$A$ closed vessel has a small hole in one face near the bottom,as shown in the figure. The velocity of water coming out from the hole at the given instant is ....... $m/s$. (Given: $P_{0} = 10^{5} \text{ N/m}^{2}$,$h = 9/4 \text{ m}$,density of water $\rho = 10^{3} \text{ kg/m}^{3}$,$g = 10 \text{ m/s}^{2}$)

$A$ vertical tank,open at the top,is filled with a liquid and rests on a smooth horizontal surface. $A$ small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is $N$ times the area of the hole,where $N$ is a large number. Neglect the mass of the tank itself. The initial acceleration of the tank is:

Water is filled in a tank up to a height of $3 \,m$. The base of the tank is at a height of $1 \,m$ above the ground. What should be the height of a hole made in the side of the tank so that water can be sprayed up to the maximum horizontal distance on the ground?

$A$ large tank filled with water to a height $h$ is to be emptied through a small hole at the bottom. The ratio of the time taken for the level to fall from $h$ to $h/2$ and that taken for the level to fall from $h/2$ to $0$ is:

$A$ vessel completely filled with water has holes $A$ and $B$ at depths $h$ and $3h$ from the top respectively. The hole $A$ is a square of side $L$ and $B$ is a circle of radius $r$. The water flowing out per second from both the holes is same. Then $L$ is equal to

The height of the water level in a tank of uniform cross-section is $5 \,m$. The volume of water leaked in $5 \,s$ through a hole of area $2.4 \,mm^2$ made at the bottom of the tank is (Assume the level of the water in the tank remains constant and acceleration due to gravity $= 10 \,ms^{-2}$).