Question

a block of wood floats in water such that half of its volume is below the water surface but in a certain liquid it Floats with One by fourth of its volume below the liquid surface find the density of the liquid?

Answer 1

Answer:

The relative density of the liquids is calculated as follows:

Let mg be the weight of the block and V its total volume V’ volume of liquid displaced by the floating object). Let d1 and d2 be the densities of the two liquids.

For floating we have to satisfy Archimedes law: mg-V’=0===>V’= mg ( where V’= V* the ratio of the immersed volume)

Hence V*d1*2/3 = mg; and V*d2*1/2 =mg

We can equate both d1*2/3=d2*1/2

and after simplifying it we get d1/d2= 1/2*3/2=3/4

The first liquid’s density is three quarters of the second

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