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 $1000\text{ }kg{{m}^{-3}}$ and density of air is $1.29\text{ }kg{{m}^{-3}}$. Find the terminal velocity.

Answer 1

The terminal velocity of an object is the maximum attainable velocity by an object falling through a fluid. It happens when the drag force is equal to the force of gravity acting on the object. Terminal velocity mainly depends on the drag coefficient and mass of the falling object. This will be helpful in finding out the answer.

Complete answer:
It has been given in the question that the radius of the water droplet be,
$r=1.5\text{ }mm=0.0015\text{ }m$
The height from which the water drop is falling down has been given as,
$h=1000\text{ }m$
Density of the air will be,
${{\rho }_{air}}=1.29\text{ }kg{{m}^{-3}}$
Density of water can be written as,
${{\rho }_{air}}=1000\text{ }kg{{m}^{-3}}$
The drag coefficient can be written as,
${{C}_{d}}=0.5$
From this given data, we have to find the mass of the water droplet and the area to find out the terminal velocity.
Terminal Velocity is given by the formula,
${{V}_{t}}=\sqrt{\dfrac{2mg}{\rho A{{C}_{d}}}}$
After finding the area and mass of the particle, we will substitute the given data in the equation to find out the final answer.
The area A will be,

& A=\pi {{r}^{2}} \\\ & \Rightarrow A=3.14\times {{(0.0015)}^{2}} \\\ & \Rightarrow A=0.000007065\text{ }{{m}^{2}} \\\ \end{aligned}$$ The mass will be, $$\begin{aligned} & m=\rho .v \\\ & \Rightarrow m=1000\times \dfrac{4}{3}\times \pi \times r \\\ & \Rightarrow m=0.000014137\text{ }kg \\\ \end{aligned}$$ Therefore, $$\begin{aligned} & {{V}_{t}}=\sqrt{\dfrac{2mg}{\rho A{{C}_{d}}}} \\\ & \Rightarrow {{V}_{t}}=\sqrt{\dfrac{2\times 0.000014137\times 9.81}{0.5\times 1.29\times 0.000007065}} \\\ & \Rightarrow {{V}_{t}}=\sqrt{60.80} \\\ & \therefore {{V}_{t}}=7.8\text{ }m{{s}^{-1}} \\\ \end{aligned}$$ $$$$ **Note:** The same formula is used by NASA to calculate the terminal velocity of a rocket falling through the earth. Objects do not fall at the same rate through the atmosphere like in vacuum because it depends on the drag coefficient and weight of the object. More streamlined objects have lesser drag coefficient and hence the terminal velocity will be higher.