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:

For spherical drops combining together: \[ n \propto R^3 \] Thus, if \(n\) identical drops merge, \[ R=n^{1/3}r \] Since terminal velocity varies as \(r^2\), \[ v_t \propto n^{2/3} \] For \(n=8\), \[ 8^{2/3}=4 \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The concept involved here is the terminal velocity of a spherical body falling through a viscous fluid, which is governed by Stokes' Law.
Terminal velocity (\(v\)) is reached when the net force on the drop becomes zero.
This occurs when the downward force of gravity is perfectly balanced by the upward buoyant force and the viscous drag force.
When several small drops coalesce to form a larger drop, the total volume remains constant, but the radius of the resulting drop increases.
Since terminal velocity is a function of the radius of the falling object, we must first determine how the radius changes and then apply the terminal velocity relationship.
Step 2: Key Formula or Approach:
The formula for the terminal velocity of a sphere of radius \(r\) and density \(\rho\) falling through a fluid of density \(\sigma\) and viscosity \(\eta\) is given by:
\[ v = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
From this expression, it is evident that for a given liquid and medium at a constant temperature, all terms except \(r^2\) are constant.
Therefore, we can establish the proportionality:
\[ v \propto r^2 \]
To find the new velocity, we need to find the ratio of the new radius (\(R\)) to the old radius (\(r\)) using the principle of conservation of volume.
Step 3: Detailed Explanation:
Let the radius of each of the eight small drops be \(r\).
The volume of one small spherical drop is given by \(V_s = \frac{4}{3}\pi r^3\).
When eight such drops coalesce, the total volume of the resulting large drop (\(V_L\)) is:
\[ V_L = 8 \times V_s \]
Let the radius of the large drop be \(R\). Its volume is \(V_L = \frac{4}{3}\pi R^3\).
Equating the two expressions for the volume of the large drop:
\[ \frac{4}{3}\pi R^3 = 8 \times \left( \frac{4}{3}\pi r^3 \right) \]
Canceling the common factor \(\frac{4}{3}\pi\) from both sides:
\[ R^3 = 8r^3 \]
Taking the cube root of both sides:
\[ R = \sqrt[3]{8r^3} = 2r \]
The radius of the large drop is exactly twice the radius of a single small drop.
Now, let the terminal velocity of a small drop be \(v\) and that of the large drop be \(V'\).
Using the proportionality \(v \propto r^2\):
\[ \frac{V'}{v} = \left( \frac{R}{r} \right)^2 \]
Substituting the value \(R = 2r\):
\[ \frac{V'}{v} = \left( \frac{2r}{r} \right)^2 = 2^2 = 4 \]
Multiplying both sides by \(v\):
\[ V' = 4v \]
This means the terminal velocity increases by a factor of four because the surface area (which determines drag) increases less rapidly than the mass (which determines gravity) as the radius grows.
Step 4: Final Answer:
The terminal velocity of the new large drop is \(4v\).
Hence, the correct option is (B).