Question

A water drop of radius 1.5 mm is falling from height 1 km having drag constant 0.5 density of water drop is 1000 kg/${m^3}$ and density of air is 1.29 kg/${m^3}$ Find the terminal velocity.
A. 8.7m/s
B. 7.8m/s
C. 5.6m/s
D. 4.3m/s

Answer 1

To solve this question, we will use the concept of terminal velocity. The terminal velocity can be calculated as $\sqrt {\dfrac{{2mg}}{{\rho CA}}}$, where C is the drag coefficient, A is the cross-section area and $\rho$ is the density.

Complete answer:
Given that,
Radius of the water drop = 1.5mm = $\dfrac{{1.5}}{{1000}}m = 0.0015m$
Height, h = 1 km.
Drag constant, C = 0.5.
Density of water drop = 1000 kg/${m^3}$
Density of air = 1.29 kg/${m^3}$

We have to find out the terminal velocity of the water drop.
So,
We know that,
Density, $\rho = \dfrac{{mass}}{{volume}}$
And,
Mass = $\rho \times volume$ ………. (i)

And we know that, a water drop is spherical in shape and volume of a sphere is given by,
$V = \dfrac{4}{3}\pi {r^3}$
Putting r = 0.0015m in this, we will get
Volume of water drop, $V = \dfrac{4}{3}\left( {3.14} \right){\left( {0.0015} \right)^3}$
Putting this value in equation (i),
Mass = $1000 \times \dfrac{4}{3}\left( {3.14} \right){\left( {0.0015} \right)^3}$
Mass = 0.000014137 kg.

Now,

Area = $\pi {r^2}$
Area = $\left( {3.14} \right) \times {\left( {0.0015} \right)^2}$
Area = 0.000007065 ${m^2}$
We know that,
Terminal velocity is given by,
${{\text{v}}_{{\text{terminal}}}} = \sqrt {\dfrac{{2mg}}{{\rho CA}}}$
Putting all the values,
${{\text{v}}_{{\text{terminal}}}} = \sqrt {\dfrac{{2\left( {0.000014137} \right)\left( {9.8} \right)}}{{\left( {1.29} \right)\left( {0.5} \right)\left( {0.000007965} \right)}}}$

Solving it, we will get

${{\text{v}}_{{\text{terminal}}}} = \sqrt {60.80}$
${{\text{v}}_{{\text{terminal}}}} = 7.795$m/s
${{\text{v}}_{{\text{terminal}}}} \simeq 7.8$m/s
Hence, we can say that the terminal velocity of the water drop is ${{\text{v}}_{{\text{terminal}}}} \simeq 7.8$m/s.

So, the correct answer is “Option B”.

Note:
As an object falls through the atmosphere, there is air resistance acting on it. Air resistance is dependent on the velocity of the falling object, so as it speeds up, the force acting on it gets larger and larger. Eventually, the force from air resistance will equal the force due to gravity. At that point the forces, acting in opposite directions, will sum to zero meaning that the object will travel at a constant velocity.