Question

Find the difference in the height of the liquid between the centre and the edge $\left(\text{in cm}\right)$, if the fluid is rotated in a container as shown in figure.
[Take ${\pi }^2=10,g=10{\text{m/s}}^2$]

• A. $0.1$
• B. $1.2$
• C. $0.4$
• D. $2.0$

Answer 1

The correct option is D $2.0$

Let the difference in the height between the liquid level at the centre and the edge be $h$.


For rotating fluid, difference in pressure $P_B-P_O$ provides centripetal force.
i.e $\Delta P=P_B-P_O=\frac{1}{2}\rho {\omega }^2{x}^2...\left(1\right)$
Also, in vertical direction, we can write:
$\Delta P=P_B-P_A=\rho gh$
[since $P_A=P_O=$ Atmospheric Pressure]
$\Rightarrow P_B-P_O=\rho gh...\left(2\right)$
From Eq. $\left(1\right)$ and $\left(2\right)$,
$\Rightarrow \rho gh=\frac{1}{2}\rho {\omega }^2{x}^2$
$\Rightarrow h=\frac{{\omega }^2{x}^2}{2g}$
$\because x=0.05\text{m}$
$\Rightarrow h=\frac{(4\pi {)}^2\times (0.05{)}^2}{2\times 10}$
$\therefore h=0.02\text{m}=2.0\text{cm}$
The difference in the height of liquid between the centre and the edge is $2.0\text{cm}$