Question

A body of uniform cross sectional area floatts in liquid of density thrice its value.The portion of exposed height will be

Answer 1

maybe 1/3 times height

Answer 2

Answer: 2/3

Explanation:

Let the density of body be = X

and volume be = v

Density of liquid be = 3X

Volume of body immersed in liquid be = n×v

Therefore,

X×v×g = 3X×n×v×g

n= 1/3

Volume of body immersed in liquid is 1/3.

Therefore, volume of body outside the liquid is 2/3.