At what speed will the velocity head of a stream of water be equal to $40 \ cm$ of $Hg$ (in $cm/sec$)?

State Bernoulli's principle in words.

At a hydroelectric power plant,the water pressure head is at a height of $300\; m$ and the water flow available is $100\; m^{3} s^{-1}.$ If the turbine generator efficiency is $60\%,$ estimate the electric power available from the plant (in $MW$) $\left(g=9.8\; m s^{-2}\right).$

$A$ closed water tank has a cross-sectional area $A$. It has a small hole at a depth of $h$ from the free surface of the water. The radius of the hole is $r$ such that $r \ll \sqrt{\frac{A}{\pi}}$. If $p_o$ is the pressure inside the tank above the water level and $p_a$ is the atmospheric pressure,the rate of flow of the water coming out of the hole is ($\rho$ is the density of water).

Obtain Bernoulli's equation for a fluid at rest.

$A$ train with cross-sectional area $S_t$ is moving with speed $v_t$ inside a long tunnel of cross-sectional area $S_0$ $(S_0 = 4S_t)$. Assume that almost all the air (density $\rho$) in front of the train flows back between its sides and the walls of the tunnel. Also,the air flow with respect to the train is steady and laminar. Take the ambient pressure and that inside the train to be $p_0$. If the pressure in the region between the sides of the train and the tunnel walls is $p$,then $p_0 - p = \frac{7}{2N} \rho v_t^2$. The value of $N$ is. . . . .