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

Step 1: Understanding the Question:
We need to calculate the terminal velocity of a metallic sphere falling through a viscous liquid (glycerine). Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is moving equals the force of gravity.
Step 2: Key Formula or Approach:
The formula for terminal velocity (\(v_t\)) is given by Stokes' law, which comes from balancing the gravitational force, buoyant force, and viscous drag force:
\[ v_t = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta} \] where:
- \(r\) = radius of the sphere
- \(g\) = acceleration due to gravity
- \(\rho\) = density of the sphere
- \(\sigma\) = density of the fluid
- \(\eta\) = viscosity of the fluid
Step 3: Detailed Explanation:
First, we must ensure all units are consistent. The options are in cm/s, so using the CGS (Centimeter-Gram-Second) system is convenient.
- Diameter = 2 mm, so radius \(r = 1 \text{ mm} = 0.1 \text{ cm}\).
- Density of sphere \(\rho = 10.5 \text{ g/cm}^3\).
- Density of glycerine \(\sigma = 1.5 \text{ g/cm}^3\).
- Viscosity of glycerine \(\eta = 10 \text{ poise}\). (Note: 1 poise = 1 g·cm\(^{-1}\)·s\(^{-1}\), which is the CGS unit).
- Acceleration due to gravity \(g = 10 \text{ m/s}^2 = 10 \times 100 \text{ cm/s}^2 = 1000 \text{ cm/s}^2\).
Now substitute these values into the terminal velocity formula:
\[ v_t = \frac{2}{9} \frac{(0.1)^2 \times 1000 \times (10.5 - 1.5)}{10} \] \[ v_t = \frac{2}{9} \frac{0.01 \times 1000 \times (9)}{10} \] \[ v_t = \frac{2}{9} \frac{10 \times 9}{10} \] \[ v_t = \frac{2}{9} \times 9 \] \[ v_t = 2 \text{ cm/s} \] Step 4: Final Answer:
The terminal velocity attained by the sphere is 2.0 cm/s.