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?

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,$d=0.8m$

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,

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

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.