JEE Challenger
More from Mechanical Properties of Fluids
PhyMedAdv 2009 • Paper 2

Fall in Height of Water Column in Sealed Cylindrical Vessel

A cylindrical vessel of height 500 mm500\text{ mm} has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm200\text{ mm}. Find the fall in height (in mm\textbf{in mm}) of water level due to opening of the orifice.
[Take atmospheric pressure =1.0×105 N/m2= 1.0 \times 10^5\text{ N/m}^2, density of water =1000 kg/m3= 1000\text{ kg/m}^3 and g=10 m/s2g = 10\text{ m/s}^2. Neglect any effect of surface tension.]

Official Numerical Answer6

Topics & Concepts

Mechanical Properties of FluidsFluid Dynamics

Step-by-Step Solution

To find the fall in the height of the water level, we consider the process of water draining out until a steady state is reached.

1. Initial State:

Let:

  • Total height of the cylindrical vessel, L=500 mm=0.5 mL = 500\text{ mm} = 0.5\text{ m}
  • Initial height of the water column =H= H
  • Cross-sectional area of the cylinder =A= A
  • Atmospheric pressure, P0=1.0×105 N/m2P_0 = 1.0 \times 10^5\text{ N/m}^2

When the vessel is sealed at the top with water at height HH, the volume of the trapped air column above the water is:
V1=A(L−H)V_1 = A(L - H)
The pressure of this trapped air is equal to the atmospheric pressure:
P1=P0P_1 = P_0

2. Final (Steady) State:

Water comes out of the bottom orifice until the pressure at the orifice equals atmospheric pressure, which stops any further flow.

  • Final height of the water column, h=200 mm=0.2 mh = 200\text{ mm} = 0.2\text{ m}
  • Final volume of the trapped air, V2=A(L−h)V_2 = A(L - h)

At the bottom of the vessel, the condition for the water level to be steady (neglecting surface tension) is:
Pbottom=P2+ρgh=P0P_{\text{bottom}} = P_2 + \rho g h = P_0
where P2P_2 is the final pressure of the trapped air.

Thus, the final pressure of the trapped air is:
P2=P0−ρghP_2 = P_0 - \rho g h

Given:

  • ρ=1000 kg/m3\rho = 1000\text{ kg/m}^3
  • g=10 m/s2g = 10\text{ m/s}^2
  • h=0.2 mh = 0.2\text{ m}

The hydrostatic pressure of the water column is:
ρgh=1000×10×0.2=2000 N/m2\rho g h = 1000 \times 10 \times 0.2 = 2000\text{ N/m}^2

3. Applying Boyle's Law:

Assuming the temperature remains constant throughout the process, the trapped air undergoes an isothermal expansion:
P1V1=P2V2P_1 V_1 = P_2 V_2

Substituting the expressions for pressure and volume:
P0⋅A(L−H)=P2⋅A(L−h)P_0 \cdot A(L - H) = P_2 \cdot A(L - h)
P0(L−H)=(P0−ρgh)(L−h)P_0 (L - H) = (P_0 - \rho g h)(L - h)
L−H=(1−ρghP0)(L−h)L - H = \left(1 - \frac{\rho g h}{P_0}\right)(L - h)

Rearranging to find the fall in height of the water level, Δh=H−h\Delta h = H - h:
Δh=H−h=(L−h)−(L−H)\Delta h = H - h = (L - h) - (L - H)
Δh=(L−h)−(1−ρghP0)(L−h)\Delta h = (L - h) - \left(1 - \frac{\rho g h}{P_0}\right)(L - h)
Δh=(L−h)(ρghP0)\Delta h = (L - h)\left(\frac{\rho g h}{P_0}\right)

4. Calculation:

Given that:

  • L−h=500 mm−200 mm=300 mmL - h = 500\text{ mm} - 200\text{ mm} = 300\text{ mm}
  • ρghP0=20001.0×105=0.02\frac{\rho g h}{P_0} = \frac{2000}{1.0 \times 10^5} = 0.02

Substitute these values into the expression for Δh\Delta h:
Δh=300 mm×0.02=6 mm\Delta h = 300\text{ mm} \times 0.02 = 6\text{ mm}

Thus, the fall in height of the water level is 6.

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