An ideal fluid flows in the pipe as shown in the figure. The pressure in the fluid at the bottom $P_2$ is the same as it is at the top $P_1.$ If the velocity of the top $v_1=2 \mathrm{m} {\mathrm{s}}^{-1}.$ Then the ratio of areas $A_1 : A_2$ is

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 :

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

A body $B$ is capable of remaining stationary inside a liquid at the position shown in Fig. (a). If the whole system is gently placed on smooth inclined plane (Fig. (b)) and is allowed to slide down, then $\left(0<\mathrm{\theta }<{90}^{\mathrm{o}}\right)$
 

Water flows through a horizontal tube as shown in the figure. If the difference of heights of water column in the vertical tubes is $h=0.02 \mathrm{m}.$ And the areas of cross-section at $A$ and $B$ are $4\times {10}^{-4} {\mathrm{m}}^2$ and $2\times {10}^{-4} {\mathrm{m}}^2,$ Respectively, then the rate of flow of water across any section is

A cylindrical vessel contains a liquid of density $\rho$ up to a height $h$. The liquid is closed by a piston of mass $m$ and area of cross-section $A$. There is a small hole at the bottom of the vessel. The speed $v$ with which the liquid comes out of the hole is