$A$ 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 $\,mm^2$)

$A$ 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 $cm$)

$A$ cylindrical vessel of cross-section $A$ contains water to a height $h$. There is a hole in the bottom of radius $a$. The time in which it will be emptied is

$A$ large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from the two holes are the same. Then the value of $R$ is

$A$ cylinder of height $h$ is filled with water and is kept on a block of height $h/2$. The level of water in the cylinder is kept constant. Four holes numbered $1, 2, 3$ and $4$ are at the side of the cylinder and at heights $0, h/4, h/2$ and $3h/4$ from the base of the cylinder, respectively. When all four holes are opened together, the hole from which water will reach the farthest distance on the plane $PQ$ is the hole number: