Consider an example shown in figure 23.1 As the water comes out from the vessel with some speed ( generally ( 2gh )^{1/2} ) , it has some momentum, which was initially almost zero , when it is in the container. This change in momentum is due to a force on water ejecting in forward direction and the reaction of this force must be experienced by the container and the liquid inside .

If the hole has a cross section ‘ a ‘ then according to **Torricelli’s theorem** and liquid is coming with a speed ( 2gh )^{1/2} .

Velocity of efflux at that instant

v = ( 2gh )^{1/2}

Change is mass = vaρ

(Change in mass per unit time)

dm/dt = vaρ

we know that ,

F = dp/dt

F = mdv/dt + vdm/dt

Velocity for that particular small instant is constant ,

mdv/dt = 0

So , F = vdm/dt

F = v( vaρ )

F = ρav²

**F = ρav² = 2ρagh**

**Note :** If the velocity of a liquid (or gas) of density p coming out through an opening of area of cross section a is v, then there will be a thrust force F = ρav² due to liquid coming out the opening. The direction of the thrust force will be just opposite to the velocity direction .

**Solved Examples**

**Problem 1 .**In a wide vertical vessel filled with water, two identical holes are opened on the opposite sides . Area of cross – section of each hole is ‘ a ‘ . Find the resultant force of reaction of water flowing out of vessel if the height difference between them is equal to h .

**Solution.** Let A and B be two openings, each of area ‘a’ the velocity of water coming out of opening are V_{A} and V_{B} , respectively .

Volume of water emerging per second through A is aV_{A}

Volume of water emerging per second through B is aV_{B}

Force of reaction due to water coming out at A is given by

FA = Rate of change of momentum per second at A

= (aV_{A}ρ)V_{A} = aρV_{A}²

Similarly, force of reaction at B is given by FB = aρV_{B}²

Therefore , net force on vessel is

**F = F _{B} – F_{A} = aρ( V_{B}² – V_{A}² ) ….( i )**

We can find value of ( V_{B}² – V_{A}² ) by direct writing the value of V_{B} and V_{A} with help of velocity of efflux ,

or we can solve by Bernoulli’s equation.

Now , Applying Bernoulli’s theorems at A ans B , we get

**P _{a} + ρV_{A}²/2 + ( h_{2 }+ h )ρg = P_{a} + ρV_{B}²/2+ ρgh_{2} ….( ii )**

Pressure at both the efflux will be equal to atmospheric pressure because both efflux is open in atmosphere .

Here Pa is atmospheric pressure in equation ( ii )

V_{B}² – V_{A}² = 2gh

Substituting this value of ( VB² – VA² ) in equation ( i ) , we get

F = aρ2gh

or **F = 2ρgah**

**Problem 2**.The side wall of a wide vertical cylindrical vessel of height h = 75 cm has a narrow vertical slit running all the way down to the bottom of the vessel. The length of the slit is = 50 cm and the width b = 1.0 mm. With the slit closed, the vessel is filled with water. Find the resultant force of reaction of water flowing out the vessel immediately after the slit is opened.

**Solution **– Consider an element of length dx of the slit as shown in Figure 23.3

Area of the slit= bdx

Discharge per sec through this area =ρ(bdx)v

Force of reaction due to element dx**dF= – ρ(bdx)v² ….. ( i )**

Negative sign is used because this is opposite to direction of v

Applying Bernoulli’s theorem at point A , we have

P_{a} = P_{a} + ρg(h – x) + ρv²/2

**v² = – 2ρg(h – x) …. ( ii )**

Substituting the value of v² from equation-( ii ) in equation-( i ) we get

dF =ρ(bdx) 2ρg (h – x)

F =2ρgb ∫(h – x) dx

[ limit of integration is from x = 0 to x = l ]

F=2ρgb [ hl – ( l²/2 )]

F = ρgbl ( 2h – l )

Substituting the given values, we get

F = (1000)(9.8)(1×10³)(0.5)[2×0.75 -0.5]

**F =5N**

**Problem 3**– A large open top container of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional area A/100 in its side wall near the bottom . The container contains a liquid of density ρ and mass Mo and kept on a smooth horizontal floor. Assuming that the liquid starts flowing

out horizontally through the hole at t=0 , calculate

(a) the acceleration of the container and

(b) its velocity when 75% of the liquid has drained out .

**Solution-** Liquid is escaping from the container. Let y be the height of the liquid at any instant t after start .

**(a)** Then velocity of escape is given by v= (2gy)^{1/2}

Mass of the liquid flowing in time dt= avρdt , where a = area of the hole = A/100.

Force is the rate of change of momentum. Hence ,

F= dp/dt = (avρdt/dt)v = av²ρ

F = 2agyρ

As F = mass of container × acceleration = Ma’

a’ = F/M

But M=mass of container = Ayρ

a’ = 2agyρ/M = [(A/100)2gyρ]/Ayρ = **g/50**

**(b)** When 75% of the liquid has drained out, the height of the liquid y is given by

Ayp =25M_{o}/50

y = M_{o}/4Aρ

But v = ( 2gy )^{1/2}

v = ( 2gM_{o}/4Aρ)^{1/2}

**v = ( M _{o}g/2Ay )^{1/2}**