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JEE Mains Physics · Mechanical Properties of Fluids

Terminal Velocity

A sphere falling through a fluid speeds up until the viscous drag 6πηrv and the buoyancy together equal its weight; that steady speed is v = 2r²(ρ − σ)g/9η, so for one material it grows as the square of the radius.

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Why this matters

Nineteen PYQs, six of them numeric, and five from 2026. Eleven use the terminal-velocity formula directly: a speed, a viscosity found from a rising bubble or a falling ball, a drop height that matches the speed in water, and the shape of the velocity-time graph; eight use only how the speed scales, for drops that merge or split, for balls of the same mass, and for the error in a measured radius.

Concept 1 of 2: The terminal-velocity formula

At first the ball accelerates. As it speeds up, the drag 6πηrv6\pi\eta r v grows, until drag plus buoyancy equals the weight. From then on it falls at a constant speed, the terminal velocity. Setting the forces equal gives the formula. An air bubble rising through a liquid is the same balance upside down: buoyancy pulls it up, drag holds it back, and its own weight is too small to count.

Definition

  • Balance: 6πηrvT+43πr3σg=43πr3ρg6\pi\eta r v_T + \tfrac{4}{3}\pi r^{3}\sigma g = \tfrac{4}{3}\pi r^{3}\rho g, with ρ the sphere's density and σ the fluid's.
  • So vT=2r2(ρ−σ)g9ηv_T = \dfrac{2r^{2}(\rho - \sigma)g}{9\eta}.
  • Air bubble rising, air's density neglected: vT=2r2σg9ηv_T = \dfrac{2r^{2}\sigma g}{9\eta}, so η=2r2σg9vT\eta = \dfrac{2r^{2}\sigma g}{9v_T}.
  • Use one system of units: SI (m, kg/m³, Pa s, g = 10 m/s²) or CGS (cm, g/cm³, poise, g = 1000 cm/s²).
  • 'Enters the water without changing speed': the free-fall speed 2gh\sqrt{2gh} equals vTv_T in water, so h=vT22gh = \dfrac{v_T^{2}}{2g}.
  • Velocity-time graph: rises from zero with a slope that keeps falling, and levels off at vTv_T.
  • In the lab: η does not depend on how fast the ball was launched, but it does depend on temperature, so the temperature must be kept steady.

Terminal velocity

vT=2r2(ρ−σ)g9ηv_T = \frac{2r^{2}(\rho - \sigma)g}{9\eta}

Worked example

A steel ball of radius 1.5 mm and density 7.8 g/cm37.8\ \text{g/cm}^{3} falls through oil of density 0.9 g/cm30.9\ \text{g/cm}^{3} and viscosity 15 poise. Find its terminal velocity. (g=1000 cm/s2g = 1000\ \text{cm/s}^{2})
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q20Moderate

Example 1 · Mechanical Properties of Fluids · Terminal Velocity

A small metallic sphere of diameter 2 mm and density 10.5 g/cm310.5\text{ }g/cm^{3} is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm31.5\text{ }g/cm^{3} respectively. The terminal velocity attained by the sphere is ____\_\_\_\_ cm/scm/s. (π=227\pi =\frac{22}{7} and g=10 m/s2g = 10\text{ }m/s^{2})

Diameter or radius

Stems usually give the diameter. Halve it before squaring, or the answer is four times too big.

Do not mix CGS and SI

With η in poise, use cm, g/cm³ and g = 1000 cm/s². With η in Pa s, use m, kg/m³ and g = 10 m/s². A g of 10 with lengths in cm gives an answer 100 times too small.

A faster launch does not change η

A ball thrown in with some speed still settles to the same terminal velocity, so the measured viscosity is the same.
Drill 10 more on the terminal-velocity formula

Concept 2 of 2: How terminal velocity scales with the radius

For one material in one fluid, everything in the formula is fixed except r2r^{2}, so vT∝r2v_T \propto r^{2}. When n equal drops merge, the volume is kept: the big radius is n1/3n^{1/3} times the small one, and the speed rises by n2/3n^{2/3}. If the MASS is kept fixed instead of the material, the answer changes completely.

Definition

  • Same material, same fluid: vT∝r2v_T \propto r^{2}.
  • n equal drops merge: R=n1/3rR = n^{1/3}r, so v′=n2/3vv' = n^{2/3}v. Eight drops give × 4, 27 give × 9, 64 give × 16.
  • One drop split into n equal droplets: each droplet falls at v/n2/3v/n^{2/3}.
  • Same MASS, different radius, fluid density negligible: the weight is fixed and must equal 6πηrv6\pi\eta r v, so v∝1/rv \propto 1/r.
  • Error in the speed from an error in the radius: Δvv=2Δrr\dfrac{\Delta v}{v} = 2\dfrac{\Delta r}{r}.

Merging drops

vT∝r2,R=n1/3r⇒v′=n2/3vv_T \propto r^{2}, \qquad R = n^{1/3}r \Rightarrow v' = n^{2/3}v

Worked example

Twenty-seven equal raindrops each fall at a terminal velocity of 6 cm/s. They merge into one drop. Find its terminal velocity.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 1 · Q25Moderate

Example 2 · Mechanical Properties of Fluids · Terminal Velocity

Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 cm/s10\text{ }cm/s coalesce to form a bigger drop. The terminal velocity of bigger drop is ____\_\_\_\_ cm/scm/s.

Same mass is not same material

For one material, v ∝ r². For one mass, a bigger radius means a lighter material and more drag: v ∝ 1/r, so doubling the radius halves the speed.

Merging multiplies by n^(2/3), not n

27 drops merging make the speed 9 times larger, not 27 times. The radius grows only by the cube root of n.

v grows as r², not inversely with r

A reason that says terminal velocity is inversely proportional to the radius is false for a given material. The squared radius is also why its error is doubled.
Drill 7 more on how terminal velocity scales with the radius

Summary — formulas & gotchas at a glance

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