A body is falling freely in a viscous liquid. If the radius of the body is doubled, its terminal velocity will become:

CONCEPT:

Terminal velocity:

• If a spherical body of radius r is dropped in a viscous fluid, it is first accelerated, and then its acceleration becomes zero and it attains a constant velocity called terminal velocity.
•  The terminal velocity is given as,

\(⇒ v=\frac{2gr^2(ρ-σ)}{9η}\)

Where v = terminal velocity, r = radius, ρ = density of the body, σ = density of liquid, g = gravitational acceleration, and η = viscosity

EXPLANATION:

Given r₁ = r, and r₂ = 2r

• We know that the terminal velocity is given as,

\(⇒ v=\frac{2gr^2(ρ-σ)}{9η}\)

If all quantities other than r is constant, then,

⇒ v ∝ r²     -----(1)

By equation 1 when r1 = r,

⇒ v₁ ∝ r2     -----(2)

By equation 1 when r2 = 2r,

⇒ v₂ ∝ (2r)2

⇒ v2 ∝ 4r2     -----(3)

By equation 2 and equation 3,

⇒ v2 = 4v₁

• Hence, option 2 is correct.

Additional Information

Stoke's law:

• When a body moves through a fluid, the fluid in contact with the body is dragged with it.
• This establishes relative motion in fluid layers near the body, due to which viscous force starts operating. The fluid exerts a viscous force on the body to oppose its motion.
• The magnitude of the viscous force depends on the shape and size of the body, its speed, and the viscosity of the fluid.
• Stokes established that if a sphere of radius r moves with velocity v through a fluid of viscosity η, the viscous force opposing the motion of the sphere is,

⇒ F = 6πηrv

• This law is called Stokes law.