There is a hole in the bottom of a tank containing water. If the total pressure at the bottom is $3 \ atm$,then the velocity of water flowing from the hole is $(1 \ atm = 10^5 \ Pa)$ :-

$A$ 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?

$A$ vessel is filled with water up to a height $h$. When a hole is made at the bottom,all the water drains out in time $t$. If the vessel is filled with water up to a height $4h$,how long will it take for the water to drain out completely when a hole is made at the bottom?

$A$ large tank is filled with water to a height $H$. $A$ small hole is made at the base of the tank. It takes $T_1$ time to decrease the height of water to $H/\eta$ (where $\eta > 1$),and it takes $T_2$ time to drain the rest of the water. If $T_1 = T_2$,then the value of $\eta$ is:

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:

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: