Question

A rain drop of radius $r$ falls in air with a terminal speed ${v_t}$. what is the terminal speed of a raindrop of radius $2\,r$ ?
A. $\dfrac{{{v_t}}}{2}$
B. ${v_t}$
C. $2{v_t}$
D. $4{v_t}$

Answer 1

Hint- We know that terminal velocity is the maximum attainable velocity of a falling body. The terminal velocity of a falling sphere is directly proportional to the square of its radius r. By comparing terminal velocity of drop with radius r and the terminal velocity of drop with radius 2r we can get the final answer.

Complete step by step answer:
When the weight of the spherical body is balanced by the buoyant force and drag force due to the fluid the object falls with constant velocity called the terminal velocity. Terminal velocity is the maximum attainable velocity of the falling body in a fluid.
Let weight of body be $w$, buoyant force provided by fluid by $B$ and drag force be ${F_d}$ .
Then, we can write
$W = B + {F_d}$
We know weight is
$w = mg$
$\Rightarrow w = V{\rho _b}g$
Where, $V$ is the volume,${\rho _b}$ is the density of the falling body and $g$ is the acceleration due to gravity. Assuming the drop to be spherical we get the volume as $\dfrac{4}{3}\pi {r^3}$
That is $W = \dfrac{4}{3}\pi {r^3} \times {\rho _b} \times g$
Buoyant force is $B = V{\rho _f}g$
Where, $V$ is the volume of fluid displaced, ${\rho _f}$ is the density of the fluid and $g$ is the acceleration due to gravity.
Since same volume of fluid is displaced, we can take volume $V = \dfrac{4}{3}\pi {r^3}$
That is $B = \dfrac{4}{3}\pi {r^3}{\rho _f}g$
From strokes we have viscous force or drag force.
${F_d} = 6\pi \eta r{v_t}$
Where, $r$ is the radius $\eta$ is the coefficient of viscosity
${v_t}$is the terminal velocity
On substituting these values in equation (1), we get
$\dfrac{4}{3}\pi {r^3} \times {\rho _b} \times g = \dfrac{4}{3}\pi {r^3} \times {\rho _f} \times g + 6\pi \eta r{v_t}$
$\Rightarrow \dfrac{4}{3}\pi {r^3}g\left( {{\rho _b} - {\rho _f}} \right) = 6\pi \eta r{v_t}$
$\Rightarrow {v_t} = \dfrac{{2{r^3}g\left( {{\rho _b} - {\rho _f}} \right)}}{{9\eta }}$
which means ${v_t} \propto {r^2}$
It is given that the radius of the sphere is initially $r$.
Let us denote terminal velocity in this case as ${v_1}$
Thus,
${v_1} \propto {r^2}$ …………………….(1)
Let us denotes terminal velocity when the raindrop has $2a$ radius be ${v_2}$
Then,
${v_2} \propto {\left( {2r} \right)^2}$ ……………..(2)
By dividing (1) and (2).
We get,
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{r^2}}}{{{{\left( {2r} \right)}^2}}}$
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{r^2}}}{{4{r^2}}}$
$\Rightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{1}{4}$
$\therefore {v_2} = 4{v_1}$
Since, terminal velocity of drop with radius r is given as ${v_t}$ we can write,
$\therefore {v_2} = 4{v_t}$
So, the terminal velocity becomes 4 times if the radius becomes $2r$

So, the correct answer is option D.

Note: Remember that for the same shape and material of the falling body the terminal velocity will increase with increase in size. Thus, the value of settling velocity or the terminal velocity will be higher when the radius is made twice the initial value. Terminal velocity is directly related to the square of radius.