A light cylindrical vessel is kept on a horizontal surface. Area of base is A. A hole of cross-sectional area 'a' is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is (a < < A) :

Solution

Physics sir · 4 steps

The jet pushes the vessel back with force ρ a v²; friction from the water's weight must match it.

1. Speed of the jet. With water height h and a ≪ A, the top surface barely moves. By Torricelli's result, the jet leaves with v² = 2gh.
2. Thrust on the vessel. Each second, mass ρ a v leaves with speed v. So the vessel feels a backward reaction F = ρ a v² = 2ρ g h a.
3. Friction available. The vessel is light, so the normal reaction is just the weight of the water, N = ρ A h g. Maximum friction is μ N = μ ρ A h g.

   μ ρ A h g ≥ 2ρ g h a ⇒ μ ≥ 2a/A

4. So. The minimum coefficient of friction is 2a/A, option (c). Notice that h cancels, so the answer holds at every water level.

The idea to keep. Thrust from an emerging jet is ρ a v², the momentum leaving per second, not pressure times area. With v² = 2gh it comes out to 2ρ g h a, twice the static pressure force on the hole.

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