Question

A drop of water of radius $0.0015mm$ is falling in air. If the coefficient of viscosity of air is $2.0\times {10}^{-5}kg{m}^{-1}{s}^{-1}$, the terminal velocity of the drop will be: (The density of water $1.0\times {10}^{3}kg{m}^{-3}$ and $g=10m{s}^{-2}$)

• A. $1.0\times {10}^{-4}m{s}^{-1}$
• B. $2.0\times {10}^{-4}m{s}^{-1}$
• C. $2.5\times {10}^{-4}m{s}^{-1}$
• D. $5.0\times {10}^{-4}m{s}^{-1}$

Answer 1

The correct option is C

$2.5\times {10}^{-4}m{s}^{-1}$

Step1: Given data

Radius, $r=0.0015mm=0.0015\times {10}^{-3}m$

Coefficient of viscosity of air, $\eta =2.0\times {10}^{-5}kg{m}^{-1}{s}^{-1}$

The density of water, $\rho =1.0\times {10}^{3}kg{m}^{-3}$

$g=10m{s}^{-2}$

Step2: Formula used

The terminal velocity of the water drop is $v_{\tau }=\frac{2r^2\rho g}{9\eta }$

Step3: Calculating the terminal velocity

When an object falls through a fluid, it achieves its highest velocity, which is known as terminal velocity. It occurs when the sum of the drag and buoyancy forces acting on the item equals the downward gravity force acting on the object.

Substituting all the given values in the terminal velocity formula

$$\begin{gathered} v_{\tau }=\frac{2\times {(0.0015\times {10}^{-3})}^2\times 1.0\times {10}^3\times 10}{9\times 2.0\times {10}^{-5}} \\ {v}_{\tau }=2.5\times {10}^{-4}m{s}^{-1} \end{gathered}$$

The terminal velocity of the drop will be $2.5\times {10}^{-4}m{s}^{-1}$

Hence, option C is the correct answer.