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?
Answers
Answered by
1
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.
=====
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.
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.
=====
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.
kvnmurty:
clic k on thanks button above please
Similar questions