$A$ 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 $\rho = 10^{3} \text{ kg/m}^{3}$,$g = 10 \text{ m/s}^{2}$)

There is a hole in the bottom of a tank containing water. If the total pressure at the bottom is $3 \ atm$,then the velocity of water flowing from the hole is $(1 \ atm = 10^5 \ Pa)$ :-

Water is filled in a tank up to a height of $20 \ cm$ from the bottom of the tank. Water flows through a hole of area $1 \ mm^2$ at its bottom. The mass of the water coming out from the hole in a time of $0.6 \ s$ is (Density of water $= 1000 \ kg \ m^{-3}$ and acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $g$)

$A$ 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:

$A$ tank filled with fresh water has a hole in its bottom and water is flowing out of it. If the size of the hole is increased,then

Two large,identical water tanks,$1$ and $2$,kept on the top of a building of height $H$,are filled with water up to height $h$ in each tank. Both the tanks contain an identical hole of small radius on their sides,close to their bottom. $A$ pipe of the same internal radius as that of the hole is connected to tank $2$,and the pipe ends at the ground level. When the water flows out of tanks $1$ and $2$ through the holes,the times taken to empty the tanks are $t_1$ and $t_2$,respectively. If $H = (16/9) h$,then the ratio $t_1 / t_2$ is: