What is filled in a cylindrical container to a height of `3 m`. The ratio of the cross-sectional area of the orifice and the beaker is `0.1`. The square of the speed of the liquid coming out from the orifice is `(g = 10 m//s^2)`.

Answer

[`7.04 m//s`]

Explanation

Let `a_(1), a_(2)` be the areas of cross-section of the beaker and the orifice respectively. `v_(1), v_(2)` be the liquid velocity for the beaker and the orifice respectively.
Here, `(a_(2))/(a_(1))=0.1, h_(1)=3 m`,
`h_(2)=52.5 cm=0.525 m`,
`h_(1)-h_(2)=3-0.525=2.475 m`.
Ac cording to equation of continnutiy
`a_(1) v_(1)=a_(2)v_(2)` or `v_(1)=(a_(2))/(a_(1))xxv_(2)`
According to Bernoulli's equation
`P_(1)+rho gh_(1)+(1)/(2)rho v_(1)^(2)=P_(2)+rho gh_(2)+(1)/(2)rho v_(2)^(2)`
Here, `P_(1)=P_(2)=` atmospheric pressure, then
`gh_(1)+(1)/(2)+(v_(1)^(2))/(2)=gh_(2)+((v_(2)^(2))/(2))`
`gh_(1)+(1)/(2)[(a_(2))/(a_(1))v_(2)]=gh_(2)+(v_(2)^(2))/(2)`
or `v_(2)^(2)[1-((a_(2))/(a_(1)))^(2)]=2g(h_(1)-h_(2))`
or `v_(2)=sqrt((2g(h_(1)-h_(2)))/ 1-(a_(2)//a_(1))^(2))=sqrt((2xx10xx2.475)/(1-(0.1)^(2)))`
`=7.04 ms^(-1)`