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

Concept: According to Stokes' law, the terminal velocity of a small spherical body moving through a viscous medium is directly proportional to the square of its radius. \[ v_t \propto r^2 \] Thus, if the radius changes, terminal velocity changes according to: \[ \frac{v_2}{v_1}=\left(\frac{R}{r}\right)^2 \] where: \[ r = \text{radius of each small drop} \] \[ R = \text{radius of the new large drop} \]

Step 1: Relating the volumes before and after coalescence.
Eight identical drops combine to form one larger drop. Volume is conserved. Volume of one small drop: \[ V=\frac{4}{3}\pi r^3 \] Volume of eight drops: \[ 8\left(\frac{4}{3}\pi r^3\right) \] Let the radius of the new large drop be \(R\). Then, \[ \frac{4}{3}\pi R^3 = 8\left(\frac{4}{3}\pi r^3\right) \] Cancel common terms: \[ R^3=8r^3 \] Taking cube root: \[ R=2r \]

Step 2: Using the terminal velocity relation.
Since: \[ v_t\propto r^2 \] therefore, \[ \frac{v_{\text{new}}}{v} = \left(\frac{R}{r}\right)^2 \] Substitute \(R=2r\): \[ \frac{v_{\text{new}}}{v} = \left(\frac{2r}{r}\right)^2 \] \[ = 2^2 = 4 \] Hence, \[ v_{\text{new}}=4v \] Therefore, the correct answer is: \[ \boxed{4v} \]