Question

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 0.980 g/cm3. The area of each base is 4.85 cm2, but in one vessel the liquid height is 0.954 m and in the other it is 1.58 m. Find the work done by the gravitational force in equalizing the levels when the two vessels are connected.

Answer 1

Answer:

Explanation:

Potential energy of vessel with liquid upto height .954 m

= mgh where m is mass , h is height of CG of the liquid and g is acceleration due to gravity

mass = volume x density

= cross sectional area x height x density

potential energy = cross sectional area x height x density x g x height / 2

because CG lies at depth of half  the height .

= 4.85 x 10⁻⁴ x .954 x 980 x .954/2 x 9.8

= 2.12  J .

Potential energy of vessel with liquid upto height 1.58 m

= 4.85 x 10⁻⁴ x 1.58 x 980 x 1.58/2 x 9.8

= 5.8 J  

Total energy = 2.12 + 5.8 = 7.92 J

when the level becomes equal , common height becomes

(1.58 + .954 )/ 2 = 1.267 m

potential energy of liquid in two cylinders = 2 x 4.85 x 10⁻⁴ x 1.267 x 980 x 1.267/2 x 9.8

= 7.477 J

Work done by gravitational force

= 7.92 - 7.477 = .443 J