Question
A spherical ball of lead . 3 cm in diameter is melted and recast into three spherical balls . the diameter of two these are 1.5 cm and 2 cm respectively . The diameter of the third ball is :
Answer: Option A
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Let he radius of the third ball be R cm . Then ,
`4/3 pi xx (3/4)^3 + 4/3 pi xx (1)^3 + 4/3 pi xx R^3 = 4/3 pi xx (3/2)^3`
`rArr 4/3pi(27/64 + 1 + R^3) = 4/3 pi xx 27/8`
`rArr 27/64 + 1 + R^3 = 27/8`
`rArr R^3 = 125/64`
`rArr R^3 = (5/4)^3`
`rArr R = 5/4.`
`:.` Diameter of the third ball = 2R = `5/2` cm = 2.5 cm.
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