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2015-04-25T02:57:57+05:30

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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

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