First slide

Fluid dynamics

Question

A U-tube filled with liquid is being rotated about an axis with angular speed ω as shown in the figure. The difference in heights of liquid in two arms is

Moderate

A

$\large \frac {\omega^2}{2g}\left ( r_1^2-r_2^2 \right )$
B

$\large \frac {\omega^2}{g}\left ( r_1^2-r_2^2 \right )$
C

$\large \frac {2\omega^2}{g}\left ( r_1^2+r_2^2 \right )$
D

$\large \frac {\omega^2}{2g}\left ( r_1^2+r_2^2 \right )$

Solution

Let us take a small element of length dr at a distance r from axis of rotation.
Mass of the element dm = (Adr.ρ) where A = cross-sectional area of the tube. p = pressure at a radial distance r.
p + dp = pressure at a radial distance (r + dr)
Net radially inward thrust on the element = (p + dp)A - p.A
$\large \therefore$ (p + dp)A - p.A = dm. ω².r
(p + dp) - p = ρω²rdr
$\large \Rightarrow p_1-p_2=\rho\omega^2\int_{-r_2}^{r_1}rdr\Rightarrow p_1-p_2=\frac {\rho \omega^2}{2}(r_1^2-r_2^2)$
$\large \Rightarrow h_1-h_2=\frac {\omega^2}{2g}(r_1^2-r_2^2)$

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