# A large open top container of negligible mass and uniform cross-sectional area A has small holes of cross-sectional area A / 100 in its side wall near the bottom. The container is kept on a  smooth horizontal floor and contains a liquid of density. Ρ and mass m0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, Calculate                         (i)  the acceleration of the container, and(ii) its velocity when 75% of the liquid has drained out.

Kevin Nash
8 years ago
Hello Student,
(i) Let at any instant of time during the flow, the height of liquid in the container is x.
The velocity of flow of liquid through small hole in the orifice by Toricelli’s theorem is
v = √2gx …(i)
The mass of liquid flowing per second through the orifice
= p x volume of liquid flowing per second
Dm / dt = ρ x √2 gx x A / 100 …(ii)
Therefore, the rate of change of momentum of the system in forward direction
= dm / dt x v = 2gx x A x ρ / 100 (from (i) and (ii))
(Alternatively you may use F = ρav2)
The rate of change of momentum of the system in the backward direction
= Force on backward direction = m x a
Where m is mass of liquid in the container at the instant t m = volume x density = A x x x ρ
∴ The rate of change of momentum of the system in the backward direction
Axρ x α
By conservation of linear momentum
A x ρ x α = 2gxAρ / 100 ⇒ a = g / 50
(ii) By Toricelli’s theorem
v' = √2g x (0.25 h)
Where h is the initial height of the liquid in the container. m0. the initial mass is
m0 = Ah x ρ ⇒ h = m0 / Aρ
∴ v’ = √2g x 0.25 x m0 / Aρ = √gm0 / 2Aρ
Thanks
Kevin Nash