Question

A small solid sphere of mass 10 g and density \( 2600 \, \text{kg/m}^3 \) is dropped into a long vertical column of glycerine. When the sphere attains terminal velocity, the magnitude of the viscous force acting on the sphere is
(Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \), Density of glycerine \( = 1300 \, \text{kg/m}^3 \))

• A. \( 20 \times 10^{-3} \, \text{N} \)
• B. \( 25 \times 10^{-3} \, \text{N} \)
• C. \( 75 \times 10^{-3} \, \text{N} \)
• D. \( 50 \times 10^{-3} \, \text{N} \)

Hint:

At terminal velocity, net force is zero. Use the balance between weight and buoyant force to calculate viscous force.

Answer 1

The Correct Option is D

At terminal velocity, the viscous force equals the apparent weight: \[ F_{\text{viscous}} = \text{Weight} - \text{Buoyant force} = mg - V \cdot \rho_{\text{gly}} \cdot g \] Mass \( m = 10 \, \text{g} = 0.01 \, \text{kg} \)
Density of sphere \( \rho_s = 2600 \, \text{kg/m}^3 \)
So volume \( V = \frac{m}{\rho_s} = \frac{0.01}{2600} \) \[ F = 0.01 \cdot 10 - \left( \frac{0.01}{2600} \cdot 1300 \cdot 10 \right) = 0.1 - 0.05 = 0.05 \, \text{N} = 50 \times 10^{-3} \, \text{N} \]