Question

. A solid of density 5000kg/m3weights 0.5 kg in air. It is completely immersed in water of density1000kg/m3

(a) Calculate the apparent weight of solid in water.(ans. 0.4 kg)

(b) What will be its apparent weight if water is replaced by a liquid of density 8000kg/m3

Answer 1

(a)

we know that

apparent weight = true weight (acting downwards) - buoyant force (acting upwards)

here

true weight, W = m.g = 0.5kg x 9.81 m/s²

or

W = 4.905 kg

and

buoyant force, F = ρgV

ρ = density of liquid = 1000 kg/m³

V is the volume of immersed object or displaced liquid = mass / density = 0.5kg / 5000 kg/m³

so, V = 0.0001 m³

thus,

F = 1000 x 9.81 x 0.0001

or

F = 0.981 N

thus, apparent weight will be

W' = W - F = 4.905 - 0.981

or

W' = 3.924 N

(b)

if water is replaced with a liquid of density 8000 kg/m³, the buoyant force will be

F = ρgV = 8000 x 9.81 x 0.0001

or

F = 7.848 N

thus, apparent weight will be

W' = W - F = 4.905 - 7.848

or

W' = -2.943 N

thus, the solid will float in liquid.