Question

The plastic tube in Figure has a cross-sectional area of$\mathbf{5}\mathbf{.}\mathbf{00}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. The tube is filled with water until the short arm (of length$\mathit{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathit{m}$) is full. Then the short arm is sealed and more water is gradually poured into the long arm. If the seal will pop off when the force on it exceeds$\mathbf{9}\mathbf{.}\mathbf{80}\mathit{N}$, what total height of water in the long arm will put the seal on the verge of popping?

Short answer

The total height of water in long arm is2.8$m$

Step-by-step solution

The given data

1. Density of water $p=1000kg/m^3$
2. Force$F=9.80N$
3. Cross sectional area,$A=5\times {10}^{-4}{m}^2$
4. Length of the short arm,$d=0.8m$

Understanding the concept of hydrostatic pressure

To solve for pressure, we can use the formula for hydrostatic pressure difference, which depends on density and height. Hydrostatic pressure is that, which is exerted by a fluid at any point of time at equilibrium due to the force applied by gravity.

Formula:

Pressure applied on a body, $p=pgh$

Pressure applied by force F on area A,

$p=\frac{F}{A}$

Calculation of the total height of water

Using equations (i) & (ii), we get

$$\begin{gathered} pgh=\frac{F}{A} \\ h=\frac{F}{pgA} \\ =\frac{9.8}{1000\times 9.8\times 5\times {10}^{-4}} \\ =2m \end{gathered}$$

So, the total height of water in long arm can be given as:

lenght of water(h) + lenght of short arm (d)=2+0.8

=$2.8m$

So, the total height of height is 2.8 m.