#### 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 `a/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

Is there an error in this question or solution?

Solution for 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 `A/2`Cm and 2 Cm, Find the Diameter of the Third Ball. concept: Volume of a Sphere. For the course CBSE