Question

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of each base is 3.89 cm2, but in one vessel the liquid height is 0.993 m and in the other it is 1.76 m. Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

Answer 1

Explanation:

Work done by gravity is given by the formula,

           W = $\rho A (h_{1} - h)g (h - h_{2})$ ......... (1)

It is known that when levels are same then height of the liquid is as follows.

           h = $\frac{h_{1} + h_{2}}{2}$ ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

       W = $\frac{1}{4} \rho gA(h_{1} - h_{2})^{2})$

           = $\frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})$

           = 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.