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:

$A$ cylindrical vessel, open at the top, contains $15 \text{ litres}$ of water. Water drains out through a small opening at the bottom. $5 \text{ litres}$ of water comes out in time $t_1$, the next $5 \text{ litres}$ in further time $t_2$, and the last $5 \text{ litres}$ in further time $t_3$. Then,

$A$ tank with a small hole at the bottom has been filled with water and kerosene (specific gravity $0.8$). The height of water is $3\,m$ and that of kerosene $2\,m$. When the hole is opened,the velocity of fluid coming out from it is nearly ........ $ms^{-1}$. (Take $g = 10\,ms^{-2}$ and density of water $= 10^3\,kg\,m^{-3}$)

The height of water in a tank is $H$. The range of the liquid emerging out from a hole in the wall of the tank at a depth $\frac{3H}{4}$ from the upper surface of water will be:

$A$ cylindrical tank is filled with water to a level of $3 \,m$. $A$ hole is opened at a height of $52.5 \,cm$ from the bottom. The ratio of the area of the hole to that of the cross-sectional area of the tank is $0.1$. The square of the speed with which water will be coming out from the orifice is $(g=10 \,ms^{-2})$.

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: