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 $(1 \ atm = 10^5 \ Pa)$ :-

$A$ cylindrical vessel filled with water up to the height $H$ becomes empty in time $t_0$ due to a small hole at the bottom of the vessel. If water is filled to a height $4H$,it will flow out in time:

$A$ cylindrical vessel is filled with water up to a height of $1 \,m$ from the base. $A$ small orifice is opened at some height in the cylinder, and the water level is reduced to the height of the orifice in $20 \,s$. If the base area of the cylinder is $100$ times the area of the orifice, then the height of the orifice from the base is (take $g = 10 \,m/s^2$): (in $\,cm$)

Tanks $A$ and $B$ open at the top contain two different liquids up to a certain height. $A$ hole is made in the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of in $A$. If the liquid mass flux through each hole is equal, then the ratio of the densities of the liquids respectively, is

In a cylindrical vessel containing liquid of density $\rho$,there are two holes in the side walls at heights of $h_1$ and $h_2$ respectively,such that the range of efflux at the bottom of the vessel is the same. The height of a hole,for which the range of efflux would be maximum,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