The velocity of a small ball of mass 0.3 g and density 8 g / cc when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is 1.3 g / cc, then the value of viscous force acting on the ball will be x × 10⁻⁴ N, The value of x is _________. [use g=10 m / s²]

Solution

Physics sir · 4 steps

Constant velocity means zero net force, so viscous drag is weight minus buoyancy.

1. Terminal velocity. Once the velocity is constant, the acceleration is zero. Weight acts down, and buoyancy and viscous force act up, so they must balance.

   Fᵥ = mg - Vσ g = mg(1 - σ/ρ)

2. Weight and buoyancy. Here m = 0.3 g = 3 × 10⁻⁴ kg, so mg = 3 × 10⁻³ N. Buoyancy is mg × 1.3/8 = 3 × 10⁻³ × 0.1625 = 4.875 × 10⁻⁴ N.
3. Subtract. Fᵥ = 30 × 10⁻⁴ - 4.875 × 10⁻⁴ = 25.125 × 10⁻⁴ N.
4. So. The viscous force is about 25 × 10⁻⁴ N, so x = 25.

The common slip: 30. Leaving out the buoyant force and setting viscous force equal to mg gives 30, not 25.

The idea to keep. At terminal velocity, viscous force equals weight minus buoyancy, which is mg(1 - σ/ρ). You do not need the radius or viscosity at all.

Answer from the NTA final answer key. Solution by PrepWiser.