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((m_0 g )/(2A rho)) `
B. `sqrt((m_0 g )/(A rho )) `
C. `sqrt((3)/(4) (m_0 g )/(A rho)) `
D. `sqrt((m_0 g )/(4 A rho )) `

Answer

A

Explanation

Velocity of efflux when 75% liquid has been drained out i.e. Height of liquid
` h = (H)/(4) = (m _ 0 ) /(4 A rho ) `
` v = sqrt (2gh ) = sqrt(2g ((m_0)/(4 A rho ))) `
` v = sqrt((m_0 g )/(2 A rho )) `