A solid sphere of mass 0.60\text{ kg} and volume 2.0 \times 10^{-4}\text{ m}^3 is released… — JAMB UTME

A solid sphere of mass $0.60\text{ kg}$ and volume $2.0 \times 10^{-4}\text{ m}^3$ is released from rest in a tall vessel filled with a viscous liquid of density $1000\text{ kg/m}^3$ . As the sphere falls, it eventually reaches a constant terminal velocity of $4.0\text{ m/s}$ . Assuming that the viscous drag force is directly proportional to the speed of the sphere, what is the magnitude of the viscous drag force acting on the sphere when its speed is $1.5\text{ m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

A
$0.75\text{ N}$
$1.50\text{ N}$ Answer
C
$2.25\text{ N}$
D
$2.50\text{ N}$

Answer

1.50\text{ N}

At terminal velocity, the sphere is in translational equilibrium under three forces: downward weight (

6.0\text{ N}

), upward upthrust (

2.0\text{ N}

), and upward viscous drag (

4.0\text{ N}

). Since viscous drag is directly proportional to velocity (

F_v = k v

), the proportionality constant

k

1.0\text{ N}\cdot\text{s/m}

. Therefore, at

1.5\text{ m/s}

, the viscous force is

1.0 \times 1.5 = 1.50\text{ N}

Step-by-Step Solution

Calculate the downward gravitational force (weight) acting on the sphere.

W = m \times g = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force pulling the sphere downward.

Calculate the buoyant force (upthrust) exerted by the liquid using Archimedes' principle.

U = \rho_l \times V \times g = 1000\text{ kg/m}^3 \times (2.0 \times 10^{-4}\text{ m}^3) \times 10\text{ m/s}^2 = 2.0\text{ N}

Upthrust equals the weight of the fluid displaced by the submerged sphere.

Determine the viscous drag force at terminal velocity (

v_t = 4.0\text{ m/s}

) using equilibrium of forces.

F_v(v_t) = W - U = 6.0\text{ N} - 2.0\text{ N} = 4.0\text{ N}

At terminal velocity, the net acceleration is zero, so downward weight is balanced by upward upthrust and viscous drag.

Find the constant of proportionality

k

for viscous drag (

F_v = k v

k = \frac{F_v(v_t)}{v_t} = \frac{4.0\text{ N}}{4.0\text{ m/s}} = 1.0\text{ N}\cdot\text{s/m}

Viscous force is given as directly proportional to speed.

Calculate the viscous drag force at a speed of

1.5\text{ m/s}

F_v(1.5) = k \times 1.5\text{ m/s} = 1.0\text{ N}\cdot\text{s/m} \times 1.5\text{ m/s} = 1.50\text{ N}

Applying the constant

k

to the specified speed.

Key Concept

Terminal velocity in viscous fluids and Archimedes' Principle

Estimated Time:2m 30s

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