Question

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

• A. $-6.1\times {10}^4{\text{N/m}}^2$
• B. $-5.2\times {10}^4{\text{N/m}}^2$
• C. $-7.2\times {10}^4{\text{N/m}}^2$
• D. $-2.4\times {10}^4{\text{N/m}}^2$

Answer 1

The correct option is C $-7.2\times {10}^4{\text{N/m}}^2$

For a rotating fluid, the pressure difference can be given as
$\Delta P=\frac{1}{2}\rho {\omega }^2{x}^2...\left(1\right)$
where $x$ is the distance of the point from the axis of rotation.
The pressure will increase in a direction away from axis of rotation.
$\Rightarrow P_C>P_B\&P_A>P_B$
At point $A$,
$P_A=P_B+\frac{1}{2}\times 1000\times (3{)}^2\times (3{)}^2$
$\Rightarrow P_A=P_B+40500...\left(1\right)$

At point $C$,
$P_C=P_B+\frac{1}{2}\times 1000\times (3{)}^2\times (5{)}^2$
$\Rightarrow P_C=P_B+112500...\left(2\right)$

On subtracting Eq.$\left(2\right)$ from $\left(1\right)$, we get
$P_A-P_C=-72000{\text{N/m}}^2$
$\therefore P_A-P_C=-7.2\times {10}^4{\text{N/m}}^2$
The pressure difference between points $A$ and $C$ is $-7.2\times {10}^4{\text{N/m}}^2$