Change of pressure with depth | ICSE Class 9 Physics Notes

This note covers thrust and pressure, units, density, pressure at different depths, the liquid-column formula, container shape, dams and diving, transmission of pressure, Pascal’s law, and calculations involving pressure and force.

What are thrust and pressure?

How does area affect the action of a force?

Thrust is the force acting perpendicular to a surface. Perpendicular means at a right angle to that surface; the word normal has the same meaning here. Pressure is thrust per unit area. A force and the pressure it produces are therefore different quantities.

Definition: Pressure is the normal force acting per unit area of a surface. For a given thrust, a smaller area experiences greater pressure.

Let p represent pressure, F the normal force or thrust, and A the area over which it acts. For uniform pressure, p = F/A. If the pressure varies across the area, this calculation gives the average pressure.

In the International System of Units (SI), the unit of thrust is the newton, symbol N. The SI unit of pressure is the pascal, symbol Pa. One pascal is the pressure produced by one newton acting normally and uniformly over one square metre: 1 Pa = 1 N/m².

A sharp needle can pierce skin whereas a blunt object with a wider contact area, such as the back of a spoon, does not do so under the same force. The smaller contact area of the needle produces greater pressure.

How can pressure be calculated from weight?

Cross-sectional area is the area of a cut perpendicular to an object’s length. Weight is the gravitational force on a body. Let m denote mass and g acceleration due to gravity. Then F = mg when the thrust equals the body’s weight. The SI unit of mass is the kilogram, symbol kg.

Worked example 1. Two thigh bones each have a cross-sectional area of 10 cm² and support an upper-body mass of 40 kg. Take g = 10 m/s². Find the average pressure. Here cm² means square centimetres and m/s² means metres per second squared.

Formula: F = mg; p = F/A. Substitute: total A = 20 cm² = 20 × 10⁻⁴ m²; F = 40 × 10 = 400 N. Answer: p = 400/(20 × 10⁻⁴) = 2 × 10⁵ Pa, equivalent to 200000 Pa.

The two areas must be added because both bones support the stated mass. Converting square centimetres to square metres before division ensures that the final pressure is expressed in pascals.

What properties and units are needed to describe liquid pressure?

What do fluid, density and depth mean?

A fluid is a substance that can flow; liquids and gases are fluids. A stationary fluid is a fluid at rest. The pressure relationships developed here concern a liquid at rest, rather than a liquid flowing through a pipe.

Density means mass per unit volume. Its symbol is the Greek letter ρ, pronounced rho. If mass m occupies volume V, then ρ = m/V. Volume is the space occupied by the substance. The SI unit of density is the kilogram per cubic metre, written kg/m³.

The SI unit of volume is the cubic metre, written m³. Density describes how much mass occupies a given volume; it does not mean the total mass of all the liquid in a vessel. A larger quantity of the same uniform liquid need not have a larger density.

Depth, represented by h, is the vertical distance below the liquid surface. The SI unit of depth is the metre, symbol m. In a formula, an italic-style m may denote mass, while m following a numerical length is the unit metre; context distinguishes them.

Why is liquid density treated as constant?

A liquid is largely incompressible: its volume changes very little under pressure. Its density is therefore nearly constant. For these calculations, density and g are treated as constant over the liquid column. Gases can show large changes of density with pressure, so this assumption cannot simply be transferred to a tall column of air.

Why does pressure increase with depth in a liquid?

How does the weight of liquid produce a pressure difference?

Liquid below the surface has liquid above it. The weight of that liquid must be supported. A deeper point therefore has greater pressure than a shallower point in the same stationary liquid. This pressure due to the weight of a liquid at rest is called hydrostatic pressure.

Consider an imaginary vertical cylinder of liquid within the surrounding liquid. Its horizontal cross-sectional area is A, its height is h, and its density is ρ. Its upper pressure P₁ and lower pressure P₂ act on equal end areas.

The upper pressure produces a downward force P₁A. The lower pressure produces an upward force P₂A. The cylinder’s weight mg acts downward. Since the liquid is at rest, these forces are in equilibrium, meaning that their resultant, or combined force, is zero.

Derivation: How is the liquid-column formula obtained?

1. The volume of the cylinder is V = Ah. Rearranging the density definition gives m = ρV, so its mass is ρAh.
2. Its weight is mg = ρAhg. The upward force at the lower end balances the downward force at the upper end together with this weight.
3. Thus P₂A = P₁A + ρAhg. Subtracting P₁A gives (P₂ − P₁)A = ρAhg.
4. Dividing by the common area A gives P₂ − P₁ = ρgh. If p denotes the pressure added by a liquid column of depth h, then p = hρg.

p = hρg expresses the pressure due to the liquid column. It is the difference between the pressure at depth h and the pressure at the top of that column.

A vertical cylindrical element is drawn inside a liquid-filled container. Its upper and lower ends are labelled 1 and 2, their vertical separation is h, and a downward arrow inside is labelled mg. Pressure-force arrows act on the element.

See Fig. 9.3 in your NCERT textbook

The area cancels because increasing the column area increases both the weight of liquid and the area supporting that weight. This cancellation explains why area is absent from the final pressure formula. It does not mean that area is irrelevant when calculating the total force on a surface.

Which factors affect pressure at a point in a stationary liquid?

What changes when depth or density changes?

The relationship p = hρg identifies three factors: depth, liquid density and acceleration due to gravity. A comparison is meaningful when the quantities not being compared remain constant. Saying that pressure increases with depth assumes the same liquid and the same value of g.

At constant density and g, liquid-column pressure is directly proportional to depth. Direct proportionality means that the ratio of pressure to depth stays constant. Increasing depth by a given factor increases this pressure by the same factor.

At constant depth and g, a denser liquid exerts greater liquid-column pressure. The reason is that a column of the same height and area then has greater mass and greater weight. The supporting area has not changed, so its pressure is greater.

For fixed depth and density, the formula also shows that pressure is proportional to g. In ordinary calculations within one liquid container, g is taken as constant. Changing a container’s width or shape does not itself alter g.

What is the difference between depth and distance?

Depth is measured vertically, not along a sloping wall or along a path through the liquid. The weight balance used in the derivation involves the vertical height of the column. A longer sloping path does not by itself imply a greater pressure difference.

Note: Compare pressures only after identifying the liquid, the vertical depth and the pressure on the surface. Equal depths in different liquids need not give equal pressures.

At the free surface, meaning the exposed upper boundary of the liquid, h is zero. The additional pressure due to a liquid column is then zero. This does not say that the total pressure at an exposed surface is zero, because the air above it also exerts pressure.

How do pressure direction and container shape affect the result?

Is pressure confined to the bottom of a vessel?

A stationary liquid exerts normal forces on surfaces touching it, including the sides of its container and the surfaces of an immersed object. Immersed means placed within the liquid. At a point in a fluid at rest, the pressure is the same in all directions.

Pressure is a scalar quantity, meaning that it has magnitude but no direction. The force resulting from pressure has a direction: it acts perpendicular to the surface concerned. On differently oriented surfaces, the forces can point in different directions even when the pressure has the same value.

A beaker contains liquid and a small immersed object. Arrows around the object indicate forces perpendicular to its surfaces. The drawing illustrates the normal action of a liquid at rest.

See Fig. 9.1a in your NCERT textbook

Why can differently shaped vessels have the same pressure?

At the same horizontal level within the same connected liquid at rest, pressure is the same. Otherwise an unbalanced horizontal force would cause flow. The same-level rule is therefore a consequence of the condition that the liquid is stationary.

The hydrostatic paradox is the apparently surprising result that vessels containing different amounts of the same liquid can have the same bottom pressure. With equal liquid depths and equal pressures at the surfaces, different shapes do not alter the pressure at a common bottom level.

Three vessels labelled A, B and C have different shapes and are connected by a horizontal passage at the bottom. The liquid surfaces are at the same height, although the amounts of liquid in the vessels differ.

See Fig. 9.4 in your NCERT textbook

Do not confuse equal pressure with equal force. For a horizontal base under uniform pressure, F = pA. If two bases have different areas, equal pressures can produce different forces. On a vertical wall, pressure varies with depth, so a single bottom-pressure value cannot represent the pressure over the whole wall.

How is liquid-column pressure different from total pressure?

What pressure is already present at the surface?

Atmospheric pressure is the pressure exerted by the atmosphere, the air surrounding Earth. An exposed liquid surface is under this pressure. Let Pₐ denote atmospheric pressure and P denote the total pressure at depth h in a liquid open to the atmosphere.

The liquid adds hρg to the surface pressure. Therefore P = Pₐ + hρg. The total pressure measured relative to a vacuum, a space without matter in the ideal case, is called absolute pressure.

The excess above atmospheric pressure is called gauge pressure. Here it equals the liquid-column pressure: p = P − Pₐ. These terms distinguish two answers that might otherwise both be called simply “pressure”. The distinction is especially useful when a question gives atmospheric pressure.

What do the two pressure values mean for a swimmer?

Worked example 2. A swimmer is 10 m below a lake surface. The water density is 1000 kg/m³ and g = 10 m/s². Calculate the pressure due to the water column alone.

Formula: p = hρg. Substitute: p = 10 × 1000 × 10. Answer: p = 100000 Pa = 1.00 × 10⁵ Pa. Atmospheric pressure is not added because the question asks for the water-column contribution.

Worked example 3. A swimmer is 10 m below a lake surface open to the atmosphere. Use water density 1000 kg/m³, g = 10 m/s² and atmospheric pressure 1.01 × 10⁵ Pa. Find the total pressure.

Formula: p = hρg; P = Pₐ + p. Substitute: p = 10 × 1000 × 10 = 1.00 × 10⁵ Pa; P = 1.01 × 10⁵ + 1.00 × 10⁵. Answer: P = 2.01 × 10⁵ Pa, equivalent to 201000 Pa.

These calculations describe the same depth and water density but answer different questions. The total pressure includes the pressure already acting at the surface. The water-column pressure describes the increase on moving from that surface to the swimmer’s depth.

Note: A zero value of hρg at the surface does not imply a vacuum. In an open vessel, the total pressure at the surface equals atmospheric pressure.

When comparing two underwater levels, the shared surface pressure cancels from the subtraction. The pressure difference depends on their vertical separation, the liquid density and g. It does not require adding atmospheric pressure separately at each step of the final difference calculation.

How does increasing pressure explain dams, diving and submarine design?

Why is a dam broader near its base?

A dam is a barrier that holds back water. The water beside its lower portions is at greater depth than the water beside its upper portions. Since pressure increases with depth, the lower portions must withstand greater water pressure.

A dam is therefore made broader near the base to provide greater strength against this greater pressure. The explanation is a consequence of p = hρg. Increasing the breadth does not reduce the water pressure at a particular depth; it helps the structure withstand the pressure acting there.

What changes as a diver descends?

A diver works or moves underwater. Greater depth means greater surrounding water pressure. Protection suitable for the pressure encountered is therefore important in diving. A pressure-resistant diving suit is an application of the need to withstand underwater pressure.

The same depth relationship explains why a submarine, a vessel designed to operate underwater, must withstand large external pressures. The pressure on a surface and the force on it are related, but the force also depends on the area and the pressure acting on the other side.

Worked example 4. At an ocean depth of 1000 m, take sea-water density as 1.03 × 10³ kg/m³, g = 10 m/s² and atmospheric pressure as 1.01 × 10⁵ Pa. Find gauge and absolute pressures.

Formula: p = hρg; P = Pₐ + p. Substitute: p = 1000 × 1.03 × 10³ × 10 = 103 × 10⁵ Pa. Answer: gauge pressure = 103 × 10⁵ Pa; absolute pressure = (103 + 1.01) × 10⁵ = 104.01 × 10⁵ Pa, equivalent to 10401000 Pa.

How does pressure produce a net force on a window?

Net force is the resultant after opposing forces are combined. If a window has pressure on both sides, subtract the opposing pressures before multiplying by area. Let ΔP mean outside pressure minus inside pressure; Δ, called delta, denotes a difference. Then the inward force is F = ΔP A.

Worked example 5. A submarine window measures 20 cm × 20 cm at an ocean depth of 1000 m. Sea-water density is 1.03 × 10³ kg/m³ and g = 10 m/s². Both the sea-surface pressure and the submarine’s internal pressure are 1.01 × 10⁵ Pa. Treat pressure as uniform over the small window. Find the net inward force.

Formula: ΔP = hρg; F = ΔP A. Substitute: A = 0.20 × 0.20 = 0.04 m²; ΔP = 1000 × 1.03 × 10³ × 10 = 103 × 10⁵ Pa. Answer: F = 103 × 10⁵ × 0.04 = 4.12 × 10⁵ N, or 412000 N, inward.

The atmospheric contributions cancel because the stated internal pressure equals the pressure at the sea surface. Using the outside absolute pressure alone would calculate the outside force without subtracting the opposing force from the air inside the submarine.

What does Pascal’s law say about transmission of pressure?

How is an applied pressure change transmitted?

Transmission of pressure means that an applied change of pressure is communicated through a fluid. It is distinct from the increase of pressure with depth caused by the fluid’s own weight. A liquid can have different initial pressures at different depths and still transmit an additional pressure equally.

Definition: Pascal’s law states that external pressure applied to a fluid contained in a vessel is transmitted undiminished and equally in all directions. Undiminished means without a reduction in the transmitted pressure change.

Consider a horizontal liquid-filled cylinder with a piston, a movable close-fitting part that presses on the liquid. Vertical tubes connected at different positions show the pressure through the heights of their liquid columns. Initially, the liquid reaches the same level in the tubes.

When the piston is pushed, the levels rise and again reach the same height. The increase is transmitted throughout the liquid. This demonstration concerns equal pressure changes; it does not mean that the original total pressures at all possible depths were identical.

A piston at the left communicates with a horizontal liquid-filled passage. Three upright vessels labelled C, A and B are connected to it. The vessels have different shapes, and their liquid surfaces are drawn at a common level.

See Fig. 9.6a in your NCERT textbook

Derivation: How are forces on two liquid-contact surfaces related?

Let F₁ be an applied normal force on area A₁. Let F₂ be the force due to its transmitted pressure change on area A₂. Subscripts 1 and 2 distinguish the two surfaces; both forces refer to the same added pressure.

1. The applied force produces an additional pressure F₁/A₁ on the confined liquid.
2. By Pascal’s law, this additional pressure reaches the second surface without diminution.
3. At the second surface, the force per unit area due to this pressure change is F₂/A₂.
4. Equating these pressure changes gives F₁/A₁ = F₂/A₂. Multiplying by A₂ gives the force relation.

F₂ = F₁A₂/A₁. Equal transmitted pressures do not require equal forces when the receiving areas differ. Hydraulic lifts and hydraulic brakes are applications of Pascal’s law.

How should simple pressure numericals be solved and checked?

Which relationship fits the question?

Begin by identifying what the question requests: normal force per area, pressure due to a liquid column, total pressure, pressure difference, or force due to transmitted pressure. Write the relationship in words if the symbols initially seem confusing.

1. List the data. Identify the depth, density, value of g, relevant area and any surface or internal pressure given.
2. Convert the units. Use metres for depth, square metres for area and kilograms per cubic metre for density before calculating pressure in pascals.
3. Choose the reference. Decide whether atmospheric pressure must be added, subtracted or cancels from the comparison.
4. Calculate and interpret. Show the substitution and state the unit. For force, also describe the direction when the physical situation specifies it.

A length conversion and an area conversion are different. Since 1 cm = 10⁻² m, squaring gives 1 cm² = 10⁻⁴ m². For a rectangular area, convert both side lengths before multiplying, or convert the completed area using the squared conversion.

How is Pascal’s law used with circular pistons?

For a circular face, let r be its radius and d its diameter, the distance across the circle through its centre. Then r = d/2 and A = πr², where π is the constant ratio of a circle’s circumference to its diameter.

Worked example 6. Two needle-free syringes filled with water are joined by a tightly fitted water-filled tube. Their piston diameters are 1.0 cm and 3.0 cm. Find the force on the larger piston when 10 N is applied to the smaller one. Consider the transmitted pressure change; atmospheric pressure is common to both pistons.

Here d₁ and d₂ denote the smaller and larger diameters. Formula: A₂/A₁ = (d₂/d₁)²; F₂ = F₁A₂/A₁. Substitute: A₂/A₁ = (3.0/1.0)² = 9. Answer: F₂ = 10 × 9 = 90 N.

The areas depend on the squares of the diameters, so using the diameter ratio alone would give the wrong force. The common units cancel in this ratio. In a direct calculation of pressure, however, the area must be expressed in square metres to obtain pascals.

A final check should compare the answer with the physical trend. Greater depth in the same liquid should increase liquid-column pressure. A larger receiving area under the same transmitted pressure should produce a larger force. These checks support, but do not replace, the calculation.

Glossary

• Thrust — The force acting perpendicular, or normal, to the surface under consideration.
• Pressure — Normal force per unit area, measured in pascals in the International System of Units.
• Pascal — Pressure produced by one newton acting normally and uniformly over one square metre.
• Fluid — A substance that can flow; both liquids and gases belong to this category.
• Stationary fluid — A fluid at rest, with no flow through the region being considered.
• Density — Mass per unit volume of a substance, measured in kilograms per cubic metre.
• Depth — The vertical distance of a point below the surface of a liquid.
• Hydrostatic pressure — Pressure associated with the weight of a fluid at rest.
• Equilibrium — The condition in which the forces considered balance and their resultant is zero.
• Atmospheric pressure — The pressure exerted by the air of the atmosphere on exposed surfaces.
• Absolute pressure — Total pressure measured relative to a vacuum, including the pressure already present at the liquid surface.
• Gauge pressure — The excess of the total pressure over the surrounding atmospheric pressure.
• Hydrostatic paradox — The result that differently shaped vessels can have equal bottom pressure despite containing different amounts of liquid.
• Pascal’s law — External pressure applied to a confined fluid is transmitted undiminished and equally in all directions.
• Scalar quantity — A physical quantity described by magnitude without an associated direction.

Common errors and misconceptions

• Misconception: Pressure and thrust are the same quantity. Correct: Thrust is a normal force, measured in newtons. Pressure is thrust per unit area, measured in pascals; changing area can change pressure without changing thrust.
• Misconception: A wider vessel must produce greater pressure at the same liquid depth. Correct: Liquid-column pressure depends on depth, density and g. Vessel width and shape do not appear in p = hρg.
• Misconception: Liquid pressure acts downwards only. Correct: A stationary liquid exerts normal forces on all contacting surfaces, including side walls. Pressure at a point is the same in all directions.
• Misconception: The distance along a sloping wall is the depth to use. Correct: Use the vertical depth below the liquid surface, because the hydrostatic relationship comes from a vertical weight balance.
• Misconception: Total pressure at an open liquid surface is zero. Correct: The liquid-column contribution is zero there, but the total pressure equals atmospheric pressure.
• Misconception: Pascal’s law makes total pressure identical at every depth. Correct: The added pressure is transmitted equally. The pressure difference caused by the weight of liquid between different levels remains.
• Misconception: Equal transmitted pressure means equal force on every piston. Correct: Force equals pressure multiplied by area. A larger area receives a larger force from the same pressure change.
• Misconception: Outside absolute pressure alone gives the net force on a submarine window. Correct: Subtract the inside pressure before multiplying by the window area, because the inside pressure produces an opposing force.

Exam-style questions with model answers

Q1. Define thrust and pressure, and give the SI unit of each. [2 marks]

1. Thrust is the force acting perpendicular to a surface. Its SI unit is the newton, N.
2. Pressure is thrust per unit area of the surface. Its SI unit is the pascal, Pa, equivalent to N/m².

Q2. For a stationary liquid of uniform density, explain how liquid-column pressure depends on depth, density and acceleration due to gravity. State what is held constant in each comparison. [3 marks]

1. At constant density and acceleration due to gravity, pressure increases directly with vertical depth. A deeper point has a greater height of liquid above it.
2. At constant depth and acceleration due to gravity, pressure increases directly with density. The same volume of a denser liquid has greater weight.
3. At constant depth and density, pressure increases directly with acceleration due to gravity. These three dependences are expressed together by p = hρg.

Q3. Derive the pressure difference between the lower and upper ends of an imaginary vertical liquid cylinder at rest. Its cross-sectional area is A, height h and uniform density ρ. Acceleration due to gravity is g; the upper and lower pressures are P₁ and P₂. [5 marks]

1. The cylinder has volume V = Ah, where V denotes its volume. Its mass is m = ρV = ρAh, using density as mass per unit volume.
2. The weight of the liquid cylinder is mg = ρAhg. This force acts vertically downwards and must be included in the force balance.
3. The upper pressure produces downward force P₁A. The lower pressure produces upward force P₂A on the cylinder’s equal-area lower face.
4. Because the liquid is at rest, the upward force equals the total downward force: P₂A = P₁A + ρAhg.
5. Subtracting P₁A and dividing by A gives P₂ − P₁ = ρgh. Thus a liquid column of height h adds pressure p = hρg.

Q4. A swimmer is 10 m below a lake surface. Water density is 1000 kg/m³, acceleration due to gravity is 10 m/s², and atmospheric pressure is 1.01 × 10⁵ Pa. Calculate the water-column pressure and total pressure, distinguishing the two. [4 marks]

1. The water-column pressure is p = hρg, using the vertical depth, water density and acceleration due to gravity supplied in the question.
2. Substituting gives p = 10 × 1000 × 10 = 100000 Pa = 1.00 × 10⁵ Pa. This is the increase due to the water.
3. Total pressure includes the atmospheric pressure already acting at the lake surface: P = Pₐ + p.
4. Therefore P = 1.01 × 10⁵ + 1.00 × 10⁵ = 2.01 × 10⁵ Pa. It exceeds the water-column pressure by atmospheric pressure.

Q5. A submarine window measures 20 cm × 20 cm at a depth of 1000 m. Sea-water density is 1.03 × 10³ kg/m³ and g = 10 m/s². Sea-surface pressure and internal pressure both equal 1.01 × 10⁵ Pa. Treat pressure as uniform across the small window. Calculate the net inward force. [4 marks]

1. Convert the side lengths to metres. The window area is A = 0.20 × 0.20 = 0.04 m².
2. The outside pressure is surface pressure plus hρg. Subtracting the equal internal and surface pressures leaves pressure difference ΔP = hρg.
3. Substitute the stated depth, density and g: ΔP = 1000 × 1.03 × 10³ × 10 = 103 × 10⁵ Pa.
4. The net force is F = ΔP A = 103 × 10⁵ × 0.04 = 4.12 × 10⁵ N, directed inward because outside pressure is greater.

Q6. State Pascal’s law and apply it to two water-filled syringes connected by a water-filled tube. The piston diameters are 1.0 cm and 3.0 cm. A force of 10 N acts on the smaller piston. Considering the transmitted pressure change, calculate the force on the larger piston and explain why the forces differ. Atmospheric pressure is common to both pistons; use circular area A = πd²/4, where d is diameter and π is the circle constant. [5 marks]

1. Pascal’s law states that external pressure applied to a confined fluid is transmitted undiminished and equally in all directions throughout it.
2. Let F₁ and A₁ be the smaller piston’s force and area, and F₂ and A₂ the larger piston’s force and area. Equal transmitted pressure gives F₁/A₁ = F₂/A₂.
3. For circular faces, the common factors π and one quarter cancel in the area ratio. Thus A₂/A₁ = (3.0/1.0)² = 9.
4. Therefore F₂ = F₁A₂/A₁ = 10 × 9 = 90 N. The common atmospheric pressure does not contribute to the added pressure being considered.
5. The forces differ because the piston areas differ. Equal pressure means equal force per unit area, so the larger area receives the larger force.

Key takeaways

• Thrust is a normal force; pressure is thrust per unit area, measured in pascals rather than newtons.
• For a stationary liquid of uniform density, liquid-column pressure is p = hρg, with h measured vertically.
• Pressure increases with depth at constant density and g, and increases with density at constant depth and g.
• Container shape and width do not change liquid-column pressure when depth, density and g remain the same.
• Equal pressure at the same horizontal level does not imply equal total force on surfaces of different areas.
• Total pressure in an open liquid equals atmospheric pressure plus the pressure due to the liquid column.
• Increasing pressure with depth explains the broader base of a dam and the need to withstand pressure underwater.
• Pascal’s law concerns equal transmission of added pressure; forces produced by that pressure depend on the receiving areas.

Test yourself

What happens to pressure if the same thrust acts over a smaller area?

Pressure increases because pressure equals thrust divided by area, with the thrust held constant.

In p = hρg, what does each symbol mean?

The symbols represent liquid-column pressure p, vertical depth h, liquid density ρ and acceleration due to gravity g.

Why does the area disappear from the liquid-column pressure formula?

The column’s weight and its supporting area both contain the same cross-sectional-area factor, which cancels during division.

Can equal depths in different liquids give different liquid-column pressures?

Yes. At the same depth and g, different densities produce different pressures according to p = hρg.

Why does a wider dam base help without reducing the water pressure?

Its greater breadth provides strength against the greater pressure near the bottom; pressure still depends on the water depth.

What is the total pressure at the free surface of water open to air?

It equals atmospheric pressure. The additional pressure due to the water column is zero at the surface.

Does Pascal’s law remove the pressure difference between different depths?

No. It transmits the added pressure equally; the pressure difference due to the intervening liquid’s weight remains.

Why must internal pressure be considered when finding net force on a submarine window?

Internal pressure exerts an opposing force. Subtract it from outside pressure before multiplying by the window area.