A solid of density 'd' has weight 'w'. It is completely immersed in a liquid of density 'd', then apparent weight of the solid is :
Answers
Answer:
apparent weight of the solid = 0
w( 1 - D/d)
Explanation:
A solid of density 'd' has weight 'w'. It is completely immersed in a liquid of density 'd', then apparent weight of the solid is :
The object suffers an apparent weight loss equal to the weight of the fluid displaced.
A solid
Density = d
Weight = w
Density = Mass / Volume
=> Volume = Mass / Density
=> V = w /d
Volume displaced of fluid = V
Density of fluid = d
Weight of Fluid Displaces = Vd = w
apparent weight loss = w
apparent weight of the solid = Weight of solid - apparent weight loss
= w - w
= 0
apparent weight of the solid = 0
if we say density of fluid = D ( capital D but not small D)
then
Volume displaced of fluid = V
Density of fluid = D
Weight of Fluid Displaces = VD = wD/d
apparent weight loss = wD/d
apparent weight of the solid = Weight of solid - apparent weight loss
= w - wD/d
= w( 1 - D/d)
apparent weight of the solid = w( 1 - D/d)