Help!!! Please help me solve this in Physics..!!?

Question

A solid of density 5000 kg/ m3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg/ m cube. Calculate the apparent weight of the solid in water. What will be its apparent weight if water is replaced by a liquid of density 8000 Kg/m3?

Guys please help me solve this.....the answers are:
1... 0.4 kgf
2...Zero

Though I know the answer I still need to know how it works....please help me!

Answer 1

Two forces act on the solid immersed in a liquid:

( 1 ) Its weight, mg, downwards and
( 2 ) Force of buoyancy, equal to the weight of liquid displaced, upwards.
The resultant downward force is the apparent weight.

In the first case, as the density of liquid is less by a factor of 5, the weight of liquid displace is also 1/5th of the weight of the body = 0.5 x 1/5 = 0.1.
Hence, apparent weight = its wt. - wt. of liquid displaced
= 0.5 - 0.1 = 0.4 kgf

In the second case, density of liquid is more than the density of solid. So if the solid is completely immersed, the upward force of buoyancy will be more than the weight which will push the body upwards in such a way that it will float and only such a part of the solid will be immersed in liquid that will displace liquid of weight equal to the weight of the solid. Hence, apparent weight = weight of the body - weight of liquid displaced = 0.5 - 0.5 kgf. = 0 kgf.

Answer 2

A volume of water equal to the volume of the solid would weigh 0.1 kgf, and the same volume of the 8000 Kg/m3 liquid would weigh 0.8 kgf.
You specify total immersion, so
1) 0.5 kgf - 0.1 kgf = 0.4 kgf
but
2) 0.5 kgf - 0.8 kgf = - 0.3 kgf, not 0
Only if the solid is permitted to float in the denser liquid would the apparent weight be 0.