Hydrostatics

Q: What is hydrostatics in physics?

Hydrostatics is the part of fluid mechanics that studies fluids at rest. It covers pressure and its variation with depth, atmospheric and gauge pressure, barometers and manometers, Pascal's law and hydraulic machines, and Archimedes' principle with floatation.

Q: Why does pressure in a liquid increase with depth?

Each layer of liquid must support the weight of all the liquid above it. Going deeper adds more liquid on top, so the pressure rises by rho g h. For a liquid of uniform density the increase is linear with depth, which is why dams are thicker at the base.

Q: Does the pressure at the bottom depend on the shape of the vessel?

No. Pressure at a point depends only on its depth below the free surface, the liquid's density and g. Vessels of any shape filled to the same height give the same bottom pressure. This result is called the hydrostatic paradox.

Q: What is the difference between gauge pressure and absolute pressure?

Absolute pressure is the actual pressure at a point. Gauge pressure is the excess over atmospheric pressure, so gauge pressure equals absolute pressure minus about 1.013 x 10^5 Pa. Tyre gauges and open manometers read gauge pressure.

Q: What hydrostatics questions come in JEE Main and JEE Advanced?

JEE Main asks U-tubes with two liquids, manometers, hydraulic lifts and floating blocks. JEE Advanced adds forces and torques on dam walls, liquids in accelerated or rotating containers, buoyancy in lifts, and spring-balance or scale readings when a body is lowered into a liquid.

Q: How does a hydraulic lift multiply force?

By Pascal's law the pressure applied on the small piston is transmitted to the large piston. Equal pressure on a bigger area gives a bigger force, so F2 equals F1 times S2 over S1. Energy is conserved: the large piston moves a proportionally smaller distance.

Q: Which hydrostatics topics are important for NEET?

NEET regularly asks pressure at a depth, the barometer and gauge pressure, two liquids in a U-tube, Pascal's law with the hydraulic lift, and floating-fraction questions such as ice in water. Most are one-line numericals based on P = P0 + rho g h and Archimedes' principle.

Q: When does a body float, sink or stay suspended in a liquid?

Compare densities. If the body is denser than the liquid it sinks; if it is less dense it floats with the fraction submerged equal to the density ratio; if the densities are equal it stays suspended wherever it is placed.

PhysicsFluid MechanicsFor NEET aspirants

Hydrostatics is the study of fluids at rest: how pressure builds up with depth ( $P = P_{0} + ρ g h$ ), how a pressure applied at one point spreads through a liquid (Pascal's law), and why bodies float or feel lighter in a liquid (Archimedes' principle). Hydrostatics questions in JEE Main, JEE Advanced and NEET test U-tubes, side-wall forces, hydraulic lifts, accelerated containers and floating blocks, all covered here with worked examples.

On this page1Density2Pressure and depth3Barometer and U-tubes4Side walls5Accelerated liquids6Pascal's law7Buoyancy and floatation

Key Formulas - Quick Reference

Density $ρ = \frac{m}{V}$ ; relative density $= \frac{ρ}{ρ _{water}}$ (no unit)
Pressure $P = \frac{F _{⊥}}{A}$ ; $1 atm = 1.013 \times 1 0^{5} Pa$ , $1 bar = 1 0^{5} Pa$
★ Must learnPressure at depth $h$ : $P = P_{0} + ρ g h$ ; gauge pressure $P - P_{0} = ρ g h$
Barometer: $P_{0} = ρ_{Hg} g h$ ( $76 cm$ of Hg at sea level); two liquids in a U-tube: $ρ_{1} h_{1} = ρ_{2} h_{2}$
★ Must learnForce on a vertical side wall of width $b$ : $F = \frac{1}{2} ρ g b h^{2}$ , acting $\frac{h}{3}$ above the bottom; torque about the bottom edge $τ = \frac{1}{6} ρ g b h^{3}$
★ Must learnHorizontal acceleration $a$ : $tan θ = \frac{a}{g}$ ; vertical acceleration $a$ (up): $P = P_{0} + ρ h (g + a)$
Rotating liquid: surface $y = \frac{ω ^{2} r ^{2}}{2 g}$
★ Must learnPascal / hydraulic lift: $\frac{F _{1}}{S _{1}} = \frac{F _{2}}{S _{2}}$
★ Must learnBuoyant force $F_{B} = ρ_{L} V_{in} g$ ; floating body: $\frac{V _{in}}{V} = \frac{ρ}{ρ _{L}}$
Apparent weight $= m g - F_{B} = V g (ρ - ρ_{L})$

1. Fluids, Density and Relative Density

A fluid is a substance that deforms continuously under a shear (tangential) stress, however small that stress is. Liquids and gases are both fluids. Fluid mechanics deals with fluids at rest (hydrostatics) and in motion (hydrodynamics).

Density is mass per unit volume. At a point, $ρ = Δ V \to 0 lim \frac{Δ m}{Δ V} = \frac{d m}{d V}$ . It is a positive scalar. SI unit $kg m^{- 3}$ , CGS unit $g cm^{- 3}$ ( $1 g cm^{- 3} = 1000 kg m^{- 3}$ ), dimensions $[ML^{- 3}]$ .

Quantity	Definition	Unit
Relative density (R.D.)	$\frac{density of substance}{density of water at 4 ^{\circ} C}$	none (same value in SI and CGS)
Specific gravity	$\frac{weight of substance}{weight of equal volume of water at 4 ^{\circ} C}$	none; numerically equal to R.D.

Example: mercury has R.D. $13.6$ , so $ρ_{Hg} = 13.6 g cm^{- 3} = 13600 kg m^{- 3}$ .

2. Pressure in a Fluid at Rest

A fluid at rest pushes perpendicular to every surface it touches: a wall, the bottom of a vessel, or a body immersed in it. The push comes from molecules colliding with the surface. On an imaginary surface inside the fluid, the fluid on the two sides pushes equally and oppositely, otherwise that part of the fluid would accelerate.

Pressure at a point is the normal force per unit area on a small surface $d A$ around that point:

P = \frac{d F}{d A} (uniform: P = \frac{F}{A})

SI unit: pascal, $1 Pa = 1 N m^{- 2}$ . Dimensions $[ML^{- 1} T^{- 2}]$ .

Pressure is a scalar: it has no direction of its own. The force it produces on a surface is a vector along the normal to that surface, whatever the orientation of the surface.

2.1 Atmospheric, absolute and gauge pressure

Atmospheric pressure $P_{0}$ is the pressure of the air at a place; it changes with weather and altitude. Its average value at sea level is

1 atm = 1.013 \times 1 0^{5} Pa = 76 cm of Hg = 760 torr

Term	Meaning	Relation
Absolute pressure	Actual pressure at the point	$P$
Gauge pressure	Excess over atmospheric pressure (what a tyre gauge reads)	$P - P_{0}$
Bar	Unit used in meteorology	$1 bar = 1 0^{5} Pa$

2.2 Variation of pressure with depth

Take a thin horizontal slab of fluid of area $A$ and height $d y$ at height $y$ above a reference level. Density $ρ$ and $g$ are uniform.

Weight of the slab: $d W = ρ g A d y$ (downward).
Upward force on the bottom face: $P A$ . Downward force on the top face: $(P + d P) A$ .
Equilibrium: $P A - (P + d P) A - ρ g A d y = 0$ , so
$\frac{d P}{d y} = - ρ g$
Pressure decreases as we go up.
Integrate between heights $y_{1}$ and $y_{2}$ :
$P_{2} - P_{1} = - ρ g (y_{2} - y_{1})$
Let point 2 be the free surface ( $P_{2} = P_{0}$ ) and point 1 lie a depth $h = y_{2} - y_{1}$ below it:
$P = P_{0} + ρ g h$

Figure 1: Pressure at depth

h

P = P_{0} + ρ g h

. The graph is a straight line: intercept

P_{0}

, slope

ρ g

Result: $P = P_{0} + ρ g h$ . Pressure grows linearly with depth. All points at the same level in the same connected liquid at rest have the same pressure.

The shape of the container does not matter. A tall narrow vessel and a wide vessel filled to the same height give the same pressure at the base, even though they hold very different amounts of liquid. This is the hydrostatic paradox (NCERT). The extra weight of liquid in a widening vessel is carried by the slanting walls, not by the base.

Figure 2: Hydrostatic paradox. Liquid filled to the same height

H

gives the same base pressure

P_{0} + ρ g H

, whatever the shape or the amount of liquid.

Liquids in layers. When immiscible liquids lie one above another, each layer adds its own column: at the bottom of a layer of depth $h_{1}$ (density $ρ_{1}$ ) resting on a layer of depth $h_{2}$ (density $ρ_{2}$ ), $P = P_{0} + ρ_{1} g h_{1} + ρ_{2} g h_{2}$ . The pressure-depth graph is made of straight pieces whose slope $ρ g$ changes at each interface.

Figure 3: Layered liquids add their columns:

P - P_{0} = ρ_{1} g h_{1} + ρ_{2} g h_{2}

. Here

ρ_{1} g h_{1} = 3.92 kPa

at the interface and

13.72 kPa

at the bottom (

g = 9.8 m s^{- 2}

). The graph bends at the interface because the slope

ρ g

changes.

2.3 Barometer and manometer

A mercury barometer measures atmospheric pressure. A long tube filled with mercury is inverted in a dish of mercury; the space above the column is a vacuum ( $P = 0$ ). Points 1 (inside the tube) and 2 (on the open surface) are at the same level, so $P_{1} = P_{2}$ :

P_{0} = 0 + ρ g h = ρ g h

Mercury is chosen because its large density keeps $h$ short. For $h = 76 cm$ : $P_{0} = (13.6 \times 1 0^{3}) (9.8) (0.760) = 1.01 \times 1 0^{5} Pa$ . With water the column would be about $10.3 m$ tall.

An open-tube manometer measures the pressure $P$ of a gas. A U-tube holding a liquid of density $ρ$ is connected to the gas at one end and open to air at the other. At the level of the lower surface, $P_{1} = P_{2}$ :

P = P_{0} + ρ g h \Rightarrow P - P_{0} = ρ g h (gauge pressure)

Figure 4: Points 1 and 2 at the same level in a connected liquid are at the same pressure. Barometer:

P_{0} = ρ g h

. Manometer: gauge pressure

P - P_{0} = ρ g h

U-tube with two immiscible liquids. Pour oil into one arm of a U-tube holding water. Choose the level of the oil-water interface: below it the same liquid (water) joins the two arms, so the pressures at A and B on that level are equal:

P_{0} + ρ_{w} g h_{w} = P_{0} + ρ_{o} g h_{o} \Rightarrow ρ_{o} h_{o} = ρ_{w} h_{w}

The lighter liquid stands higher. Points higher up at one level (C and D) are not at equal pressure, because different liquids lie between them.

Figure 5: Equate pressures only at the level of the lower interface (A and B), where the same liquid joins the two points. Drawn to scale:

h_{o} = 10 cm

of oil balances

h_{w} = 8 cm

of water, so

ρ_{o} = 1000 \times \frac{8}{10} = 800 kg m^{- 3}

Any U-tube or manometer problem can be solved by walking through the liquid from one open end to the other and adding or subtracting $ρ g h$ at each step.

Figure 6: The pressure walk solves any U-tube, manometer or layered-liquid question in one line. Down:

+ ρ g h

; up:

- ρ g h

; sideways only within one liquid. For the oil-water U-tube:

P_{0} + ρ_{o} g h_{o} - ρ_{w} g h_{w} = P_{0}

2.4 Force, average pressure and torque on a side wall

Pressure on a vertical wall differs at different depths, so we add up strips. Let the wall have width $b$ and the liquid depth be $h$ . Use gauge pressure (air pushes on both sides of the wall, so $P_{0}$ cancels).

Strip at depth $x$ with height $d x$ : $d F = ρ g x \cdot b d x$ , normal to the wall.
Total force:
$F = \int_{0}^{h} ρ g b x d x = \frac{1}{2} ρ g b h^{2}$
Average pressure: $⟨ P ⟩ = \frac{F}{bh} = \frac{1}{2} ρ g h$ , half the gauge pressure at the bottom.
Torque about the bottom edge (lever arm $h - x$ ):
$τ = \int_{0}^{h} ρ g b x (h - x) d x = ρ g b [\frac{h ^{3}}{2} - \frac{h ^{3}}{3}] = \frac{1}{6} ρ g b h^{3}$
Point of action: $\frac{τ}{F} = \frac{h}{3}$ above the bottom (the centroid of the pressure triangle).

Figure 7: Pressure on a side wall grows linearly with depth, so the push on the wall is a triangle. The total force

F = \frac{1}{2} ρ g b h^{2}

pushes the wall outward and acts

h /3

above the bottom.

Exam Trick

Force on a flat wall = pressure at its centroid × area. For a vertical rectangle the centroid is at depth $h /2$ , so $F = ρ g \frac{h}{2} \cdot bh$ . The same shortcut works for any submerged plane surface.

Key idea

Pressure depends only on depth below the free surface:

P = P_{0} + ρ g h

. Points at one level in one connected liquid share the same pressure, whatever the shape of the vessel.

Quick Recall: tap to check

A tank holds

0.5 m

of oil (

800 kg m^{- 3}

) on

1 m

of water. Gauge pressure at the bottom?

ρ_{1} g h_{1} + ρ_{2} g h_{2} = 3920 + 9800 = 13720 Pa

(

g = 9.8 m s^{- 2}

Why can we not equate pressures at points C and D of the oil-water U-tube?

Different liquids lie between them; equate only at the interface level, where the same liquid joins both arms.

Where does the total force of water on a vertical dam act?

\frac{h}{3}

above the base, the centroid of the pressure triangle.

3. Pressure in Accelerated and Rotating Liquids

3.1 Container with horizontal acceleration

In a container accelerating with $a$ , a liquid at rest relative to the container feels gravity $g$ (down) and a pseudo force $a$ per unit mass (backward). Its free surface sets itself perpendicular to the effective gravity $g_{eff} = g^{2} + a^{2}$ , so it tilts, rising at the back wall:

tan θ = \frac{a}{g}

Every layer parallel to the tilted free surface is at one pressure. For a point $A$ at perpendicular depth $h$ below the free surface:

P_{A} = P_{0} + ρ h g^{2} + a^{2}

The same answer comes from the vertical or the horizontal direction. If $l_{1}$ is the vertical depth of $A$ below the surface and $l_{2}$ its horizontal distance from the surface, then $P_{A} = P_{0} + ρ g l_{1} = P_{0} + ρ a l_{2}$ , with $l_{1} = h sec θ = \frac{h a ^{2} + g ^{2}}{g}$ and $l_{2} = h cosec θ = \frac{h a ^{2} + g ^{2}}{a}$ . Along a vertical line only gravity changes the pressure; along a horizontal line only the pseudo force does.

Figure 8: Tank accelerating at

a

(here

a = 0.5 g

θ = 26. 6^{\circ}

). The surface is perpendicular to

g_{eff}

, so

tan θ = \frac{a}{g}

and

P_{A} = P_{0} + ρ h g^{2} + a^{2}

3.2 Container with vertical acceleration

If the container accelerates upward with $a$ , $g_{eff} = g + a$ and the surface stays horizontal: $P = P_{0} + ρ h (g + a)$ . Downward acceleration gives $P = P_{0} + ρ h (g - a)$ . In free fall ( $a = g$ ) the liquid pressure equals $P_{0}$ everywhere.

Horizontal acceleration

a

Surface tilts, rising at the back wall: $tan θ = \frac{a}{g}$ . Pressure at perpendicular depth $h$ : $P_{0} + ρ h g^{2} + a^{2}$ .

Vertical acceleration

a

Surface stays flat. Only $g$ changes: $P = P_{0} + ρ h (g \pm a)$ , $+$ for upward and $-$ for downward acceleration; zero gauge pressure in free fall.

JEE Advanced

Rotating liquid. In a cylinder rotating at $ω$ about its vertical axis, a liquid element at radius $x$ needs a centripetal force; in the rotating frame it feels $m ω^{2} x$ outward and $m g$ down. The free surface is perpendicular to the resultant, so its slope is $\frac{d y}{d x} = \frac{ω ^{2} x}{g}$ . Integrating from the axis:

y = \frac{ω ^{2} x ^{2}}{2 g} (paraboloid of revolution)

Radially, $\frac{d P}{d x} = ρ ω^{2} x$ , so pressure at the level of the lowest surface point, at distance $r$ from the axis, is $P (r) = P_{0} + \frac{1}{2} ρ ω^{2} r^{2}$ .

Figure 9: Rotating liquid. The surface rises by

y = \frac{ω ^{2} r ^{2}}{2 g}

at distance

r

from the axis (curve plotted from this equation).

4. Pascal's Law and Hydraulic Machines

Pascal's law: a pressure applied at one point of an enclosed incompressible fluid is transmitted undiminished to every part of the fluid and to the walls of its container.

Two connected cylinders of cross-sections $S_{1}$ and $S_{2}$ are fitted with pistons at the same level. A force $F_{1}$ on piston 1 raises the pressure everywhere by $p_{1} = F_{1} / S_{1}$ . Piston 2 then feels $F_{2} = p_{1} S_{2}$ :

F_{2} = F_{1} \frac{S _{2}}{S _{1}}

If the pistons were at different heights, $p_{2} = p_{B} - ρ g h$ and $p_{B} = p_{1} + ρ g h$ still give $p_{2} = p_{1}$ : the depth terms cancel. So when $S_{2} ≫ S_{1}$ , $F_{2} ≫ F_{1}$ . This is the hydraulic lift (and hydraulic press). The force is multiplied, not the work: piston 2 moves up only $\frac{S _{1}}{S _{2}}$ times as far as piston 1 moves down, because the liquid volume is fixed.

Figure 10: Hydraulic lift. With both pistons at one level,

\frac{F _{1}}{S _{1}} = \frac{F _{2}}{S _{2}}

, so

F_{2} = F_{1} \frac{S _{2}}{S _{1}}

: a small force lifts a heavy load.

Hydraulic brakes (NCERT) use the same idea: a small push on the brake pedal raises the pressure in the brake fluid, and the pressure acting on larger pistons at every wheel presses the brake pads with a much larger force, equally on all wheels.

Key idea

Pascal's law multiplies force, not energy:

F_{2} = F_{1} \frac{S _{2}}{S _{1}}

, while the large piston moves only

\frac{S _{1}}{S _{2}}

as far.

5. Archimedes' Principle, Buoyancy and Floatation

Archimedes' principle: a body wholly or partly immersed in a fluid is pushed up by a force equal to the weight of the fluid it displaces:

F_{B} = ρ_{L} V_{in} g

$V_{in}$ = immersed volume of the body, $ρ_{L}$ = density of the liquid. The buoyant force acts at the centre of the displaced liquid (the centre of buoyancy).

5.1 Why the buoyant force exists

For a cylinder of face area $A$ and length $L$ standing in a liquid, the top face at depth $h_{1}$ is pushed down by $P_{1} A$ and the bottom face at depth $h_{2}$ is pushed up by $P_{2} A$ . The side forces cancel in pairs. Since $P_{2} - P_{1} = ρ_{L} g (h_{2} - h_{1})$ , the net upward force is $ρ_{L} g A L = ρ_{L} g V$ .

Figure 11: The bottom face is deeper, so

P_{2} > P_{1}

. Net upward force

F_{B} = ρ_{L} g A (h_{2} - h_{1}) = ρ_{L} g V

, the weight of liquid displaced (Archimedes' principle).

General proof. Before the body is placed, the region it will occupy is filled with liquid of weight $V ρ_{L} g$ , which is in equilibrium; so the surrounding liquid pushes that region up with exactly $V ρ_{L} g$ . Pressure at each point depends only on depth, so when the body replaces the liquid, the surrounding liquid pushes on the body with the same net force: $F_{B} = V ρ_{L} g$ . Full immersion is not needed for the argument.

5.2 Law of floatation and apparent weight

A body floats when the buoyant force on its immersed part balances its weight: $ρ V g = ρ_{L} V_{in} g$ . Hence

\frac{V _{in}}{V} = \frac{ρ}{ρ _{L}}

A body sinks if $ρ > ρ_{L}$ , floats partly submerged if $ρ < ρ_{L}$ , and stays wherever it is placed inside the liquid if $ρ = ρ_{L}$ . A fully immersed body that does not float has apparent weight $W_{app} = m g - F_{B} = V g (ρ - ρ_{L})$ ; the loss of weight equals the weight of liquid displaced.

Figure 12: Law of floatation.

ρ V g = ρ_{L} V_{in} g

, so

\frac{V _{in}}{V} = \frac{ρ}{ρ _{L}}

. Drawn to scale for ratios 0.25, 0.5 and 0.8.

A spring balance shows this directly. When a stone hanging from it is lowered into water, the balance reading drops by $F_{B}$ ; if the beaker stands on a weighing scale, the scale reading rises by the same $F_{B}$ , because the stone pushes the water down as hard as the water pushes it up.

Figure 13: Stone:

m = 0.5 kg

V = 200 cm^{3}

g = 10 m s^{- 2}

. Buoyant force

F_{B} = ρ_{w} V g = 2.0 N

, so the balance reads

W - F_{B} = 3.0 N

. By Newton's third law the stone pushes the water down with

2.0 N

, so the scale reads

22.0 N

instead of

20.0 N

Floating body

$ρ < ρ_{L}$ . Weight = buoyant force on the immersed part: $\frac{V _{in}}{V} = \frac{ρ}{ρ _{L}}$ . Apparent weight is zero.

Fully immersed, denser body

$ρ > ρ_{L}$ . $F_{B} = ρ_{L} V g$ on the whole volume; apparent weight $= V g (ρ - ρ_{L})$ , the reading on a spring balance.

Exam Trick

Floating fraction = density ratio. Ice ( $ρ = 0.9 g cm^{- 3}$ ) floats with $90%$ under water; wood of R.D. $0.6$ floats $60%$ submerged. For "loss of weight" problems: loss of weight in water (in gf) = volume in $cm^{3}$ .

JEE Advanced

Buoyancy in an accelerated frame. In a lift accelerating up with $a$ , $F_{B} = ρ_{L} V (g + a)$ and the weight is effectively $m (g + a)$ too, so the floating fraction $\frac{ρ}{ρ _{L}}$ does not change. In a horizontally accelerated tank the buoyant force also has a horizontal part $ρ_{L} V a$ , and $F_{B} = ρ_{L} V g^{2} + a^{2}$ along $g_{eff}$ .

Key idea

Buoyant force = weight of liquid displaced,

F_{B} = ρ_{L} V_{in} g

. A floating body displaces its own weight; a sunk body loses weight equal to

F_{B}

Quick Recall: tap to check

An ice cube floats in water. What fraction is under water?

\frac{ρ _{ice}}{ρ _{w}} = 0.9

, i.e.

90%

5 N

stone of volume

200 cm^{3}

hangs in water. Spring balance reading?

5 - 1000 \times 2 \times 1 0^{- 4} \times 10 = 3 N

(

g = 10 m s^{- 2}

Does the floating fraction change in a lift accelerating upward?

No: both the weight and the buoyant force scale with

g + a

Figure 14: Revision map of hydrostatics: every result follows from

P = P_{0} + ρ g h

and the balance of forces on a piece of fluid.

6. Solved Examples

Solved Example 1

Two liquids that do not mix are placed in an L-shaped tube: the vertical left arm (length

l

) holds liquid of density

2 ρ

, and the horizontal part plus the vertical right arm (each of length

l

) hold liquid of density

ρ

. Both arms are open. Find the displacement

x

of the liquid surfaces at equilibrium.

Solution:

At equilibrium the heavier liquid drops by $x$ in the left arm and the lighter one rises by $x$ in the right arm. Pressure at the level of the horizontal tube must be the same from both sides:

2 ρ g (l - x) = ρ g (l + x) \Rightarrow 2 l - 2 x = l + x

Answer: $x = \frac{l}{3}$ .

Solved Example 2

Three immiscible liquids of densities

ρ

2 ρ

and

3 ρ

fill a U-tube:

ρ

in the left arm (length

l

2 ρ

in the horizontal part (length

l

) and

3 ρ

in the right arm (length

l

). Find the displacement

x

at equilibrium.

Solution:

The densest liquid ( $3 ρ$ ) sinks by $x$ , pushing a length $x$ of the $2 ρ$ liquid up into the left arm under the $ρ$ column. Equate pressures at the bottom level:

ρ g l + 2 ρ g x = 3 ρ g (l - x) \Rightarrow l + 2 x = 3 l - 3 x

Answer: $x = \frac{2 l}{5}$ .

Solved Example 3

A U-tube with a horizontal part of length

l

contains liquid only in its horizontal part. The tube is rotated with angular speed

ω

about the vertical axis through arm A. How high (

x_{0}

) does the liquid rise in the far arm B, and what is the pressure difference between the two ends of the liquid in the horizontal part?

Solution:

The liquid is flung outward: a length $x_{0}$ near the axis empties and the liquid rises $x_{0}$ in arm B. Take an element of length $d x$ at distance $x$ from the axis. The net inward pressure force supplies the centripetal force: $A d P = (ρ A d x) ω^{2} x$ , so $d P = ρ ω^{2} x d x$ .

From the inner free end ( $x = x_{0}$ , pressure $P_{0}$ ) to the bottom of arm B ( $x = l$ ):

Δ P = ρ ω^{2} \int_{x_{0}}^{l} x d x = \frac{1}{2} ρ ω^{2} (l^{2} - x_{0}^{2})

This excess supports the column of height $x_{0}$ in arm B: $Δ P = ρ g x_{0}$ . Therefore

ω^{2} x_{0}^{2} + 2 g x_{0} - ω^{2} l^{2} = 0 \Rightarrow x_{0} = \frac{g ^{2} + ω ^{4} l ^{2} - g}{ω ^{2}}

Answer: pressure difference $Δ P = ρ g x_{0} = \frac{1}{2} ρ ω^{2} (l^{2} - x_{0}^{2})$ , with $x_{0}$ as above (neglecting the tube's width).

Solved Example 4

A liquid of density

ρ

in a bucket spins with angular velocity

ω

about the bucket's vertical axis. Show that the pressure at radial distance

r

from the axis, at the level of the lowest point of the free surface, is

P = P_{0} + \frac{1}{2} ρ ω^{2} r^{2}

Solution:

In the rotating frame a surface element at $(x, y)$ feels the pseudo force $m ω^{2} x$ (outward) and its weight $m g$ . In equilibrium the resultant is perpendicular to the surface:

tan θ = \frac{d y}{d x} = \frac{ω ^{2} x}{g} \Rightarrow y = \frac{ω ^{2} x ^{2}}{2 g}

The surface is a parabola. At $x = r$ the liquid surface stands $y = \frac{ω ^{2} r ^{2}}{2 g}$ above the lowest level, so the pressure at that lowest level, directly below, is

P (r) = P_{0} + ρ g y = P_{0} + \frac{1}{2} ρ ω^{2} r^{2}

Hence shown.

Solved Example 5

A beaker of circular cross-section of radius

4 cm

is filled with mercury up to a height of

10 cm

. Find the force exerted by the mercury on the bottom of the beaker. Atmospheric pressure

= 1 0^{5} N m^{- 2}

, density of mercury

= 13600 kg m^{- 3}

g = 10 m s^{- 2}

Solution:

Pressure at the bottom: $P = P_{0} + ρ g h = 1 0^{5} + (0.1) (13600) (10) = 1 0^{5} + 13600 = 1.136 \times 1 0^{5} N m^{- 2}$ .

Area $= π (0.04)^{2} = 5.03 \times 1 0^{- 3} m^{2}$ . Force $= (1.136 \times 1 0^{5}) (5.03 \times 1 0^{- 3})$ .

Answer: $F \approx 571 N$ .

Solved Example 6

A cubical iron block of side

5 cm

floats on mercury. (i) What height of the block is above the mercury? (ii) Water is poured into the vessel until it just covers the block. What is the height of the water column? Density of mercury

13.6 g cm^{- 3}

, iron

7.2 g cm^{- 3}

Solution:

Mass of block $= 125 \times 7.2 = 900 g$ .

(i) Let $h$ be the height above mercury. Floatation: $5 \times 5 \times (5 - h) \times 13.6 = 900 \Rightarrow 5 - h = 2.65$ , so $h = 2.35 cm$ .

(ii) Let $x$ be the height of the block in water, so $(5 - x)$ is in mercury. Weight of water displaced + weight of mercury displaced = weight of block:

25 x (1) + 25 (5 - x) (13.6) = 900 \Rightarrow x + 68 - 13.6 x = 36 \Rightarrow x = \frac{32}{12.6}

Answer: (i) $2.35 cm$ ; (ii) $x = 2.54 cm$ .

Solved Example 7

A tank of water is placed on a spring balance. A stone of weight

w

hanging from a thread is lowered into the water without touching the sides or the bottom. How does the balance reading change?

Solution:

The water pushes the stone up with buoyant force $B$ ; by Newton's third law the stone pushes the water down with the same $B$ . For the "water + tank" system: weight $W$ down, $B$ down, spring force $R$ up. Equilibrium gives $R = W + B$ .

Answer: the reading increases by $B$ , the weight of the water displaced by the stone (the thread carries the rest, $w - B$ ).

Solved Example 8

A cylindrical vessel containing a liquid is closed by a smooth piston of mass

m

and area

A

. Atmospheric pressure is

P_{0}

. Find the pressure of the liquid just below the piston.

Solution:

Forces on the piston: weight $m g$ (down), air force $P_{0} A$ (down), liquid force $P A$ (up). Equilibrium: $P A = P_{0} A + m g$ .

Answer: $P = P_{0} + \frac{m g}{A}$ .

Solved Example 9

A rubber ball of mass

m

and radius

r

is held under water at depth

h

and released. Neglecting water and air resistance, how high above the surface will it jump?

Solution:

Let it rise $x$ above the surface. While rising through depth $h$ the buoyant force $\frac{4}{3} π r^{3} ρ g$ does work $\frac{4}{3} π r^{3} ρ g h$ ; gravity does work $- m g (h + x)$ . Kinetic energy is zero at both ends:

\frac{4}{3} π r^{3} ρ g h = m g (h + x)

Answer: $x = \frac{( \frac{4}{3} π r ^{3} ρ - m ) h}{m}$ , where $ρ$ is the density of water.

Solved Example 10

A wooden cube supporting a

200 g

mass just floats in water (its top level with the surface). When the mass is removed the cube rises by

2 cm

. Find the side of the cube.

Solution:

Let side $= l$ and density of wood $= ρ$ (water $= 1 g cm^{- 3}$ ). With the mass the cube is fully submerged: $ρ l^{3} + 200 = l^{3}$ .

Without the mass it rises $2 cm$ , so the immersed height is $l - 2$ : $ρ l^{3} = l^{2} (l - 2)$ .

Subtract: $200 = l^{3} - l^{2} (l - 2) = 2 l^{2}$ .

Answer: $l = 10 cm$ .

Solved Example 11

A boat floating in a water tank carries a number of large stones. If the stones are unloaded into the water, what happens to the water level?

Solution:

Let the boat weigh $W$ and the stones $w$ (in gf; water $= 1 g cm^{- 3}$ ). While in the boat the stones are floated, so the water displaced is $(W + w) cm^{3}$ .

In the water, the stones sink and displace only their own volume $\frac{w}{ρ} cm^{3}$ , where $ρ > 1$ is their density. Total displaced $= W + \frac{w}{ρ} < W + w$ .

Answer: the water level falls.

Solved Example 12

Two solid uniform spheres, each of radius

5 cm

and specific gravities

0.5

and

2

, are joined by a light string and fully immersed in a tank of water. Find the tension in the string and the contact force between the heavier sphere and the bottom. (

g = 9.8 m s^{- 2}

)

Solution:

Let the volume of each sphere be $V$ and water density $ρ$ .

Lighter sphere (held down by the string): $T + 0.5 V ρ g = V ρ g \Rightarrow T = 0.5 V ρ g$ .

Heavier sphere: $T + R + V ρ g = 2 V ρ g \Rightarrow R = V ρ g - T = 0.5 V ρ g$ .

$V = \frac{4}{3} π (0.05)^{3} = 5.24 \times 1 0^{- 4} m^{3}$ , so $0.5 V ρ g = 0.5 (5.24 \times 1 0^{- 4}) (1000) (9.8)$ .

Answer: $T = R \approx 2.57 N$ .

Solved Example 13

A rod of length

6 m

and mass

12 kg

(specific gravity

0.5

) is hinged at one end

3 m

below the water surface. (i) What weight must be attached to the other end so that

5 m

of the rod is submerged? (ii) Find the magnitude and direction of the force exerted by the hinge.

Solution:

Let A be the hinge, B the free end, AC $= 5 m$ the submerged part. Mass of AC $= \frac{5}{6} \times 12 = 10 kg$ ; its volume displaces water of mass $\frac{10}{0.5} = 20 kg$ . So $F_{B} = 20 kg wt$ , acting at the middle of AC ( $2.5 m$ from A). The rod's weight acts at $3 m$ from A.

(i) Torques about A (the common factor $cos θ$ of the inclination cancels): $12 \times 3 + x \times 6 = 20 \times 2.5 \Rightarrow 6 x = 14$ , so $x = 2.33 kg wt$ .

(ii) Vertical balance with hinge force $R$ taken upward: $12 + 2.33 = 20 + R \Rightarrow R = - 5.67 kg wt$ .

Answer: (i) $2.33 kg wt$ ; (ii) $5.67 kg wt$ acting downward on the rod.

Solved Example 14

A cylinder of cross-section

300 cm^{2}

and length

10 cm

, made of material of specific gravity

0.8

, floats in water with its axis vertical. Find the work done in pushing it down until it is just fully immersed. (

g = 9.8 m s^{- 2}

)

Solution:

Weight $= (300 \times 1 0^{- 4}) (0.1) (800) g = 2.4 g N$ . Floating depth: $2.4 g = (300 \times 1 0^{- 4}) x (1000) g \Rightarrow x = 0.08 m$ . It must be pushed a further $0.02 m$ .

The extra upthrust grows linearly from $0$ to $ρ g A (0.02) = (1000) (9.8) (0.03) (0.02) = 5.88 N$ ( $= 0.6 g$ ), so the work is the area of this triangle:

W = \frac{1}{2} (5.88) (0.02)

Answer: $W = 0.0588 J$ .

Solved Example 15

96 g

piece of alloy is made of two metals of specific gravities

11.4

and

7.4

. Its weight in water is

86 g wt

. Find the mass of each metal.

Solution:

Let the mass of the first metal be $m$ ; the second is $96 - m$ . Loss of weight in water $= 96 - 86 = 10 g wt$ $=$ total volume in $cm^{3}$ :

\frac{m}{11.4} + \frac{96 - m}{7.4} = 10

Solving: $m (\frac{1}{11.4} - \frac{1}{7.4}) = 10 - 12.97$ , so $m = 62.7 g$ .

Answer: $62.7 g$ of the metal with S.G. $11.4$ and $33.3 g$ of the other.

Solved Example 16

A dam holds water to a depth of

10 m

along a width of

20 m

. Find the total force of the water on the dam and the torque about its base. (

g = 9.8 m s^{- 2}

, ignore atmospheric pressure, which acts on both sides.)

Solution:

$F = \frac{1}{2} ρ g b h^{2} = \frac{1}{2} (1000) (9.8) (20) (10)^{2} = 9.8 \times 1 0^{6} N$ .

$τ = \frac{1}{6} ρ g b h^{3} = \frac{1}{6} (1000) (9.8) (20) (1000) = 3.27 \times 1 0^{7} N m$ , consistent with $F$ acting $\frac{h}{3} = 3.33 m$ above the base.

Answer: $9.8 \times 1 0^{6} N$ ; $3.27 \times 1 0^{7} N m$ .

Solved Example 17

A mercury manometer connected to a gas cylinder shows the mercury

20 cm

higher in the open arm. Find the gauge and absolute pressure of the gas. (

P_{0} = 1.013 \times 1 0^{5} Pa

g = 9.8 m s^{- 2}

)

Solution:

Gauge pressure $= ρ g h = (13600) (9.8) (0.20) = 2.67 \times 1 0^{4} Pa$ .

Absolute pressure $= 1.013 \times 1 0^{5} + 0.267 \times 1 0^{5} = 1.28 \times 1 0^{5} Pa$ .

Answer: gauge $2.67 \times 1 0^{4} Pa$ ; absolute $1.28 \times 1 0^{5} Pa$ .

Solved Example 18

A beaker of water is in a lift accelerating upward at

\frac{g}{2}

. The gauge pressure at a depth of

0.5 m

is (take

g = 9.8 m s^{- 2}

)
(A)

2450 Pa

(B)

4900 Pa

(C)

7350 Pa

(D)

9800 Pa

Solution:

Answer: (C). $g_{eff} = g + a = 1.5 g$ , so $P - P_{0} = ρ h g_{eff} = (1000) (0.5) (14.7) = 7350 Pa$ .

Solved Example 19

In a hydraulic lift the small piston has area

10 cm^{2}

and the large piston

0.50 m^{2}

. What force on the small piston lifts a

1500 kg

car? (

g = 9.8 m s^{- 2}

)

Solution:

$F_{1} = F_{2} \frac{S _{1}}{S _{2}} = (1500) (9.8) \frac{10 \times 1 0 ^{- 4}}{0.50} = 14700 \times 2 \times 1 0^{- 3}$ .

Answer: $F_{1} = 29.4 N$ (a force about 500 times smaller than the car's weight).

Solved Example 20

A tank holds a

0.5 m

layer of oil (density

800 kg m^{- 3}

) floating on

1.0 m

of water. Its base has area

2.0 m^{2}

. Find the gauge pressure at the base and the force on the base due to the liquids. (

g = 9.8 m s^{- 2}

)

Solution:

Each layer adds its column: $P - P_{0} = ρ_{1} g h_{1} + ρ_{2} g h_{2} = (800) (9.8) (0.5) + (1000) (9.8) (1.0) = 3920 + 9800 = 13720 Pa$ .

Force $= (P - P_{0}) A = 13720 \times 2.0 = 27440 N$ .

Answer: $1.37 \times 1 0^{4} Pa$ ; $2.74 \times 1 0^{4} N$ .

Solved Example 21

A U-tube contains water. Oil of density

900 kg m^{- 3}

is poured into one arm until the oil column is

12 cm

long. How high does the water in the other arm stand above the oil-water interface?
(A)

10.8 cm

(B)

12.0 cm

(C)

13.3 cm

(D)

9.0 cm

Solution:

Answer: (A). At the interface level the same liquid (water) joins both arms, so $ρ_{o} h_{o} = ρ_{w} h_{w}$ : $h_{w} = \frac{900 \times 12}{1000} = 10.8 cm$ . The lighter oil stands higher.

Solved Example 22

An ice cube floats in a glass filled with water to the brim. When the ice melts completely, the water
(A) overflows
(B) level falls
(C) level stays the same
(D) level first rises, then falls

Solution:

Answer: (C). Floating ice of mass $m$ displaces water of mass $m$ . On melting it becomes water of the same mass $m$ , which exactly fills the volume it was displacing, so the level does not change.

Practice Questions

At what depth in water is the absolute pressure double the atmospheric pressure? ( $P_{0} = 1.013 \times 1 0^{5} Pa$ , $g = 9.8 m s^{- 2}$ )Answer: $h = \frac{P _{0}}{ρ g} \approx 10.3 m$
A block of wood floats in water with $\frac{3}{5}$ of its volume submerged and in oil with $\frac{3}{4}$ submerged. Find the densities of wood and oil.Answer: wood $600 kg m^{- 3}$ , oil $800 kg m^{- 3}$
A tank $2 m$ wide holds water $3 m$ deep. Find the force on one $2 m$ wide side wall. ( $g = 10 m s^{- 2}$ )Answer: $F = \frac{1}{2} ρ g b h^{2} = 9 \times 1 0^{4} N$
A truck carrying a tank of water accelerates at $2.45 m s^{- 2}$ . At what angle to the horizontal does the water surface settle? ( $g = 9.8 m s^{- 2}$ )Answer: $tan θ = 0.25$ , $θ \approx 1 4^{\circ}$
A metal piece weighs $210 g wt$ in air and $180 g wt$ in water. Find its relative density.Answer: $\frac{210}{30} = 7$
The pistons of a hydraulic press have diameters $2 cm$ and $20 cm$ . What force on the small piston produces $1 0^{4} N$ on the large one?Answer: $100 N$

Common Mistakes to Avoid

Watch out

Adding $P_{0}$ when the question asks for gauge pressure, or leaving it out when it asks for absolute pressure.
Taking $h$ as the distance from the bottom. In $P = P_{0} + ρ g h$ , $h$ is the depth below the free surface.
Thinking a wider vessel gives more pressure at the base: pressure depends on depth only, not on shape or amount of liquid.
Using $ρ g h \times$ area with the bottom pressure for a side wall. Use the average pressure $\frac{1}{2} ρ g h$ (pressure at the centroid).
Using the total volume instead of the immersed volume in $F_{B} = ρ_{L} V_{in} g$ , and using the body's density instead of the liquid's.
In an accelerated tank, tilting the surface the wrong way: the surface rises at the back wall (opposite to $a$ ).
Mixing units: $g cm^{- 3}$ with SI. Convert $1 g cm^{- 3} = 1000 kg m^{- 3}$ before substituting.

Frequently Asked Questions

What is hydrostatics in physics?

Hydrostatics is the part of fluid mechanics that studies fluids at rest. It covers pressure and its variation with depth, atmospheric and gauge pressure, barometers and manometers, Pascal's law and hydraulic machines, and Archimedes' principle with floatation.

Why does pressure in a liquid increase with depth?

Each layer of liquid must support the weight of all the liquid above it. Going deeper adds more liquid on top, so the pressure rises by rho g h. For a liquid of uniform density the increase is linear with depth, which is why dams are thicker at the base.

Does the pressure at the bottom depend on the shape of the vessel?

No. Pressure at a point depends only on its depth below the free surface, the liquid's density and g. Vessels of any shape filled to the same height give the same bottom pressure. This result is called the hydrostatic paradox.

What is the difference between gauge pressure and absolute pressure?

Absolute pressure is the actual pressure at a point. Gauge pressure is the excess over atmospheric pressure, so gauge pressure equals absolute pressure minus about 1.013 x 10^5 Pa. Tyre gauges and open manometers read gauge pressure.

What hydrostatics questions come in JEE Main and JEE Advanced?

JEE Main asks U-tubes with two liquids, manometers, hydraulic lifts and floating blocks. JEE Advanced adds forces and torques on dam walls, liquids in accelerated or rotating containers, buoyancy in lifts, and spring-balance or scale readings when a body is lowered into a liquid.

How does a hydraulic lift multiply force?

By Pascal's law the pressure applied on the small piston is transmitted to the large piston. Equal pressure on a bigger area gives a bigger force, so F2 equals F1 times S2 over S1. Energy is conserved: the large piston moves a proportionally smaller distance.

Which hydrostatics topics are important for NEET?

NEET regularly asks pressure at a depth, the barometer and gauge pressure, two liquids in a U-tube, Pascal's law with the hydraulic lift, and floating-fraction questions such as ice in water. Most are one-line numericals based on P = P0 + rho g h and Archimedes' principle.

When does a body float, sink or stay suspended in a liquid?

Compare densities. If the body is denser than the liquid it sinks; if it is less dense it floats with the fraction submerged equal to the density ratio; if the densities are equal it stays suspended wherever it is placed.

Previous year questions on Hydrostatics

6 questions from past papers, each with a step-by-step solution.

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