Figure shows a metal ball suspended by thread of negligible mass from an upright cylinder that floats partially submerged in water. The cylinder has height 6 cm, face area $11 cm^2$ on the top and bottom and density $0.5 g/cm^3, 4 cm$ of cylinder's height is inside the water surface. If density of the metal ball is $8 gm/cm^3$ then its radius is equal to $\left(\dfrac{3}{x} \right)^{1/3}$ cm. calculate x. ($\rho_w = 1 gm/cm^3$)

The mass of the cylinder is its volume times its density, giving 11 * 6 * 0.5 = 33 g. With the metal ball attached, the total floating mass is balanced by the buoyant force from the 4 cm submerged height of the cylinder, so (33 + mass of ball) * g = 11 * 4 * 1 * g. This gives the mass of the ball as 11 g. Using the ball's density of 8 g/cm^3, its volume is 11/8 cm^3, and applying the sphere volume formula 4/3 * pi * r^3 = 11/8 yields r^3 = 33/(8*pi). Equating this to the given form (3/x)^(1/3) implies x is 8.

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