There is a hole of area $A_0$ at the bottom of a cylindrical vessel of cross-sectional area $A$. Water is filled up to a height $h$ and the water flows out in time $t$. If water is filled to a height $4h$,it will flow out in time equal to:

$A$ hole is made at the bottom of a tank filled with water (density $1000 \, kg/m^3$). If the total pressure at the bottom of the tank is $3 \, atm$ $(1 \, atm = 10^5 \, N/m^2)$,then the velocity of efflux is:

The height of the water level in a tank of uniform cross-section is $5 \,m$. The volume of water leaked in $5 \,s$ through a hole of area $2.4 \,mm^2$ made at the bottom of the tank is (Assume the level of the water in the tank remains constant and acceleration due to gravity $= 10 \,ms^{-2}$).

$A$ water tank kept on the ground has an orifice of $2 \,mm$ diameter on the vertical side. What is the minimum height of the water above the orifice for which the output flow of water is found to be turbulent (in $\,cm$)? (Assume, $g=10 \,m/s^2, \rho_{\text{water}}=10^3 \,kg/m^3$, viscosity $=1$ centi-poise)

$A$ 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 having a large diameter is filled with water to a height $H$. $A$ hole of cross-sectional area $5 \,cm^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 \times 10^{-3} \,m^3 \,s^{-1}$, then the value of $H$ is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,cm$)