Given :

By melting two solid spheres of 1CM and 6cm a hollow spheres of 1cm thick is formed.

To find :

• Find the area of outer curve surface of new SPHERE.

Solution :

We need to remember some points before SOLVING such problems or QUESTIONS

Recasting, melting, reformed & transformation, if these words are in any questions, then it means we have to find out volume of given dimension.

Volume of sphere → 4/3 πr³

• According to the given condition

★ Thickness of hollow sphere after melting = 1cm

★ Radius of the FIRST sphere (x) = 1cm

★ Radius of the second sphere (y) = 6cm

Consider internal radius be x

★ Internal radius (r) = x

★ External radius (R)

→ internal radius + thickness = x + 1

★ Volume of two sphere = Volume of hollow sphere

→ 4/3 π r³ + 4/3 πr³ = 4/3πR³ - 4/3πr³

→ 4/3πx³ + 4/3πy³ = 4/3π(R³ - r³)

→ 4/3π(x³ + y³) = 4/3π(R³ - r³)

→ 4/3π(x³ + y³) = 4/3π{(x + 1)³ - x³}

• Cancel 4/3π & apply identity
• (a + b)³ = a³ + b³ + 3ab(a + b)

→ (6)³ + 1 = {x³ + 1 + 3*x*1(x + 1)} - x³

→ 216 + 1 = {x³ + 1 + 3x(x + 1)} - x³

→ 216 + 1 = x³ + 1 + 3x² + 3x - x³

→ 216 = 3x² + 3x + 1 - 1 + x³ - x

→ 216 = 3x² + 3x

→ 3x² + 3x - 216 = 0

• Take 3 as a common

→ 3(x² + x - 72) = 0

→ x² + x - 72 = 0

• Splitting middle term

→ x² + 9x - 8x - 72 = 0

→ x(x + 9) - 8(x + 9) = 0

→ (x + 9)(x - 8) = 0

•°• x = -9 or x = 8

★ Length never be in negative

→ Take radius = 8cm = x = Internal radius

→ External radius = x + 1 = 9cm

★ Outer curved surface area of new sphere

→ 4πR²

• Put the value of external radius

→ 4π(9)²

→ 4π × 81

→ 4 × 22/7 × 81

→ 88 × 81/7

→ 1018.28cm²

•°• Outer curved surface area of new hollow sphere is 1018.28 cm²

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