Water flows into a cylindrical vessel of large cross-sectional area at a rate of $10^{-4} \; m^3/s$. It flows out from a hole of area $10^{-4} \; m^2$, which has been punched through the base. How high does the water rise in the vessel (in $; m$)?

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:

$A$ tank is filled with water and two holes $A$ and $B$ are made in it. For getting the same range,the ratio of $h^{\prime}$ to $h$ is ..........

$A$ cylindrical vessel is filled with a liquid up to height $H$. $A$ small hole is made in the vessel at a distance $y$ below the liquid surface as shown in the figure. The liquid emerging from the hole strikes the ground at a distance $x$.

$A$ 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

Equal volumes of two immiscible liquids of densities $\rho$ and $2\rho$ are filled in a vessel as shown in the figure. Two small holes are punched at depths $h/2$ and $3h/2$ from the surface of the lighter liquid. If $v_1$ and $v_2$ are the velocities of efflux at these two holes,then $v_1/v_2$ is: