Answers

2015-01-23T20:47:58+05:30
Consider a uniform cylinder immersed in a liquid!.


Force on the upper face of the cylinder = hρgA
Force on the lower face of the cylinder = [h + L]ρgA
Difference in force = LρgA

But LA is the volume of liquid displaced by the cylinder, and LrgA is the weight of the liquid displaced by the cylinder.

Therefore there is a net upward force on the cylinder equal to the weight of the fluid displaced by it.

The same result will be obtained for a body of any shape, regular or not by taking into account the vertical and horirontal components of the forces on the object.


If a sphere of radius r made of material of density σ is fully immersed in a liquid of density ρ the apparent weight of the sphere is given by:

Apparent weight = actual weight - upthrust = 4/3 πr3g(ρ - s)

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2015-01-24T21:44:41+05:30
Suspend an  object  in  air  using  a  spring  balance.Let  its  weight  in  air  be  W1.Fill  the  measuring  cylinder  with  water  upto  a  certain  level  and  let  its  volume  be  V1.Immerse  the  object  in  the  measuring  cylinder .Note  its  weight  from  the  spring  balance  and  let   it  be  W2  and  let  the  volume  of the  water  left  in  the  cylinder be V2.Loss  of  the  weight  of  the  object in water =W1  -  W2
Mass=  Volume×Density
W=  V×D
W1-W2=(V1-V2)×D   (Density  of  water =1g/cm³)
=V2-V1×1
=V2-V1
Weight  of  loss  of  an  object=Weight  of  water  displaced
Archimedes'  principle  is  verified.








































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