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 . Now the top is completely sealed with a cap and the orifice at the bottom is opened. [2009]

Question

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 = $1.0 \times 10^{5} {N/m}^{2}$ , density of water = $1000 {kg/m}^{3}$ and $g = 10 {m/s}^{2}$ . Neglect any effect of surface tension.] [2009]

Smart Achievers · Accepted Answer

(6)

$Initially, pressure of air column above water$

$P_{1} = 10^{5} {N m}^{- 2}$ $and$ $V_{1} = (500 - H) A,$

$where A is the area of cross-section of the vessel.$

$Finally, the volume of air column above water$ $V_{2} = (500 - 200) A = 300 A .$

$If P_{2} is the pressure of air then$ $P_{2} + ρ g h = P_{1}$

$∴$ $P_{2} + 10^{3} \times 10 \times \frac{200}{1000} = 10^{5}$

$∴$ $P_{2} = 9.8 \times 10^{4} {N/m}^{2}$

$Assuming the temperature remains constant, according to Boyle's law$

$P_{1} V_{1} = P_{2} V_{2}$

$∴$ $10^{5} \times (500 - H) A = (9.8 \times 10^{4}) \times 300 A$ $\Rightarrow H = 206 mm$

$∴$ $Fall in height of water level due to the opening of orifice$ $= 206 - 200 = 6 mm$

Solution

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