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A cylindrical tank with cross sectional area. At any time 't' it contains water with mass 'm' and density 'p'. The tank has cylindrical hole at the bottom of area AO. If the fluid drains from the tank through the hole at volumetric flow rate 'q'. If [q = C.h]; where C is constant, and h is the water level in the tank. Derive an expression describing the case relating the changing variable with time.

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To derive the expression relating the changing variable with time, let's consider the given information and apply some principles of fluid mechanics.

Given:
- Cross-sectional area of the tank: A
- Mass of water in the tank: m
- Density of water: ρ
- Area of the hole at the bottom: A₀
- Volumetric flow rate: q = C⋅h, where C is a constant and h is the water level in the tank.

We can start by relating the mass of water in the tank to its volume using the density:

m = ρ⋅V

The volume V can be calculated using the cross-sectional area A and the water level h:

V = A⋅h

Now, let's express the rate of change of mass with respect to time:

dm/dt = d(ρ⋅V)/dt

Using the product rule of differentiation, we can expand this expression:

dm/dt = ρ⋅dV/dt + V⋅dρ/dt

Next, let's consider how the volume V changes with time. Since water is draining out of the tank through the hole at the bottom, the volumetric flow rate q is equal to the cross-sectional area of the hole A₀ multiplied by the velocity v of the water draining out:

q = A₀⋅v

The velocity v can be related to the water level h by applying the principle of Torricelli's law for flow through an orifice:

v = √(2⋅g⋅h)

Where g is the acceleration due to gravity. Substituting this expression for v into the equation for q, we have:

q = A₀⋅√(2⋅g⋅h)

Now, let's differentiate the equation q = A₀⋅√(2⋅g⋅h) with respect to time t:

dq/dt = d(A₀⋅√(2⋅g⋅h))/dt

Using the chain rule of differentiation, we can calculate this:

dq/dt = A₀⋅(1/2)⋅(2⋅g/h)⋅(dh/dt)

Simplifying further, we have:

dq/dt = A₀⋅g/√h⋅(dh/dt)

Since we know that q = C⋅h, we can substitute this into the equation:

C⋅dh/dt = A₀⋅g/√h⋅(dh/dt)

Now, rearranging the equation to isolate the changing variable, we get:

C⋅dh/dt - A₀⋅g/√h⋅(dh/dt) = 0

Combining the terms on the left-hand side and factoring out the common factor of dh/dt, we have:

(dh/dt)⋅(C - A₀⋅g/√h) = 0

Since dh/dt cannot be zero (as the water level is changing), the expression in parentheses must be zero:

C - A₀⋅g/√h = 0

Solving for h, we get:

C = A₀⋅g/√h

Now, we can solve this equation to obtain an expression relating the changing variable (h) with time. By manipulating the equation further, we can isolate h:

√h = A₀⋅g/C

Squaring both sides:

h = (A₀⋅g/C)
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