Question

Consider a viscous fluid having uniform density \( \rho \) and coefficient of viscosity \( \eta \). Two solid spheres with radius \( R_1 \) and \( R_2 \) fall into this liquid and acquire terminal velocities of \( V \) and \( 8V \) respectively. If the densities of the two spheres are \( 2\rho \) and \( 3\rho \) respectively, then \( R_1 \) is equal to \( R_2 \)?

• A. 2
• B. \( \frac{1}{2} \)
• C. 8
• D. \( \frac{1}{8} \)

Hint:

When comparing the terminal velocities and radii of spheres in a viscous fluid, remember to use Stokes' law and carefully analyze the ratios of densities and velocities to find the relationship between the radii.

Answer 1

The Correct Option is D

Given that the terminal velocity \( V \) of a sphere falling in a viscous fluid is given by Stokes' law: \[ V = \frac{2}{9} \frac{R^2 (\rho_s - \rho_f) g}{\eta} \] Where: - \( V \) is the terminal velocity, - \( R \) is the radius of the sphere, - \( \rho_s \) and \( \rho_f \) are the densities of the sphere and the fluid respectively, - \( g \) is the acceleration due to gravity, - \( \eta \) is the coefficient of viscosity of the fluid. ### Step 1: Set up the equation for both spheres. For sphere 1: \[ V = \frac{2}{9} \frac{R_1^2 (2\rho - \rho)}{\eta} \] For sphere 2, with terminal velocity \( 8V \): \[ 8V = \frac{2}{9} \frac{R_2^2 (3\rho - \rho)}{\eta} \] ### Step 2: Relate the two equations. Dividing the second equation by the first equation: \[ \frac{8V}{V} = \frac{R_2^2 (3\rho - \rho)}{R_1^2 (2\rho - \rho)} \] This simplifies to: \[ 8 = \frac{R_2^2 (2\rho)}{R_1^2 ( \rho)} \] \[ 8 = \frac{R_2^2}{R_1^2} \cdot 2 \] Thus, \[ \frac{R_2^2}{R_1^2} = 4 \] Taking the square root of both sides: \[ \frac{R_2}{R_1} = 2 \] Therefore, \[ R_1 = \frac{R_2}{8} \] Thus, \( R_1 \) is \( \frac{1}{8} \) of \( R_2 \).