A solid of density 5000 kg/m^3 weighs 0.5 kgf in air.It is completely immersed in water of density 1000 kg/m^3. Calculate the apparent weight of the solid in water.

Answer 1

volume=.5kg/5000kg/m3= 100ml
mwass water displaced=100g

apparent mass=origianal-displaced=400g

Answer 2

We have,

Density of solid, d = 5000 kg/m3

Weight in air, W = 0.5 kgf = 0.5g N

Density of water, q = 1000 kg/m3

Let, V be the volume of the solid.

So, W = Vdg

=> 0.5g = V × 5000 × g

=> V = 10-4 m-3

So, Buoyant force is, B = Vqg = 10-4 × 1000 × g = 0.1 g N = 0.1 kgf

Thus, apparent weight = W – B = 0.5 – 0.1 = 0.4 kgf

Answer 3

45kg

Answer 4

A solid of density of 5000kg m^3 weighs 0.5 kg f in air is completely immersed in the liquid of density 8000 kg m^3 calculate the apparent weight of solid in liquid

Answer 5

0.5g

Answer 6

0.4kgf