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 comes out of the hole, the top surface of the liquid accelerates (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))`