**Force of
Reaction due to Ejection of Liquid**

A cylindrical vessel has an opening of cross-sectional area “a” near the bottom. A disc is held against the opening to prevent liquid of density ρ from coming out. If the height of the liquid above opening is h. Let us analyse the force on the disc in this situation.

The disc experiences a hydrostatic pressure from the liquid inside the vessel. Pressure at the level of the disc is:

P₁ = P_{atm} + ρgh

The air pressure on the outside of
the disc is P₂ = P_{atm}.

The net outward force = (P₂ – P₁) a = ρgha.

Now the disc is moved a short distance away in horizontal direction. The liquid comes out, strikes the disc inelastically and drops vertically downward. The water in this case will impart impulsive force on the disc.

When the disc is moved away, the liquid moves out with speed (v) = √2gh.

The mass per second, i.e. the rate of mass coming out of the opening is given by:

\(\frac{dm}{dt}=\rho av=\rho a\sqrt{2gh}\).

Momentum per second imparted by water rightward is given by:

\(\frac{dm}{dt}v=\left( \rho av \right)v=\rho a{{v}^{2}}=2\rho gha\).

The change in momentum per second after striking the disc is:

\(\Delta \overrightarrow{P}=\overrightarrow{{{P}_{f}}}-\overrightarrow{{{P}_{i}}}\).

\(\Delta \overrightarrow{P}=0-\left( \rho a{{v}^{2}} \right)=-\rho a{{v}^{2}}=-2\rho gha\).

Taking the rightward direction is positive, the force on the liquid is towards the left and its reaction with disc is towards the right. The force obtained is twice the hydrostatic force. The force due to atmosphere pressure is cancelled out on the both sides.