# A solid of density 500kg/m3 weighs 0.5kg in air. It is completely immersed in water of density 1000kg/m3 . Calculate the apparent weight of this object in water. How will its apparent weight change if water is replaced by a liquid of density 8000kg/m3?

1
by rabindraKunda805

2015-08-21T03:11:05+05:30

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Mass m = 0.50 kg  in air
Density d = 500 kg/m^3        volume  = V = 10^-3 m^3

Water density d_w = 1000 kg/m^3
Apparent weight  in water =     m g -  (m/d) d_w g      as the solid is completely immersed.

W_app = m g [1 - d_w / d] = 0.5 * 10 * (-1) = - 5 Newtons

Apparent weight is negative.
There is actually  an external force holding object down against the buoyancy force in water.
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water replaced by  a liquid of density  8000 kg/m^3  and the solid is completely immersed by some external force.
W_app = m g [ 1 - d_l / d]  = 0.5 * 10 * (-15) = - 75 Newtons

Apparent weight is negative.  so external force makes the object immersed and be there.

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