# Maths!!!! A spherical ball of lead 3cm in radius is melted and recasted in 3 spherical balls if the radius of the two balls are 1.5cm and 2cm respectively . Find the radius of third ball

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Radius of ball which is melted = 3cm

Radius of first formed ball = 1.5 cm

Radius of second formed ball = 2 cm

Let Radius of third formed ball = r cm

Volume of large ball = sum of volume of 3 formed balls

⇒4π/3 × 3³ = 4π/3(1.5³ + 2³ + r³)

⇒ 3³ =1.5³ + 2³ + r³

⇒ 27 = 3.375 + 8 + r³

⇒ r³ = 27 - 8 - 3.375 = 15.625

⇒ r = ∛15.625

⇒ r = 2.5cm

∴Radius of third ball is 2.5cm

Hope this helps!

Radius of first formed ball = 1.5 cm

Radius of second formed ball = 2 cm

Let Radius of third formed ball = r cm

Volume of large ball = sum of volume of 3 formed balls

⇒4π/3 × 3³ = 4π/3(1.5³ + 2³ + r³)

⇒ 3³ =1.5³ + 2³ + r³

⇒ 27 = 3.375 + 8 + r³

⇒ r³ = 27 - 8 - 3.375 = 15.625

⇒ r = ∛15.625

⇒ r = 2.5cm

∴Radius of third ball is 2.5cm

Hope this helps!

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Conservation of mass principle, as the density is constant:

volume of 3 cm ball = sum of volumes of small ones.

4π/3 * 3³ = 4π/3 (1.5³ + 2³ + R³)

Solving that R = 2.5 cm

volume of 3 cm ball = sum of volumes of small ones.

4π/3 * 3³ = 4π/3 (1.5³ + 2³ + R³)

Solving that R = 2.5 cm