A U-tube of narrow, uniform bore has a horizontal part 30 cm long between its two vertical limbs.

Step 1. Along the horizontal part (a). A short piece of mercury of length $dr$ at a distance $r$ from the axis moves in a circle. The pressure difference across it gives its centripetal force: $A\,dp = (\rho A\,dr)\,\omega^2 r$. Adding up from $0$ to $L$:

or, in centimetres of mercury, $\dfrac{\omega^2 L^2}{2g}$.

Step 2. Round the tube. Let the mercury rise by $x$ in the closed limb, so it falls by $x$ in the open limb. Start at the open surface, at the air's pressure $P_0 = 76$. Down the open limb, $+\,(h - x)$. Along the bottom, $+\,\omega^2L^2/2g$. Up the closed limb, $-\,(h + x)$. So the pressure at the closed surface is

This must equal the pressure of the trapped air, which by Boyle's law is $P_0\,\dfrac{l_0}{l_0 - x}$.

Step 3. A rise of $4.0\ \text{cm}$ (b). The air is squeezed from $20\ \text{cm}$ to $16\ \text{cm}$, so its pressure is $76 \times \dfrac{20}{16} = 95.0\ \text{cm}$ of mercury. Then

Step 4. The open limb empties (c). Now $x = h = 10\ \text{cm}$. The air is squeezed to $10\ \text{cm}$, so its pressure is $152\ \text{cm}$ of mercury, and