A cube is floating on mercury with (1//5)^(th)of its volume submerged. If suficient water Is added to cover the cube, what fraction of its volume will remain immersed in mercury ? Density of mercury = 13600 kg//m^(3) .

A cube is floating on mercury with (1//5)^(th)of its volume submerged.

1.	A cube is floating on mercury with (1//5)^(th)of its volume submerged. If suficient water Is added to cover the cube, what fraction of its volume will remain immersed in mercury ? Density of mercury = 13600 kg//m^(3) .
Answer» SOLUTION :When the cube floats on MERCURY, Volume of cube `xx` density of cube D = Volume of immersed portion of the cube `xx` density of mercury D. ` "" V xx D= (1)/(5)V xx 13600` ` ""D = (13600)/( 5)=2720 kg//m^(3)` When water is poured let the fraction of cube immersed in mercury be x. Then the fraction immersed in water will be `(1-x)` ` "" VD = xVd+(1-x) Vrho ` where` RHO` is the density of water ,`rho = 1000 kg //m^(3)` ` 2720 = 13600 x+( 1-x) 1000` `"" x=0.14 m `