Who devised for first time, a method for measuring atmospheric pressure ?

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 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 $3.$

A $U-$tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with $10.0\; cm$ of water in one arm and $12.5\; cm $ of spirit in the other. if $15.0\; cm$ of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels  (in $cm$) of mercury in the two arms?

(the specific gravity of spirit is $0.8.)$

Consider the wall of a dam to be straight with height $H$ and length $L$. It holds a lake of water of height $h (h < H)$ on one side. Let the density of water be $\rho_{ w }$. Denote the torque about the axis along the bottom length of the wall by $\tau_1$. Denote also a similar torque due to the water up to height $h / 2$ and wall length $L / 2$ by $\tau_2$. Then $\tau_1 / \tau_2$ (ignore atmospheric pressure) is

An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius $0.1 \mathrm{~mm}$ is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume $V_0$ to $\frac{100}{101} V_0$ with the movable piston. Considering air as an ideal gas, the height $(h)$ of the liquid column in the capillary above the liquid level in $\mathrm{cm}$ is. . . . . . .

[Given: Surface tension of the liquid is $0.075 \mathrm{Nm}^{-1}$, atmospheric pressure is $10^5 \mathrm{~N} \mathrm{~m}^{-2}$, acceleration due to gravity $(g)$ is $10 \mathrm{~m} \mathrm{~s}^{-2}$, density of the liquid is $10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and contact angle of capillary surface with the liquid is zero]