Two cylindrical vessels of equal crosssectional area 'A' contain water upto heights 'H1' and 'H2'. The vessels are interconnected so that the levels in them become equal. Calculate th work done by the force of gravity during the process. The density of water is P
2007-02-07 21:50:20 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Work is calculated by the product of force and displacement.
In the case of liquids it is calculated by the product of pressure and volume displaced.
In the given problem,
A volume of water = (H2 - H1) A /2 is displaced from one cylinder to the other so that the level becomes equal.
The average pressure acting in this volume of water is
(H2 - H1) ρ g /2, the density of water is ρ.
Work done = (H2 - H1) ρ g /2 * (H2 - H1) A /2
(H2 - H1)^2 * A * ρ *g / 4.
2007-02-07 23:24:56 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
When the two vessels are interconnected, the level of fall or rise respectively in the two cylinders will be (H1-H2)/2, since their diameters are equal. The potential energy in the cylinder of higher level is used to do work against gravity in the cylinder of lower level. This energy is given by formula W= F* D
= A*(H1-H2)/2* P*g * (H1-H2)/2 = A*P*[(H1-H2)/2]^2 *g
2007-02-08 07:09:14 · answer #2 · answered by Paleologus 3 · 0⤊ 0⤋
W = PA(H2 - H1)g/2
2007-02-08 05:56:42 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
THE DIAMETER OR THE AREA OF CROSS-SECTION OF THE PIPE WILL ALSO BE NEEDED.
2007-02-08 05:57:07 · answer #4 · answered by krissh 3 · 0⤊ 0⤋