an object whose 2/3 part inside water dipped,while remaining 1/3 outside the water ,what will be the desity of this object,plz explain mathematical method to use this process???
2006-09-12 22:50:46 · 4 answers · asked by science125 1 in Science & Mathematics ➔ Physics
If I read your question right, the density of the object is 2/3 that of water, or 0.67 gm/cm3. That's because it is floating in equilibrium, with it's entire weight displacing 2/3 of its volume in water.
Look at it this way. The density difference between the object and water, working over 2/3 the object volume, lifts the other 1/3 of the object in the air.
1/3 rho = 2/3 (1.0 gm/cm3 - rho)
for which rho = 0.67 gm/cm3
2006-09-12 23:17:57 · answer #1 · answered by SAN 5 · 0⤊ 0⤋
The density of the object remains unaltered, if there is no chemical or osmotic process between it and the water.
Simply dipping an object into water does not change its density. You're probably looking for some sort of buoyancy equation.
2006-09-12 23:02:27 · answer #2 · answered by Jeff L 3 · 0⤊ 0⤋
If that particle attracts water then the volume of that object will expand so the density will also expand with its 2/3 portion.
2006-09-12 22:58:44 · answer #3 · answered by Arka 1 · 0⤊ 0⤋
SAN is correct.
2006-09-13 06:34:24 · answer #4 · answered by dwarf 3 · 0⤊ 0⤋