Question

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

• A. $\frac{79}{72}$
• B. $\frac{19}{36}$
• C. $\frac{39}{72}$
• D. $\frac{79}{36}$

Answer 1

The Correct Option is D

To find the ratio of the terminal velocities of the two spherical balls, we need to analyze the forces acting on each ball as they move through the viscous medium. The terminal velocity, \( v_t \), is attained when the forces of gravity and buoyancy become equal to the viscous drag force. According to Stoke's Law, the expression for the terminal velocity \( v_t \) of a small spherical object moving through a viscous fluid is given by:

v_t = \frac{2}{9} \cdot \frac{(r^2) \cdot g \cdot (\rho - \rho_m)}{\eta}

where:

• r is the radius of the sphere.
• g is the acceleration due to gravity.
• \rho is the density of the material of the sphere.
• \rho_m is the density of the medium.
• \eta is the coefficient of viscosity of the medium.

Given that the two spheres have radii \( r_1 = 1 \, \text{mm} = 0.001 \, \text{m} \) and \( r_2 = 2 \, \text{mm} = 0.002 \, \text{m} \), and densities \( \rho_1 \) and \( \rho_2 \) with the condition \( \rho_1 = 8 \rho_2 \). The density of the medium is \( 0.1 \rho_2 \).

For the first sphere:

v_{t1} = \frac{2}{9} \cdot \frac{(r_1^2) \cdot g \cdot (\rho_1 - \rho_m)}{\eta} = \frac{2}{9} \cdot \frac{(0.001^2) \cdot g \cdot (\rho_1 - 0.1 \rho_2)}{\eta}

For the second sphere:

v_{t2} = \frac{2}{9} \cdot \frac{(r_2^2) \cdot g \cdot (\rho_2 - \rho_m)}{\eta} = \frac{2}{9} \cdot \frac{(0.002^2) \cdot g \cdot (\rho_2 - 0.1 \rho_2)}{\eta}

Simplifying the expressions for each terminal velocity, we use the fact that:

\rho_1 = 8 \rho_2

Now, substituting the values:

For sphere 1:

v_{t1} = \frac{2}{9} \cdot \frac{0.000001 \cdot g \cdot (8 \rho_2 - 0.1 \rho_2)}{\eta} = \frac{2}{9} \cdot \frac{0.000001 \cdot g \cdot 7.9 \rho_2}{\eta}

For sphere 2:

v_{t2} = \frac{2}{9} \cdot \frac{0.000004 \cdot g \cdot 0.9 \rho_2}{\eta}

The ratio of their terminal velocities is:

\frac{v_{t1}}{v_{t2}} = \frac{0.000001 \cdot 7.9 \rho_2}{0.000004 \cdot 0.9 \rho_2} = \frac{7.9}{4 \times 0.9} = \frac{7.9}{3.6} = \frac{79}{36}

Thus, the ratio of their terminal velocities is \frac{79}{36}.