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 $\rho$):

$A$ 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

$A$ large vessel of height $H$ is filled with a liquid of density $\rho$ up to the brim. $A$ small hole of radius $r$ is made at the side vertical face,close to the base. The horizontal force required to stop the gushing of liquid is ...........

$A$ tube of length $50\,cm$ is filled completely with an incompressible liquid of mass $250\,g$ and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega = x \sqrt{F} \text{ rad } s^{-1}$. If $F$ is the force exerted by the liquid at the other end,then the value of $x$ will be:

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

$A$ 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 ........ $N$.