$A$ cylindrical tank has a hole of $1 \ cm^2$ in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of $70 \ cm^3/s$,then the maximum height up to which water can rise in the tank is: $(g = 9.8 \ m/s^2)$ (in $cm$)

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:

$A$ cylinder containing water up to a height of $25 \ cm$ has a hole of cross-section $\frac{1}{4} \ cm^2$ in its bottom. It is counterpoised in a balance. What is the initial change in the balancing weight when water begins to flow out?

$A$ large open tank has two holes in the wall. One is a square hole of side $L$ at a depth $y$ from the top and the other is a circular hole of radius $R$ at a depth $4y$ from the top. When the tank is completely filled with water, the quantities of water flowing out per second from the two holes are the same. Then the value of $R$ is

$A$ container of large uniform cross-sectional area $A$ is resting on a horizontal surface and holds two immiscible,non-viscous,and incompressible liquids of densities $d$ and $2d$,each of height $\frac{H}{2}$,as shown. The upper surface of the liquid with lower density is open to the atmosphere. $A$ small hole is made on the wall of the container at height $h$ $(h < \frac{H}{2})$. The initial speed of efflux of the liquid at the hole is:

$A$ vessel of area of cross-section $A$ has liquid to a height $H$. There is a hole at the bottom of the vessel having area of cross-section $a$. The time taken to decrease the level from $H_1$ to $H_2$ will be: