Gauge pressure is defined as:

$A$ spherical marble of radius $1 \, cm$ is stuck in a circular hole of radius slightly smaller than its own radius (for calculation purpose,both can be taken same) at the bottom of a bucket filled with water up to a height of $10 \, cm$. If the mass of the marble is $20 \, g$,then the net force on the marble due to water is close to:

$A$ tank $5 \,m$ high is half filled with water and then is filled to the top with oil of density $0.85 \,g/cm^3$. The pressure at the bottom of the tank,due to these liquids is ........ $g/cm^2$.

Two vessels $A$ and $B$ have the same base area and contain water to the same height,but the mass of water in $A$ is four times that in $B$. The ratio of the liquid thrust at the base of $A$ to that at the base of $B$ is (in $:1$)

$A$ cubical block of wood, of length $10 \,cm$, floats at the interface between oil of density $800 \,kg/m^3$ and water. The lower surface of the block is $1.5 \,cm$ below the interface. If the depth of water is $10 \,cm$ below the interface and oil is up to $10 \,cm$ above the interface, then the difference in pressure at the lower and the upper face of the wooden block is:
(Assume density of water, $\rho_w = 1000 \,kg/m^3$ and acceleration due to gravity, $g = 10 \,m/s^2$) (in $\,Pa$)

Consider a vessel filled with a liquid up to height $H$. The bottom of the vessel lies in the $X-Y$ plane passing through the origin. The density of the liquid varies with the $Z$-axis as $\rho(z) = \rho_0 \left[ 2 - \left( \frac{z}{H} \right)^2 \right]$. If $P_1$ and $P_2$ are the pressures at the bottom surface and top surface of the liquid respectively, the magnitude of $(P_1 - P_2)$ is: