A hollow sphere of inner and outer diameter of 12 cm and 16 cm respectively floats half submerged in water. The specific gravity of the material of the sphere is

Options

A. 8.65
B. 6.85
C. 0.865
D. 0.685

Answer

C

Explanation

For equilibrium, `F_g = F_("buoy") implies mg = F_("buoy")`, So `V rho g = V_w rho_w g`
This implies `[4/3 pi (R^3 - r^3)] rho = (2/3 pi R^3) rho_w`
Solving this, `(rho)/(rho_w) = 1/(2(1 - (r^3)/(R^3))) = 1/(2[1 - (6/8)^3]) = 0.865`