Question

A rain drop of diameter 1 mm falls with a terminal velocity of 0.7 ms$^{-1}$ in air. If the coefficient of viscosity of air is $2\times10^{-5}$ Pas, the viscous force on the rain drop is

• A. $13.2\times10^{-8}$ N
• B. $6.6\times10^{-8}$ N
• C. $26.4\times10^{-8}$ N
• D. $10.4\times10^{-8}$ N

Hint:

Stokes' Law, $F_v = 6\pi\eta r v$, describes the drag force on a sphere moving slowly through a viscous fluid. Remember that at terminal velocity, this viscous drag force exactly balances the net downward force (gravity minus buoyancy).

Answer 1

The Correct Option is A

When a spherical object moves through a viscous fluid at terminal velocity, the net force on it is zero. This means the downward force of gravity (minus buoyancy) is balanced by the upward viscous drag force.
The viscous force on a spherical object is given by Stokes' Law:
$F_v = 6\pi\eta r v_t$.
Where:
$\eta$ is the coefficient of viscosity.
$r$ is the radius of the sphere.
$v_t$ is the terminal velocity.
We are given the following values:
Diameter = 1 mm, so radius $r = 0.5$ mm = $0.5 \times 10^{-3}$ m.
Terminal velocity $v_t = 0.7$ m/s.
Coefficient of viscosity $\eta = 2 \times 10^{-5}$ Pa$\cdot$s.
Now, substitute these values into Stokes' Law:
$F_v = 6 \pi (2 \times 10^{-5}) (0.5 \times 10^{-3}) (0.7)$.
$F_v = 6 \pi (2 \times 0.5 \times 0.7) \times 10^{-5} \times 10^{-3}$.
$F_v = 6 \pi (0.7) \times 10^{-8}$.
$F_v = 4.2 \pi \times 10^{-8}$.
Using the approximation $\pi \approx 3.14$:
$F_v = 4.2 \times 3.14 \times 10^{-8} = 13.188 \times 10^{-8}$ N.
This value is approximately $13.2 \times 10^{-8}$ N.