Step 1: Given Variables & Conversion to SI Units

To ensure dimensional consistency, all values are converted to the International System of Units (SI).

• Mass of the load, $M = 3000\text{ kg}$
• Acceleration due to gravity, $g = 9.8\text{ m/s}^2$
• Area of the output piston, $A = 425\text{ cm}^2 = 425 \times 10^{-4}\text{ m}^2 = 0.0425\text{ m}^2$

Step 2: Calculating the Force Exerted by the Load

The force exerted by the weight of the car on the output piston is determined by Newton's Second Law of Motion [Per $F = mg$].

$F = M \times g$

$F = 3000\text{ kg} \times 9.8\text{ m/s}^2 = 29,400\text{ N}$

Step 3: Application of Pascal's Law

According to Pascal's Law, the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. Therefore, the pressure at the output piston ($P$) must be equal to the pressure exerted on the smaller input piston [Per Pascal's Principle of Pressure Transmission].

$P = \frac{F}{A}$

Step 4: Calculation of Maximum Pressure

Substituting the calculated force and the given area into the pressure formula:

$P = \frac{29400\text{ N}}{0.0425\text{ m}^2}$

$P \approx 691,764.7\text{ Pa}$

Step 5: Final Result Analysis

Rounding to significant figures consistent with the input data ($3000$ kg and $425$ cm$^2$):

$P \approx 6.92 \times 10^5\text{ Pa}$

Final Solution: The maximum pressure the smaller piston must bear is $6.92 \times 10^5\text{ Pa}$ (or approx $6.92 \times 10^5\text{ N/m}^2$).

Step 1: Theoretical Foundation of Bernoulli's Equation

Bernoulli’s equation is derived based on the Work-Energy Theorem applied to a streamline flow of an ideal fluid [Per the Law of Conservation of Energy]. The derivation assumes that the fluid is incompressible and non-viscous (laminar flow with zero internal friction). The total energy per unit mass is conserved because dissipative forces, such as viscosity and turbulence, are neglected.

Step 2: Role of Dissipative Forces

Dissipative forces are primarily internal frictional forces caused by the viscosity of the fluid. When a fluid flows, different layers move with different velocities, creating a velocity gradient $\frac{dv}{dx}$. According to Newton’s Law of Viscosity, the viscous force $F$ is given by:

$F = \eta A \frac{dv}{dx}$

Where $\eta$ is the coefficient of viscosity and $A$ is the area of contact. As fluid velocity increases, the velocity gradient $\frac{dv}{dx}$ typically increases, leading to a significant rise in viscous drag.

Step 3: Transition to Turbulence

The importance of dissipative forces is dictated by the Reynolds Number ($R_e$):

$R_e = \frac{\rho v D}{\eta}$

Where $\rho$ is the density, $v$ is the average velocity, $D$ is the hydraulic diameter, and $\eta$ is the dynamic viscosity. As velocity $v$ increases, $R_e$ increases. When $R_e$ exceeds a critical threshold (typically $2000–4000$), the flow regime shifts from laminar to turbulent.

Step 4: Qualitative Analysis of Dissipation

As the velocity increases, the following physical phenomena occur:

• Increased Viscous Work: Higher velocity gradients translate to higher rates of mechanical energy conversion into thermal energy [Per the First Law of Thermodynamics].
• Turbulent Eddies: At high velocities, kinetic energy is transferred into smaller, chaotic eddies (vortices). This chaotic motion consumes energy, which is ultimately dissipated as heat via internal friction.
• Validity of Bernoulli’s Principle: As $v$ increases, the assumption of "ideal fluid flow" becomes invalid because the "non-viscous" constraint is violated. The pressure drop across a length of pipe becomes greater than what Bernoulli’s equation predicts because energy is lost to dissipation rather than just potential/kinetic exchange.

Final Solution: Yes, dissipative forces become significantly more important as fluid velocity increases. High velocities induce higher velocity gradients, increasing viscous losses, and promote the transition from laminar flow to turbulent flow, where kinetic energy is dissipated into thermal energy via vortex formation, rendering the ideal-flow assumptions of Bernoulli’s equation increasingly inaccurate.

Step 1: Defining Surface Tension and Interfacial Forces

The shape of a liquid surface at the point of contact with a solid is determined by the balance of three interfacial surface tensions acting at the junction. Let:

• $\gamma_{sl}$ = Interfacial tension between solid and liquid.
• $\gamma_{sg}$ = Interfacial tension between solid and gas (air).
• $\gamma_{lg}$ = Interfacial tension between liquid and gas (air).

The angle of contact ($\theta$) is defined as the angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid. [Per Young-Dupré equation]:

$\cos \theta = \frac{\gamma_{sg} - \gamma_{sl}}{\gamma_{lg}}$

Step 2: Analytical Comparison of Force Magnitudes

The nature of the angle $\theta$ depends on the relative magnitudes of the cohesive forces (between liquid molecules) and adhesive forces (between liquid and solid molecules).

Step 3: Visual Representation of Contact Angles

Step 4: Theoretical Justification

For Water-Glass: The adhesive force between water molecules and glass molecules is significantly stronger than the cohesive force between water molecules. To minimize the system's potential energy, the water spreads against the glass, pulling the liquid upward and resulting in an acute angle ($\theta < 90^\circ$).

For Mercury-Glass: The cohesive force between mercury molecules is vastly stronger than the adhesive force between mercury and glass. The mercury minimizes its contact area with the glass to maintain its spherical shape, causing the liquid to depress and form an obtuse angle ($\theta > 90^\circ$).

Final Solution: The angle of contact is acute for water-glass because adhesive forces exceed cohesive forces, drawing the liquid against the surface; it is obtuse for mercury-glass because cohesive forces exceed adhesive forces, causing the liquid to recede from the surface.

Theoretical Foundation: Surface Tension and Molecular Dynamics

Surface tension ($\gamma$) is defined as the force per unit length acting in the plane of the interface between a liquid and another medium. At the molecular level, this property arises from the net inward cohesive force experienced by molecules at the liquid surface, as they lack an equal number of neighboring molecules compared to those in the bulk liquid [Per the Kinetic Molecular Theory].

Step 1: Thermal Energy and Cohesive Forces

As the temperature ($T$) of a liquid increases, the average kinetic energy of its molecules increases. This rise in thermal agitation results in the following physical consequences:

• Increased molecular separation, which weakens the effective range of intermolecular attractive forces (van der Waals or hydrogen bonding).
• A reduction in the net cohesive force acting on the surface layer.

Step 2: Analysis of the Relationship

Mathematically, the empirical relation for surface tension variation with temperature is often approximated by the Eötvös rule or the van der Waals relation, which shows an inverse correlation:

$\gamma(T) \approx \gamma_0 \left( 1 - \frac{T}{T_c} \right)^n$

where $T_c$ is the critical temperature and $n$ is a constant [By the Thermodynamic Definition of Surface Energy]. Since $\gamma$ decreases as $T$ approaches $T_c$ (where surface tension becomes zero), it is established that the thermal energy works against the cohesive forces that sustain surface tension.

Step 3: Visual Representation of Intermolecular Force Decay

Step 4: Conclusion

Given the inverse relationship between thermal kinetic energy and the strength of cohesive intermolecular forces, an increase in temperature invariably leads to a reduction in the energy required to create a new unit of surface area. Therefore, the surface tension of liquids generally decreases with an increase in temperature.

Final Solution: Surface tension of liquids generally decreases with temperatures.

Step 1: Analyzing the Physical Mechanism

In a thin liquid film, surface tension acts along the boundaries where the liquid interfaces with the solid frame. The force of surface tension ($F_s$) acting on a length $L$ of the film is given by $F_s = \gamma \times L \times n$, where $\gamma$ is the surface tension of the liquid and $n$ is the number of free surfaces. Since a soap film has two surfaces (top and bottom), the total force supporting the weight $W$ is $F_s = 2 \gamma L$ [Per the definition of surface tension force on a film].

Step 2: Evaluating the Initial Configuration (a)

In Figure (a), the weight $W_a = 4.5 \times 10^{-2} \, \text{N}$ is supported by the film of length $L$. The balance of forces is: $W_a = 2 \gamma L$ Given $W_a = 4.5 \times 10^{-2} \, \text{N}$.

Step 3: Calculating Configuration (b)

In Figure (b), the slider length $L$ remains identical to Figure (a). Although the frame shape is different, the effective length of the liquid film supporting the weight is still $L$, and the liquid film still possesses two surfaces. Therefore, the force supported is: $W_b = 2 \gamma L$ Since $2 \gamma L = W_a$, then $W_b = W_a$. $W_b = 4.5 \times 10^{-2} \, \text{N}$.

Step 4: Calculating Configuration (c)

In Figure (c), the slider length is the same as in (a) and (b), which is $L$. The orientation or shape of the wire frame does not alter the surface tension force magnitude as long as the length of the contact line $L$ and the number of surfaces ($n=2$) remain unchanged [By the Principle of Surface Tension proportionality]. Therefore: $W_c = 2 \gamma L = W_a$ $W_c = 4.5 \times 10^{-2} \, \text{N}$.

Visual Representation of Force Balance

Physical Explanation

The weight supported by a liquid film depends exclusively on the surface tension of the liquid ($\gamma$), the length of the slider ($L$), and the number of free surfaces of the film.

• Surface tension acts tangentially to the liquid surface.
• For a film, there are two surfaces (front and back), hence the multiplier of 2.
• Changes in the shape of the frame do not affect the total force $F = 2\gamma L$, provided the length of the slider in contact with the liquid film remains constant and the temperature (which determines $\gamma$) remains constant.

Final Solution: For both Figures (b) and (c), the weight supported is 4.5 × 10⁻² N, because the effective length of the film and the surface tension remain unchanged.

Step 1: Identifying the Governing Fluid Dynamic Principles

To analyze a fluid in a steady flow undergoing a change in cross-sectional area (a constriction), we must distinguish between the principle governing mass flow and the principle governing energy transformation.

• Conservation of Mass (Equation of Continuity): States that for an incompressible fluid in steady flow, the product of cross-sectional area ($A$) and flow speed ($v$) is constant: $A_1v_1 = A_2v_2$. This determines the kinematic change in speed. [Per the Continuity Equation derived from the Law of Conservation of Mass].
• Bernoulli’s Principle: Relates pressure ($P$), velocity ($v$), and elevation ($h$) along a streamline: $P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$. This determines the dynamic pressure changes resulting from velocity variations. [Per the Work-Energy Theorem for fluids].

Step 2: Analyzing the Constriction Phenomenon

When a fluid enters a constriction, the cross-sectional area decreases ($A_2 < A_1$). By the Equation of Continuity, since $A_1v_1 = A_2v_2$, it follows that $v_2 = v_1(A_1/A_2)$. Because $A_1/A_2 > 1$, the velocity must increase ($v_2 > v_1$).

Step 3: Determining the Correct Contextual Application

The question asks specifically for the principle followed by the increase in flow speed. While Bernoulli's principle describes the pressure drop that *accompanies* the increase in speed, the increase in speed *itself* is a direct manifestation of the requirement that mass is conserved across a flow channel.

Therefore, the increase in flow speed at a constriction is dictated by the requirement that the mass flux remains constant throughout the pipe.

Final Solution: For a fluid in a steady flow, the increase in flow speed at a constriction follows conservation of mass.

Given Physical Principles & Initial Conditions

The phenomenon is governed by Bernoulli's Principle, which states that for an incompressible, non-viscous, steady flow of fluid, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline [Per Bernoulli’s Equation].

The equation is expressed as:

$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$

Where:

• $P$: Static pressure of the fluid
• $\rho$: Density of the fluid (air)
• $v$: Velocity of the fluid
• $h$: Elevation (constant in this horizontal scenario)

Step 1: Analyzing the Forces on the Paper

Consider a light piece of paper held horizontally. Let $P_{atm}$ be the atmospheric pressure. There are two surfaces involved: the top surface and the bottom surface.

• Force due to pressure on top ($F_{top}$): $P_{top} \times A$
• Force due to pressure on bottom ($F_{bot}$): $P_{bot} \times A$
• Weight of paper ($W$): $mg$ acting downwards.

Step 2: Bernoulli’s Application to Airflow

When you blow over the paper, you impart velocity $v$ to the air above it, while the air beneath the paper remains relatively stationary ($v \approx 0$).

By Bernoulli's Principle ($P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$):

Since $v_{top} > v_{bot}$, it follows that $P_{top} < P_{bot}$.

Step 3: Calculating Net Lift

The pressure difference ($\Delta P = P_{bot} - P_{top}$) creates an upward force called Dynamic Lift.

$F_{lift} = (P_{bot} - P_{top}) \times A = \frac{1}{2}\rho v^2 A$

This upward force offsets the force of gravity ($mg$), effectively keeping the paper horizontal or causing it to lift.

Conclusion

Blowing over the paper creates a region of high velocity and low pressure above the paper. Because the pressure below the paper is higher than the pressure above it, a net upward force is generated, which counteracts the gravitational weight of the paper, keeping it elevated and horizontal.

Final Solution: Blowing over the paper creates a pressure difference (lower pressure above, higher pressure below) according to Bernoulli's Principle, generating an upward lift force that supports the paper against gravity.

1. Theoretical Foundation of Bernoulli’s Equation

Bernoulli’s equation is derived from the work-energy theorem applied to a streamline flow of an incompressible, non-viscous fluid. The equation is represented as:

$P + \frac{1}{2}\rho v^2 + \rho gh = \text{Constant}$

Where:

• $P$ is the fluid pressure at a given point.
• $\rho$ is the density of the fluid.
• $v$ is the fluid velocity.
• $h$ is the height above a reference level.

2. Comparative Analysis: Absolute vs. Gauge Pressure

The relationship between Absolute Pressure ($P_a$) and Gauge Pressure ($P_g$) is defined by the atmospheric pressure ($P_{atm}$):

$P_a = P_g + P_{atm} \implies P_g = P_a - P_{atm}$

[Per the definition of pressure measurement standards in fluid mechanics].

3. Mathematical Substitution in Bernoulli’s Equation

Consider two points (1 and 2) along a streamline. Applying the equation using absolute pressures:

$P_{a1} + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_{a2} + \frac{1}{2}\rho v_2^2 + \rho gh_2$

Substituting $P_{a} = P_{g} + P_{atm}$ into the equation:

$(P_{g1} + P_{atm}) + \frac{1}{2}\rho v_1^2 + \rho gh_1 = (P_{g2} + P_{atm}) + \frac{1}{2}\rho v_2^2 + \rho gh_2$

[By algebraic substitution property].

4. Analytical Deduction

Since $P_{atm}$ is constant at both points in an open-system flow, it appears on both sides of the equation. Subtracting $P_{atm}$ from both sides yields:

$P_{g1} + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_{g2} + \frac{1}{2}\rho v_2^2 + \rho gh_2$

This demonstrates that the pressure terms effectively cancel out, leaving the structure of the equation invariant. Therefore, the use of gauge pressure instead of absolute pressure does not alter the physical validity of the results, provided that gauge pressure is used consistently across all pressure terms in the equation.

5. Visual Representation of Pressure Differentials

Conclusion on Applicability

Final Solution: No, it does not matter if one uses gauge instead of absolute pressure, provided that the same pressure scale (either gauge or absolute) is used for all pressure terms in the Bernoulli equation. The atmospheric pressure component effectively cancels out from both sides of the equation, leaving the relative pressure differences—which govern flow dynamics—identical.

Step 1: Identifying Given Variables

• Weight supported by the film ($W$): $1.5 \times 10^{-2} \text{ N}$
• Length of the slider ($l$): $30 \text{ cm} = 0.3 \text{ m}$
• Surface tension ($\gamma$): To be determined

Step 2: Understanding Physical Principles

A soap film has two surfaces in contact with air (one on the front and one on the back). According to the principle of surface tension, the surface tension force acts along the length of the film. Since the film is thin and has two sides, the total force due to surface tension ($F$) is exerted on both sides of the slider.

Total effective length ($L$) = $2 \times l$ [Due to the two free surfaces of the soap film].

The force due to surface tension is defined as $F = \gamma \times L = \gamma \times (2l)$.

Step 3: Establishing the Equilibrium Condition

At equilibrium, the force exerted by the surface tension must balance the weight ($W$) supported by the slider.

$F = W$

$\gamma \times 2l = W$ [By Newton's First Law of Motion, as the slider is in static equilibrium].

Step 4: Algebraic Derivation and Calculation

Rearranging the equation to solve for surface tension ($\gamma$):

$\gamma = \frac{W}{2l}$

Substitute the given values into the equation:

$\gamma = \frac{1.5 \times 10^{-2} \text{ N}}{2 \times 0.3 \text{ m}}$

$\gamma = \frac{1.5 \times 10^{-2}}{0.6} \text{ N/m}$

$\gamma = 2.5 \times 10^{-2} \text{ N/m}$

Visual Representation of Forces

Data Summary

Final Solution: The surface tension of the soap film is $2.5 \times 10^{-2} \text{ N/m}$.

Given Variables & Initial Setup

To determine the terminal velocity ($v_t$) and the viscous force ($F_v$) acting on the spherical oil drop, we identify the following parameters provided in the problem statement:

• Radius of the drop ($r$): $2.0 \times 10^{-5} \, \text{m}$
• Density of oil ($\rho$): $1.2 \times 10^3 \, \text{kg m}^{-3}$
• Viscosity of air ($\eta$): $1.8 \times 10^{-5} \, \text{Pa s}$
• Acceleration due to gravity ($g$): $9.8 \, \text{m s}^{-2}$
• Density of air ($\sigma$): Negligible (as per instructions to neglect buoyancy)

Step 1: Theoretical Foundation for Terminal Velocity

A spherical body falling through a viscous medium attains terminal velocity when the net force acting on it is zero. At this state, the gravitational force ($F_g$) is balanced by the sum of the viscous drag force ($F_v$) and the buoyant force ($F_b$). Since buoyancy is neglected:

$F_g = F_v$

By Stokes' Law [Reference: Stokes' Law of Viscosity], the viscous force is given by $F_v = 6\pi\eta r v_t$. The gravitational force is given by $F_g = mg = \rho (\frac{4}{3}\pi r^3) g$.

Step 2: Deriving the Expression for Terminal Velocity

Equating the forces:

$6\pi\eta r v_t = \frac{4}{3}\pi r^3 \rho g$

Solving for $v_t$:

$v_t = \frac{2 r^2 \rho g}{9 \eta}$

Step 3: Calculating Terminal Speed

Substituting the given values into the derived formula:

$v_t = \frac{2 \times (2.0 \times 10^{-5} \, \text{m})^2 \times (1.2 \times 10^3 \, \text{kg m}^{-3}) \times 9.8 \, \text{m s}^{-2}}{9 \times (1.8 \times 10^{-5} \, \text{Pa s})}$

$v_t = \frac{2 \times (4.0 \times 10^{-10}) \times 1.2 \times 10^3 \times 9.8}{16.2 \times 10^{-5}}$

$v_t = \frac{9.408 \times 10^{-6}}{16.2 \times 10^{-5}} \approx 0.05807 \, \text{m s}^{-1}$

Step 4: Calculating the Viscous Force

At terminal velocity, the viscous force $F_v$ is equal to the weight of the drop ($mg$):

$F_v = \rho \cdot V \cdot g = \rho \cdot (\frac{4}{3}\pi r^3) \cdot g$

$F_v = (1.2 \times 10^3) \times \frac{4}{3} \times 3.14159 \times (2.0 \times 10^{-5})^3 \times 9.8$

$F_v = 1.2 \times 10^3 \times 4.1888 \times 8 \times 10^{-15} \times 9.8$

$F_v \approx 3.94 \times 10^{-10} \, \text{N}$

Final Solution: The terminal speed is approximately $5.8 \times 10^{-2} \, \text{m s}^{-1}$ and the viscous force on the drop is approximately $3.94 \times 10^{-10} \, \text{N}$.

Step 1: Theoretical Foundation of Elasticity in Solids

For an ideal solid subject to a tangential (shearing) force $F$ applied over an area $A$, the material undergoes a deformation characterized by a displacement $\Delta x$. This deformation is quantified as shear strain, denoted by $\theta$ or $\gamma$. According to Hooke’s Law for shear, the shearing stress is directly proportional to the shearing strain within the elastic limit.

The relationship is expressed as:

$\sigma_s = G \cdot \gamma$

[Where $\sigma_s$ is the shearing stress ($F/A$), $G$ is the shear modulus (modulus of rigidity), and $\gamma$ is the shear strain ($\Delta x / L$)]

Step 2: Theoretical Foundation of Fluid Dynamics

Fluids (liquids and gases) do not possess a modulus of rigidity; they cannot sustain a static shear stress. When a tangential force is applied to a fluid, it results in continuous deformation rather than a fixed state of strain. This is governed by Newton’s Law of Viscosity, which states that the shearing stress is proportional to the velocity gradient, also known as the rate of shear strain.

The relationship is expressed as:

$F/A = \eta \cdot \frac{dv}{dx}$

[Where $\eta$ is the coefficient of viscosity and $dv/dx$ represents the rate of shear strain, $\frac{d\gamma}{dt}$]

Step 3: Comparative Analysis

Step 4: Filling the Blanks

• For solids, the shearing force is proportional to shear strain. [Justification: The restoration force depends on the displacement relative to the original dimensions.]
• For fluids, the shearing force is proportional to rate of shear strain. [Justification: Fluids flow continuously under shear; the resistive force depends on the speed of relative motion.]

Final Solution: For solids with elastic modulus of rigidity, the shearing force is proportional to shear strain, while for fluids it is proportional to rate of shear strain.

Step 1: Identifying the Principle of Hydrostatic Equilibrium

In a U-tube containing fluids in equilibrium, the pressure at any two points at the same horizontal level in a continuous fluid must be equal. Since the mercury columns in both arms are at the same level, the pressure exerted by the water column in the first arm must balance the pressure exerted by the methylated spirit column in the second arm at the interface with the mercury.

Step 2: Defining Physical Variables

Let the variables be defined as follows:

• Height of water column ($h_w$) = $10.0 \text{ cm} = 0.10 \text{ m}$
• Height of spirit column ($h_s$) = $12.5 \text{ cm} = 0.125 \text{ m}$
• Density of water ($\rho_w$) = $1.0 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$
• Density of spirit ($\rho_s$) = Unknown
• Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$

Step 3: Formulating the Pressure Balance Equation

[Per the Hydrostatic Pressure Law $P = \rho gh$], at the level of the mercury interface:

$P_{\text{water}} = P_{\text{spirit}}$

$\rho_w \cdot g \cdot h_w = \rho_s \cdot g \cdot h_s$

Step 4: Algebraic Derivation for Density

Canceling the gravitational constant ($g$) from both sides of the equation:

$\rho_w \cdot h_w = \rho_s \cdot h_s$

$\rho_s = \rho_w \cdot \left( \frac{h_w}{h_s} \right)$

Step 5: Calculating Specific Gravity

Specific gravity (relative density) is defined as the ratio of the density of the substance to the density of water at standard temperature: $SG = \frac{\rho_s}{\rho_w}$.

Substituting the density ratio from Step 4:

$SG = \frac{h_w}{h_s}$

$SG = \frac{10.0 \text{ cm}}{12.5 \text{ cm}}$

$SG = 0.8$

Visual Representation of the Equilibrium

Step 6: Final Tabulation of Data

Final Solution: The specific gravity of the methylated spirit is 0.8.

Theoretical Foundation: The Molecular Basis of Surface Tension

Surface tension ($\gamma$) is defined as the force per unit length acting perpendicular to an imaginary line drawn on the liquid surface. It arises due to the unbalanced cohesive forces experienced by molecules at the surface interface [Molecular Theory of Matter].

Step 1: Analyzing Potential Energy at the Interface

Inside the bulk of a liquid, a molecule is surrounded by neighbors on all sides, resulting in a net cohesive force of zero. Conversely, a molecule at the surface experiences a net inward attractive force because there are no liquid molecules above it to balance the downward pull [Newton’s Third Law of Motion applied to Intermolecular Forces].

To move a molecule from the bulk to the surface, work must be done against these inward cohesive forces. This work is stored as potential energy. Thus, the surface of a liquid possesses an excess surface energy.

Step 2: Derivation of the Surface Tension Constant

Consider a rectangular film of liquid of length $l$ being extended by a small distance $\Delta x$.

• The total work done ($W$) to increase the surface area ($\Delta A$) is given by: $W = \gamma \cdot \Delta A$.
• Since the surface area has two sides, $\Delta A = 2 \cdot l \cdot \Delta x$.
• The work done per unit area is: $\gamma = \frac{W}{\Delta A} = \frac{F \cdot \Delta x}{2 \cdot l \cdot \Delta x} = \frac{F}{2l}$.

Step 3: Independence from Surface Area

The surface tension ($\gamma$) is a property that depends solely on the nature of the liquid and the surrounding medium (temperature and impurities).

• If the area of the surface increases, the number of molecules at the surface increases proportionally.
• While the total surface energy ($U = \gamma A$) increases, the energy per unit area ($\gamma = \frac{dU}{dA}$) remains constant.
• Because $\gamma$ is a scalar property defined by the net inward force per unit length—which is a local molecular phenomenon—it does not depend on the macroscopic geometric extent of the surface.

Visual Representation of Surface Molecules

Conclusion

Surface tension is an intensive property of the substance. Just as density ($\rho = \frac{m}{V}$) is independent of the total volume of the substance, surface tension is independent of the surface area because it is defined by the average molecular interactions at the boundary layer, which are uniform regardless of the scale of the surface.

Final Solution: Surface tension is an intensive property determined by the cohesive forces between liquid molecules at the interface; therefore, it remains constant irrespective of the total macroscopic area of the surface.

Theoretical Foundation: The Continuity Equation

The phenomenon is governed by the Equation of Continuity, which is a manifestation of the Principle of Conservation of Mass in fluid dynamics. For an incompressible, non-viscous fluid in steady flow, the product of the cross-sectional area ($A$) and the velocity of the fluid ($v$) remains constant at any point along the streamline:

$A_1v_1 = A_2v_2 = \text{Constant}$ [Per the Law of Conservation of Mass for steady flow]

Step 1: Identifying the System Variables

• Let $A_1$ represent the initial cross-sectional area of the water stream coming out of the tap.
• Let $v_1$ represent the initial velocity of the water as it exits the tap.
• When fingers are placed to close the tap, the available area for the water to escape is restricted to the gaps between the fingers. Let this smaller area be $A_2$.
• Since the fingers effectively block the majority of the original outlet, it is established that $A_2 \ll A_1$.

Step 2: Analytical Derivation of Velocity

Based on the relation derived from the continuity equation, the exit velocity $v_2$ can be expressed as:

$v_2 = v_1 \left( \frac{A_1}{A_2} \right)$

Because $A_2$ is significantly smaller than $A_1$, the ratio $\frac{A_1}{A_2}$ becomes a large value greater than 1. Consequently, $v_2$ must increase proportionally to maintain the flow rate [By the Principle of Continuity].

Step 3: Visual Representation of Flow Restriction

Step 4: Physical Synthesis

The total volume of water exiting the tap per unit time (volume flux) must remain constant to satisfy mass conservation [Equation of Continuity]. When the fingers obstruct the opening, the cross-sectional area ($A$) is drastically reduced. To ensure that the same volume of water per second passes through the restricted openings, the velocity ($v$) of the water particles must increase significantly.

This forced acceleration of fluid through the narrow channels creates the high-speed "jets" observed.

Final Solution: The reduction in the cross-sectional area of the water outlet, caused by the fingers, necessitates an increase in the velocity of the water flow to conserve mass, as dictated by the Equation of Continuity ($Av = \text{constant}$). Therefore, the water gushes out at a higher speed through the remaining gaps.

Step 1: Conversion of Units and Given Variables

First, we convert the speeds of the air from $km/h$ to $m/s$ to maintain consistency with the SI units of density ($kg/m^3$) and area ($m^2$).

• Speed over lower wing ($v_1$): $180 \times \frac{5}{18} = 50 \, m/s$
• Speed over upper wing ($v_2$): $234 \times \frac{5}{18} = 65 \, m/s$
• Area of each wing ($A$): $25 \, m^2$
• Total area of both wings ($A_{total}$): $2 \times 25 = 50 \, m^2$
• Density of air ($\rho$): $1 \, kg/m^3$
• Acceleration due to gravity ($g$): $9.8 \, m/s^2$

Step 2: Applying Bernoulli’s Principle

According to Bernoulli's principle for horizontal flow [Per Bernoulli's Equation: $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$], the pressure difference ($\Delta P$) between the lower and upper surfaces is given by:

$P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Step 3: Calculating the Pressure Difference

Substitute the given values into the equation:

$\Delta P = \frac{1}{2} \times 1 \times (65^2 - 50^2)$

$\Delta P = 0.5 \times (4225 - 2500)$

$\Delta P = 0.5 \times 1725 = 862.5 \, Pa$

Step 4: Determining the Lift Force and Mass

The upward lift force ($F$) generated by the pressure difference must equal the weight of the plane ($Mg$) to maintain level flight [Per Newton's Second Law of Motion].

$F = \Delta P \times A_{total}$

$F = 862.5 \times 50 = 43125 \, N$

Since $F = Mg$, we solve for mass ($M$):

$M = \frac{F}{g} = \frac{43125}{9.8}$

$M \approx 4400.51 \, kg$

Final Solution: The mass of the plane is approximately 4400.51 kg.

Step 1: Theoretical Foundation of Reynolds Number

The transition from laminar flow to turbulent flow in a fluid is governed by the dimensionless quantity known as the Reynolds number ($Re$), defined by the expression:

$Re = \frac{\rho v D}{\eta}$

Where:

• $\rho$ = Density of the fluid [kg/m³]
• $v$ = Flow velocity of the fluid [m/s]
• $D$ = Characteristic linear dimension (characteristic length of the object) [m]
• $\eta$ = Dynamic viscosity of the fluid [Pa·s]

[Per the principle of dynamic similarity in fluid mechanics, the flow regime transition occurs at a specific critical Reynolds number ($Re_c$).]

Step 2: Comparison of Model and Actual Plane

For both the scale model in the wind tunnel and the actual plane in flight, the fluid medium (air) remains the same, implying $\rho$ and $\eta$ are constant across both scenarios. To maintain dynamic similarity (i.e., to have the same $Re$ for the model and the actual plane), the following relationship must hold:

$v_m D_m = v_a D_a$

Where subscript '$m$' denotes the model and '$a$' denotes the actual plane.

Step 3: Analytical Derivation of Velocity

Rearranging the equation to solve for the velocity $v$:

$v = \frac{Re_c \cdot \eta}{\rho \cdot D}$

From this equation, it is evident that velocity $v$ is inversely proportional to the characteristic dimension $D$ ($v \propto \frac{1}{D}$).

Since the scale model of a plane is significantly smaller than the actual plane ($D_m \ll D_a$), it mathematically follows that the velocity required to reach the critical Reynolds number for the model must be significantly higher than the velocity required for the actual plane.

Step 4: Visualizing Fluid Flow Regimes

Step 5: Final Logical Deduction

Because the dimension $D$ for the model is smaller, the denominator in the proportionality $v \propto \frac{1}{D}$ decreases, forcing $v$ to increase to maintain the same critical $Re$ at which turbulence initiates. Therefore, the model reaches turbulence at a greater speed compared to the speed at which the actual plane would transition to turbulence under identical fluid conditions.

Final Solution: For the model of a plane in a wind tunnel, turbulence occurs at a greater speed for turbulence for an actual plane.

Step 1: Theoretical Foundation of the Angle of Contact

The angle of contact ($\theta$) is defined as the angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid. This angle is determined by the balance of three surface tension forces acting at the junction of the liquid, solid, and gas (air) phases [Ref: Young-Dupré Equation].

The equilibrium condition at the triple interface is given by the relation:

$\cos \theta = \frac{S_{sg} - S_{sl}}{S_{lg}}$

Where:

• $S_{sg}$ = Surface tension between solid and gas
• $S_{sl}$ = Surface tension between solid and liquid
• $S_{lg}$ = Surface tension between liquid and gas

Step 2: Analysis of Detergent Influence

When detergent is dissolved in water, it acts as a surfactant (surface-active agent). The primary physical changes induced by the detergent are as follows:

1. Reduction of $S_{lg}$ (Liquid-Gas Interfacial Tension): Detergent molecules accumulate at the water-air interface, disrupting the cohesive hydrogen bonding of water molecules. This significantly lowers the surface tension ($S_{lg}$).
2. Alteration of $S_{sl}$ (Solid-Liquid Interfacial Tension): Detergents increase the wetting ability of the liquid on the solid surface (often fabric or container walls). This effectively decreases the interfacial tension ($S_{sl}$) between the solid and the liquid.

Step 3: Mathematical Derivation for Angle Modification

Per the equation $\cos \theta = \frac{S_{sg} - S_{sl}}{S_{lg}}$, for the angle of contact to decrease (meaning $\theta$ becomes smaller), $\cos \theta$ must increase. This requires:

• A significant reduction in the denominator ($S_{lg}$).
• A relative decrease in the difference $(S_{sg} - S_{sl})$.

Since surfactants are engineered to maximize surface wetting, the reduction in $S_{sl}$ is usually more pronounced relative to $S_{lg}$, leading to an increase in the value of $\cos \theta$. As $\theta$ approaches $0^\circ$, $\cos \theta$ approaches $1$, indicating enhanced wetting capabilities [Per the principles of capillarity and wetting].

Step 4: Visual Representation of Contact Angle

Conclusion

Detergents lower the surface tension of water ($S_{lg}$) and reduce the interfacial tension between the solid and the liquid ($S_{sl}$). This shift in the force balance at the triple interface increases the $\cos \theta$ value, which geometrically results in a smaller angle of contact. This property is crucial for cleaning, as it allows the detergent-laden water to penetrate porous materials and displace dirt effectively.

Final Solution: The addition of detergent reduces the liquid-gas surface tension and the solid-liquid interfacial tension, forcing the liquid to spread more readily over the solid surface, thereby decreasing the angle of contact to facilitate wetting.

Step 1: Identifying the Physical Principle

The problem relies on the principle of hydrostatic equilibrium in a U-tube. For a fluid in equilibrium, the pressure at any two points at the same horizontal level in the same continuous liquid must be equal [Per Pascal’s Law of Hydrostatics: $P = P_0 + \rho gh$].

Step 2: Defining Given Variables

• Density of water ($\rho_w$): $1.0 \, \text{g/cm}^3$
• Density of spirit ($\rho_s$): $0.8 \, \text{g/cm}^3$
• Density of mercury ($\rho_m$): $13.6 \, \text{g/cm}^3$
• Height of water column ($h_w$): $15.0 \, \text{cm}$
• Height of spirit column ($h_s$): $15.0 \, \text{cm}$

Step 3: Analytical Derivation of the Level Difference

Let the initial height of mercury be equal in both arms. When water is added to one arm and spirit to the other, the mercury level shifts. Let the mercury in the water-arm side drop by $x$ and rise by $x$ in the spirit-arm side. The total difference in mercury levels becomes $2x$.

Pressure balance at the interface of mercury and the other liquids:

$P_{\text{right}} = P_{\text{left}}$

$\rho_w g h_w = \rho_s g h_s + \rho_m g (2x)$

Since $g$ is common to all terms, we simplify:

$\rho_w h_w = \rho_s h_s + \rho_m (2x)$

Step 4: Algebraic Substitution & Calculation

Substitute the known values into the equation:

$(1.0)(15.0) = (0.8)(15.0) + (13.6)(2x)$

$15.0 = 12.0 + 27.2x$

$15.0 - 12.0 = 27.2x$

$3.0 = 27.2x$

$x = \frac{3.0}{27.2} \approx 0.1103 \, \text{cm}$

The difference in the levels of mercury is $2x$:

Difference $= 2 \times 0.1103 = 0.2206 \, \text{cm}$

Step 5: Visual Representation

Final Solution: The difference in the levels of mercury in the two arms is 0.221 cm.

1. Theoretical Foundation of Bernoulli's Principle

Bernoulli's equation, given by $P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$, is derived under a specific set of idealized constraints [Derived from the Work-Energy Theorem for steady, incompressible, and non-viscous flow]. For this equation to hold, the fluid must satisfy the following criteria:

• Incompressibility: The density ($\rho$) of the fluid remains constant throughout the flow.
• Steady Flow: The velocity at any point in the fluid remains constant over time.
• Non-viscous (Inviscid): Internal friction between fluid layers (viscosity) is negligible.
• Irrotational: Fluid elements do not rotate about their centers of mass.

2. Analysis of River Rapids

In a river rapid, the physical conditions deviate significantly from the idealizations required for Bernoulli's theorem:

1. Turbulence: Rapids are characterized by highly irregular, chaotic motion. The velocity vectors at any point fluctuate rapidly in both magnitude and direction. This violates the Steady Flow condition [Per Fluid Dynamics continuity and momentum conservation].
2. Viscous Dissipation: The violent mixing, collisions with boulders, and high-velocity gradients induce significant internal friction. Energy is dissipated as heat, violating the assumption of a Non-viscous (Inviscid) fluid [Per the Navier-Stokes equations for real fluids].
3. Energy Loss: Bernoulli's equation represents the conservation of mechanical energy along a streamline. In a rapid, mechanical energy is converted into thermal energy due to turbulence and viscosity, meaning $\Delta E \neq 0$ [Per the Law of Conservation of Energy].

3. Comparative Summary

4. Graphical Representation of Flow

Final Solution: No, Bernoulli’s equation cannot be used to describe the flow of water through a rapid. The flow in a river rapid is highly turbulent and viscous; Bernoulli's equation assumes laminar, non-viscous, steady flow. In rapids, mechanical energy is dissipated into internal energy (heat) due to internal friction and turbulent mixing, rendering the assumption of constant total energy along a streamline invalid.

1. Theoretical Foundation: The Barometric Law

The variation of atmospheric pressure $P$ with height $h$ is governed by the hydrostatic equilibrium of a fluid column. In a static atmosphere, the pressure gradient is defined by the weight of the air column per unit area [Per the Hydrostatic Equilibrium Principle]:

$$\frac{dP}{dh} = -\rho g$$

where $\rho$ is the density of air and $g$ is the acceleration due to gravity. Assuming the atmosphere behaves as an ideal gas [Per the Ideal Gas Law $PV = nRT$], the density $\rho$ is directly proportional to the pressure $P$ at a constant temperature $T$:

$$\rho = \frac{PM}{RT}$$

Substituting this into the pressure gradient equation yields the exponential decay function:

$$P(h) = P_0 e^{-\frac{Mgh}{RT}}$$

2. Analysis of the Exponential Decay

The relationship indicates that atmospheric pressure does not decrease linearly with height. Instead, it follows a negative exponential curve. The constant $H = \frac{RT}{Mg}$ (the scale height) represents the height at which the pressure decreases by a factor of $1/e \approx 0.37$.

3. Data Comparison: Density vs. Height

The rapid decrease in pressure at lower altitudes is primarily due to the high compressibility of air. Unlike liquids, which have nearly constant density, the density of air drops sharply as we move away from the Earth's surface.

4. Logical Synthesis

The assertion that the atmosphere extends beyond $100\text{ km}$ does not contradict the pressure drop at $6\text{ km}$ because:

• Compressibility: The lower atmosphere (troposphere) contains approximately $80\%$ of the total atmospheric mass within the first $10\text{--}15\text{ km}$ due to gravitational compaction [Per Newton’s Law of Universal Gravitation].
• Non-Linearity: Because $P(h)$ decreases exponentially, the majority of the pressure drop occurs where the density is highest. By the time one reaches the thermosphere ($>80\text{ km}$), the air is so rarefied that the change in pressure per unit height is negligible, even though the geometric height is large.
• Conclusion: The "half-pressure" point is reached early because the vast majority of the air molecules are concentrated near the surface. The remaining $94\text{ km}$ of atmosphere consists of a very diffuse gas that exerts minimal additional pressure on the Earth's surface.

Final Solution: Atmospheric pressure decreases exponentially with height because air is highly compressible. Since the density of air is greatest near the Earth's surface, more than $50\%$ of the atmospheric mass is concentrated in the lowest $6\text{ km}$, causing the pressure to drop by half in that region, while the remaining atmosphere up to $100\text{ km}$ is extremely rarefied.