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

Substiuting the proper limits iin Eq. `(i)`, derived in the theroy, we have
`int_(0)^(t_(1)) dt = (A)/(asqrt(2g)) int_(H)^(H//2) y^(-1//2)dy`
or `t_(1) = (2A)/(asqrt(2g))[sqrt(y)]_(H2)^(H)` or `t_(1) = (2A)/(asqrt(2g))[sqrt(H) - sqrt((H)/(2))]` or `t_(1) = (A)/(a) sqrt((H)/(g))(sqrt(2) - 1)`
Similarly `int_(0)^(t_(2))dt = -(A)/(asqrt(2g))int_(H//2)^(0)y^(-1//2) = dy`
or `t_(2) = (A)/(a) sqrt((H)/(g))`
We get
`(t_(1))/(t_(2)) = sqrt(2) - 1` or `(t_(1))/(t_(2)) = 0.414`