Question

When an iron ball is moving through a fluid, its terminal velocity is proportional to

• A. $r^2$
• B. $\frac{1}{r}$
• C. $r^{-2}$
• D. $\sqrt{r}$

Hint:

For spherical objects falling in viscous media, always remember: $v_t \propto r^2$. A larger raindrop falls significantly faster than a smaller one!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a spherical body falls through a viscous fluid, it initially accelerates due to gravity. As its velocity increases, the viscous drag force opposing its motion also increases (according to Stokes' Law). Eventually, a state is reached where the upward forces (buoyant force + viscous drag) perfectly balance the downward gravitational force. At this point, net force is zero, and the body falls with a constant maximum velocity known as the terminal velocity.
Step 2: Key Formula or Approach:
At terminal velocity (\(v_t\)):
\[ \text{Weight} = \text{Buoyant Force} + \text{Viscous Drag} \]
\[ \frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v_t \]
Where:
\(r\) = radius of the spherical ball
\(\rho\) = density of the ball material (iron)
\(\sigma\) = density of the fluid
\(\eta\) = coefficient of viscosity of the fluid
\(g\) = acceleration due to gravity
Step 3: Detailed Explanation:
Rearranging the balance equation to solve for \(v_t\):
\[ 6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho - \sigma) g \]
\[ v_t = \frac{\frac{4}{3}\pi r^3 (\rho - \sigma) g}{6\pi \eta r} \]
\[ v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
Looking at the final expression, for a given fluid (constant \(\sigma\), constant \(\eta\)) and a given material (constant \(\rho\)), all terms are constant except the radius \(r\).
Therefore, the relationship is:
\[ v_t \propto r^2 \]
The terminal velocity is directly proportional to the square of the radius of the ball.
Step 4: Final Answer:
Terminal velocity is proportional to \(r^2\).