# A ball with a portion of it under water was floating in a lake,when the lake froze.The ball was removed without breaking the ice,leaving a hole 24 cm across at the top and 8 cm deep.the radius of the ball was.(Please give proper explanation.)

1
by Neeraj2308

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