Question

A spherical liquid drop of radius \(R\) acquires the terminal velocity \(v_1\) when falls through a gas of viscosity \(\eta\). Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity \(v_2\) falling through the same gas. The ratio of terminal velocities \(v_1/v_2\) is ________.

• A. 4
• B. 0.25
• C. 32
• D. 16

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Terminal velocity depends on the square of the radius of the sphere. When a large drop breaks into smaller droplets, the total volume remains constant, allowing us to find the relationship between the old and new radii.
Step 2: Key Formula or Approach:
1. Terminal velocity \(v \propto r^2\).
2. Volume conservation: \(\frac{4}{3}\pi R^3 = n \left(\frac{4}{3}\pi r^3\right)\), where \(n\) is the number of droplets.
Step 3: Detailed Explanation:
1. Find new radius \(r\): \(R^3 = 64 r^3 \implies R = \sqrt[3]{64} r \implies R = 4r\). 2. Find ratio of velocities: Since \(v_1 = k R^2\) and \(v_2 = k r^2\): \[ \frac{v_1}{v_2} = \left(\frac{R}{r}\right)^2 \] Substituting \(R/r = 4\): \[ \frac{v_1}{v_2} = (4)^2 = 16 \]
Step 4: Final Answer:
The ratio \(v_1/v_2\) is 16.