A cylindrical vessel of height 500mm 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 200mm. Find the fall in height(in mm) of water level due to opening of the orifice.
[Take atmospheric pressure `=1.0xx10^5N//m^2`, density of water=1000kg//m^3` and `g=10m//s^2`. Neglect any effect of surface tension.]

Options

A. 6mm
B. 8mm
C. 10mm
D. 12mm

Answer

A

Explanation

Let A in the area of crosa section of cylinder.
Initial volume of air at the top: `V_1 =A(500 - H) xx 10^(-3) m^(3)`
Final volume of air at the top: `V_2 = A(500 - 200) xx 10^(-3) m^(3)=300A xx 10^(-3)m`
Initial pressure of air at the top: `P=1 xx 10^(5) N//m^2`
If `P_2` is final pressure of air at the top then pressure near the orifice must be equal to atmospheric pressure as water is not coming out from the orifice.
`P_(2)+pgh=P_(0)`
`rArr P_(2)+1000 xx 10 xx 0.2 =1 xx 10^(5) N//m^(2)`
`rArr P_(2)=0.98 xx 10^(5) N//m^(2)`
`P_(1)V_(1)=P_(2)V_(2)`
`10^(5) xx A (500-H) xx 10^(-3) =0.98 xx 10^(5) xx 300A xx 10^(-3)`
`rArr (500-H) =0.98 xx 300`
`rArr 500-H=294`
`rArr H=206mm`
Since final level is 200 mm hence fall in water level is 6mm.