see the diagram.
let the solid ball have a radius of R cm.
From the diagram: CD = 8 cm and AB = 24 cm given.
Hence, DB = 24/2 = 12 cm.
From the triangle ODB, we get: OD² + DB² = OB² -- Pythagoras theorem.
=> (R-8)² + 12² = R²
=> 16 R = 208 => R = 13 cm
Radius of the ball = 13 cm.
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we can also find the relative density of the ball.
Let the ball have a density of d gm/cm³.
mass of the ball = 4/3 π 13³ * d = 9,202.78 * d gms
Volume of the ball immersed in water is found using integral calculus with limits of y = OD to OC. Let h = OD = 5 cm.
Volume of water displaced = 2 π R³ / 3 - π h [ R² - h² / 3 ]
= 2077.64 cm³
weight of water displaced = volume * density = 2077.64 cm³ * 1 gm/cm³ = 2077.64 gms
weight of the ball = weight of the water displaced , Archimedes principle.
2077.64 = 9, 202.78 * d
d = density of the ball = 0.226 gm/cm³
