Question

A spherical ball of diameter 1 cm and density \( 5 \times 10^3 \, \text{kg m}^{-3} \) is dropped gently in a large tank containing viscous liquid of density \( 3 \times 10^3 \, \text{kg m}^{-3} \) and coefficient of viscosity \( 0.1 \, \text{Ns m}^{-2} \). The distance the ball moves in 1 s after attaining terminal velocity is (\( g = 10 \, \text{m s}^{-2} \))

• A. $\frac{10}{9}$ m
• B. $\frac{2}{3}$ m
• C. $\frac{4}{9}$ m
• D. $\frac{4}{5}$ m
• E. $\frac{9}{10}$ m

Hint:

Always use Stokes’ law for small spheres moving in viscous fluids.

Answer 1

The Correct Option is A

Concept: At terminal velocity, net force becomes zero. The viscous drag balances effective weight: \[ 6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho_s - \rho_l) g \]

Step 1: Given: \[ r = 0.5 \text{ cm} = 0.005 \text{ m} \] \[ \rho_s = 5 \times 10^3, \quad \rho_l = 3 \times 10^3 \] \[ \eta = 0.1, \quad g = 10 \]

Step 2: Terminal velocity formula: \[ v_t = \frac{2 r^2 (\rho_s - \rho_l) g}{9 \eta} \]

Step 3: Substitute values: \[ v_t = \frac{2 \times (0.005)^2 \times (2000) \times 10}{9 \times 0.1} \] \[ = \frac{2 \times 25 \times 10^{-6} \times 2000 \times 10}{0.9} \] \[ = \frac{1}{0.9} = \frac{10}{9} \]

Step 4: Distance in 1 sec: \[ s = v_t \times t = \frac{10}{9} \times 1 = \frac{10}{9} \] Final Answer: \[ s = \frac{10}{9} \text{ m} \]