All JEE Main FUILDS PYQs (2002-2025) | Complete Problem Analysis & Solutions

TLDR;

Alright guys, so this session is all about cracking Fluid Mechanics for JEE Main! The aim is to tackle those tricky problems by understanding the concepts and applying the right formulas. We'll be covering everything from pressure and density to fluid flow and surface tension, with a mix of topic-wise and type-wise questions. Plus, there's a special Unacademy Plus subscription offer and a promise of some "bhutiya kahani" later on.

Fluid Mechanics is formula-heavy but needs conceptual clarity.
JEE Main PYQs will be solved with key concept discussions.
Topics include pressure, fluid flow, and surface tension.

Intro and Session Overview [0:05]

The session is focused on Fluid Mechanics, a chapter that many students find challenging due to its formula-based nature and the difficulty in choosing the correct formula for a given problem. The goal is to improve conceptual clarity and the application of formulas. The session will cover past year questions (PYQs) for JEE Main, along with key concepts. The presenter also mentions that the session is structured topic-wise and type-wise, based on feedback from previous sessions.

Unacademy Plus Subscription Offer [2:59]

The presenter highlights a limited-time offer for the Unacademy Plus subscription, available for ₹4,999 for one year using the code SBJEE. This subscription provides access to various batches and resources for JEE preparation. The presenter encourages viewers to explore the platform and take advantage of the offer.

Session Topics and Question Types [4:07]

The session will cover three main topics: pressure, density, and Pascal's law; fluid flow and Bernoulli's equation; and surface tension with capillary rise. The questions will be arranged by type, including problems where bodies are partially submerged, direct pressure calculations, and pressure calculations for flowing fluids. The presenter mentions that some miscellaneous questions will also be included.

Key Concepts: Submerged Bodies [6:17]

When a body is submerged in a fluid, two forces act on it: gravity (mg) and buoyancy (Fb). For a stationary body, these forces are equal. The buoyancy force is calculated as Vρg, where V is the volume of fluid displaced, ρ is the density of the fluid, and g is the acceleration due to gravity. The density used for calculating the gravitational force is the density of the body, while the density used for calculating the buoyancy force is the density of the fluid.

Problem 1: Partially Submerged Cube [8:58]

A 400g solid cube with a 10cm edge is floating in water. The problem asks for the volume of the cube that is outside the water. The solution equates the gravitational force to the buoyancy force. The volume of displaced water is calculated as 10 * 10 * x cm³, where x is the height of the cube submerged in water. The height of the cube above the water is then found to be 6 cm, and the volume above the water is 600 cm³.

Problem 2: Ice Cube in Water and Kerosene [12:29]

An ice cube floats partially in water and partially in kerosene. The ratio of the volume of ice submerged in water to that in kerosene is to be determined. Specific gravities of kerosene and ice are given as 0.8 and 0.9, respectively. The solution equates the gravitational force on the ice cube to the sum of the buoyancy forces due to water and kerosene. The ratio of the volumes is found to be 1:1.

Problem 3: Sphere with Concentric Cavity [17:02]

A sphere with relative density σ and diameter D has a concentric cavity of diameter d. The ratio of d/D is to be found if the sphere just floats on water. The solution equates the gravitational force on the sphere (considering the cavity) to the buoyancy force. The ratio d/D is found to be (σ/(σ-1))^(1/3).

Problem 4: Floating Liquid Drop [21:12]

A liquid drop of density ρ is floating half-immersed in a liquid of density σ. The surface tension is given as 7.5 x 10^-4 N/m. The radius of the drop in centimeters is to be determined. The solution balances the gravitational force, buoyancy force, and the force due to surface tension. The force due to surface tension is calculated as 2πrT.

Problem 5: Ball in Oil and Water Mixture [25:45]

A ball's density lies between that of oil and water. The question asks which picture correctly represents the equilibrium position of the ball in a mixture of oil and water. The correct picture shows the oil on top, the ball in the middle, and the water at the bottom.

Problem 6: Ball in Non-Mixing Liquids [27:28]

A solid ball of density ρ3 is dropped in a jar containing two non-mixing liquids with densities ρ1 and ρ2. The ball comes to equilibrium as shown in the figure. The correct relationship between the densities is ρ1 < ρ3 < ρ2.

Problem 7: Hollow Spherical Shell [28:09]

A hollow spherical shell of outer radius R floats just submerged under water. The inner radius is r, and the specific gravity of the material is 27/8. The value of r is to be determined. The solution equates the gravitational force on the shell to the buoyancy force. The inner radius r is found to be 8/9 R.

Problem 8: Wooden Block in Water and Oil [30:12]

A wooden block floats on water with 4/5 of its volume submerged. When oil is poured in, the block is just under the oil surface, with half its volume in water and half in oil. The relative density of the oil to water is to be determined. The solution first finds the density of the block in terms of the density of water. Then, it equates the gravitational force on the block to the sum of the buoyancy forces due to water and oil. The relative density of oil is found to be 0.6.

Problem 9: Maximum Weight on Floating Block [34:08]

A cubical block of side 0.5 meters floats on water with 30% of its volume submerged. The maximum weight that can be put on the block without fully submerging it is to be determined. The solution first finds the density of the block in terms of the density of water. Then, it equates the gravitational force on the block plus the additional weight to the buoyancy force when the block is fully submerged.

Key Concepts: Direct Pressure Calculation [38:13]

The pressure at a depth h in a fluid is given by P = P₀ + ρgh, where P₀ is the atmospheric pressure, ρ is the density of the fluid, and g is the acceleration due to gravity. The pressure at the bottom of a container depends on the height of the fluid, not the shape or radius of the container.

Problem 10: Mercury in a Tube [39:07]

Mercury is filled in a tube of radius 2 cm up to a height of 30 cm. The force exerted by the mercury on the bottom of the tube is to be determined. The solution uses the formula P = P₀ + ρgh to calculate the pressure at the bottom of the tube. The force is then calculated as pressure times area.

Problem 11: Force on a Square Wall [40:48]

Two liquids of densities ρ1 and ρ2 are filled behind a square wall of side 10 meters, each liquid having a height of 5 meters. The ratio of the forces due to these liquids exerted on the upper part to that of the lower part is to be determined. The solution involves integrating the pressure over the area of the wall. The force is calculated by integrating ρgh dA, where dA is a small area element.

Problem 12: Cylinder Suspended by Spring in Liquid [51:23]

A uniform cylinder of length L and mass M is suspended vertically from a fixed point by a massless spring. It is half-submerged in a liquid of density σ at equilibrium. The extension x₀ in the spring is to be determined. The solution balances the gravitational force, buoyancy force, and the spring force.

Problem 13: Submarine Pressure Increase [54:53]

A submarine experiences a pressure of 3 x 10^5 Pascals at a certain depth. If the depth is doubled, the percentage increase in pressure is to be determined. The solution uses the formula P = P₀ + ρgh to calculate the pressure at both depths. The percentage increase is then calculated as (P₂ - P₁)/P₁ * 100.

Key Concepts: Hydraulic Lift [57:51]

In a hydraulic lift, the pressure is the same throughout the fluid. Therefore, F₁/A₁ = F₂/A₂, where F₁ and F₂ are the forces applied on the smaller and larger pistons, respectively, and A₁ and A₂ are the areas of the smaller and larger pistons, respectively.

Problem 14: Hydraulic Press [58:55]

A hydraulic press has two arms with diameters 1.4 cm and 14 cm. A force of 10 N is applied on the thinner arm. The force required to be applied on the thicker arm to maintain equilibrium is to be determined. The solution uses the formula F₁/A₁ = F₂/A₂ to calculate the force on the thicker arm.

Problem 15: Hydraulic Lift with Mass on Piston [1:00:32]

A hydraulic press can lift 100 kg. When a mass m is placed on the smaller piston, it can lift a certain weight. The diameter of the larger piston is increased by four times, and the diameter of the smaller piston is decreased by four times, keeping the mass on the smaller piston the same. The new weight that can be lifted is to be determined. The solution uses the formula F₁/A₁ = F₂/A₂ to relate the forces and areas.

Key Concepts: Rotating Tube [1:04:46]

When a tube filled with liquid is rotated, the liquid experiences a centrifugal force. The mass distribution is not uniform, so integration is required to calculate the total force. The force on a small element dm at a distance x from the axis of rotation is dmω²x.

Problem 16: Rotating Tube with Liquid [1:09:03]

A tube of length 50 cm is filled completely with an incompressible liquid of mass 250 g. The tube is rotated in a horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is F. The value of ω is to be determined. The solution integrates the centrifugal force over the length of the tube.

Key Concepts: Pressure of Flowing Water [1:11:49]

When water flows out of a tube and hits a wall, the force exerted on the wall is given by F = ρAv², where ρ is the density of the water, A is the area of the tube, and v is the velocity of the water. This formula applies when the water hits a stationary object.

Problem 17: Pressure Pump on Vertical Wall [1:14:44]

A pressure pump has a horizontal tube of cross-sectional area 10 cm² for the outflow of water with a speed of 20 m/s. The force exerted on a vertical wall in front of the tube is to be determined. The solution uses the formula F = ρAv² to calculate the force.

Problem 18: Sliding Vessel Due to Water Impact [1:15:31]

A light cylindrical vessel is kept on a horizontal surface. A hole of cross-sectional area a is made at the bottom. The minimum coefficient of friction necessary to prevent sliding of the vessel due to the impact of the water force is to be determined. The solution equates the frictional force to the force exerted by the water.

Key Concepts: Rising Air Bubble [1:19:48]

As an air bubble rises in a liquid, the pressure decreases, and the volume increases due to Boyle's law (P₁V₁ = P₂V₂). The pressure at a depth h is P = P₀ + ρgh, where P₀ is the atmospheric pressure, ρ is the density of the liquid, and g is the acceleration due to gravity.

Problem 19: Air Bubble Rising in a Lake [1:23:12]

An air bubble of radius r rises from the bottom of a lake to the surface. Its radius becomes 5r/4. The depth of the lake is to be determined, given that the atmospheric pressure is equal to 10 meters of water height. The solution uses Boyle's law to relate the pressure and volume at the bottom and top of the lake.

Remaining Questions in Section 1 [1:25:44]

The presenter notes that the remaining questions in the first section are mixed and not arranged by type.

Problem 20: Assertion and Reason on Toothpaste [1:26:14]

Assertion: Squeezing a toothpaste tube demonstrates Pascal's principle. Reason: Pascal's law states that pressure changes in an enclosed incompressible fluid are transmitted undiminished. Both are true, and the reason correctly explains the assertion.

Problem 21: Pressure in a Reservoir [1:27:16]

Statement 1: Pressure in a water reservoir is the same at all points at the same level. Statement 2: Pressure applied to an enclosed water is transmitted equally in all directions. Both statements are true.

Problem 22: Bicycle Tire Pressure [1:28:05]

A bicycle tire is filled with air at 270 kPa at 20°C. The approximate pressure at 36°C is to be determined. The solution uses the relation P₁/T₁ = P₂/T₂.

Problem 23: Viscous Force on a Ball in Glycerin [1:30:08]

A small ball of mass 0.3 g and density 8 g/cm³ is dropped in glycerin. The velocity becomes constant after some time. The density of glycerin is 1.3 g/cm³. The viscous force acting on the ball is to be determined. The solution equates the gravitational force to the buoyancy force plus the viscous force.

Problem 24: Interconnected Cylindrical Vessels [1:34:18]

Two cylindrical vessels of equal cross-sectional area contain water up to heights of 100 cm and 150 cm. The vessels are interconnected, so the water levels become equal. The work done by gravity is to be determined. The solution calculates the initial and final potential energies of the water.

Problem 25: Solid Sphere Floating in Liquid [1:38:40]

A solid sphere of radius R has a mass density ρ(r) = ρ₀(1 - r²/R²). The minimum density of a liquid in which it will float is to be determined. The solution integrates the density over the volume of the sphere to find the total mass. Then, it equates the gravitational force to the buoyancy force.

Problem 26: Leaking Cylinder in Water [1:41:51]

A leak-proof cylinder of length 1 meter is floating vertically in water at 0°C, with 20 cm above the surface. When the water temperature increases to 4°C, 21 cm is above the surface. The density of water at 4°C relative to 0°C is to be determined. The solution equates the weight of the cylinder to the buoyant force at both temperatures.

Problem 27: Air Bubble with Upward Acceleration [1:45:58]

An air bubble of radius 1 cm in water has an upward acceleration of 9.8 cm/s². The density of water is 1 g/cm³, and the drag force is negligible. The mass of the bubble is to be determined. The solution uses Newton's second law to relate the forces and acceleration.

Problem 28: Soap Bubble Volume Increase [1:48:09]

A soap bubble blown by a mechanical pump increases in volume at a constant rate. The graph that correctly represents the time dependence of the pressure inside the bubble is to be determined. The solution relates the volume to the radius and then uses the formula for excess pressure inside a soap bubble.

Problem 29: Simple Pendulum in Liquid [1:49:48]

A simple pendulum oscillating in air has a time period T. The bob is completely immersed in a non-viscous liquid with a density 1/16 of the bob's material. The new period of oscillation is to be determined. The solution calculates the effective gravitational acceleration in the liquid.

Problem 30: Submarine Depth and Pressure [1:52:15]

A submarine experiences a pressure of 3 x 10^6 Pa at depth d1. At depth d2, it experiences 8.08 x 10^6 Pa. The difference d2 - d1 is to be determined. The solution uses the formula P = P₀ + ρgh to relate pressure and depth.

Problem 31: Non-Mixing Liquids in a Circular Tube [1:53:18]

A uniform tube is bent into a circle of radius R in a vertical plane. Equal volumes of two immiscible liquids with densities ρ1 and ρ2 are filled, occupying half the circle. The angle between the radius vector passing through the common interface and the vertical is to be determined. The solution equates the pressures at the bottom of the tube due to both liquids.

End of Section 1 and Start of Section 2 [2:04:00]

The presenter announces the end of the first section and a 10-minute break before starting the second section on fluid flow and Bernoulli's principle.

Key Concepts: Torricelli's Law [2:15:18]

When a hole is made in a container filled with a liquid, the velocity of efflux is given by v = √(2gh), where h is the height of the liquid above the hole. In a liquid, the buoyancy force and viscosity are also important.

Problem 32: Efflux from a Tank [2:15:43]

A tank has a hole at a certain height. The problem involves finding the relationship between the height and the velocity of the water coming out of the hole.

Problem 33: Water Tank with Hole [2:21:16]

A water tank has a cross-sectional area of 750 cm² on top of a house, h meters above a tap. The speed of water coming out of the tap with a cross-section of 30 cm/s is 30 cm/s. The rate of change of height (dh/dt) is to be determined. The solution uses the equation of continuity (A₁v₁ = A₂v₂) to relate the velocities and areas.

Problem 34: Fluid Flow Through a Bent Pipe [2:22:35]

An ideal fluid flows smoothly through a bent pipe. The cross-sectional area changes from A to A/2. The pressure difference between the wide and narrow sections is given. The velocity at the wider section is to be determined. The solution uses both the equation of continuity and Bernoulli's equation.

Problem 35: Water Flowing from a Tap [2:24:49]

Water flows from a tap with an initial speed of 1 m/s. The cross-sectional area of the tap is 10⁻⁴ m². The cross-sectional area of the stream 0.15 m below the tap is to be determined. The solution uses the equation of continuity and kinematic equations.

Problem 36: Water Tank with a Hole [2:26:18]

A cylinder of height 20 meters is completely filled with water. The velocity of efflux from a small hole on the side of the cylinder near the bottom is to be determined. The solution uses Torricelli's law (v = √(2gh)).

Problem 37: Pressure Meter Attached to a Closed Pipe [2:27:17]

A pressure meter attached to a closed pipe reads 3.5 x 10^5 N/m². On opening the valve, the reading falls to 3.0 x 10^5 N/m². The speed of the water is to be determined. The solution uses Bernoulli's equation.

Problem 38: Water Tank with Hole and Applied Force [2:30:48]

A cylindrical tank of radius 1 meter is filled with water. The top surface is 15 meters from the bottom. A hole is made at a height of 5 meters from the bottom. A force of 5 x 10^5 N is applied on the top surface using a piston. The speed of efflux from the hole is to be determined. The solution uses Bernoulli's equation.

Problem 39: Water Tank with Hole and Applied Load [2:32:56]

A water tank has a large cross-sectional area. A hole near the bottom has an area of 1 cm². A load of 25 kg is applied on the water. The velocity of the water coming out is to be determined. The solution uses Bernoulli's equation.

Problem 40: Fluid Flow Through a Horizontal Pipe [2:34:11]

A fluid flows through a horizontal pipe with varying cross-section. The pressure changes from P to P/2. The rate of flow is to be determined. The solution uses both the equation of continuity and Bernoulli's equation.

Problem 41: Water Flowing Continuously from a Tap [2:35:58]

Water flows continuously from a tap. The diameter of the tap is given. The diameter of the water stream at a certain distance below the tap is to be determined. The solution uses the equation of continuity and kinematic equations.

Problem 42: Water Tank with Hole - Range [2:38:08]

A water tank is filled to a height of 12 meters. A hole is made at a depth h below the water level. The value of h for which the emerging stream strikes the ground at the maximum range is to be determined. The solution uses the concept of maximum range for projectile motion.

Problem 43: Fluid Flowing Through a Horizontal Pipe [2:40:34]

A fluid is flowing through a horizontal pipe with varying cross-section. The speed and pressure at one point are given. The speed at another point where the pressure is different is to be determined. The solution uses Bernoulli's equation.

Key Concepts: Terminal Velocity [2:41:51]

When an object falls through a viscous fluid, it reaches a terminal velocity when the gravitational force is balanced by the buoyancy force and the viscous force. The viscous force is given by Stokes' law: F = 6πηrv, where η is the viscosity, r is the radius, and v is the velocity.

Problem 44: Velocity-Time Graph for a Ball in Glycerin [2:42:58]

A small steel ball is dropped in a long cylinder containing glycerin. The correct representation of the velocity-time graph is to be determined. The graph shows the velocity increasing initially and then becoming constant.

Problem 45: Terminal Velocity with Negligible Density [2:43:38]

A small spherical ball of radius r is falling through a viscous medium of negligible density and has a terminal velocity v. Another ball of the same mass but radius 2r is falling through the same medium. The new terminal velocity is to be determined. The solution uses Stokes' law and the formula for terminal velocity.

Problem 46: Raindrop Terminal Velocity [2:48:11]

A raindrop of radius r is falling from a height. The force of buoyancy is negligible. The terminal velocity depends on r. The solution uses Stokes' law and the formula for terminal velocity.

Problem 47: Coalescing Water Droplets [2:49:14]

Small water droplets of radius r are formed and fall with terminal velocity v. Eight droplets coalesce to form a larger drop. The new terminal velocity is to be determined. The solution relates the volumes and radii of the droplets.

Problem 48: Viscous Force and Terminal Velocity [2:50:28]

A spherical ball of volume V is made of material with density ρ1. It falls through a liquid with density ρ2. The viscous force is proportional to the square of the speed (-kv²). The terminal speed is to be determined. The solution balances the forces and solves for the terminal velocity.

Problem 49: Terminal Speed of Gold and Silver Spheres [2:52:38]

The terminal speed of a gold sphere in a viscous liquid is given. The terminal speed of a silver sphere of the same size in the same liquid is to be determined. The solution uses the formula for terminal velocity.

Problem 50: Stokes' Law Verification [2:54:41]

In an experiment to verify Stokes' law, a spherical ball falls through a distance h in air. The terminal velocity is the same as the velocity just before entering the liquid. The relationship between h and r is to be determined. The solution uses Stokes' law and the formula for terminal velocity.

Problem 51: Air Bubble in Water - Acceleration [2:56:47]

An air bubble of radius 1 cm in water has an upward acceleration. The density of the water is given. The mass of the bubble is to be determined. The solution uses Newton's second law to relate the forces and acceleration.

Problem 52: Assertion and Reason on Fluid Flow [2:59:08]

Assertion: When you squeeze one end of a tube to get toothpaste out, Pascal's principle is observed. Reason: Change in pressure applied to an enclosed incompressible fluid is transmitted undiminished. Both are true, and the reason correctly explains the assertion.

Problem 53: Assertion and Reason on Fluid Properties [3:00:20]

Assertion: Pressure in a reservoir is the same at all points at the same level. Reason: Pressure applied to an enclosed water is transmitted equally in all directions. Both statements are true.

Problem 54: Bicycle Tire Pressure and Temperature [3:00:59]