The diagram shows a venturimeter through which water is flowing. The speed of water at $X$ is $2 \, cm/s$. The speed of water at $Y$ (taking $g = 1000 \, cm/s^2$) is ........ $cm/s$.

$A$ liquid enters at point $A_{1}$ with a speed of $3.5 \ m/s$ and leaves at point $A_{2}$. Find the height attained by the liquid above point $A_{2}$ (in $cm$). (in $.25$)

$A$ 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. . . . .

An $L$-shaped glass tube is just immersed in flowing water such that its opening is pointing against the flowing water. If the speed of the water current is $v$,then

$A$ 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).

Assertion $(A)$: The upper surface of the wing of an aeroplane is made convex and the lower surface is made concave.
Reason $(R)$: The air currents at the top have smaller velocity and thus less pressure at the bottom than at the top.