Question

By melting two solid spheres of 1cm and 6cm a hollow spheres of 1cm thick is formed. Find the area of outer curve surface of new sphere.​

Answer 1

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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Answer 2

${\bf{Given\::}}$

• Two solid metallic spheres of radii 1 cm and 6 cm, which were melting to formed a hollow sphere of thickness 1 cm.

$\\ {\bf{To\: Find\::}}$

• The area of outer curved surface area of new sphere.

$\\ {\bf{Solution\::}}$

As we know that,

Volume of a metallic sphere is,

$\orange\bigstar\:{\Large\mid}\:\bf\purple{Volume\:=\:\dfrac{4}{3}\:\pi\:r^3\:}\:{\Large\mid}\:\green\bigstar$

Given that,

• Radius of two solid metallic spheres is 1 cm and 6 cm.

Let,

• r = 1 cm

• R = 6 cm

Then,

Sum of volume of these two spheres are,

$\dashrightarrow\:\tt{Volume\:=\:\dfrac{4}{3}\:\pi\:r^3\:+\:4\:\pi\:R^3\:}$

$\dashrightarrow\:\tt{Volume\:=\:\dfrac{4}{3}\:\pi\:(r^3\:+\:R^3)\:}$

$\dashrightarrow\:\tt{Volume\:=\:\dfrac{4}{3}\:\pi\:(1^3\:+\:6^3)\:}$

$\dashrightarrow\:\tt{Volume\:=\:\dfrac{4}{3}\:\pi\:(1\:+\:216)\:}$

$\dashrightarrow\:\bf{Volume\:=\:\dfrac{4}{3}\:\pi\times{(217)}\:}$

Now,

Let,

• Internal radius of the hollow sphere is x cm.

• External radius of the hollow sphere is (x + 1) cm [due 1 cm thick].

Volume of the hollow sphere is,

$\dashrightarrow\:\bf{Volume\:=\:\dfrac{4}{3}\:\pi\:\left((x\:+\:1)^3\:-\:x^3\right)\:}$

According to the question,

• Melting of two metallic spheres are formed a hollow sphere.

That means,

➣ Sum of volume of two metallic spheres is equal to the volume of hollow sphere.

$:\implies\:\tt{\dfrac{4}{3}\:pi\times{(217)}\:=\:\dfrac{4}{3}\:\pi\:\left((x\:+\:1)^3\:-\:x^3\right)\:}$

$:\implies\:\tt{217\:=\:\left(x^3\:+\:1\:+\:3x(x\:+\:1)\right)\:-\:x^3\:}$

$:\implies\:\tt{217\:=\:x^3\:+\:1\:+\:3x(x\:+\:1)\:-\:x^3\:}$

$:\implies\:\tt{217\:=\:1\:+\:3x^2\:+\:3x\:}$

$:\implies\:\tt{3x^2\:+\:3x\:=\:217\:-\:1\:}$

$:\implies\:\tt{3x^2\:+\:3x\:=\:216\:}$

$:\implies\:\tt{3\:(x^2\:+\:x)\:=\:216\:}$

$:\implies\:\tt{x^2\:+\:x\:=\:\dfrac{216}{3}\:}$

$:\implies\:\tt{x^2\:+\:x\:=\:72\:}$

$:\implies\:\tt{x^2\:+\:x\:-\:72\:=\:0\:}$

$:\implies\:\tt{x^2\:+\:9x\:-\:8x\:-\:72\:=\:0\:}$

$:\implies\:\tt{x\:(x\:+\:9)\:-\:8\:(x\:+\:9)\:=\:0\:}$

$:\implies\:\tt{(x\:-\:8)\:(x\:+\:9)\:=\:0\:}$

$:\implies\:\tt{x\:-\:8\:=\:0\:~~~or~~~\:x\:+\:9\:=\:0\:}$

$:\implies\:\tt{x\:=\:8\:{\red{\checkmark}}~~~or~~~\:x\:=\:-9\:(impossible)}$

$:\implies\:\bf{x\:=\:8\:cm\:}$

Hence,

• Internal radius of hollow sphere is 8 cm.

• External radius of hollow sphere is 9 cm.

As we know that,

➣ External curved surface area of a hollow sphere is,

$\orange\bigstar\:{\Large\mid}\:\bf\blue{Volume\:=\:4\pi\:r^2\:}\:{\Large\mid}\:\green\bigstar$

➠ 4 × π × 9²

➠ 4π × 81

➠ 324π

➠ 324 × $\tt{\dfrac{22}{7}}$

➠ 46.285 × 22

➠ $\bf\pink{1018.285 \:cm^2}$

∴ The area of outer curved surface area of the hollow sphere is 1018.285 cm².