The reading of a pressure meter attached to a closed pipe is $4.5 \times 10^4 \ N/m^2$. On opening the valve,water starts flowing and the reading of the pressure meter falls to $2.0 \times 10^4 \ N/m^2$. The velocity of water is found to be $\sqrt{V} \ m/s$. The value of $V$ is . . . . . .

$A$ 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).

Water from a tap emerges vertically downwards with an initial speed of $1.0\,ms^{-1}.$ The cross-sectional area of the tap is $10^{-4}\,m^2.$ Assume that the pressure is constant throughout the stream of water and that flow is streamlined. The cross-sectional area of the stream,$0.15\,m$ below the tap would be: (take $g = 10\,ms^{-2}$ )

$A$ large cylindrical tank of cross-sectional area $1\ m^2$ is filled with water. It has a small hole at a height of $1\ m$ from the bottom. $A$ movable piston of mass $5\ kg$ is fitted on the top of the tank such that it can slide in the tank freely without friction. $A$ load of $45\ kg$ is applied on the top of the water by the piston,as shown in the figure. The value of $v$ when the piston is $7\ m$ above the bottom is $(g = 10\ m/s^2)$ ....... $m/s$

$A$ tank of height $15 \ m$ and cross-section area $10 \ m^2$ is filled with water. There is a small hole of cross-section area $a$ which is much smaller than the container, located at a height of $12 \ m$ from the base of the container. How much force should be applied with a piston at the top level, so that the water coming out of the hole hits the ground at a distance of $16 \ m$ (in $kN$)? (Take, density of water $\rho = 1000 \ kg \ m^{-3}$ and $g = 10 \ m/s^2$)

What is the Magnus effect?