A spherical particle of radius 3.0 \times 10^{-3}\text{ m} and density 8.0 \times 10^3\tex… — JAMB UTME

A spherical particle of radius $3.0 \times 10^{-3}\text{ m}$ and density $8.0 \times 10^3\text{ kg/m}^3$ is released from rest and falls vertically through a tall column of a viscous liquid of density $2.0 \times 10^3\text{ kg/m}^3$ . If the coefficient of viscosity of the fluid is $0.40\text{ Pa}\cdot\text{s}$ and the acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the magnitude of its terminal velocity in $\text{m/s}$ .

Answer: 0.3 m/s

Answer

The magnitude of the terminal velocity of the falling sphere is

0.3\text{ m/s}

When a body falls at terminal velocity through a viscous medium, its weight is balanced by the sum of buoyancy upthrust and Stokes' viscous drag force. Applying

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

with sphere radius

r = 0.003\text{ m}

, sphere density

\rho_s = 8000\text{ kg/m}^3

, fluid density

\rho_f = 2000\text{ kg/m}^3

, viscosity

\eta = 0.40\text{ Pa}\cdot\text{s}

, and

g = 10\text{ m/s}^2

yields

v_t = 0.3\text{ m/s}

Step-by-Step Solution

Formulate the dynamic equilibrium condition at terminal velocity.

At terminal velocity, the net acceleration is zero, leading to the force balance equation

W = U + F_v

, where

W

is the gravitational weight of the sphere,

U

is the buoyant upthrust, and

F_v

is the retarding viscous force.

Terminal velocity occurs when the downward force of gravity is precisely balanced by the sum of upward resistive and buoyancy forces.

Substitute algebraic expressions for weight, upthrust, and Stokes' viscous drag.

W = \frac{4}{3}\pi r^3 \rho_s g

U = \frac{4}{3}\pi r^3 \rho_f g

, and

F_v = 6\pi \eta r v_t

Archimedes' principle defines the upthrust force equal to the weight of displaced liquid, while Stokes' law governs viscous resistance on spherical bodies.

Solve the equilibrium equation for terminal velocity

v_t

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

Equating

\frac{4}{3}\pi r^3 (\rho_s - \rho_f) g = 6\pi \eta r v_t

and simplifying cancels common factors of

\pi

and

r

Substitute the specified numerical parameters into the derived expression.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 2000) \times 10}{9 \times 0.40} = \frac{2 \times 9.0 \times 10^{-6} \times 6000 \times 10}{3.6} = \frac{1.08}{3.6} = 0.3\text{ m/s}

Direct calculation yields the exact value of terminal velocity.

Key Concept

Terminal Velocity, Stokes' Law, and Archimedes' Principle

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