JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Fluid Mechanics

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A cylindrical vessel of height 500 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 mm. Find the fall in height (in mm) of water level due to opening of the orifice.
[Take atmospheric pressure

= 1.0 \times 1 0^{5} N / m^{2}

, density of water

= 1000 kg / m^{3}

and

g = 10 m / s^{2}

. Neglect any effect of surface tension.]

Answer: 6

Solution

When water flows out, the pressure in the sealed air space decreases. At equilibrium, the pressure at the bottom equals atmospheric pressure:

P + ρ g h = P_{0}

, where

h = 200

mm. Thus

P = P_{0} - ρ g h = 1 0^{5} - 1000 \times 10 \times 0.2 = 98000

N/m². Using the ideal gas law for the trapped air:

P_{0} V_{0} = P V

, where

V_{0} = A (500 - H)

and

V = A (500 - 200) = 300 A

. Thus

1 0^{5} (500 - H) = 98000 \times 300

, giving

H = 206

mm. The fall in height is

206 - 200 = 6

mm.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Integer answer

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JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Fluid Mechanics

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