91Ó°ÊÓ

1. Density of water $p=1000kg/m^3$
2. Force$F=9.80N$
3. Cross sectional area,$A=5×{10}^{-4}{m}^2$
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

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

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.