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Question
A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be `3/2`cm and 2 cm, find the diameter of the third ball.
Solution
Volume of lead ball = `4/3 πr^3`
=`4/3× 22/7×(3/2)^3`
∴ According to question,
Volume of lead ball = `4/3×π(3/4)^3+4/3π(2/2)^3+4/3π(d/2)^3`
⇒`4/3π(3/2)^3=4/3π(3/4)^3+4/3[π(2/2)^3+(d/2)^3]`
⇒`4/3π[(3/2)^3]=4/3π[(3/4)^3+(2/2)^3+(d/2)^3]`
⇒`27/8=27/64+8/8+(d^3)/8`
⇒ `[27/8-27/64-1]8=d^3`
⇒` d^3/8=125/64`
⇒`d/2=5/4`
⇒ d=`10/4`
⇒ d=2.5cm
Therefore, the diameter of the third ball is 2.5cm .
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