Question

A vertical U-tube is spinned about the vertical axis passing through the left hand limb, with an angular speed $\omega$ such that the liquid columns in the limbs have heights $h$ and $3h$, as shown in the figure. Find the value of $\omega$

• A. $\sqrt{\frac{5g}{41h}}$
• B. $\sqrt{\frac{3g}{41h}}$
• C. $\sqrt{\frac{5g}{14h}}$
• D. $\sqrt{\frac{3g}{14h}}$

Answer 1

The correct option is A $\sqrt{\frac{5g}{41h}}$

Let $m_1$ and $m_2$ be the masses of the two liquids in the horizontal pipe of cross-section area $S$.

$\therefore$ $m_1=\rho \left(S4h\right)=4\rho Sh$

Also $m_2=\left(2\rho \right)\left(S3h\right)=6\rho Sh$

From figure, $r_1=2h$ $r_2=2h+\frac{3h}{2}=\frac{11h}{2}$

Point C be the centre of mass of the liquids in the horizontal pipe.

$\therefore$ Position of centre of mass $x=\frac{m_1{r}_1+m_2{r}_2}{m_1+m_2}$

OR $x=\frac{4\rho Sh\left(2h\right)+6\rho Sh\left(11h/2\right)}{4\rho Sh+6\rho Sh}=\frac{41}{10}h$

Pressure at A $P_A=P_o+\rho gh$

Pressure at B $P_B=P_o+\left(2\rho \right)g\left(3h\right)$

$⟹$ $P_B-P_A=5\rho gh$

Using $(P_B-P_A)S=(m_1+m_2)x{w}^2$

$\therefore$ $5\rho ghS=\left(10\rho Sh\right)\times \frac{41}{10}h{w}^2$ $⟹w^2=\frac{5g}{41h}$

$⟹$ $w=\sqrt{\frac{5g}{41h}}$