A solid sphere of radius ‘R’ floats in a liquid of density '$\sigma$'  such that its diameter x-x is below distance ‘h’ from free surface as shown. The density of sphere is  $\rho$. If 2h=R then the value of  $\frac{\sigma }{\rho }$

Since plane angle between h and R is ${60}^0$ solid angle is $\pi$ steradians. 

So the volume outside is $\frac{\pi R^3}{3}-\frac{1}{3}\pi \times \frac{3R^2}{4}\times \frac{R}{2}$ and volume inside is total volume minus outside volume which comes out to be $\frac{27}{24}\pi R^3$ using Buoyant force equal to its weight we get   $\frac{\sigma }{\rho }=\frac{32}{27}$