Question

A rectangular block of wood floats in water with two-third of its volume immersed. When placed in another liquid, it floats with half of its volume immersed. Calculate the relative density of the liquid. (5)

Answer 1

2/3/1/2=4/3 is the answer

Answer 2

Answer:

The first liquid’s density is 3/4 of the second.

Explanation:

The calculation of relative density of the liquids is below :

Weight of the block - mg

total volume - V1

Volume of liquid displaced by the floating object- V2

Densities of the two liquids - d1, d2

For floating as per Archimedes law:

mg-V2=0

V2= mg  

Hence V1*d1*2/3 = mg; and V1*d2*1/2 =mg

equating both d1*2/3=d2*1/2

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