Question

A metallic sphere of diameter 2mm and density 10.5 g/cc is dropped in glycerine having viscosity 10 poise and density 1.5 g/cc. The terminal velocity attained by sphere is ______ cm/s. [\(\pi = \frac{22}{7}, g = 10 \text{ m/s}^2\)]

• A. 2.0
• B. 1.0
• C. 1.5
• D. 3.0

Hint:

Unit consistency is absolutely critical in fluid mechanics problems. When densities are in g/cc and viscosity in poise, the CGS system is usually the easiest to work with.
Don't forget to convert acceleration due to gravity (\(g\)) from m/s\(^2\) to cm/s\(^2\) if you are using CGS units.

Answer 1

The Correct Option is A

We need to calculate the terminal velocity of a metallic sphere in glycerine using the known properties and conditions. The terminal velocity can be determined using Stokes' Law for a sphere falling under gravity in a fluid.

The formula for terminal velocity \( v_t \) for a sphere is given by: 

\(v_t = \frac{2}{9} \frac{r^2 ( \rho_s - \rho_f ) g}{\eta}\)

Where:

• \(r\) is the radius of the sphere.
• \(\rho_s\) is the density of the sphere.
• \(\rho_f\) is the density of the fluid.
• \(g\) is the acceleration due to gravity.
• \(\eta\) is the viscosity of the fluid.

Given:

• Diameter of the sphere = 2 mm, so radius \(r = 1 \, \text{mm} = 0.1 \, \text{cm}\)
• Density of the sphere \(\rho_s = 10.5 \, \text{g/cc}\)
• Density of glycerine \(\rho_f = 1.5 \, \text{g/cc}\)
• Viscosity of glycerine \(\eta = 10 \, \text{poise}\)
• Acceleration due to gravity \(g = 10 \, \text{m/s}^2 = 1000 \, \text{cm/s}^2\) (as we need the answer in cm/s)

Substituting these values into the formula for terminal velocity:

\(v_t = \frac{2}{9} \frac{(0.1)^2 (10.5 - 1.5) \times 1000}{10}\)

Simplify the calculation:

\(v_t = \frac{2}{9} \frac{0.01 \times 9 \times 1000}{10} = \frac{2}{9} \times \frac{90}{10} = \frac{2}{9} \times 9 = 2 \, \text{cm/s}\)

Therefore, the terminal velocity attained by the sphere is 2.0 cm/s.