Question

A vertical $U$-tube is rotated about the vertical axis passing through the left limb, with an angular speed $\omega$ such that the liquid columns in the limbs have height $h$ and $3h$ where $h$ is $205mm$, as shown in figure. The value of $\omega$ (in $rad/s$) is (Take $g=10$ $m{s}^{-2}$)

• A. $5.55$
• B. $2.44$
• C. $1.55$
• D. $3.55$

Answer 1

The correct option is B $2.44$


Taking an element $dr$ at distance $r$ from the axis as shown in figure.
$P_2-P_1=Po+6\rho gh-(P_o+\rho gh)$
$=5\rho gh$
The pressure difference across the element,
$dP=\rho {\omega }^{2}rdr$
On integrating (from $0$ to $4h$ for liquid of density $\rho$ and from $4h$ to $7h$ for the liquid of density $2\rho$), we get
${\int }_{P_1}^{P_2}dP={\int }_0^{4h}\rho {\omega }^{2}rdr+{\int }_{4h}^{7h}2\rho {\omega }^{2}rdr$
$5\rho gh=\frac{\rho {\omega }^2}{2}[r^2{]}_0^{4h}+\frac{2\rho {\omega }^2}{2}[r^2{]}_{4h}^{7h}$
⇒$5gh={\omega }^2\left[\right(8h^2)+(49h^2-16h^2\left)\right]$
⇒$5gh=41{\omega }^2{h}^2$
$\omega =\sqrt{\frac{5g}{41h}}=2.44rad/s$