A small spherical ball of radius 0.1 mm and density 10⁴ kg m⁻³ falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of h will be ___________ m.
(Given g = 10 ms⁻², viscosity of water = 1.0 × 10⁻⁵ N-sm⁻²).

Solution

Physics sir · 4 steps

The speed it gains in free fall must already equal its terminal speed in water. Find that, then the height.

1. What constant speed means. Inside the water the ball's speed does not change, so the forces on it are balanced from the very first moment. That means it enters the water already moving at its terminal velocity.

   vₜ = 2r²(ρ - σ)g/9η

2. Put in numbers. Here r = 10⁻⁴ m, so r² = 10⁻⁸ m². The density difference is ρ - σ = 10⁴ - 10³ = 9 × 10³ kg m⁻³. Then vₜ = (2 × 10⁻⁸ × 9 × 10³ × 10)/(9 × 10⁻⁵) = 20 m s⁻¹.
3. Free fall height. Falling from rest through h, the ball reaches v = √2gh. Set this equal to vₜ: h = vₜ²/2g = 20²/(2 × 10) = 400/20.
4. So. h = 20 m.

The common slip: 24.7. Leaving out buoyancy (using ρ instead of ρ - σ) gives vₜ ≈ 22.2 m s⁻¹ and h ≈ 24.7 m, not 20.

The idea to keep. If a body moves at constant speed in a fluid right after entering it, it entered at terminal velocity. Match √2gh to vₜ, and always subtract the fluid's density for buoyancy.

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