If drop of radius R breaks into 64 drops, find v₁/v₂.

16. r=R/4 from volume conservation. v∝r² so v₁/v₂=(R/r)²=16.

What is the terminal velocity formula?

v_t=2r²(ρ_drop−ρ_fluid)g/(9η). It's proportional to r².

Spherical drop radius R terminal velocity v₁ broken into 64 drops terminal velocity v₂ ratio v₁/v₂

JEERANKERS

PhysicsFluid Mechanics

QMCQFluid Mechanics

A spherical liquid drop of radius $R$ acquires terminal velocity $v_1$ through a gas of viscosity $\eta$. The drop breaks into 64 identical droplets each acquiring terminal velocity $v_2$. The ratio $v_1/v_2$ is:

A) 4 B) 0.25 C) 32 D) 16

✅ Correct Answer

D) 16

✓

Solution

Find radius of each small drop

$\tfrac{4}{3}\pi R^3=64\times\tfrac{4}{3}\pi r^3\Rightarrow r=\dfrac{R}{4}$

Apply Stokes' law: v ∝ r²

$v_t=\dfrac{2r^2(\rho_1-\rho_2)g}{9\eta}\propto r^2$

$\dfrac{v_1}{v_2}=\left(\dfrac{R}{r}\right)^2=\left(\dfrac{R}{R/4}\right)^2=16$

📘 Stokes' Law: v_t ∝ r²

Volume conservation: $R^3=64r^3$ gives $r=R/4$. By Stokes' law $v_t\propto r^2$: $v_1/v_2=(R/r)^2=16$.

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Important Concepts

Volume Conservation4/3πR³=64×4/3πr³ → r=R/4. The 64 small drops have same total volume.

Stokes' Lawv_t=2r²(Δρ)g/(9η)∝r². Smaller radius → much smaller terminal velocity.

Ratio Formulav₁/v₂=(R/r)²=4²=16.

Physical MeaningLarge drop (R) falls 16× faster than small drops (R/4). This explains why rain drops reach terminal velocity quickly but mist drops stay suspended.

FAQs

How to use volume conservation?

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Total volume unchanged: R³=64r³ → r=R/4.

Why v∝r²?

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Stokes drag∝r, gravity∝r³. At terminal: v∝r³/r=r².

What is Stokes' drag force?

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F_drag=6πηrv. At terminal velocity: F_drag=F_gravity→v_t=2r²Δρg/(9η).

What if 27 drops?

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r=R/3, v₁/v₂=9.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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