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

Let at some instant of time the level of liquid in the tank is y. Velocity of efflux at this instant of time `v=sqrt(2gy)`
Now, at this instant volume of liquid coming out the hole per second is `((dV_(1))/(dt))`
Volume of liquid coming down in the tank per second is `((dV_(2))/(dt))`
`(dV_(1))/(dt)=(dV_(2))/(dt)impliesav=A((-dy)/(dt))thereforeasqrt(2gy)=A(-(dy)/(dt))` ...(i)
(Here area of cross-section of hole and tank are respectively a and A)
Substituting the proper limits in equation (i).
`int_(0)^(t_(1))dt=-(A)/(asqrt(2g))int_(H)^(H//2)y^(-1//2)dyimpliest_(1)=(2A)/(asqrt(2g))[sqrt(y)]_(H//2)^(H)=(2A)/(asqrt(2g))[sqrt(H)-sqrt((H)/(2))]=(A)/(a)sqrt((H)/(g))(sqrt(2)-1)` ....(ii)
Similarly `int_(0)^(t_(2))dt=-(A)/(asqrt(2g))int_(H//2)^(0)y^(-1//2)dyimpliest_(2)=(A)/(a)sqrt((H)/(g))` ....(iii)
From equation (ii) and (iii), `(t_(1))/(t_(2))=sqrt(2)-1=0.414`