Physics Fluids and Thermal - How to Solve: Fluid Statics (Pressure, Pascal’s Law, Archimedes’ Principle, Buoyancy) – IIT JEE Guide

How to Solve: Fluid Statics (Pressure, Pascal’s Law, Archimedes’ Principle, Buoyancy) – IIT JEE Guide

Introduction

Mastering fluid statics unlocks 5-7 marks in IIT JEE (Main + Advanced) every year—enough to push you into the top 1% if you solve them fast and error-free. These concepts also explain why ships float, how hydraulic brakes work, and why deep-sea divers need pressure suits.

WHAT YOU NEED TO KNOW FIRST

1. Newton’s Laws of Motion – Forces in equilibrium.
2. Density & Mass-Volume Relationships – ρ = m/V.
3. Basic Calculus (for pressure variation with depth) – dP/dh = ρg.

KEY TERMS & FORMULAS

1. Pressure (P)

• Definition: Force per unit area.
• Formula: [ P = \frac{F}{A} ]
• P = Pressure (Pa or N/m²)
• F = Force (N)
• A = Area (m²)
• MEMORISE THIS

2. Pressure in a Fluid (Hydrostatic Pressure)

• Formula: [ P = P_0 + \rho g h ]
• P = Pressure at depth h (Pa)
• P₀ = Atmospheric pressure (1.01 × 10⁵ Pa at sea level)
• ρ = Density of fluid (kg/m³)
• g = Acceleration due to gravity (9.8 m/s²)
• h = Depth (m)
• MEMORISE THIS

3. Pascal’s Law

• Statement: Pressure applied to a confined fluid is transmitted undiminished in all directions.
• Formula: [ \frac{F_1}{A_1} = \frac{F_2}{A_2} ]
• F₁, F₂ = Forces on pistons (N)
• A₁, A₂ = Areas of pistons (m²)
• MEMORISE THIS

4. Archimedes’ Principle (Buoyancy)

• Statement: A body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.
• Formula: [ F_b = \rho_f V_d g ]
• F_b = Buoyant force (N)
• ρ_f = Density of fluid (kg/m³)
• V_d = Volume of fluid displaced (m³)
• g = Acceleration due to gravity (9.8 m/s²)
• MEMORISE THIS

5. Apparent Weight in Fluid

• Formula: [ W_{app} = W_{actual} - F_b ]
• W_app = Apparent weight (N)
• W_actual = Actual weight (mg)
• F_b = Buoyant force (N)
• MEMORISE THIS

6. Floatation Condition

• Formula: [ \rho_{body} V_{body} g = \rho_{fluid} V_{submerged} g ]
• Simplifies to: [ \frac{V_{submerged}}{V_{body}} = \frac{\rho_{body}}{\rho_{fluid}} ]
• MEMORISE THIS

STEP-BY-STEP METHOD

Step 1: Identify the Problem Type

• Pressure in a fluid? → Use P = P₀ + ρgh.
• Hydraulic system? → Use Pascal’s Law (F₁/A₁ = F₂/A₂).
• Floating/sinking object? → Use Archimedes’ Principle (F_b = ρ_f V_d g).
• Apparent weight? → Use W_app = W_actual - F_b.

Step 2: Draw a Free-Body Diagram (FBD)

• For buoyancy problems, always draw:
• Weight (mg) acting downward.
• Buoyant force (F_b) acting upward.
• Normal force (if object is on a surface).

Step 3: Write Equilibrium Conditions

• Floating object: F_b = mg.
• Sinking object: F_b < mg.
• Rising object (balloon): F_b > mg.

Step 4: Calculate Volume Displaced

• If object is fully submerged, V_d = V_body.
• If partially submerged, V_d = V_submerged.

Step 5: Solve for Unknown

• Plug values into the correct formula.
• Check units (kg/m³ for density, m³ for volume, N for force).

Step 6: Verify Answer

• Does the answer make physical sense?
• Buoyant force cannot exceed weight of displaced fluid.
• Pressure increases with depth, not height.

WORKED EXAMPLES

Example 1 – Basic (Pressure at Depth)

Question: What is the pressure at a depth of 10 m in water? (Take ρ_water = 1000 kg/m³, P₀ = 1.01 × 10⁵ Pa, g = 9.8 m/s²)

Solution:
1. Identify formula: P = P₀ + ρgh.
2. Plug in values: [ P = 1.01 \times 10^5 + (1000)(9.8)(10) ]
3. Calculate: [ P = 1.01 \times 10^5 + 98000 = 1.99 \times 10^5 \text{ Pa} ]
4. Final answer: 1.99 × 10⁵ Pa

What we did and why: - Used hydrostatic pressure formula because we needed pressure at a depth. - Added atmospheric pressure because it acts on the surface.

Example 2 – Medium (Pascal’s Law – Hydraulic Lift)

Question: A hydraulic lift has two pistons: A₁ = 0.01 m² and A₂ = 0.1 m². If a force of 50 N is applied on the smaller piston, what is the maximum weight that can be lifted on the larger piston?

Solution:
1. Identify formula: Pascal’s Law (F₁/A₁ = F₂/A₂).
2. Rearrange for F₂: [ F_2 = F_1 \times \frac{A_2}{A_1} ]
3. Plug in values: [ F_2 = 50 \times \frac{0.1}{0.01} = 50 \times 10 = 500 \text{ N} ]
4. Final answer: 500 N

What we did and why: - Used Pascal’s Law because pressure is transmitted equally in a confined fluid. - Area ratio determines force multiplication.

Example 3 – Exam-Style (Buoyancy & Apparent Weight)

Question: A solid cube of side 0.2 m and density 800 kg/m³ is submerged in water (ρ_water = 1000 kg/m³). What is the apparent weight of the cube? (g = 10 m/s²)

Solution:
1. Calculate actual weight (W_actual): [ V_{cube} = (0.2)^3 = 0.008 \text{ m}^3 ] [ m = \rho V = 800 \times 0.008 = 6.4 \text{ kg} ] [ W_{actual} = mg = 6.4 \times 10 = 64 \text{ N} ]
2. Calculate buoyant force (F_b): - Since cube is fully submerged, V_d = V_cube = 0.008 m³. [ F_b = \rho_{water} V_d g = 1000 \times 0.008 \times 10 = 80 \text{ N} ]
3. Calculate apparent weight (W_app): [ W_{app} = W_{actual} - F_b = 64 - 80 = -16 \text{ N} ] - Negative sign means the cube floats upward (net upward force = 16 N).
4. Final answer: 16 N (upward)

What we did and why: - Used Archimedes’ Principle to find buoyant force. - Compared actual weight vs. buoyant force to determine if the object sinks/floats. - Negative apparent weight means the object is less dense than the fluid.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—fluid statics is all about pressure and forces in fluids at rest. Here’s the crash course:

1. Pressure at depth? P = P₀ + ρgh. Don’t forget P₀!
2. Pascal’s Law? F₁/A₁ = F₂/A₂. Bigger area = bigger force.
3. Buoyancy? F_b = ρ_fluid V_d g. If F_b > mg, it floats.
4. Apparent weight? W_app = mg - F_b. Negative? It’s rising!
5. Floatation? V_submerged / V_total = ρ_object / ρ_fluid.

Common traps? - Forgetting P₀ in pressure questions. - Using wrong volume in buoyancy. - Mixing up densities (fluid vs. object).

You’ve got this. Now go solve those 5-7 marks like a pro!