volume=.5kg/5000kg/m3= 100ml
mwass water displaced=100g
apparent mass=origianal-displaced=400g
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.
6 answers
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
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
45kg
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
0.5g
0.4kgf