Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $d.$ The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x_{1}$ and in the other, $x_{2}$. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is

A liquid is kept in a cylindrical vessel. When the vessel is rotated about its axis, the liquid rises at its sides. If the radius of the vessel is $0.05\,\, m$ and the speed of rotation is $2$ revolutions per second, the difference in the heights of the liquid at the centre and at the sides of the vessels will be ...... $cm.$ $($ take $g = 10\,\, ms^{-2}$ and $\pi^2 = 10)$

A fixed thermally conducting cylinder has a radius $\mathrm{R}$ and height $\mathrm{L}_0$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass $M$ is held at a distance $L$ from the top surface, as shown in the figure. The atmospheric pressure is $\mathrm{P}_0$.

$1.$  The piston is now pulled out slowly and held at a distance $2 \mathrm{~L}$ from the top. The pressure in the cylinder between its top and the piston will then be

$(A)$ $\mathrm{P}_0$ $(B)$ $\frac{\mathrm{P}_0}{2}$  $(C)$ $\frac{P_0}{2}+\frac{M g}{\pi R^2}$  $(D)$ $\frac{\mathrm{P}_0}{2}-\frac{\mathrm{Mg}}{\pi \mathrm{R}^2}$

$2.$  While the piston is at a distance $2 \mathrm{~L}$ from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

$(A)$ $\left(\frac{2 \mathrm{P}_0 \pi \mathrm{R}^2}{\pi \mathrm{R}^2 \mathrm{P}_0+\mathrm{Mg}}\right)(2 \mathrm{~L})$  $(B)$ $\left(\frac{\mathrm{P}_0 \pi R^2-\mathrm{Mg}}{\pi R^2 \mathrm{P}_0}\right)(2 \mathrm{~L})$ 

$(C)$ $\left(\frac{\mathrm{P}_0 \pi \mathrm{R}^2+\mathrm{Mg}}{\pi \mathrm{R}^2 \mathrm{P}_0}\right)(2 \mathrm{~L})$  $(D)$ $\left(\frac{\mathrm{P}_0 \pi \mathrm{R}^2}{\pi \mathrm{R}^2 \mathrm{P}_0-\mathrm{Mg}}\right)(2 \mathrm{~L})$

$3.$  The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium, the height $\mathrm{H}$ of the water column in the cylinder satisfies

$(A)$ $\rho g\left(\mathrm{~L}_0-\mathrm{H}\right)^2+\mathrm{P}_0\left(\mathrm{~L}_0-\mathrm{H}\right)+\mathrm{L}_0 \mathrm{P}_0=0$

$(B)$ $\rho \mathrm{g}\left(\mathrm{L}_0-\mathrm{H}\right)^2-\mathrm{P}_0\left(\mathrm{~L}_0-\mathrm{H}\right)-\mathrm{L}_0 \mathrm{P}_0=0$

$(C)$ $\rho g\left(\mathrm{~L}_0-\mathrm{H}\right)^2+\mathrm{P}_0\left(\mathrm{~L}_0-\mathrm{H}\right)-\mathrm{L}_0 \mathrm{P}_0=0$

$(D)$ $\rho \mathrm{g}\left(\mathrm{L}_0-\mathrm{H}\right)^2-\mathrm{P}_0\left(\mathrm{~L}_0-\mathrm{H}\right)+\mathrm{L}_0 \mathrm{P}_0=0$

Give the answer question $1,2$ and $3.$

Figure shows two containers $P$ and $Q$ with same base area $A$ and each filled upto same height with same liquid. Select the correct alternative .............

A vertical U-tube of uniform cross-section contains water in both the arms. A $10 \,cm$ glycerine column ($R.D$. $=1.2$ ) is added to one of the limbs. The level difference between the two free surfaces in the two limbs will be ...... $cm$

The pressure of confined air is $p$. If the atmospheric pressure is $P$, then