There is a small hole in the bottom of a fixed container containing a liquid upto height `'h'`. The top of the liquid as well as the hole at the bottom are exposed to atmosphere. As the liquid come out of the hole. (Area of the hole is `'a'` and that of the top surface is `'A'`) :

Options

A. the top surface of the liquid accelerates with acceleration `= g`
B. the top surface of the liquid accelertes with acceleration `= g'(a^(2))/(A^(2))`
C. the top surface of the liquid retards with retardation `= g'(a)/(A)`
D. the top surface of the liquid retards with retardation `= (ga^(2))/(A^(2))`

Answer

D

Explanation

The velocity of fluid at the hole is `V_(2) = sqrt((2gh)/(1 + (a^(2)//A^(2)))`
Using continuity equation at the two cross-sections `(1)` and `(2)` :
`V_(1)A = V_(2)a rArr V_(1) = (a)/(A)V_(2)`
`rArr` acceleration (of top surface) `= -V_(1)(dV_(1))/(dh)`
`= -(a)/(A)V_(2)(d)/(dh)((a)/(A)V_(2))`
`a_(10) = -(a^(2))/A^(2)V_(2)(dV_(2))/(dh) = -(a^(2))/A^(2)sqrt(2gh) sqrt(2g).(1)/(2sqrt(h)) rArr a_(1) = (-ga^(2))/(A^(2))`