ICSE Class 9 Answered
Sorry everybody ! Please answer the 19th one not the 20th one.
Asked by nishabr7 | 11 Sep, 2018, 08:19: PM
Expert Answer
Let us assume an object of mass m floating on a liquid of density ρ and it is partially immersed to an depth h as shown in figure.
The floating object is subjected to a pressure difference between top surface and bottom surface as shown in figure.
Top surface is under atmospheri pressure p0. Pressure p at bottom surface directed upwards is increased by height h of liquid
column and this pressure p = p0+ρgh. Pressure also acting on the side surface of the object, but pressure at any point on the side surface
will cancel due to equal magnitude pressure on a point at opposite side. Hence the net pressure on side is zero.
Upward thrust is the force acting on the object upwards and this force is product of pressure difference (p-p0) and cross section area A.
Up-thrust = (p-p0)×A = (ρgh)×A = V×ρ×g .................(2)
where V is the immersed volume of the object
when the object is weighed when it is partially or wholly immersed, we have to consider this upward thrust which is acting against
the gravitational force mg.
hence weight of the object when it is partially immersed = ( m - V×ρ)×g ......................(3)
when the object is completely immersed, volume of the object and immersed volume are one and the same.
If we write m = V×ρs , where ρs is density of solid object, then eqn.(3) can be written as
weight of the object when it is fully immersed = V×(ρs - ρ)×g .....................(4)
weight loss of the object when it is partially/fully immersed is weight of the displaced liquid volume.
weight loss = V×ρ×g .................(5)
Answered by Thiyagarajan K | 12 Sep, 2018, 12:12: PM
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