Question

A $U$-shaped tube contains a liquid of density $\rho$ and it is rotated about the left dotted line as shown in the figure. Find the difference in the levels of the liquid column.

• A. $\frac{{\omega }^2{L}^2}{2\sqrt{2g}}$
• B. $\frac{{\omega }^2{L}^2}{2g}$
• C. $\frac{{2\omega }^2{L}^2}{2g}$
• D. $\frac{{2\sqrt{2}\omega }^2{L}^2}{g}$

Answer 1

The correct option is B $\frac{{\omega }^2{L}^2}{2g}$

Weight of liquid of height $H=\frac{\pi d^2}{4}\times H\rho \times g\to \left(i\right)$

Let us consider a mass $dm$ situated at a distance $x$ from $A$ as shown in the figure. The centripetal force required for the mass to rotate $=\left(dm\right){\omega }^2$

$\therefore$ The total centripetal force required for the mass of length $L$ to rotate

$={\int }_0^l\left(\rho x\frac{\pi d^2}{4}\right)xdx\times \left(x{\omega }^2\right)=\rho \times \frac{\pi d^2}{4}\times {\omega }^2\times \frac{l^2}{2}\to \left(ii\right)$

This centripetal force required is provided by the weight of liquid of height "H'

From $\left(i\right)$ and $\left(ii\right)$$

$\frac{\pi d^2}{4}\times H\times \rho \times g=\rho \times \frac{\pi d^2}{4}\times {\omega }^2\times \frac{l^2}{2}\Rightarrow H=\frac{{\omega }^2{l}^2}{2g}$