A large open top container of negligible mass and uniform cross sectional area `A` has a small uniform cross sectional area `a` in its side wall near the bottom. The container is kept over a smooth horizontal floor and contains a liquid of density `rho` and mass `m_(0)`. Assuming that the liqud starts flowing through the hole A the acceleration of the container will be

Options

A. g
B. `g//50 `
C. `g//100 `
D. `g//10`

Answer

B

Explanation

Mass of water = (Volume ) (density )
` therefore m _ 0 = (AH) rho `
` therefore H = (m_0 )/(A rho ) " " `…(i)
Velocity of efflux , ` v = sqrt (2 gH ) = sqrt (2g (m_0)/(Arho)) = sqrt((2m_0 g )/(A rho)) `
Thrust force on the container due to draining out of liquid from the bottom is given by
F = (density of liquid) (area of hole ) (velocity of efflux ) ` ""^2 `
` F = rho a v ^ 2 `
` F = rho (A//100 ) v ^ 2 = rho (A//100) ((2m_0 g )/(A rho)) `
` F = (m_0 g )/(50) `
` therefore ` Acceleration of the container ` a = F//m_0 = g //50 , a = g//50 `