A cube floats both in water and in a liquid of specific gravity $0.8 .$ Therefore,

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The cube has displaced equal volume of water and the liquid while floating.

If some weights are placed on the top surface of the cube to make it just sink, the load in case of water will be 0.8 times of that to be used in case of the liquid.

The cube has displaced equal weight of water and the liquid while floating.

Apparent weight of the cube is the same in water and in the liquid.

answer is A, C.

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Detailed Solution

$W_{a p} = 0 i n b o t h$

$m g = ρ V g$

$I n w a t e r, m g + m^{'} g = ρ V_{T} g$

In the liquid , $m g + m^{''} g = 0.8 ρ V_{T} g = 0.8 (m g + m^{'} g)$

$0.8 m^{'} = 0.2 m + m^{''}$

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