Answer:
a. The time required for the tank to empty halfway is presented as follows;

b. The time it takes for the tank to empty the remaining half is presented as follows;

The total time 't', is presented as follows;

Step-by-step explanation:
a. The diameter of the tank = D₀
The height of the tank = H
The diameter of the orifice at the bottom = D
The equation for the flow through an orifice is given as follows;
v = √(2·g·h)
Therefore, we have;


Where;
P₁ = P₂ = The atmospheric pressure
z₁ - z₂ = dh (The height of eater in the tank)
A₁·v₁ = A₂·v₂
v₂ = (A₁/A₂)·v₁
A₁ = π·D₀²/4
A₂ = π·D²/4
A₁/A₂ = D₀²/(D²) = v₂/v₁
v₂ = (D₀²/(D²))·v₁ = √(2·g·h)
The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;
dt = dh/v₁ = (v₂/v₁)/v₂·dh = (D₀²/(D²))/v₂·dh = (D₀²/(D²))/√(2·g·h)·dh
We have;

The time for the tank to drop halfway is given as follows;

![t_1 =\left[{ (D_0^2)/(D\cdot √(2\cdot g) ) \cdot\frac{h^{-(1)/(2) +1}}{-(1)/(2) +1 } \right]_{(H)/(2) }^(H) =\left[ { (D_0^2 \cdot 2\cdot √(h) )/(D\cdot √(2\cdot g) ) \right]_{(H)/(2) }^(H) = { (2 \cdot D_0^2 )/(D\cdot √(2\cdot g) ) \cdot \left(√(H) - \sqrt{(H)/(2) } \right)](https://img.qammunity.org/2022/formulas/engineering/college/b0eot0d8bsqdfpisauyc711s4pgd8x1cj7.png)

The time required for the tank to empty halfway, t₁, is given as follows;

(b) The time it takes for the tank to empty completely, t₂, is given as follows;

![t_2 =\left[{ (D_0^2)/(D\cdot √(2\cdot g) ) \cdot\frac{h^{-(1)/(2) +1}}{-(1)/(2) +1 } \right]_(0)^{(H)/(2) } =\left[ { (D_0^2 \cdot 2\cdot √(h) )/(D\cdot √(2\cdot g) ) \right]_(0 )^{(H)/(2) } = { (2 \cdot D_0^2 )/(D\cdot √(2\cdot g) ) \cdot \left( \sqrt{(H)/(2) } -0\right)](https://img.qammunity.org/2022/formulas/engineering/college/m2flhtdmnwkvzei0d52kg98c3qh9z5v3z7.png)

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

The total time, t, to empty the tank is given as follows;

