A spherical ball of radius 1 × 10⁻⁴ m and density 10⁵ kg / m³ falls freely under gravity through a distance h before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of h is approximately:
(The coefficient of viscosity of water is 9.8 × 10⁻⁶ N s / m²)

Solution

Physics sir · 4 steps

The ball must hit the water already at terminal velocity, so set √2gh equal to it.

1. Why no change. Inside the water the speed stays the same only if the viscous drag balances the weight from the start. That happens only when the ball enters at its terminal velocity vₜ. Free fall through h gives it a speed v = √2gh, so v must equal vₜ.

   √2gh = vₜ = (2r²ρ g)/9η

2. Terminal velocity. The ball is 100 times denser than water, so the options are built without the buoyant force. Then vₜ = (2 × 10⁻⁸ × 10⁵ × 9.8)/(9 × 9.8 × 10⁻⁶) = 2000/9 ≈ 222.2 m s⁻¹. The 9.8 in g cancels the 9.8 in η.
3. Find h. h = vₜ²/2g = 49383/19.6 ≈ 2519 m. Keeping buoyancy would use ρ - σ = 99000 kg m⁻³ and give about 2469 m, which is not an option. So the question means for you to ignore it.
4. So. h ≈ 2518 m, which is option (a).

The idea to keep. Velocity does not change in a fluid only at terminal velocity. Match the entry speed √2gh to vₜ = 2r²(ρ-σ)g/9η and solve for h.

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