Question

A drop of liquid having radius 2mm has a terminal velocity $$20\text{cm}{\text{s}}^{-1}$$, the terminal velocity of the drop 1 mm radius will be:
A. $$40\text{cm}{\text{s}}^{-1}$$
B. $$20\text{cm}{\text{s}}^{-1}$$
C. $$10\text{cm}{\text{s}}^{-1}$$
D. $$5\text{cm}{\text{s}}^{-1}$$

Answer 1

Hint: Recall the expression for the terminal velocity. The terminal velocity is proportional to the square of the radius of the drop. Solve the two equations by keeping the other quantities the same.

Formula used:
Terminal velocity, $$v_t={\displaystyle \frac{2r^2\left(\rho -\sigma \right)g}{9\eta }}$$
Here, r is the radius of the drop, $$\rho$$ is the density of the drop, $$\sigma$$ is the density of the medium through which the drop is falling, g is the acceleration due to gravity and $$\eta$$ is the coefficient of viscosity.

Complete step by step answer:
We have given that the terminal velocity of the first drop is $$v=20\text{cm}{\text{s}}^{-1}$$. The radius of the second drop is given $$r_2=1\text{mm}$$.We have the expression for the terminal velocity,
$$v_t={\displaystyle \frac{2r^2\left(\rho -\sigma \right)g}{9\eta }}$$
Here, r is the radius of the drop, $$\rho$$ is the density of the drop, $$\sigma$$ is the density of the medium through which the drop is falling, g is the acceleration due to gravity and $$\eta$$ is the coefficient of viscosity.

Let’ express the terminal velocity for the drop of radius $$r_1=2\text{mm}$$ as follows,
$$v_t={\displaystyle \frac{2r_1^2\left(\rho -\sigma \right)g}{9\eta }}$$
Substituting $$r_1=2\text{mm}$$ in the above equation, we get,
$$v_t={\displaystyle \frac{2{\left(2\right)}^2\left(\rho -\sigma \right)g}{9\eta }}$$
$$\Rightarrow v_t=4\left({\displaystyle \frac{2\left(\rho -\sigma \right)g}{9\eta }}\right)$$ …… (1)
Let’ express the terminal velocity for the drop of radius $$r_1=2\text{mm}$$ as follows,
$$v_t^{\prime }={\displaystyle \frac{2r_1^2\left(\rho -\sigma \right)g}{9\eta }}$$
Substituting $$r_1=2\text{mm}$$ in the above equation, we get,
$$v_t^{\prime }={\displaystyle \frac{2{\left(1\right)}^2\left(\rho -\sigma \right)g}{9\eta }}$$
$$\Rightarrow v_t^{\prime }={\displaystyle \frac{2\left(\rho -\sigma \right)g}{9\eta }}$$ …… (2)

Using equation (1) in the above equation, we get,
$$v_t^{\prime }={\displaystyle \frac{v_t}{4}}$$
Substituting 20 cm/s for $$v_t$$ in the above equation, we get,
$$v_t^{\prime }={\displaystyle \frac{20}{4}}$$
$$\therefore v_t^{\prime }=5\text{cm}{\text{s}}^{-1}$$
Therefore, the terminal velocity of the drop of radius 1 mm will be $$5\text{cm}{\text{s}}^{-1}$$.

So, the correct answer is option D.

Note: The terminal velocity of the liquid drop is the velocity at which the viscous force and buoyant force is balanced by the gravitational force. When the object sinks with terminal velocity, the net force acting on the object becomes zero. In the solution, we have taken the density of the drop and density of the medium the same since the same drop of smaller size descends in the air medium in both situations.