Given

Initial height of air trapped =h_o

Initial pressure=2P_o

P_o=atmospheric pressure = 10⁵Pa

Depth at which there is hole in tank =h₁

Let the density of water be

From the diagram we can see that,

Pressure of water above the water level of the bigger tank is given by

P=(h₂+h_o)g

Let the atmospheric pressure above the tube be P_o.

Total pressure above the tube=P_o+P=(h₂+h_o)g+P_o

This pressure initially is balanced by pressure above the tank 2P_o (from diagram).

Therefore,

2P_o=(h₂+h_o)g+P_o

P_o=(h₂-h₀)g

The height h₂ of the water in the long tube above the top initially is given by

(b)

Efflux means water flowing out from tube or outlet.

Velocity of efflux out of the outlet depends upon the total pressure above the outlet.

Total pressure above the outlet=2P_o+(h₁-h_o)g

Let the velocity of efflux be v₁ and the velocity with which the level of tank falls be v₂.

Pressure outside the outlet =P_o

Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

Mathematically Bernoulli’s theorem can be written as

Where v=velocity of fluid at a point

z= elevation of that point from a reference level

P=pressure of fluid at that point

=density of that fluid

G=acceleration due to gravity

Applying Bernoulli’s theorem, at points outside the outlet and above the outlet, we get

Consider the difference in elevation of both the points very small, so that we can ignore ‘gz’ term on both the sides.

Again, the speed with which the water level of the tank goes out is very less compared to the velocity of the efflux. Thus, v₂=0

The speed with which water comes out of the hole is given by

(c) Water maintains its own level, so height of the water of the tank will be h₁ when water will stop flowing out.

Thus, height of water in the tube below the tank height will be =h₁

Hence height if the water above the tank will be =-h₁