Physics, asked by loswithan, 1 year ago

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)

Answers

Answered by s123450
3

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


s123450: is it correct
loswithan: how did you get the answer? any formula step by step?
Answered by ankurbadani84
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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