Question

At what speed, the velocity head of water is equal to the pressure head of $40\text{cm}$ of mercury?

Answer 1

Step 1: Given parameter

The velocity head of water is equal to the pressure head of $40\text{cm}$ of mercury

Step 2: Formula used

$Velocityhead=\frac{1}{2}\rho v^2$

$p=\rho gh$

Here,

$\rho$ is the density of water.

$h$ is the height.

$v$ is the velocity.

$p$ is the pressure.

$g$ is the gravitational acceleration.

Step 3: Calculation

$h=40\text{cm}$ of mercury (Given) and $g={10}^3\frac{\text{cm}}{{\text{s}}^2}$

Hence $\frac{1}{2}\rho v^2=\rho gh$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 1000\times 40}=\sqrt{80000}$

$Thereforev=282.84cm/s$

Hence, at $v=282.84cm/s$, the velocity head of water is equal to the pressure head of $40\text{cm}$ of mercury.