$A$ 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

Two large,identical water tanks,$1$ and $2$,kept on the top of a building of height $H$,are filled with water up to height $h$ in each tank. Both the tanks contain an identical hole of small radius on their sides,close to their bottom. $A$ pipe of the same internal radius as that of the hole is connected to tank $2$,and the pipe ends at the ground level. When the water flows out of tanks $1$ and $2$ through the holes,the times taken to empty the tanks are $t_1$ and $t_2$,respectively. If $H = (16/9) h$,then the ratio $t_1 / t_2$ 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 zero is

$A$ cylinder containing water stands on a table of height $H$. $A$ small hole is punched in the side of the cylinder at its base. The stream of water strikes the ground at a horizontal distance $R$ from the table. Then the depth of water in the cylinder is ............

$A$ 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

$A$ 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