But

Application of Bernoulli’s theorem

Weight of block
= Weight of displaced oil + Weight of displaced water
Þ
Þ = 760 gm

When cross-section of duct is decreased, the velocity of water increased and in accordance with Bernoulli’s theorem, the pressure P decreased at that place

When we move from centre to circumference, the velocity of liquid goes on decreasing and finally becomes zero

If the level in narrow tube goes down by h₁ then in wider tube goes up to h₂,
Now, Þ
Now, pressure at point A = pressure at point B

Þ h = Þ

Net force (reaction) =

\ …(i)
According to Bernoulli's theorem

Þ Þ
From equation (i),

Let A = The area of cross section of the hole
v = Initial velocity of efflux
d = Density of water,
Initial volume of water flowing out per second = Av
Initial mass of water flowing out per second = Avd
Rate of change of momentum = Adv²
Initial downward force on the flowing out water = Adv²
So equal amount of reaction acts upwards on the cylinder.
\ Initial upward reaction = [As ]
Initial decrease in weight
gm-wt.

Let A = cross-section of tank
a = cross-section hole
V = velocity with which level decreases
v = velocity of efflux

From equation of continuity
By using Bernoulli's theorem for energy per unit volume
Energy per unit volume at point A
= Energy per unit volume at point B

Þ