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

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

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

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

Application of Bernoulli’s theorem

But

Upthrust
where, a = downward acceleration,
V = volume of liquid displaced
But for free fall a = g Upthrust = 0

As the block moves up with the fall of coin, l decreases, similarly h will also decrease because when the coin is in water, it displaces water equal to its own volume only.

Weight of cylinder = upthrust due to both liquids

ÞÞ

If the rise of level in the right limb be x cm. the fall of level of mercury in left limb be 4x cm because the area of cross section of right limb is 4 times as that of left limb.
Level of water in left limb is (36 + 4x) cm.

Now equating pressure at interface of Hg and water (at A' B')

By solving we get x = 0.56 cm.