A spherical ball of lead 3cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be `3/2c m` and 2cm, find the diameter of the third ball.

Explanation

In this case the volume of the initial spherical ball will be the sum of the volumes of the new three balls.
Volume of a sphere` =4/3pir^3`
` r=3/2=1.5 cm`(As the diameter of the given sphere is 3cm)
r of other two balls,`=.75 \and\ 1cm`
Let the radius of the third new ball be `Rcm.`
Volume of the three new sphere = Volume of old sphere
`4/3pi(.75)^3+4/3pi1^3+4/3piR^3=4/3pi(1.5)^3`
`(.75^3+1^3+R^3=1.5^3`
`.421875+1+R^3=3.375`
`R^3=3.375-1.421875`
`R^3=1.953125`
`R=sqrt3(1.953125)`
`R=1.25`
So, the radius of the third new ball is `1.25cm .`
Therefore, the diameter `=2.5 cm`

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