$A$ submarine is designed to withstand an absolute pressure of $100 \text{ atm}$. How deep can it go below the water surface (in $\text{ m}$)? (Consider the density of water = $1000 \text{ kg/m}^3$, $1 \text{ atm} = 1 \times 10^5 \text{ Pa}$, and gravitational acceleration $g = 10 \text{ m/s}^2$)

$A$ barometer kept in an elevator reads $76 \, cm$ when the elevator is accelerating upwards. The most likely pressure inside the elevator (in $cm$ of $Hg$) is ........

The pressure at a point $10 \ m$ below the surface of water is (take atmospheric pressure $P_0 = 1.01 \times 10^5 \ Nm^{-2}$ and $g = 10 \ ms^{-2}$):

Water stands up to height $h$ behind a dam as shown in the figure. The front view of the dam gate is also shown in the adjoining figure. The density of water is $\rho$ and the acceleration due to gravity is $g$. If the atmospheric pressure force is also considered,the point of application of the total force acting on the dam due to water above $O$ is:

$A$ tank of oil has a height of $4 \,m$ and a density of $850 \,kg \,m^{-3}$. The gauge pressure at the bottom of the tank is (Given: $1 \,atm = 10^5 \,Pa$,Acceleration due to gravity $g = 10 \,m \,s^{-2}$) (in $\,kPa$)

$A$ $U$-shaped tube is partially filled with an incompressible liquid of density $1.2 \ g \ cm^{-3}$. Oil, which does not mix with the liquid, is poured into the left side of the $U$-tube until the liquid rises by $15 \ cm$ on the right side. If the density of the oil is $0.9 \ g \ cm^{-3}$, the oil level will stand higher than the liquid level on the right side of the $U$-tube by: (in $cm$)