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. upwards at g
B. upwards at ` g (a ^ 2 )/( A^2 ) `
C. downwards at ` g (a)/(A)`
D. downwards at ` g (a ^ 2 )/(A^2 ) `

Answer

D

Explanation

The velocity of fluid at the hole is `V_(2)=sqrt(2gh)`
Using continuity equation at the two cross-ssections A and B:
`V_(1)A=V_(2)a," "Rightarrow" "V_(1)=(a)/(A)V_(2)`
`Rightarrow" "` acceleration of top surface
`=-V_(1)(dV_1)/(dh)=-((a)/(A))V_(2)(d)/(dh)((a)/(A)V_(2))`
`a_(1)=(a^2)/(A^2)sqrt(2gh)sqrt(2g).(1)/(2sqrth)," "Rightarrow" "a_(1)=-(ga^2)/(A^2)`