$A$ horizontal right-angle pipe bend has a cross-sectional area of $10 \ cm^2$ and water flows through it at a speed of $20 \ m/s$. The force on the pipe bend due to the turning of water is ........ $N$.

$A$ 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?

Water flowing through a pipe of area of cross-section $2 \times 10^{-3} \,m^2$ hits a vertical wall horizontally with a velocity of $12 \,m \,s^{-1}$. If the water does not rebound after hitting the wall, then the force acting on the wall due to water is (in $\,N$)

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 $\rho$):

Some liquid is filled in a cylindrical vessel of radius $R$. Let $F_1$ be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side $R$. Let $F_2$ be the force applied by the liquid on the bottom of this new vessel. Then:

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 $\rho = 1.36 \times 10^4 \,kg/m^3$,$g = 10 \,m/s^2$,$\pi = \frac{22}{7}$)