If a small orifice is made at a height of $0.25\, m$ from the ground in a tank filled with water up to a height of $1\, m$,the horizontal range of the water stream will be (in $cm$):

$A$ container is filled with water up to a height of $20 \ m$. If a hole is made at the bottom,what will be the velocity of the water coming out in $m/s$?

$A$ 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 $(a \ll A)$.

$A$ large open tank containing water has two holes in its wall. $A$ square hole of side $a$ is made at a depth $y$ and a circular hole of radius $r$ is made at a depth $16y$ from the surface of the water. If an equal amount of water comes out through both the holes per second,then the relation between $r$ and $a$ will be:

$A$ large tank open to the atmosphere at the top and filled with water develops a small hole in the side at a point $20 \,m$ below the water level. If the rate of flow of water from the hole is $3 \times 10^{-3} \,m^3/min$, then the area of the hole is (Acceleration due to gravity $= 10 \,m/s^2$) (in $\,mm^2$)

$A$ tank is filled up to a height $h$ with a liquid and is placed on a platform of height $h$ from the ground. To get the maximum range $x_m$,a small hole is punched at a distance of $y$ from the free surface of the liquid. Then: