$A$ cylindrical tank of height $0.4\,m$ is open at the top and has a diameter $0.16\,m$. Water is filled in it up to a height of $0.16\,m$. How long will it take to empty the tank through a hole of radius $5 \times 10^{-3}\,m$ in its bottom? (in seconds)

$A$ tank is filled with water up to a height of $1 \, m$. $A$ hole is made at a distance of $20 \, cm$ from the top. Find the horizontal distance from the base of the tank where the water strikes the ground in $cm$.

$A$ large vessel completely filled with water has two holes '$A$' and '$B$' at depths '$h$' and '$4h$' from the top. Hole '$A$' is a square of side '$L$' and hole '$B$' is a circle of radius '$R$'. If the same quantity of water is flowing per second from both holes,then the side of the square hole is:

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

$A$ 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 are two identical small holes of cross-sectional area $a$ on the opposite sides of a tank containing a liquid of density $\rho$. The difference in height between the holes is $h$. The tank is resting on a smooth horizontal surface. The horizontal force that must be applied to the tank to keep it in equilibrium is