A large open top container of negligible mass and uniform cross-sectional area `A` has a small hole of croos-sectional area `(A)/(100)` in its side wall near the bottom.The container is kept on a smooth horizontal floor and contains a liquid of density `rho` and mass `m_(0)`. Assuming that the liquid starts flowing out horizontally through the hole at `t = 0`, The acceleration of the container is `(x)/(10) m//s^(2)` than `x` is -

Answer

`(i)0.2m//s^(2),(ii)sqrt(2g.(m_(0))/(Arho))`

Explanation

`m_(0)=Ahrho`
`h=(m_(0))/(Arho)`
velocity of efflux
`v=sqrt(2gh)impliesv=sqrt(2g m_(0)/(Arho))`
If m is mass coming out per sec.then
`F=mv=((A)/(100)xx v xxrho)v=(A)/(100)sqrt(2gm_(0)/(Arho))rhosqrt(2gm_(0)/(Arho))=m_(0)a`
`(2gm_(0))/(100)=m_(0)a implies " "a = 0.2 m//s^(2)`

(ii) Let at any intermediate stage height of liquid above hole is x.
First we have to get time taken
Velocity of efflux `v=sqrt(2gx)`
volume comingout per/sec. `(dV )/(dt)=A_(0)sqrt(2gx)`
`((-Adx))/(dt)=A_(0)sqrt(2gx)" "implies-Adx=A_(0)sqrt(2gx) dt`
`-(A)/(A_(0))underset(h)overset(x)(int)(dx)/(sqrt(x))=sqrt(2g)underset(0)overset(t)(int)dtimplies -(A)/(A_(0))[sqrt(2x)]_(h)^(x)=sqrt(2g)t`
`-(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(x)-sqrt(h)rfloor=t`
`t=(A)/(A_(0))sqrt((2)/(g))lfloorsqrt(h)-sqrt(x)rfloor" "....(1)`
Since `75%` volume drain out so volume remain `25%`
`Ax=25%` of Ah
`x=(h)/(4)` put in equation (1) we set
`t=(A)/(A_(0))sqrt((2)/(g))xx(1)/(2)sqrt(h)implies t=(A)/(A_(0))sqrt((2)/(g))xx(1)/(2)sqrt((m_(0))/(Arho))=100sqrt((2)/(g)xx(1)/(2)) sqrt(m_(0)/(Arho))`
For velocity
Change is linear at momentum `=m(dv)/(dt)`
`(A)/(100)sqrt(2gx)rho sqrt(2gx)=Axrho (dv)/(dt) implies (2g)/(100)=(dv)/(dt)`
`underset(0)overset(v)(int)dv =(2g)/(100) underset(o)overset(t)(int)dt`
`v=(2g)/(100)xxt" "implies v=(2g)/(100)xx100sqrt((2)/(g))xx(1)/(2)sqrt((m_(0))/(AP))impliesv=sqrt(2hm_(0)/(AP))`