Question

Difference in height between the liquid columns in the vertical arms is - [Take $g=10{\text{m/s}}^2$]

• A. $15\text{cm}$
• B. $25\text{cm}$
• C. $20\text{cm}$
• D. $30\text{cm}$

Answer 1

The correct option is A $15\text{cm}$

Let the difference in height between the liquid columns in the vertical arms be $x$ and atmosphere pressure be $P_0$.
$\because (l_2>l_1)$, the centrifugal force acting on the right limb will be greater, thus a higher liquid rise is obtained in this limb.


In case of rotating fluid, the pressure difference is $\Delta P=\frac{1}{2}\rho {\omega }^2{x}^2$, with pressure increasing opposite to the direction of centripetal acceleration.

For pressure at $A$,
$P_A=P_B+\frac{1}{2}\rho {\omega }^2{l}_1^2$
$\Rightarrow P_A=P_B+\frac{1}{2}\rho \times 2^2\times (0.5{)}^2$
Or, $P_A=P_B+\frac{\rho }{2}...\left(1\right)$
For pressure at $C$,
$P_C=P_B+\frac{1}{2}\rho {\omega }^2{l}_2^2$
$\Rightarrow P_A=P_B+\frac{1}{2}\rho \left(2{)}^2\right(1{)}^2$
$\Rightarrow P_C=P_B+2\rho ...\left(2\right)$
On subtracting Eq.$\left(1\right)$ from $\left(2\right)$,
$P_C-P_A=\frac{3\rho }{2}...\left(3\right)$

For pressure variation due to vertical height for both limbs,
$P_A=P_0+\rho g{h}_1...\left(4\right)$
Similarly
$P_C=P_0+\rho g{h}_2...\left(5\right)$

On subtracting Eq.$\left(4\right)$ from $\left(5\right)$,
$P_C-P_A=\rho g(h_2-h_1)=\rho gx...\left(6\right)$
On comparing Eq.$\left(3\right)$ and $\left(6\right)$, we get
$\frac{3\rho }{2}=\rho gx$
$\Rightarrow x=\frac{3}{2g}\text{m}$
$\therefore x=15\text{cm}$
Difference in the heights of the liquid columns in the limbs is $15\text{cm}$