Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_1$ and $\rho_

Question

Two small spherical metal balls, having equal masses, are made from materials of densities $ho_1$ and $ho_2$ such that $ho_1 = 8ho_2$ and having radii of $1 	ext{ mm}$ and $2 	ext{ mm}$, respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1ho_2$. The ratio of their terminal velocities would be:

Accepted Answer

1. **Terminal Velocity Formula:** The terminal velocity ($v_t$) of a spherical body falling through a viscous fluid is given by Stokes' Law:
   $$ v_t = \frac{2r^2 (ho - \sigma)g}{9\eta} $$
   where $r$ is the radius of the sphere, $ho$ is the density of the sphere, $\sigma$ is the density of the fluid, and $\eta$ is the coefficient of viscosity .
2. **Parameters for Ball 1:**
   - Radius $r_1 = 1 	ext{ mm}$
   - Density $ho_1 = 8ho_2$
   - Fluid density $\sigma = 0.1ho_2$
   $$ v_1 \propto r_1^2 (ho_1 - \sigma) = (1)^2 (8ho_2 - 0.1ho_2) = 7.9ho_2 $$
3. **Parameters for Ball 2:**
   - Radius $r_2 = 2 	ext{ mm}$
   - Density $ho_2 = ho_2$
   - Fluid density $\sigma = 0.1ho_2$
   $$ v_2 \propto r_2^2 (ho_2 - \sigma) = (2)^2 (ho_2 - 0.1ho_2) = 4(0.9ho_2) = 3.6ho_2 $$
4. **Calculate Ratio:**
   $$ \frac{v_1}{v_2} = \frac{7.9ho_2}{3.6ho_2} = \frac{79}{36} $$

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