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?

$A$ 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 $\,cm$)

$A$ cylindrical tank has a hole of area $2 \,cm^2$ at its bottom. If water is poured into the tank from a tube above it at the rate of $100 \,cm^3 \,s^{-1}$, then the maximum height up to which water can rise in the tank is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $cm$)

$A$ tank of height $5\, m$ is full of water. There is a hole of cross-sectional area $1\, cm^2$ in its bottom. The initial volume of water that will come out from this hole per second is

$A$ light cylindrical vessel is kept on a horizontal surface. The area of the base is $A$. $A$ hole of cross-sectional area $a$ is made just at its bottom side. The minimum coefficient of friction necessary to prevent the vessel from sliding due to the impact force of the emerging liquid is $(a \ll A)$.

Water is filled in a cylindrical vessel of height $H$. $A$ hole is made at height $z$ from the bottom, as shown in the figure. The value of $z$ for which the range $(R)$ of the emerging water through the hole will be maximum is: