An object is fitted in a hole at the base of a container as shown in the figure. The force due to the liquid on the object is (Assume no leakage of water,the volume of the object inside the container is $V$,and the density of the liquid is $\rho$):

$A$ stream of water flowing horizontally with a speed of $15\; m s^{-1}$ gushes out of a tube of cross-sectional area $10^{-2}\; m^2$,and hits a vertical wall nearby. What is the force in $N$ exerted on the wall by the impact of water,assuming it does not rebound?

$A$ jet of liquid of cross-sectional area $a$ strikes a wall making an angle $\theta$ with the wall. The liquid strikes the wall with velocity $v$ and rebounds elastically. If the density of the liquid is $\rho$,the normal force on the wall is:

Mercury is filled in a tube of radius $2 \,cm$ up to a height of $30 \,cm$. The force exerted by mercury on the bottom of the tube is . . . . . . $N$.
(Given: atmospheric pressure $P_0 = 10^5 \,N/m^2$,density of mercury $\rho = 1.36 \times 10^4 \,kg/m^3$,$g = 10 \,m/s^2$,$\pi = \frac{22}{7}$)

$A$ tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. The force exerted by the liquid at the other end is

Water is flowing at a speed of $1.5\, ms^{-1}$ through a horizontal tube of cross-sectional area $10^{-2}\, m^2$ and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm,the minimum force that you must exert should be ......... $N$ (density of water $= 10^3\, kgm^{-3}$)