Question

A steel ball of radius 0.05 cm and density \( 7.8 \, \text{g/cm}^3 \) is dropped into water. What is its terminal velocity?
(Density of water \( = 1 \, \text{g/cm}^3 \), Viscosity of water \( = 0.001 \, \text{Pas} \))

• A. \( 3.42 \, \text{ms}^{-1} \)
• B. \( 1.81 \, \text{ms}^{-1} \)
• C. \( 5.11 \, \text{ms}^{-1} \)
• D. \( 3.77 \, \text{ms}^{-1} \)

Hint:

In fluid mechanics, use Stokes' law and convert all units to SI before calculation.

Answer 1

The Correct Option is D

Use Stokes’ law: \( v_t = \frac{2r^2 g (\rho - \sigma)}{9\eta} \)
Convert radius: \( r = 0.05 \, \text{cm} = 0.0005 \, \text{m} \), \( \rho = 7800 \), \( \sigma = 1000 \), \( \eta = 0.001 \)
\[ v_t = \frac{2 \cdot (0.0005)^2 \cdot 9.8 \cdot (7800 - 1000)}{9 \cdot 0.001} = \frac{2 \cdot 2.5 \cdot 10^{-7} \cdot 9.8 \cdot 6800}{0.009} \approx 3.77 \, \text{ms}^{-1} \]