A large open top container of negligible mass and uniform cross sectional area `A` a has a small hole of cross 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`, calculate a the acceleration of the container and b its velocity when `75%` of the liquid has drained out.

Options

A. ` sqrt((3m g )/( 4 A rho )) `
B. ` sqrt ((2m g )/(A rho )) `
C. `2 sqrt ((mg )/(A rho)) `
D. `sqrt((mg )/(2 A rho )) `

Answer

D

Explanation

When 75% of the liquid has drained out, the height of the liquid in the container will be ` h_1 = (h )/(4) ` for this height, velocity of liquid flowing out ` v_ 1 = sqrt(2gh _ 1 ) = sqrt (2g xx (h)/(4)), = sqrt((gh)/(2) ) `
Now, ` h = (m)/(A rho ) `, Hence ` v_1 = sqrt((gm)/((2 A rho )) `