MHT-CET Physics · Mechanical Properties of Fluids

Pressure with Depth and Buoyancy

Pressure in a liquid rises by ρgh with depth, so a gas bubble grows as it rises (Boyle's law at constant temperature), and a body in a liquid feels an upthrust equal to the weight of liquid it displaces — which can decelerate a light body that falls in until it stops and floats back up.

All Mechanical Properties of Fluids notes MHT-CET Physics strategy

Why this matters

6 PYQs, five of them HARD — a small page with the chapter's hardest questions. Two are a bubble rising from a lake bed whose radius or diameter doubles, one the pressure rise at the centre of a spinning drum, one the force on the bottom of a vessel holding a suspended body, and two how deep a light body sinks when dropped into a denser liquid. Two cards.

Concept 1 of 2: Pressure with Depth, and the Rising Bubble

Every metre of water adds ρg of pressure. A bubble at the bottom of a lake carries the atmosphere plus the water above it; at the surface, only the atmosphere. At constant temperature PV stays fixed, so if the volume grows eight times (radius or diameter doubles) the pressure at the bottom was eight times the surface pressure — the water supplied seven of those eight parts. Write the atmosphere as a column of water H (or of mercury h, relative density ρ, which is hρ of water), and the depth is 7H.

Definition

Pressure at depth $d$ : $P = P_0 + \rho g d$ .
Rising bubble, constant temperature: $(P_0 + \rho g d)V = P_0 \cdot kV$ . Radius (or diameter) doubled $\Rightarrow k = 8 \Rightarrow \rho g d = 7P_0$ : depth $= 7H$ with $P_0$ as H of water, or $7h\rho$ with $P_0$ as h of mercury of relative density ρ.
Spinning drum of radius R at ω: the liquid at the rim moves at ωR, so the pressure difference between rim and centre is $\tfrac{1}{2}d\,\omega^2R^2$ .
A body hanging in the liquid pushes down on the liquid with the same force the liquid pushes up on it. For the cylinder with a hemisphere cut from its base, top face at depth h, the paper keys $\rho g(V + \pi R^2 h)$ : the weight of liquid filling the body's volume V plus the column of height h standing on its top face.

Rising bubble

(P_0 + \rho g d)\,V_1 = P_0\,V_2 \;\Rightarrow\; \rho g d = P_0\left(\frac{V_2}{V_1} - 1\right)

Worked example

A bubble's radius triples as it rises to the surface at constant temperature. The atmosphere is 10 m of water. How deep is the lake?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift II · Q24Hard

Example 1 · Mechanical Properties of Fluids · Pressure, Buoyancy, and Archimedes

The temperature of an air bubble while rising from bottom to surface of a lake remains constant but diameter is doubled. If pressure on the surface is h meter of mercury column and relative density of mercury is '

\rho

' then the depth of the lake is

Using 8H instead of 7H

The bottom pressure is 8 atmospheres, but one of them is the atmosphere itself. The water column is the other seven: depth 7H.

Drill 3 more on pressure with depth, and the rising bubble

Concept 2 of 2: Upthrust, and How Deep a Light Body Sinks

A body dropped from height h enters the liquid at √(2gh). Inside, the upthrust σVg beats its weight ρVg, so the net upward force (σ − ρ)Vg decelerates it at (σ − ρ)g/ρ. It sinks until that deceleration has used up its speed: depth = v²/2a = hρ/(σ − ρ). Then it rises and floats. The body only goes deep when its density is close to the liquid's.

Definition

Upthrust = weight of liquid displaced = $\sigma V g$ .
In the liquid, deceleration $a = \dfrac{(\sigma - \rho)g}{\rho}$ for $\sigma > \rho$ .
Maximum depth after a drop from height h (no damping): $\dfrac{v^2}{2a} = \dfrac{h\rho}{\sigma - \rho}$ .

Maximum depth

d_{\max} = \frac{2gh}{2\,(\sigma - \rho)g/\rho} = \frac{h\rho}{\sigma - \rho}

Worked example

A ball of density 800 kg/m³ is dropped from 1 m into water. How deep does it go?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q16Hard

Example 2 · Mechanical Properties of Fluids · Pressure, Buoyancy, and Archimedes

A small metal sphere of density

\rho

is dropped from height

h

into a jar containing liquid of density

\sigma(\sigma>\rho)

. The maximum depth up to which the sphere sinks is (Neglect damping forces)

Writing the deceleration as (σ − ρ)g

The net force (σ − ρ)Vg acts on the body's own mass ρV, so the deceleration is (σ − ρ)g/ρ. Dropping the ρ loses the h ρ in the answer.

Drill 1 more on upthrust, and how deep a light body sinks

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Pressure with Depth, and the Rising Bubble
Rising bubble
$(P_0 + \rho g d)\,V_1 = P_0\,V_2 \;\Rightarrow\; \rho g d = P_0\left(\frac{V_2}{V_1} - 1\right)$
Upthrust, and How Deep a Light Body Sinks
Maximum depth
$d_{\max} = \frac{2gh}{2\,(\sigma - \rho)g/\rho} = \frac{h\rho}{\sigma - \rho}$

Watch out for (2)

Using 8H instead of 7H
→ Pressure with Depth, and the Rising Bubble
Writing the deceleration as (σ − ρ)g
→ Upthrust, and How Deep a Light Body Sinks

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