Question

A spherical rain drop of mass \(m\) falls vertically through air with a terminal velocity of \(0.2\ \mathrm{ms^{-1}}\). If \(27\) such identical rain drops combine to form a bigger spherical drop, then the terminal velocity of the bigger drop if it falls vertically through air is

• A. \(0.9\ \mathrm{ms^{-1}}\)
• B. \(3.6\ \mathrm{ms^{-1}}\)
• C. \(5.4\ \mathrm{ms^{-1}}\)
• D. \(1.8\ \mathrm{ms^{-1}}\)

Hint:

According to Stokes' law, \[ \boxed{v_t\propto r^2.} \] If \(n\) identical drops combine, \[ \boxed{R=n^{1/3}r.} \] Hence, \[ \boxed{v_t\propto n^{2/3}.} \]

Answer 1

The Correct Option is D

Step 1: Use the relation for terminal velocity. For a spherical drop moving through a viscous medium, \[ v_t\propto r^2. \]

Step 2: Find the radius of the new drop. When \[ 27 \] identical drops combine, \[ R^3 = 27r^3. \] Hence, \[ R = 3r. \] Therefore, \[ \frac{v_2}{v_1} = \left(\frac{R}{r}\right)^2 = 3^2 = 9. \]

Step 3: Calculate the terminal velocity. Given, \[ v_1 = 0.2\ \mathrm{ms^{-1}}. \] Thus, \[ v_2 = 9\times0.2 = 1.8\ \mathrm{ms^{-1}}. \] Hence, \[ \boxed{1.8\ \mathrm{ms^{-1}}}. \] Thus, \[ \boxed{(D)} \] is the correct answer.