The cubical container $\mathrm{ABCDEFGH}$ which is completely filled with an ideal (nonviscous and incompressible) fluid, moves in a gravity free space with $a$ acceleration of $a=a_0\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)$, where $a_0$ is a positive constant. Then the pressure will be minimum at point 

A cylindrical tank of height $0.4 \mathrm{m}$ is open at the top and has a diameter $0.16 \mathrm{m}$. Water is filled in it up to a height of $0.16 \mathrm{m}$. How long it will take to empty the tank through a hole of radius $5\times {10}^{-3} \mathrm{m}$ in its bottom.

A narrow tube completely filled with a liquid is lying on a series of cylinders as shown in figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by,

A $\mathrm{U}$-tube of base length $I$ filled with 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

A given shaped glass tube having uniform cross-section is filled with water and is mounted on a rotatable shaft as shown in figure. If the tube is rotated with a constant angular velocity $\omega$ then

A candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D(D>>d)$ as shown in figure. If it is burning at the rate of $2 \mathrm{cm} {\mathrm{s}}^{-1}$ hour then the top of the candle will

There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is $h$. As the liquid comes out of the two holes, the tank will experience a net horizontal force proportional to

Assertion: Any pressure increase at one point of a static connected fluid passed to each point is undiminished.

Reason: Fluid is assumed to be incompressible.