Question

Eight identical spherical liquid drops, each falling with a terminal velocity $v$, coalesce to form a single large drop. The terminal velocity of the new large drop is:

• A. $2v$
• B. $4v$
• C. $8v$
• D. $16v$

Hint:

When identical drops combine: \[ R = n^{1/3} r \] and since terminal velocity follows: \[ v \propto r^2 \] therefore: \[ v' = n^{2/3} v \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem investigates the physics of fluid dynamics, specifically focusing on terminal velocity and the conservation of mass/volume during the coalescence of liquid droplets.
Terminal velocity is reached when an object falling through a viscous fluid experiences zero net force.
This occurs because the downward gravitational force is perfectly balanced by the upward buoyancy and the viscous drag force (Stokes' Law).
When multiple small drops merge into one large drop, the total volume remains constant, but the surface area changes.
The change in radius directly affects the terminal velocity because the gravitational force (proportional to mass/volume, hence \(r^3\)) and the drag force (proportional to velocity and radius, \(rv\)) scale differently.
Key Formula or Approach:
According to Stokes' Law, the terminal velocity \(v\) of a spherical object of radius \(r\) is given by:
\[ v = \frac{2r^2 (\rho - \sigma) g}{9\eta} \] Where:
- \(\rho\) is the density of the drop.
- \(\sigma\) is the density of the medium.
- \(g\) is the acceleration due to gravity.
- \(\eta\) is the coefficient of viscosity.
From this expression, we see that for the same liquid and medium, the terminal velocity is directly proportional to the square of the radius:
\[ v \propto r^2 \] Step 2: Detailed Explanation:
Let the radius of each of the eight identical small drops be \(r\).
Let the radius of the resulting large drop be \(R\).

1. Volume Conservation:
The volume of 8 small drops equals the volume of the single large drop.
\[ 8 \times \left( \frac{4}{3}\pi r^3 \right) = \frac{4}{3}\pi R^3 \]
By canceling out common terms \(\frac{4}{3}\pi\), we get:
\[ 8r^3 = R^3 \]
Taking the cube root on both sides:
\[ R = \sqrt[3]{8r^3} = 2r \]
This tells us the radius of the new drop is exactly double the radius of the individual small drops.

2. Relating Terminal Velocities:
Let \(v\) be the terminal velocity of the small drop and \(V'\) be the terminal velocity of the large drop.
Using the proportionality \(v \propto r^2\):
\[ \frac{V'}{v} = \left( \frac{R}{r} \right)^2 \]
Substitute \(R = 2r\) into the ratio:
\[ \frac{V'}{v} = \left( \frac{2r}{r} \right)^2 \]
\[ \frac{V'}{v} = (2)^2 = 4 \]
\[ V' = 4v \]
The increase in terminal velocity happens because the mass of the drop increases by a factor of 8 (increasing the downward pull), while the drag-related radius only increases by a factor of 2.
This results in a net four-fold increase in the speed required to reach force equilibrium.
Step 3: Final Answer:
The terminal velocity of the new large drop is \(4v\).