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

Concept:
According to Stokes' law, terminal velocity of a spherical drop is proportional to the square of its radius. \[ v \propto r^2 \]

Step 1: Relate volume of drops. If 8 identical drops combine to form one large drop: \[ \frac{4}{3}\pi R^3 = 8 \left(\frac{4}{3}\pi r^3\right) \] \[ R^3 = 8r^3 \] \[ R = 2r \]

Step 2: Use relation of terminal velocity. Since: \[ v \propto r^2 \] then for the new drop: \[ \frac{v'}{v} = \left(\frac{R}{r}\right)^2 \] Substitute: \[ \frac{v'}{v} = \left(\frac{2r}{r}\right)^2 \] \[ \frac{v'}{v} = 4 \] \[ v' = 4v \]

Step 3: Conclusion. Hence, the terminal velocity of the large drop is: \[ \boxed{4v} \]