$A$ cylindrical tank having a large diameter is filled with water to a height $H$. $A$ hole of cross-sectional area $5 \,cm^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 \times 10^{-3} \,m^3 \,s^{-1}$, then the value of $H$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,cm$)

There are two holes, one each along the opposite sides of a wide rectangular tank. The cross-section of each hole is $0.01 \,m^2$ and the vertical distance between the holes is $1 \,m$. The tank is filled with water. The net force on the tank in newton when the water flows out of the holes is (Density of water $= 1000 \,kg/m^3$, $g = 10 \,m/s^2$)

$A$ cylindrical vessel filled with water up to a height of $H$ stands on a horizontal plane. The side wall of the vessel has a plugged circular hole touching the bottom. The coefficient of friction between the bottom of the vessel and the plane is $\mu$ and the total mass of the water plus the vessel is $M$. What should be the minimum diameter of the hole so that the vessel begins to move on the floor if the plug is removed? (Here,the density of water is $\rho$)

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

$A$ 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

For a container open to the atmosphere,derive the velocity of the liquid exiting a narrow hole in the container wall using Bernoulli's equation and obtain Torricelli's law.