Answer: An incompressible fluid is one which does not undergo a change in volume for a large range of pressures. Thus, its density has a constant value throughout the fluid. In most cases, all liquids are considered incompressible.
In simple words: It's a fluid that you can't "squish" into a smaller space, no matter how much pressure you apply. Its density stays the same.

📝 Teacher's Note: Mention that while gases are highly compressible, liquids like water only compress by tiny fractions even under massive pressure.

🎯 Exam Tip: The key phrases are "no change in volume" and "constant density."

Question 3. Why two or more mercury drops form a single drop when brought in contact with each other?
Answer: A spherical shape has the minimum surface area-to-volume ratio of all geometric forms. When two drops of a liquid are brought in contact, the cohesive forces between their molecules coalesce the drops into a single larger drop. This happens because, for the same total volume of liquid, the surface area of one large resulting drop is less than the combined surface area of the smaller individual drops. This decrease in surface area leads to a decrease in surface energy, which is a more stable state. The excess energy is released as heat.

Proof: Let \( n \) droplets each of radius \( r \) coalesce to form a single drop of radius \( R \).
As volume remains constant:
Volume of the drop = volume of \( n \) droplets
\( \implies \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \)
\( \implies R^3 = nr^3 \)
\( \implies R = n^{1/3}r \)
Surface area of \( n \) droplets = \( n \times 4\pi r^2 \)
Surface area of the final drop = \( 4\pi R^2 = 4\pi (n^{1/3}r)^2 = n^{2/3} \times 4\pi r^2 \)
\( \implies \) Change in surface area = \( 4\pi r^2 (n^{2/3} - n) \)
Since \( n > n^{2/3} \) for \( n > 1 \), the term is negative, indicating a decrease in surface area.
In simple words: Nature likes to save energy. Since one big drop has less surface than two small ones, mercury drops join together to minimize their surface and get to a lower energy state.

📝 Teacher's Note: This is a great way to introduce the concept of "minimizing potential energy." Liquids always try to reach the configuration with the least surface area.

🎯 Exam Tip: In the explanation, use the term "cohesive forces" and "surface energy minimization." These are the technical terms examiners look for.

Question 4. Why does velocity increase when water flowing in broader pipe enters a narrow pipe?
Answer: According to the Equation of Continuity, for a steady flow of an incompressible fluid, the product of cross-sectional area \( A \) and velocity \( v \) remains constant (\( Av = \text{constant} \)). When a tube narrows, the cross-sectional area \( A \) decreases. To keep the product \( Av \) constant, the velocity \( v \) must increase. This ensures that the same volume of fluid that enters the broad part in a given time must exit the narrow part in the same amount of time.
In simple words: Think of a crowd of people in a wide hallway suddenly entering a narrow door; they have to move much faster through the door to keep the line from backing up.

📝 Teacher's Note: Use the formula \( A_1v_1 = A_2v_2 \) and show that if \( A_2 < A_1 \), then \( v_2 > v_1 \).

🎯 Exam Tip: Always mention the "Equation of Continuity" by name when answering questions about fluid speed and pipe width.

Question 5. Why does the speed of a liquid increase and its pressure decrease when a liquid passes through constriction in a horizontal pipe?
Answer: Consider a horizontal constricted tube. By the equation of continuity, \( v_1 A_1 = v_2 A_2 \). Since at the constriction the area \( A_2 \) is less than \( A_1 \), the velocity \( v_2 \) must be greater than \( v_1 \).
According to Bernoulli's equation for a horizontal pipe:
\( p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2 \)
\( \implies p_1 - p_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) \)
Since \( v_2 > v_1 \), the term \( (v_2^2 - v_1^2) \) is positive, meaning \( p_1 - p_2 > 0 \), or \( p_1 > p_2 \). Thus, as speed increases at the constriction, the pressure must decrease to keep the total energy constant.
In simple words: When liquid speeds up to squeeze through a narrow gap, it uses up some of its "pressure energy" to gain "motion energy" (kinetic energy). So, the pressure drops where the speed is highest.

📝 Teacher's Note: This is the Venturi effect. Use the example of how a perfume sprayer works or how air moves over an airplane wing to illustrate this.

🎯 Exam Tip: Be careful to state that the pipe is "horizontal" so that potential energy (\( h\rho g \)) can be ignored in the equation.

Question 6. Derive an expression of excess pressure inside a liquid drop.
Answer: Consider a small spherical liquid drop with a radius \( R \). The pressure \( p \) inside is greater than the pressure \( p_0 \) outside.
Initial surface area \( A = 4\pi R^2 \).
Imagine the radius increases by a tiny amount \( dR \).
New surface area \( = 4\pi (R + dR)^2 \approx 4\pi (R^2 + 2RdR) \).
\( \implies \) Increase in area \( dA = 8\pi RdR \).
The increase in surface energy is \( dW = T \cdot dA = 8\pi TR dR \). ... (1)
The work done by the excess pressure \( (p - p_0) \) is:
\( dW = \text{Force} \times \text{distance} = (\text{excess pressure} \times \text{area}) \times dR \)
\( \implies dW = (p - p_0) \times 4\pi R^2 \times dR \). ... (2)
Equating (1) and (2):
\( (p - p_0) \times 4\pi R^2 \times dR = 8\pi TR dR \)
\( \implies p - p_0 = \frac{2T}{R} \)
This is Laplace’s law for a liquid drop.
In simple words: Because surface tension acts like a tight skin pulling inward, the liquid inside is squeezed. This "squeeze" creates extra pressure inside the drop compared to the air outside.

📝 Teacher's Note: Clarify the difference between a drop (one surface: \( 2T/R \)) and a soap bubble (two surfaces: \( 4T/R \)).

🎯 Exam Tip: This is a standard 3-mark derivation. Draw a neat diagram showing the drop and the infinitesimal increase \( dR \) to score full marks.

Question 7. Obtain an expression for conservation of mass starting from the equation of continuity.
Answer: Consider a fluid in steady flow through a tube with varying cross-sections \( A_1 \) and \( A_2 \). Let \( v_1 \) and \( v_2 \) be the velocities at these points.
The volume flux (volume per unit time) is given by:
\( \frac{dV_1}{dt} = A_1v_1 \) and \( \frac{dV_2}{dt} = A_2v_2 \)
By the equation of continuity for an incompressible fluid, volume flux is constant: \( A_1v_1 = A_2v_2 \).
If \( \rho_1 \) and \( \rho_2 \) are the densities at points 1 and 2, the mass flux (mass per unit time) is:
\( \frac{dm_1}{dt} = \rho_1 A_1 v_1 \) and \( \frac{dm_2}{dt} = \rho_2 A_2 v_2 \)
Since mass cannot be created or destroyed within the flow tube (steady flow), mass entering must equal mass leaving:
\( \frac{dm_1}{dt} = \frac{dm_2}{dt} \)
\( \implies \rho_1 A_1 v_1 = \rho_2 A_2 v_2 \)
For an incompressible fluid (\( \rho_1 = \rho_2 \)), this reduces to \( Av = \text{constant} \).
In simple words: If you put 1 kg of water into a pipe every second, 1 kg must come out every second. If the pipe gets narrow, that same 1 kg just has to move faster.

📝 Teacher's Note: Remind students that mass flux is simply density multiplied by volume flux (\( \rho \times Av \)).

🎯 Exam Tip: Clearly state the assumption of "steady flow" and "conservation of mass" at the beginning of the derivation.

Question 8. Explain the capillary action.
Answer: Capillary action is the phenomenon of rise or fall of a liquid inside a capillary tube when it is dipped in the liquid.

(1) Capillary Rise (Wetting Liquid): When a tube is dipped in a liquid like water, the meniscus is concave. The pressure on the concave side (point A, above the meniscus) is atmospheric (\( p_A \)). The pressure at point B just below the meniscus is \( p_B < p_A \) due to surface tension. Outside the tube at the same level, pressure is atmospheric (\( p_D = p_A \)). Since \( p_D > p_B \), the liquid is pushed up the tube until the hydrostatic pressure of the liquid column balances the pressure difference.

(2) Capillary Depression (Non-wetting Liquid): For liquids like mercury, the meniscus is convex. The pressure just below the meniscus (\( p_B \)) is greater than the atmospheric pressure (\( p_A \)). Since \( p_B > p_D \) (where \( p_D \) is atmospheric pressure at the same level outside), the liquid flows out of the tube, causing the level inside the tube to fall below the outside level.
In simple words: Surface tension creates a pressure difference at the curved surface. Depending on whether the curve points up or down, this pressure difference either "sucks" the liquid up or "pushes" it down the thin tube.

📝 Teacher's Note: Use a sponge or a piece of chalk dipping in ink as a real-world example of capillary action in tiny "tubes."

🎯 Exam Tip: Mention the pressure difference across the curved surface as the fundamental reason. Using diagrams for rise and fall is essential.

Question 9. Derive an expression for capillary rise for a liquid having a concave meniscus.
Answer: Consider a capillary tube of radius \( r \) dipped in a liquid of density \( \rho \) and surface tension \( T \). Let \( h \) be the capillary rise and \( \theta \) be the angle of contact. The radius of curvature of the meniscus is \( R = \frac{r}{\cos \theta} \).
The gauge pressure (pressure difference) across the meniscus is given by Laplace's Law:
\( p - p_0 = \frac{2T}{R} \) ... (1)
Also, the pressure due to the liquid column of height \( h \) is:
\( p - p_0 = h\rho g \) ... (2)
Equating (1) and (2):
\( h\rho g = \frac{2T}{R} \)
Substituting \( R = \frac{r}{\cos \theta} \):
\( h\rho g = \frac{2T \cos \theta}{r} \)
\( \implies h = \frac{2T \cos \theta}{r\rho g} \)
This is the expression for capillary rise.
In simple words: The upward pull from surface tension at the edges of the tube has to support the weight of the liquid column. When these two forces balance, we get the final height of the liquid.

📝 Teacher's Note: Explain that for water in clean glass, \( \theta \approx 0 \), so \( \cos \theta = 1 \), simplifying the formula to \( h = \frac{2T}{r\rho g} \).

🎯 Exam Tip: This formula is called the "ascent formula." It's very common in numericals, so remember to convert all units (radius, surface tension, density) to SI before calculating.

Question 10. Find the pressure 200 m below the surface of the ocean if pressure on the free surface of liquid is one atmosphere. (Density of sea water = \( 1060 \text{ kg/m}^3 \))
Answer:
Data: \( h = 200 \text{ m} \), \( \rho = 1060 \text{ kg/m}^3 \), \( p_0 = 1.013 \times 10^5 \text{ Pa} \), \( g = 9.8 \text{ m/s}^2 \)
Absolute pressure, \( p = p_0 + h\rho g \)
\( \implies p = (1.013 \times 10^5) + (200 \times 1060 \times 9.8) \)
\( \implies p = (1.013 \times 10^5) + (20.776 \times 10^5) \)
\( \implies p = 21.789 \times 10^5 \text{ N/m}^2 \) (or \( 2.1789 \text{ MPa} \))
In simple words: Deep underwater, you feel the weight of all the water above you plus the air pressure from the atmosphere. Adding these together gives the total pressure at that depth.

📝 Teacher's Note: Distinguish between gauge pressure (\( h\rho g \)) and absolute pressure (\( p_0 + h\rho g \)) for the students.

🎯 Exam Tip: Note the units. Pa (Pascals) and \( \text{N/m}^2 \) are the same thing. Express final large answers in scientific notation.

Question 11. In a hydraulic lift, the input piston had surface area \( 30 \text{ cm}^2 \) and the output piston has surface area of \( 1500 \text{ cm}^2 \). If a force of 25 N is applied to the input piston, calculate weight on output piston.
Answer:
Data: \( A_1 = 30 \text{ cm}^2 = 3 \times 10^{-3} \text{ m}^2 \), \( A_2 = 1500 \text{ cm}^2 = 0.15 \text{ m}^2 \), \( F_1 = 25 \text{ N} \)
By Pascal’s law, pressure is constant:
\( \frac{F_1}{A_1} = \frac{F_2}{A_2} \)
\( \implies \) The force on the output piston, \( F_2 = F_1 \times \frac{A_2}{A_1} \)
\( \implies F_2 = 25 \times \frac{0.15}{3 \times 10^{-3}} = 25 \times 50 = 1250 \text{ N} \)
In simple words: A hydraulic lift works like a "force multiplier." Because the output piston is 50 times larger than the input one, the force you apply gets multiplied by 50.

📝 Teacher's Note: Point out that the ratio of areas is the same whether you use \( \text{cm}^2 \) or \( \text{m}^2 \), but converting to SI is a safer habit for all physics problems.

🎯 Exam Tip: Clearly write "By Pascal's Law" as the starting point of your solution.

Question 12. Calculate the viscous force acting on a rain drop of diameter 1 mm, falling with a uniform velocity 2 m/s through air. The coefficient of viscosity of air is \( 1.8 \times 10^{-5} \text{ Ns/m}^2 \).
Answer:
Data: \( d = 1 \text{ mm} \implies r = 0.5 \text{ mm} = 5 \times 10^{-4} \text{ m} \), \( v = 2 \text{ m/s} \), \( \eta = 1.8 \times 10^{-5} \text{ Ns/m}^2 \)
By Stokes’ law, viscous force \( f = 6\pi \eta rv \)
\( \implies f = 6 \times 3.142 \times (1.8 \times 10^{-5}) \times (5 \times 10^{-4}) \times 2 \)
\( \implies f = 3.394 \times 10^{-7} \text{ N} \)
In simple words: As the raindrop falls, the air "rubs" against it, trying to slow it down. This "air friction" is the viscous force calculated here.

📝 Teacher's Note: Remind students that Stokes' law only applies to small spherical bodies moving through a fluid at low speeds (laminar flow).

🎯 Exam Tip: Don't forget the \( 6\pi \) factor in Stokes' law. Many students accidentally use \( 4\pi \).

Question 13. A horizontal force of 1 N is required to move a metal plate of area \( 10^{-2} \text{ m}^2 \) with a velocity of \( 2 \times 10^{-2} \text{ m/s} \), when it rests on a layer of oil \( 1.5 \times 10^{-3} \text{ m} \) thick. Find the coefficient of viscosity of oil.
Answer:
Data: \( F = 1 \text{ N} \), \( A = 10^{-2} \text{ m}^2 \), \( v = 2 \times 10^{-2} \text{ m/s} \), \( y = 1.5 \times 10^{-3} \text{ m} \)
Velocity gradient \( = \frac{dv}{dy} = \frac{2 \times 10^{-2}}{1.5 \times 10^{-3}} = \frac{40}{3} \text{ s}^{-1} \)
Viscous force \( F = \eta A \frac{dv}{dy} \)
\( \implies \eta = \frac{F}{A(dv/dy)} = \frac{1}{(10^{-2}) \times (40/3)} = \frac{300}{40} = 7.5 \text{ Pa}\cdot\text{s} \) (or \( \text{Ns/m}^2 \))
In simple words: Viscosity is a measure of how "thick" or "sticky" a liquid is. Here, we calculate how much resistance the oil layer offers as we slide a plate over it.

📝 Teacher's Note: Explain that the velocity gradient is the change in speed divided by the thickness of the liquid layer.

🎯 Exam Tip: Unit for viscosity is \( \text{Ns/m}^2 \) or \( \text{Pa}\cdot\text{s} \). Writing the correct unit is vital for the last half-mark.

Question 14. With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity \( 0.1 \text{ Ns/m}^2 \) and specific gravity 0.9? Density of air is \( 1.29 \text{ kg/m}^3 \).
Answer:
Data: \( r = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m} \), \( \eta = 0.1 \text{ Pa}\cdot\text{s} \), \( \rho_L = 900 \text{ kg/m}^3 \), \( \rho_a = 1.29 \text{ kg/m}^3 \), \( g = 9.8 \text{ m/s}^2 \)
Terminal velocity \( v_t = \frac{2r^2g(\rho_a - \rho_L)}{9\eta} \)
\( \implies v_t = \frac{2 \times (2 \times 10^{-4})^2 \times 9.8 \times (1.29 - 900)}{9 \times 0.1} \)
\( \implies v_t = \frac{2 \times 4 \times 10^{-8} \times 9.8 \times (-898.71)}{0.9} \)
\( \implies v_t \approx -0.782 \times 10^{-3} \text{ m/s} \) (negative sign means rising upward)
In simple words: Since air is much lighter than the liquid, the buoyant force pushes it up. The "friction" from the liquid eventually balances this push, leading to a steady rising speed.

📝 Teacher's Note: Highlight that the same formula works for falling drops and rising bubbles; the difference is just the sign of \( (\rho_{\text{body}} - \rho_{\text{medium}}) \).

🎯 Exam Tip: Always specify the direction of movement (upward or downward) especially when a negative sign appears in your result.

Question 15. The speed of water is 2m/s through a pipe of internal diameter 10 cm. What should be the internal diameter of nozzle of the pipe if the speed of water at nozzle is 4 m/s?
Answer:
Data: \( d_1 = 10 \text{ cm} = 0.1 \text{ m} \), \( v_1 = 2 \text{ m/s} \), \( v_2 = 4 \text{ m/s} \)
By Equation of Continuity: \( A_1v_1 = A_2v_2 \)
\( \implies \frac{\pi d_1^2}{4} v_1 = \frac{\pi d_2^2}{4} v_2 \)
\( \implies d_2^2 = d_1^2 \times \frac{v_1}{v_2} = (0.1)^2 \times \frac{2}{4} = 0.01 \times 0.5 = 0.005 \)
\( \implies d_2 = \sqrt{0.005} \approx 0.0707 \text{ m} \) (or 7.07 cm)
In simple words: To double the speed of water, you have to make the pipe opening narrower. This calculation tells us exactly how much smaller that nozzle needs to be.

📝 Teacher's Note: Show students that for circular pipes, \( \frac{v_1}{v_2} = (\frac{d_2}{d_1})^2 \). This shortcut is useful for MCQs.

🎯 Exam Tip: You can keep the diameter in cm if the velocities are in the same units, as it's a ratio. But conversion to meters is always standard.

Question 16. With what velocity does water flow out of an orifice in a tank with gauge pressure \( 4 \times 10^5 \text{ N/m}^2 \) before the flow starts? Density of water = \( 1000 \text{ kg/m}^3 \).
Answer:
Data: \( p - p_0 = 4 \times 10^5 \text{ Pa} \), \( \rho = 10^3 \text{ kg/m}^3 \)
The gauge pressure is \( p - p_0 = h\rho g \).
According to Toricelli’s law, velocity of efflux \( v = \sqrt{2gh} \).
From pressure: \( hg = \frac{p - p_0}{\rho} \)
\( \implies v = \sqrt{2 \times \frac{p - p_0}{\rho}} \)
\( \implies v = \sqrt{\frac{2 \times 4 \times 10^5}{10^3}} = \sqrt{800} \approx 28.28 \text{ m/s} \)
In simple words: The high pressure inside the tank "shoots" the water out. The more pressure there is, the faster the water sprays out of the hole.

📝 Teacher's Note: This is a combination of Bernoulli's principle and Toricelli's theorem. Explain that depth \( h \) is just another way of expressing gauge pressure.

🎯 Exam Tip: The formula \( v = \sqrt{\frac{2\Delta p}{\rho}} \) is a very useful derived form to remember for efflux problems.

Question 17. The pressure of water inside the closed pipe is \( 3 \times 10^5 \text{ N/m}^2 \). This pressure reduces to \( 2 \times 10^5 \text{ N/m}^2 \) on opening the valve of the pipe. Calculate the speed of water flowing through the pipe. (Density of water = \( 1000 \text{ kg/m}^3 \))
Answer:
Data: \( p_1 = 3 \times 10^5 \text{ Pa} \), \( v_1 = 0 \) (static water), \( p_2 = 2 \times 10^5 \text{ Pa} \), \( \rho = 10^3 \text{ kg/m}^3 \)
Using Bernoulli's equation for a horizontal pipe:
\( p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2 \)
As \( v_1 = 0 \):
\( \frac{1}{2}\rho v_2^2 = p_1 - p_2 \)
\( \implies v_2^2 = \frac{2(p_1 - p_2)}{\rho} = \frac{2(3 - 2) \times 10^5}{10^3} = 200 \)
\( \implies v_2 = \sqrt{200} = 14.14 \text{ m/s} \)
In simple words: When the water is still, it has high pressure. As soon as you open the valve and it starts moving, some of that pressure energy turns into speed (kinetic energy).

📝 Teacher's Note: Emphasize that in a horizontal pipe, potential energy is the same at both points and cancels out.

🎯 Exam Tip: Pay attention to the "closed pipe" phrase; it tells you that the initial velocity of the fluid is zero.

Question 18. Calculate the rise of water inside a clean glass capillary tube of radius 0.1 mm, when immersed in water of surface tension \( 7 \times 10^{-2} \text{ N/m} \). The angle of contact between water and glass is zero, density of water = \( 1000 \text{ kg/m}^3 \), \( g = 9.8 \text{ m/s}^2 \).
Answer:
Data: \( r = 0.1 \text{ mm} = 10^{-4} \text{ m} \), \( T = 7 \times 10^{-2} \text{ N/m} \), \( \theta = 0^\circ \), \( \rho = 10^3 \text{ kg/m}^3 \), \( g = 9.8 \text{ m/s}^2 \)
Capillary rise, \( h = \frac{2T \cos \theta}{r\rho g} \)
\( \implies h = \frac{2 \times 7 \times 10^{-2} \times \cos 0^\circ}{10^{-4} \times 10^3 \times 9.8} \)
\( \implies h = \frac{14 \times 10^{-2}}{0.98} \approx 0.1429 \text{ m} \) (or 14.29 cm)
In simple words: Surface tension pulls the water up the tiny tube. Using the formula, we find that in such a thin tube, water will climb up by about 14 centimeters.

📝 Teacher's Note: Remind students that for "pure water and clean glass," the angle of contact is always assumed to be zero (\( \cos 0 = 1 \)).

🎯 Exam Tip: State the formula clearly before substituting values. Using \( \cos 0^\circ = 1 \) is a specific technical point that should be shown in the steps.

Question 19. An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the gauge pressure. Surface tension of water = \( 7.2 \times 10^{-2} \text{ N/m} \). [Ans. 720 \( \text{N/m}^2 \)]
Answer: Data : \( R = 2 \times 10^{-4} \text{ m} \), \( T = 7.2 \times 10^{-2} \text{ N/m} \), \( p = 10^3 \text{ kg/m}^3 \)
The gauge pressure inside the bubble = \( \frac{2T}{R} \)
= \( \frac{2(7.2 \times 10^{-2})}{2 \times 10^{-4}} \)
= \( 7.2 \times 10^2 = 720 \text{ Pa} \)
In simple words: Inside a tiny air bubble underwater, the surface of the water acts like a tight skin that squeezes the air inside. This "squeeze" creates extra pressure, which we calculate using the surface tension and the size of the bubble.

📝 Teacher's Note: Remind students that an air bubble inside a liquid has only one free surface, so the formula is \( 2T/R \). If it were a soap bubble in air, it would have two surfaces, making it \( 4T/R \).

🎯 Exam Tip: Always convert units to SI (like mm to m) before starting calculations to avoid power-of-ten errors.

Question 20. Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m. [Ans. \( 1.628 \times 10^{-7} \text{ J} = 1.628 \text{ erg} \)]
Answer: Data : \( r = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \), \( T = 0.472 \text{ J/m}^2 \)
Let \( R \) be the radius of the single drop formed due to the coalescence of 8 droplets of mercury.
Volume of 8 droplets = volume of the single drop as the volume of the liquid remains constant.
\( \therefore 8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \)
\( \therefore 8r^3 = R^3 \)
\( \therefore 2r = R \)
Surface area of 8 droplets = \( 8 \times 4\pi r^2 \)
Surface area of single drop = \( 4\pi R^2 \)
\( \therefore \) Decrease in surface area = \( 8 \times 4\pi r^2 - 4\pi R^2 \)
= \( 4\pi(8r^2 - R^2) \)
= \( 4\pi[8r^2 - (2r)^2] \)
= \( 4\pi \times 4r^2 \)
\( \therefore \) The energy released = surface tension \( \times \) decrease in surface area = \( T \times 4\pi \times 4r^2 \)
= \( 0.472 \times 4 \times 3.142 \times 4 \times (1 \times 10^{-3})^2 \)
= \( 2.373 \times 10^{-5} \text{ J} \)
The decrease in surface energy = \( 0.072 \times 4 \times 3.142 \times 18 \times (1 \times 10^{-4})^2 \)
= \( 1.628 \times 10^{-7} \text{ J} \)
In simple words: When many small droplets join to form one big drop, the total surface area decreases. Since it takes energy to maintain a surface, losing area means energy is released.

📝 Teacher's Note: This solution contains inconsistent data (Question mentions 27 water droplets, but solution solves for 8 mercury droplets). Explain to students that the method remains the same: equate volumes to find the new radius, then find the change in area.

🎯 Exam Tip: In coalescence problems, the surface area always decreases, meaning energy is always released (the process is exothermic).

Question 21. A drop of mercury of radius 0.2 cm is broken into 8 identical droplets. Find the work done if the surface tension of mercury is 435.5 dyne/cm. [Ans. \( 2.189 \times 10^{-5} \text{ J} \)]
Answer: Let \( R \) be the radius of the drop and \( r \) be the radius of each droplet.
Data : \( R = 0.2 \text{ cm} \), \( n = 8 \), \( T = 435.5 \text{ dyn/cm} \)
As the volume of the liquid remains constant, volume of \( n \) droplets = volume of the drop
\( \therefore n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \)
\( \therefore r^3 = \frac{R^3}{n} \)
\( \implies r = \frac{R}{\sqrt[3]{n}} = \frac{R}{\sqrt[3]{8}} = \frac{R}{2} \)
Surface area of the drop = \( 4\pi R^2 \)
Surface area of \( n \) droplets = \( n \times 4\pi r^2 \)
\( \therefore \) The increase in the surface area = surface area of \( n \) droplets - surface area of drop
= \( 4\pi(nr^2 - R^2) = 4\pi(8 \times \frac{R^2}{4} - R^2) \)
= \( 4\pi(2 - 1)R^2 = 4\pi R^2 \)
\( \therefore \) The work done
= surface tension \( \times \) increase in surface area
= \( T \times 4\pi R^2 = 435.5 \times 4 \times 3.142 \times (0.2)^2 \)
= \( 2.19 \times 10^2 \text{ ergs} = 2.19 \times 10^{-5} \text{ J} \)
In simple words: To break a big drop into smaller ones, you have to create more surface area. Creating this surface requires physical effort or "work," which depends on how "sticky" the liquid's surface is (surface tension).

📝 Teacher's Note: Use the analogy of blowing a bubble—you have to use your breath (energy) to create the surface. Breaking a drop is similar.

🎯 Exam Tip: Remember to convert the final answer from CGS (ergs) to SI (Joules) by using the conversion factor \( 1 \text{ J} = 10^7 \text{ ergs} \).

Question 22. How much work is required to form a bubble of 2 cm radius from the soap solution having surface tension 0.07 N/m. [Ans. \( 0.7038 \times 10^{-3} \text{ J} \)]
Answer: Data : \( r = 4 \text{ cm} = 4 \times 10^{-2} \text{ m} \), \( T = 25 \times 10^{-3} \text{ N/m} \)
Initial surface area of soap bubble = 0
Final surface area of soap bubble = \( 2 \times 4\pi r^2 \)
\( \therefore \) Increase in surface area = \( 2 \times 4\pi r^2 \)
The work done
= surface tension \( \times \) increase in surface area
= \( T \times 2 \times 4\pi r^2 \)
= \( 25 \times 10^{-3} \times 2 \times 4 \times 3.142 \times (4 \times 10^{-2})^2 \)
= \( 1.005 \times 10^{-3} \text{ J} \)
The work done = \( 0.07 \times 8 \times 3.142 \times (2 \times 10^{-2})^2 \)
= \( 7.038 \times 10^{-4} \text{ J} \)
In simple words: A soap bubble is a hollow shell with two surfaces: one on the inside and one on the outside. Because there are two surfaces, we have to double the area when calculating the work needed to make it.

📝 Teacher's Note: The source text contains two different sets of data and solutions. Point out that a soap bubble always has 2 free surfaces, whereas a liquid drop has only 1.

🎯 Exam Tip: Double-check if the question says "bubble" (2 surfaces) or "drop" (1 surface). This is the most common trap in surface tension problems.

Question 23. A rectangular wire frame of size 2 cm \( \times \) 2 cm, is dipped in a soap solution and taken out. A soap film is formed, if the size of the film is changed to 3 cm \( \times \) 3 cm, calculate the work done in the process. The surface tension of soap film is \( 3 \times 10^{-2} \text{ N/m} \). [Ans. \( 3 \times 10^{-5} \text{ J} \)]
Answer: Data : \( A_1 = 2 \times 2 \text{ cm}^2 = 4 \times 10^{-4} \text{ m}^2 \),
\( A_2 = 3 \times 3 \text{ cm}^2 = 9 \times 10^{-4} \text{ m}^2 \), \( T = 3 \times 10^{-2} \text{ N/m} \)
As the film has two surfaces, the work done is \( W = 2T(A_2 - A_1) \)
= \( 2(3 \times 10^{-2})(9 \times 10^{-4} - 4 \times 10^{-4}) \)
= \( 3.0 \times 10^{-5} \text{ J} = 30 \mu\text{J} \)
In simple words: When you stretch a soap film on a frame, you are pulling against the surface tension of both sides of the film. The work you do is stored as energy in that extra surface area.

📝 Teacher's Note: Use the "rubber sheet" analogy. Stretching a film is like stretching a balloon—it requires force and does work.

🎯 Exam Tip: For any film problem involving a wire frame, always include the factor of 2 for the two surfaces of the film.

Question. Why does a knife have a sharp edge or a needle has a sharp tip?
Answer: For a given force, the pressure over which the force is exerted depends inversely on the area of contact; smaller the area, greater the pressure. For instance, a force applied to an area of \( 1 \text{ mm}^2 \) applies a pressure that is 100 times as great as the same force applied to an area of \( 1 \text{ cm}^2 \). The edge of a knife or the tip of a needle has a small area of contact. That is why a sharp needle is able to puncture the skin when a small force is exerted, but applying the same force with a finger does not.
In simple words: Pressure is force divided by area. Because a sharp point has a tiny area, even a small push creates a huge pressure, allowing it to cut or pierce easily.

📝 Teacher's Note: Demonstrate this using the same weight on a flat block vs. a pointed object to show how pressure changes while force remains constant.

🎯 Exam Tip: Mention the inverse relationship between pressure and area (\( P \propto 1/A \)) to secure full marks.

Question. A student of mass 50 kg is standing on both feet. Estimate the pressure exerted by the student on the Earth. Assume reasonable value to any quantity you need; justify your assumption. You may use \( g = 10 \text{ m/s}^2 \), By what factor will it change if the student lies on back?
Answer: Assume area of each foot = area of a \( 6 \text{ cm} \times 25 \text{ cm} \) rectangle.
\( \therefore \) Area of both feet = \( 0.03 \text{ m}^2 \)
\( \therefore \) The pressure due to the student’s weight
= \( \frac{mg}{A} = \frac{50 \times 10}{0.03} = 16.7 \text{ kPa} \)
According to the most widely used Du Bois formula for body surface area (BSA), the student’s BSA = \( 1.5 \text{ m}^2 \), so that the area of his back is less than half his BSA, i.e., \( < 0.75 \text{ m}^2 \). When the student lies on his back, his area of contact is much larger than his feet. So, estimating the area of contact to be \( 0.3 \text{ m}^2 \), i.e., 10 times more than the area of his feet, the pressure will be less by a factor of 10 or more. [Du Bois formula : \( BSA = 0.2025 \times W^{0.425} \times H^{0.725} \), where W is weight in kilogram and H is height in metre.]
In simple words: When standing, your weight is concentrated on the small area of your feet, creating high pressure. When lying down, that same weight is spread over your whole back, so the pressure on any single spot is much lower.

📝 Teacher's Note: This is a great exercise for Fermi estimation. Encourage students to actually measure their foot area using graph paper.

🎯 Exam Tip: Clearly state your assumptions (like the dimensions of the feet) when a question asks you to "estimate."

Question. The figures show three containers filled with the same oil. How will the pressures at the reference compare?
Answer: Filled to the same level, the pressure is the same at the bottom of each vessel.

In simple words: Pressure at the bottom of a liquid depends only on how deep the liquid is, not on the shape or width of the container. This is called the "Hydrostatic Paradox."

📝 Teacher's Note: This is a classic conceptual trap. Students often think the wider vessel has more pressure because it holds more liquid. Remind them that \( P = h\rho g \).

🎯 Exam Tip: Use the formula \( P = h\rho g \) to justify why pressure only depends on height \( h \) and density \( \rho \).

Question. Prove that equivalent SI unit of surface tension is \( \text{J/m}^2 \).
Answer: The SI unit of surface tension =
\( \frac{\text{newton (N)}}{\text{metre (m)}} = \frac{\text{N} \cdot \text{m}}{\text{m} \cdot \text{m}} = \frac{\text{J}}{\text{m}^2} \)
In simple words: Surface tension can be thought of as force per length, but it is also the amount of energy stored in every square meter of a liquid's surface.

📝 Teacher's Note: Show students that Joules (Work) = Newton \( \times \) metre. Multiplying the numerator and denominator by 'metre' is a quick algebraic trick to convert units.

🎯 Exam Tip: Remember: Surface Tension \( = \text{Force}/\text{Length} = \text{Energy}/\text{Area} \). Both definitions are physically equivalent.

Question. Take a ring of about 5 cm in diameter. Tie a thread slightly loose at two diametrically opposite points on the ring. Dip the ring into a soap solution and take it out. Break the film on any one side of the thread. Discuss what happens.
Answer: On taking the ring out, there is a soap film stretched over the ring, in which the thread moves about quite freely. Now, if the film is punctured with a pin on one side—side A in below figure—then immediately the thread is pulled taut by the film on the other side as far as it can go. The thread is now part of a perfect circle, because the surface tension on the side F of the film acts everywhere perpendicular to the thread, and minimizes the surface area of the film to as small as possible.
In simple words: The soap film wants to be as small as possible. When you pop one side, the other side pulls on the thread with equal force from all directions, stretching it into a perfect curve to minimize the remaining film's area.

📝 Teacher's Note: This experiment perfectly demonstrates that surface tension is a force that acts to minimize surface area.

🎯 Exam Tip: Mention that the thread takes the shape of a circular arc because a circle encloses the maximum area for a given perimeter, leaving the minimum area for the film.

Question. How does a waterproofing agent work?
Answer: Wettability of a surface, and thus its propensity for penetration of water, depends upon the affinity between the water and the surface. A liquid wets a surface when its contact angle with the surface is acute. A waterproofing coating has angle of contact obtuse and thus makes the surface hydrophobic.
In simple words: Waterproofing makes a surface "scared" of water. It changes the angle where water meets the surface so that water beads up into balls and rolls off instead of soaking in.

📝 Teacher's Note: Use the example of water on a lotus leaf vs. water on a cotton shirt to explain hydrophobic and hydrophilic surfaces.

🎯 Exam Tip: Use keywords like "angle of contact," "acute," "obtuse," and "hydrophobic" to score high.

Question. What would happen if two streamlines intersect?
Answer: The velocity of a fluid molecule is always tangential to the streamline. If two streamlines intersect, the velocity at that point will not be constant or unique.
In simple words: A streamline is like a one-way track for water. If two tracks crossed, a water molecule wouldn't know which way to go, which would mean the flow is messy (turbulent) rather than smooth.

📝 Teacher's Note: Compare streamlines to lanes on a highway. If cars (fluid particles) cross paths, it causes a collision (turbulence).

🎯 Exam Tip: State that intersection would imply two different directions of velocity at the same point, which is physically impossible in steady flow.

Question. Identify some examples of streamline flow and turbulent flow in everyday life. How would you explain them? When would you prefer a streamline flow?
Answer: Smoke rising from an incense stick inside a wind-less room, air flow around a car or aeroplane in motion are some examples of streamline flow. Fish, dolphins, and even massive whales are streamlined in shape to reduce drag. Migratory birds species that fly long distances often have particular features such as long necks, and flocks of birds fly in the shape of a spearhead as that forms a streamlined pattern. Turbulence results in wasted energy. Cars and aeroplanes are painstakingly streamlined to reduce fluid friction, and thus the fuel consumption. Turbulence is commonly seen in washing machines and kitchen mixers. Turbulence in these devices is desirable because it causes mixing.
In simple words: Streamline flow is smooth and orderly (like cars in lanes), while turbulent flow is swirling and chaotic (like white water rapids). We want streamline flow for speed and efficiency, but turbulence is great for mixing things like soap in a washer.

📝 Teacher's Note: Explain that "streamlining" an object (like a car) is done specifically to keep the air flow smooth and reduce the energy needed to push through it.

🎯 Exam Tip: Differentiate based on energy: Streamline flow conserves energy, while turbulence dissipates it as heat/noise.

Question. The CGS unit of viscosity is the poise. Find the relation between the poise and the SI unit of viscosity.
Answer: By Newton’s law of viscosity, \( \frac{F}{A} = \eta \frac{dv}{dy} \)
where \( \frac{F}{A} \) is the viscous drag per unit area, \( \frac{dv}{dy} \) is the velocity gradient and \( \eta \) is the coefficient of viscosity of the fluid. Rewriting the above equation as
\( \eta = \frac{(F/A)}{(dv/dy)} \)
\( [\eta] = \frac{[FA^{-1}]}{[dv/dy]} = [ML^{-1}T^{-2}][T^1] = [ML^{-1}T^{-1}] \)
SI unit : the pascal second (abbreviated Pa.s), \( 1 \text{ Pa.s} = 1 \text{ N.m}^{-2}.\text{s} \)
CGS unit: \( \text{dyne.cm}^{-2}.\text{s} \), called the poise.
\( 1 \text{ Pa.s} = 10 \text{ poise} \)
In simple words: Viscosity measures how "thick" or "sticky" a fluid is. Poise and Pascal-seconds are just different ways to measure this thickness, similar to how inches and centimeters measure length.

📝 Teacher's Note: Help students derive the conversion \( 1 \text{ Pa.s} = 10 \text{ poise} \) by converting 1 Newton to dynes and 1 \( \text{m}^2 \) to \( \text{cm}^2 \).

🎯 Exam Tip: Dimensions of viscosity \( [ML^{-1}T^{-1}] \) are frequently asked in competitive exams. Memorize them!

Question. A water pipe with a diameter of 5.0 cm is connected to another pipe of diameter 2.5 cm. How would the speeds of the water flow compare?
Answer: Water is an incompressible fluid (almost). Then, by the equation of continuity, the ratio of the speeds, is
\( \frac{v_1}{v_2} = \frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{2.5}{5}\right)^2 = \frac{1}{4} \)
\( \implies v_2 = 4v_1 \)
In simple words: When water moves from a wide pipe to a narrow one, it must speed up to get the same amount of water through in the same time. If the pipe is half as wide, the water flows four times faster.

📝 Teacher's Note: Use a garden hose analogy—if you put your thumb over the end (narrowing the opening), the water squirts out much faster.

🎯 Exam Tip: The speed is inversely proportional to the square of the diameter (\( v \propto 1/d^2 \)). Don't forget to square the ratio!

Question. Have you experienced a sideways jerk while driving a two wheeler when a heavy vehicle overtakes you?
Answer: Suppose a truck passes a two-wheeler or car on a highway. Air passing between the vehicles flows in a narrower channel and must increase its speed according to Bernoulli’s principle causing the pressure between them to drop. Due to greater pressure on the outside, the two-wheeler or car veers towards the truck.
In simple words: Fast-moving air has lower pressure. When a truck zooms past, the air between you and the truck moves very fast, creating a low-pressure "vacuum" that sucks you toward the truck.

📝 Teacher's Note: This is a dangerous real-world application of Bernoulli's principle. Explain that "suction" is actually high pressure on the outside pushing you in.

🎯 Exam Tip: Always link "increased velocity" with "decreased pressure" when citing Bernoulli's principle.

Question. Why does dust get deposited only on one side of the blades of a fan?
Answer: Blades of a ceiling/table fan have uniform thickness (unlike that of an aerofoil) but are angled (cambered) at 8° to 12° from their plane. When they are set rotating, this camber causes the streamlines above/behind a fan blade to detach away from the surface of the blade creating a very low pressure on that side. The lower/front streamlines however follow the blade surface. Dust particles stick to a blade when it is at rest as well as when in motion both by intermolecular force of adhesion and due to static charges. However, they are not dislodged from the top/behind surface because of complete detachment of the streamlines. The lower/front surface retains some of the dust because during motion, a thin layer of air remains stationary relative to the blade.
In simple words: On one side of the blade, the air flow is so smooth that it stays stuck to the surface, protecting the dust. On the other side, the air swirls away, but the "dead zone" it leaves behind allows dust to sit undisturbed.

📝 Teacher's Note: This explains the boundary layer concept. Even in fast flow, the air right at the surface is effectively still.

🎯 Exam Tip: Focus your answer on the detachment of streamlines and the stationary layer of air relative to the blade.

Question. Does Bernoulli’s equation change when the fluid is at rest? How?
Answer: Bernoulli’s principle is for fluids in motion. Hence, it is pointless to apply it to a fluid at rest. Nevertheless, for a fluid is at rest, the Bernoulli equation gives the pressure difference due to a liquid column. For a static fluid, \( v_1 = v_2 = 0 \). Bernoulli’s equation in that case is \( p_1 + \rho gh_1 = p_2 + \rho gh_2 \)
Further, taking \( h_2 \) as the reference height of zero, i.e., by setting \( h_2 = 0 \), we get \( p_2 = p_1 + \rho gh_1 \)
This equation tells us that in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by \( h_1 \) and consequently, \( p_2 \) is greater than \( p_1 \) by an amount \( \rho gh_1 \). In the case, \( p_1 = p_0 \), the atmospheric pressure at the top of the fluid, we get the familiar gauge pressure at a depth \( h_1 = \rho gh_1 \). Thus, Bernoulli’s equation confirms the fact that the pressure change due to the weight of a fluid column of length h is \( \rho gh \).
In simple words: Bernoulli's equation is a master formula. When things stop moving, the "speed" parts drop out, leaving us with the basic rule that pressure simply increases the deeper you go into a liquid.

📝 Teacher's Note: Show how the speed terms \( \frac{1}{2}\rho v^2 \) vanish to zero, reducing the complex equation back to the basic hydrostatic pressure formula.

🎯 Exam Tip: Note that Bernoulli's equation is essentially a statement of the Law of Conservation of Energy applied to fluids.