Explain why you cannot make a decent cup of tea on a mountain.

$(a)$ Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$,what is the change in pressure $dp$ over differential height $dh$?
$(b)$ Considering the pressure $P$ to be proportional to the density,find the pressure $P$ at a height $h$ if the pressure on the surface of the earth is $P_{0}$.
$(c)$ If $P_{0} = 1.03 \times 10^5 \text{ N/m}^2$,$\rho_0 = 1.29 \text{ kg/m}^3$,and $g = 9.8 \text{ m/s}^2$,at what height will the pressure drop to $\frac{1}{10}$ of the value at the surface of the earth?
$(d)$ This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

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

Write the formula of pressure at depth $h$ in a liquid.

The heights of mercury in a barometer at the base and at the top of a mountain are $75 \, cm$ and $50 \, cm$,respectively. If the ratio of the density of mercury to the density of air is $10^4$,what is the height of the mountain?

In a cylindrical container open to the atmosphere from the top,a liquid is filled up to a $10\, m$ depth. The density of the liquid varies with depth $h$ from the surface as $\rho(h) = 100 + 6h^2$,where $h$ is in meters and $\rho$ is in $kg/m^3$. The pressure at the bottom of the container will be: (Atmospheric pressure $P_0 = 10^5\, Pa$,$g = 10\, m/s^2$)