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

A, D

Explanation

Substituting the proper limits in Eq. (i), derived in the theory, we have
`int_(0)^(t_1) dt= -(A)/(a sqrt(2g)) int_(H)^(H//2) y_(-1//2) dy`
or `t_(1)=(2A)/(asqrt(2g))[sqrt(H)-sqrt((H)/(2))]`
or `t_(1) = (A)/(a) sqrt((H)/(g)) (sqrt(2)-1)` ...(ii)
Similarly, `int_(0)^(t_(2)) dt =- (A)/(a sqrt(2g)) int_(H//2)^(0) t^(-1//2) dy`
or `t_(2) = (A)/(a) sqrt((H)/(g))` ...(iii)
From Eqs. (ii) and (iii), we get
`(t_(1))/(t_(2)) = sqrt(2) - 1`
or `(t_(1))/(t_(2)) = 0.414`