Terminal Velocity Variation with Density Ratio

In a terminal velocity experiment using spherical balls falling through a viscous liquid, the net force on the ball determines its motion. Terminal velocity ($v$) is reached when the gravitational force ($F_g$) is balanced by the sum of the buoyant force ($F_b$) and the viscous drag force ($F_d$).

Forces Acting on the Spherical Ball

• Gravitational Force ($F_g$): Acts downwards. $F_g = \frac{4}{3}\pi r^3 \sigma g$, where $r$ is the radius, $\sigma$ is the ball's density, and $g$ is acceleration due to gravity.
• Buoyant Force ($F_b$): Acts upwards. $F_b = \frac{4}{3}\pi r^3 \rho g$, where $\rho$ is the liquid's density.
• Viscous Drag Force ($F_d$): Acts upwards, opposing motion. According to Stokes' Law, $F_d = 6\pi \eta r v$, where $\eta$ is the liquid's viscosity and $v$ is the velocity.

Deriving Terminal Velocity

At terminal velocity, $F_g = F_b + F_d$. Substituting the expressions:

$ \frac{4}{3}\pi r^3 \sigma g = \frac{4}{3}\pi r^3 \rho g + 6\pi \eta r v $

Rearranging to solve for $v$:

$ 6\pi \eta r v = \frac{4}{3}\pi r^3 g (\sigma - \rho) $

$ v = \frac{2r^2 g}{9\eta} (\sigma - \rho) $

Relationship with Density Ratio ($\sigma/\rho$)

To analyze the variation with the ratio $\sigma/\rho$, we can rewrite the equation:

$ v = \left( \frac{2r^2 g}{9\eta} \rho \right) \left( \frac{\sigma}{\rho} - 1 \right) $

Let $ K = \frac{2r^2 g}{9\eta} \rho $. This constant $K$ depends on the liquid's properties and the ball's size. The relationship simplifies to:

$ v = K \left( \frac{\sigma}{\rho} - 1 \right) $

This equation represents a linear relationship between terminal velocity ($v$) and the density ratio ($\sigma/\rho$).

Graphical Interpretation

The equation $v = K(\frac{\sigma}{\rho} - 1)$ implies:

• The graph of $v$ versus $\sigma/\rho$ is a straight line.
• The slope of the line is $K$, which is positive.
• The line intercepts the $\sigma/\rho$ axis at $\frac{\sigma}{\rho} = 1$ (where $v = 0$).
• When $\sigma > \rho$ (i.e., $\sigma/\rho > 1$), $v$ is positive, indicating downward motion (sinking).
• When $\sigma < \rho$ (i.e., $\sigma/\rho < 1$), $v$ is negative, indicating upward motion (rising).

Option D correctly depicts this linear relationship, showing $v = 0$ at $\sigma/\rho = 1$, positive $v$ for $\sigma/\rho > 1$, and negative $v$ for $\sigma/\rho < 1$, all with a constant positive slope.