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)
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2/3/1/2=4/3 is the answer
s123450:
is it correct
how did you get the answer? any formula step by step?
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0
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
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