A hollow cylindrical container floats in water with half its length immersed. A liquid of specific gravity `sigma` is slowly poured into the container until the levels of the liquid inside the container and of water outside are the same. It is observed that the container is submerged to two-thirds its length. Then the value of `sigma` is

Answer

`0.25`

Explanation

Intially:
`mg=rhoA(L)/(2)g" "...(i)`
Finally:
`mg+sigmarhoA(2L)/(3)g=rhoA(2L)/(3)g" "...(ii)`
From (i) and (ii) :
`rhoA(L)/(2)g=rhoA(2L)/(3)g-sigmarhoA(2L)/(3)g" "Rightarrow" "(1)/(2)=(2)/(3)-(2)/(3)sigma" "Rightarrow" "sigma=(1)/(4)=0.25`