A U tube is rotated about one of it's limbs with an angular velocity $\omega$. Find the difference in height $H$ of the liquid (density $\rho$) level, where diameter of the tube $d << L$.

Step 1: Set up the pressure gradient equation.

In the rotating frame of reference, the centrifugal force creates a radial pressure gradient along the horizontal section of the tube.

$$\frac{dP}{dr} = \rho \omega^2 r$$

Step 2: Integrate to find the pressure difference between the limbs.

Integrate the pressure gradient from the axis of rotation at $r = 0$ to the outer limb at $r = L$.

$$\Delta P = \int_0^{L} \rho \omega^2 r \, dr$$ $$\Delta P = \frac{1}{2} \rho \omega^2 L^2$$

Step 3: Relate the pressure difference to the height difference.

The pressure difference $\Delta P$ is balanced by the hydrostatic pressure of the liquid column of height $H$.

$$\Delta P = \rho g H$$

Step 4: Solve for the height difference $H$.

Equate the two expressions for the pressure difference.

$$\rho g H = \frac{1}{2} \rho \omega^2 L^2$$ $$H = \frac{\omega^2 L^2}{2g}$$

Answer: The difference in height is $H = \dfrac{\omega^2 L^2}{2g}$.