The bottom of a cylindrical vessel has a hole of area $A$. If water is filled up to a height $h$,it flows out in $t$ seconds. If water is filled to a height $4h$,it will flow out in time:

$A$ 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:

Water is filled in a tank up to a height of $20 \ cm$ from the bottom of the tank. Water flows through a hole of area $1 \ mm^2$ at its bottom. The mass of the water coming out from the hole in a time of $0.6 \ s$ is (Density of water $= 1000 \ kg \ m^{-3}$ and acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $g$)

There is a hole in the bottom of a tank containing water. If the total pressure at the bottom is $3 \text{ atm}$ $(1 \text{ atm} = 10^5 \text{ N/m}^2)$,then the velocity of water flowing from the hole is:

Tanks $A$ and $B$ open at the top contain two different liquids up to a certain height. $A$ hole is made in the wall of each tank at a depth $h$ from the surface of the liquid. The area of the hole in $B$ is twice that of the hole in $A$. If the liquid mass flux through each hole is equal,then the ratio of the densities of the liquids,$\rho_A / \rho_B$,is:

$A$ large tank is filled with water (density $= 10^3 \ kg/m^3$). $A$ small hole is made at a depth $10 \ m$ below the water surface. The range of water issuing out of the hole is $R$ on the ground. What extra pressure must be applied on the water surface so that the range becomes $2R$? (Take $1 \ atm = 10^5 \ Pa$ and $g = 10 \ m/s^2$)