91Ó°ÊÓ

1. Density of liquid in the cylinder,$p=1.30×{10}^3\frac{kg}{m^2}$
2. Area of each base,$A=4.00c{m}^{2}or4.00×{10}^{-4}{m}^2$
3. Height of liquid in one vessel, $h_1=0.854m$and in the other vessel$h_2=1.560m$

Using the formula of pressure, we can find the force. Using this force, we can find the work done by the gravitational force in equalizing the levels when the two vessels are connected. Work done by a body is the product of the force applied to it in the direction of its displacement and the displacement of the body.

Formulae:

Pressure applied by force F on area A,$p=\frac{F}{A}$ (i)

Work done by body in terms of force, W = F.d (ii)

Pressure applied on a fluid surface, p = pgh (iii)

Difference in the height of vessels,

$$\begin{gathered} h=h_2-h_1 \\ =1.560m-0.854m \\ =0.706m \end{gathered}$$

Therefore, half of this height is given by: (0.706 /2) m = 0.353 m

Pressure of the liquid atusing equation (iii) and given values:

$$\begin{gathered} p=\left(1.30×{10}^3\right)\left(9.8\frac{m}{s^2}\right)(0.353m) \\ =4.50×{10}^{3}Pa \end{gathered}$$

As, using equation $p=4.50×{10}^{3}Pa\mathrm{and}A=4.00×{10}^{-4}{m}^2$and (i), we get

$$\begin{gathered} F=(4.50×{10}^{3}Pa)×(4.00×{10}^{-4}{m}^2) \\ =1.80N \end{gathered}$$

Now using equation (ii), we get the work done by the gravitational force:

$$\begin{gathered} W=(1.80N).(0.353m) \\ =0.635J \end{gathered}$$

Therefore, the work done by the gravitational force in equalizing the levels when the two vessels are connected is $0.635J$.