Question

A closed tube filled with water is rotating about an axis as shown in figure. The pressure difference $P_A-P_C$ is

• A. $-12\times {10}^3{\text{N/m}}^2$
• B. $-24\times {10}^3{\text{N/m}}^2$
• C. $6\times {10}^3{\text{N/m}}^2$
• D. $20\times {10}^3{\text{N/m}}^2$

Answer 1

The correct option is B $-24\times {10}^3{\text{N/m}}^2$

Pressure at any point, $x$ distance away from the axis in a rotating fluid is given by:
$P=P_0+\frac{1}{2}\rho {\omega }^2{x}^2$
where, $P_0\to$ Pressure at the point on the axis at the same level i.e at point $B$ in this case

At point $A$,
$P_A=P_0+\frac{1}{2}\times 1000\times (2{)}^2\times 2^2$
$\because \omega =2\text{rad/s},\rho =1000{\text{kg/m}}^3,x=2\text{m}$
$\Rightarrow P_A=P_0+8\times {10}^3...\left(1\right)$
Similarly at point $C$,
$P_C=P_0+\frac{1}{2}\times 1000\times 2^2\times 4^2$
$\because x=4\text{m}$
$P_C=P_0+32\times {10}^3...\left(2\right)$
On subtracting Eq.$\left(2\right)$ from Eq.$\left(1\right)$
$P_A-P_C=-24\times {10}^3{\text{N/m}}^2$
The pressure difference between point $A$ and $C$ is negative, hence $P_C>P_A$.