Question

A tank is filled with a liquid upto a height H, A small hole is made at the bottom of this tank Let `t_(1)` be the time taken to empty first half of the tank and `t_(2)` time taken to empty rest half of the tank then find `(t_(1))/(t_(2))`

Answer 1

As we know that, the time in which level of liquid in a tank falls from `H_(1) " to " H_(2)` is
`t=(A)/(a)sqrt((2)/(g))sqrt(H_(1)-sqrt(H_(2))`
where, A=area of cross-section of the tank,
a=area of hole
It is given that `H_(1)=H " and " H_(2)=(H)/(2)`
`therefore " " t_(1)=(A)/(a)sqrt((2)/(g))(sqrt(H)-sqrt((H)/(2)))`
Similarly time taken to empty the rest half of the tank
`t_(2)=(A)/(a)sqrt((2(H//2))/(g))=(A)/(a)sqrt((H)/(g))`
From Eqs. (i) and (ii), we get
`(t_(1))/(t_(2))=sqrt(2)-1`
or `(t_(1))/(t_(2))=0.414`