Question

The tube is open at end $A$. Then, absolute pressure at end $B$ is - (Take atmospheric pressure to be ${10}^5\text{Pa}$)

• A. $2\times {10}^5{\text{N/m}}^2$
• B. $1.5\times {10}^5{\text{N/m}}^2$
• C. $1\times {10}^5{\text{N/m}}^2$
• D. $3\times {10}^5{\text{N/m}}^2$

Answer 1

The correct option is D $3\times {10}^5{\text{N/m}}^2$

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 between two points in a liquid, measured $\perp$ from the axis of rotation.
The pressure will increase in a direction away from axis of rotation.
$\therefore P_B>P_A$
From eq.$\left(1\right)$ we can write,
$P_B-P_A=\frac{1}{2}\times 1000\times {10}^2\times 2^2$
$\Rightarrow P_B=P_A+2\times {10}^5$
Since tube is open at end $A$, i.e $P_A=P_{\text{atm}}={10}^5{\text{N/m}}^2$
$\Rightarrow P_B=3\times {10}^5{\text{N/m}}^2$