$A$ 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

$A$ 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:

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

$A$ tube of length $L = 1 \ m$ is filled completely with an ideal liquid of mass $M_{total} = 2M$,and closed at both ends. The tube is rotated uniformly in a horizontal plane about one of its ends. If the force exerted by the liquid at the other end is $F$,then the angular velocity of the tube is $\sqrt{\frac{F}{\alpha M}}$ in $SI$ units. The value of $\alpha$ is . . . . . . .

$A$ fire hydrant delivers water of density $\rho$ at a volume rate $L$. The water travels vertically upward through the hydrant and then makes a $90^{\circ}$ turn to emerge horizontally at speed $V$. The pipe and nozzle have a uniform cross-section throughout. The force exerted by the water on the corner of the hydrant is

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 $\rho$):