A liquid of density $\rho$ is filled in tube having three vertical arms each of length $L$ as shown. Initially height of liquid in each arm in $\frac{L}{2}$2$. If left vertical arm is closed and whole system is rotated with angular velocity $\omega$ about axis through middle arm, height of liquid in left arm becomes $\frac{5L}{8}$. Given
$\left(L=\frac{40}{3} \mathrm{m},\rho ={10}^3 \mathrm{kg}/{\mathrm{m}}^3, g=10 \mathrm{m}/{\mathrm{s}}^2,P_{atm}={10}^5 \mathrm{N}/{\mathrm{m}}^2\right)$ (Assume air is ideal gas and tube is perfectly conducting)

 

Water stands upto height $\text{'}h\text{'}$ behind the dam as shown in the figure. The front view of the dam gate is also shown in the adjoining figure. Density of water is $\text{'}\rho \text{'}$ and acceleration due to gravity is $\text{'}g\text{'}$. If atmospheric pressure force is also considered, the point of application of total force acting on the dam due to water above $\text{'}o\text{'}$ is_____

A cylindrical tube, with its base as shown in the figure., is filled with water. It is moving down with a constant acceleration $a$ along a fixed inclined plane with angle $\theta =45°$. $P_1$ and $P_2$ are pressures at point $1$ and $2$, respectively, located at the base of the tube. Let $\beta =\frac{\left(P_1-P_2\right)}{\left(\rho gd\right)}$, where $\rho$ is density of water, $d$ is the inner diameter of the tube and $g$ is the acceleration due to gravity. Which of the following statement(s) is (are) correct?

The liquid shown in the figure in the two arms is mercury (specific gravity $13.6$) and water. If the difference of heights of the mercury column is $2 \mathrm{cm}$, the height $h$ of the water column is

Two liquids of densities ${\rho }_1$ and ${\rho }_2\left({\rho }_2=2{\rho }_1\right)$ are filled up behind a square wall of side $10 \mathrm{m}$ as shown in figure. Each liquid has a height of $5 \mathrm{m}.$ The ratio of the forces due to these liquids exerted on upper part $MN$ to that at the lower part $NO$ is (Assume that the liquids are not mixing):

An open-ended U-tube of uniform cross-sectional area contains water (density ${10}^3 \mathrm{kg} {\mathrm{m}}^{-3}$ ). Initially the water level stands at $0.29 \mathrm{m}$ from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density $800 \mathrm{kg} {\mathrm{m}}^{-3}$ is added to the left arm until its length is $0.1 \mathrm{m}$, as shown in the schematic figure below. The ratio $\left(\frac{h_1}{h_2}\right)$ of the heights of the liquid in the two arms is

A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is $5 \mathrm{cm}$ and the angular speed of rotation is $\mathrm{\omega } \mathrm{rad} {\mathrm{s}}^{-1}$. The difference in the height, $\mathrm{h} \left(\mathrm{in} \mathrm{cm}\right)$ of liquid at the Centre of vessel and at the sides of the vessel will be :

Water flows out of a big tank along a tube bent at right angels the inside of the tube is of radius $r.$ The length of the horizontal portion is $l.$ the rate of water flow is $Q.$ Find the moment of reaction forces of flowing water, acting on the tube walls relative to point $O.$

A $U$-tube of base length filled with the same volume of two liquids of densities $\rho$ and $2\rho$ is moving with an acceleration $a$ on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height $h$ is given by :

An open cubical tank completely filled with water is kept on a horizontal surface. Its acceleration is then slowly increased to $2 \mathrm{m} {\mathrm{s}}^{-2}$ as shown in the figure. The side of the tank is $1\mathrm{m}$. Find the mass of water that would spill out of the tank.