Question

A piece of cork floats in water such that 2/5th of its volume remains above the surface of water. If the cork is made to float in kerosene of density 0.8 g.cm-3, what fraction of its volume will remain above the surface of kerosene?

Answer 1

Answer:

Let the fraction f of volume V be submerged under water. We have

weight of cork=weight of water displaced

⇒V×0.5×g=f×V×1×g

⇒f=0.5 or 50%

Answered By