Posted by Mayank on Thursday, September 20, 2012 at 1:46pm.
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.

Physics  bobpursley, Thursday, September 20, 2012 at 2:36pm
volume=.5kg/5000kg/m3= 100ml
mwass water displaced=100g
apparent mass=origianaldisplaced=400g

Physics  Sarthak, Sunday, December 18, 2016 at 2:33am
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 = 104 m3
So, Buoyant force is, B = Vqg = 104 × 1000 × g = 0.1 g N = 0.1 kgf
Thus, apparent weight = W – B = 0.5 – 0.1 = 0.4 kgf
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