Question

Find the terminal velocity if a water drop of radius $1.5mm$ is falling from height of $1km$. It is given that it has drag constant $0.5$ and density of water drop is $1000kg/{m^3}$ density of air $12.5km/{m^3}$.
(A) $2.53m/s$
(B) $60m/s$
(C) $7m/s$
(D) $3.47m/s$

Answer 1

Hint At first, convert radius of water drop from millimetre to metre. Because, the water drop is spherical we will calculate its area by formula-
$A = \pi {R^2}$ (where, $R$ is the radius of sphere and $\pi = 3.14$)
As the drop is spherical, the volume will be given by-
$V = \dfrac{4}{3}\pi {R^3}$
Now, to calculate mass of drop use formula of density which is-
$D = \dfrac{m}{V}$ (where, $V$ is the volume of drop, $D$ is the density and $m$ is the mass)
The terminal velocity is given by-
${V_t} = \sqrt {\dfrac{{2mg}}{{{C_d}A\rho }}}$
where, ${V_t}$ is the terminal velocity
$g$ is the acceleration due to gravity
$A$ is the area projected by object
${C_d}$ is the drag coefficient
$\rho$ is density of fluid through which object is falling

Complete step by step answer:
According to question, it is given that
Radius of water drop is $R = 1.5mm = 0.0015m$ $(\because 1m = 1000m)$
So, now we will calculate the area of water drop as it is spherical the area is given by-
$A = \pi {R^2}$
Now, putting values in above equation, we get
$A = 3.14 \times {(0.0015)^2} \\\ A = 3.14 \times 0.00000225 \\\ A = 7.06 \times {10^{ - 6}}{m^2} \\\$
As shown above, we calculated the value of area for drops.
Let the density of water drop be $D$. Therefore, according to the question it is given that the density of water drops $= D = 1000kg/{m^3}$. The volume for sphere is given by
$V = \dfrac{4}{3}\pi {R^3}$
The density is given by-
$D = \dfrac{m}{V}$
To calculate the mass, we have to do transposition of above formula therefore,
$m = D \times V$
$m = 1000 \times \dfrac{4}{3} \times 3.14 \times {(0.0015)^3}$
Doing the further calculations, we get
$m = 1.413 \times {10^{ - 5}}kg$
Therefore, we got the mass of water drop
Let the drag coefficient be ${C_d}$ and $\rho$ be the density of air
Terminal velocity is given by formula,
${V_t} = \sqrt {\dfrac{{2mg}}{{{C_d}A\rho }}}$
Now, putting the values in above formula
$\Rightarrow {V_t} = \sqrt {\dfrac{{2 \times 1.413 \times {{10}^{ - 5}} \times 10}}{{0.5 \times 12.5 \times 7.06 \times {{10}^{ - 6}}}}} \\\ \Rightarrow {V_t} = \sqrt {\dfrac{{282.6}}{{44.125}}} = \sqrt {6.406} \\\ \Rightarrow {V_t} = 2.53m/s(approx) \\\$

Hence, option (A) is the correct answer.

Note The highest velocity attained by an object when falling through a fluid is called terminal velocity. It occurs when the addition of drag force and buoyancy is equal to downward gravity force which is acting on the object.