$A$ liquid is filled in a tank up to a height of $5 \,m$ and the bottom of the tank is at $5 \,m$ height from the ground. $A$ hole is punched in the tank so that the liquid coming out touches the ground at maximum horizontal distance. Then, the maximum horizontal distance is (in $\,m$)

$A$ 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 $\,mm^2$)

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:

Water flows out of a hole on the side of a bucket and follows a parabolic path. If the bucket falls freely under gravity, ignoring air resistance, the water flow

$A$ vertical tank,open at the top,is filled with a liquid and rests on a smooth horizontal surface. $A$ small hole is opened at the centre of one side of the tank. The area of cross-section of the tank is $N$ times the area of the hole,where $N$ is a large number. Neglect the mass of the tank itself. The initial acceleration of the tank is:

$A$ large tank is filled with water to a height $H$. $A$ small hole is made at the base of the tank. It takes $T_1$ time to decrease the height of water to $H/\eta$ (where $\eta > 1$),and it takes $T_2$ time to drain the rest of the water. If $T_1 = T_2$,then the value of $\eta$ is: