A cylinder containing water up to a height of `25cm` has a hole of cross-section `1/4 cm^(2)` in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begin to flow out?

Explanation

Here, `a = 0.25 cm^(2) = 0.25 xx 10^(-4)m^(-2)`,
. `h = 25 cm = 0.25 m`
Initial velocity of water flowing out of hole
`upsilon = sqrt(2gh)`
When wate emerges out from the hole. The weight of water decrease. The decrease in weight is equal to the rate of change of linear momentum. Initially velocity of liquid flowing out `upsilon` be taken constant for a small interval of time `Delta t`. then
`F = (Delta p)/(Delta t) = (Delta (m upsilon))/(Delta t) = upsilon (Delta m)/(Delta t)`
`=(upsilonDelta(V rho))/(Delta t) = upsilon rho (Delta V)/(Delta t)`
Where, `(Delta V)/(Delta t)` = volume flowing out per second
`=aupsilon`
`F = upsilon rho.(a upsilon) = a upsilon^(2) rho`
`=a(2gh) rho = 2 a gh rho`
`= 2 xx (0.25 xx 10^(-4)) xx 9.8 xx 0.25 xx 10^(3)` ltrgt `=12.5 xx 9.8 xx 10^(-3) N = 12.5 xx 10^(-3)kg f`.