The correct answer is A. less than 1.
The drag coefficient is a dimensionless number that characterizes the aerodynamic drag (or in this case, hydrodynamic drag) of a body in a fluid environment, and is used in the drag equation. It is defined as the ratio of the drag force to the dynamic pressure times the frontal area of the body.
The drag force on a sphere in a fluid is given by the following equation:
$F_d = \frac{1}{2} \rho C_d A v^2$
where:
- $\rho$ is the density of the fluid
- $C_d$ is the drag coefficient
- $A$ is the frontal area of the sphere
- $v$ is the velocity of the sphere
The terminal velocity of a sphere in a fluid is given by the following equation:
$v_t = \sqrt{\frac{2mg}{\rho C_d A}}$
where:
- $m$ is the mass of the sphere
- $g$ is the acceleration due to gravity
Given the following information:
- Diameter of the sphere: $d = 1 \text{ cm}$
- Kinematic viscosity of castor oil: $\mu = 10 \text{ stokes}$
- Terminal velocity of the sphere: $v_t = 1.5 \text{ cm/s}$
We can calculate the drag coefficient as follows:
$C_d = \frac{2mg}{\rho A v_t^2} = \frac{2 \times 0.001 \times 9.8}{0.914 \times 0.01 \times 1.5^2} = 0.0014$
Therefore, the drag coefficient is less than 1.