Question

a solid sphere is melted and recasted into a hollow cylindee of uniform thickness.If external radius of base of cylinder is 4cm,it's height is 24cm and thickness 2cm,find the radius of Sphere.

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Answer 1

Answer:

Radius is 6 cm.

Diagram :

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{2cm}}\put(9,17.5){\sf{24cm}}\end{picture}$

Solution :

Given,

• Radius of solid sphere = $\sf R_1$.

• Thickness of hollow cylinder = 2cm.

• Height of hollow cylinder, h = 24cm.

• External radius of base of cylinder, $\sf R_2 \ = \ 4cm$.

Now,

Inner radius of hollow cylinder = Enternal radius(R) - thickness of cylinder.

• Inner radius, $\sf R_1 \ = \ (4-2)cm.$
• $\sf R_1 \ = \ 2cm.$

Hence,

$\sf Volume \ of \ solid \ sphere, V \ = \ \dfrac {4}{3} \pi r^3$

$\sf Volume \ of \ hollow \ cylinder, V \ = \ \pi h (R_2^2 \ - \ R_1^2)$

Volume of sphere = Volume of hollow cylinder.

$\implies \sf \dfrac {4}{3} \pi r^3 \ = \ \pi h (R_2^2 \ - \ R_1^2)$

$\implies \sf r^3 \ = \ \dfrac {3 \pi h (R_2^2 \ - \ R_1^2)}{4 \pi}$

$\implies \sf r^3 \ = \ \dfrac {(3h (R_2^2 \ - \ R_1^2)}{4}$

$\implies \sf r^3 \ = \ \dfrac {3 \times 24 \times (4^2 - 2^2)}{4} cm^3$

$\implies \sf r^3 \ = \ \dfrac {3 \times 24 \times (16-4)}{4} cm^3$

$\implies \sf r^3 \ = \ \dfrac {2 \times 24 \times 12}{4} cm^3$

$\implies \sf r^3 \ = \ \dfrac {864}{4} cm^3$

$\implies \sf r^3 \ = \ 216 cm^3$

$\implies \sf r^3 \ = \ (6)^3 cm$

$\implies Radius, r \ = \ 6cm$

$\therefore$ Radius of the sphere is 6cm.

Answer 2

$\large\underline{\red{\sf \orange{\bigstar} .}}$Answer$\large\underline{\red{\sf \pink{\bigstar} .}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

•The radius 'r'  of a solid sphere is 6 cm.

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$\large\underline{\pink{\sf \orange{\bigstar} .}}$Diagram$\large\underline{\red{\sf \blue{\bigstar} .}}$

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$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{2cm}}\put(9,17.5){\sf{24cm}}\end{picture}$

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$\large\underline{\red{\sf \green{\bigstar} .}}$Given :-

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• External radius of a hollow cylinder, R = 4 cm

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Height of a cylinder, h = 24 cm ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ 
• Thickness of a cylinder = 2 cm

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Inner radius of a hollow cylinder ,r1 = External radius -  Thickness

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• r1 = 4 - 2 = 2 cm

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Inner radius of a hollow cylinder ,r1 = 2 cm  

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Here, solid sphere of Radius 'r' is melted and recast into a hollow cylinder.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\large\underline{\red{\sf \blue{\bigstar} .}}$Volume of hollow cylinder = volume of sphere  

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• π(R² - r1²) h = 4/3 πr³

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• (R² - r1²) h = 4/3 × r³

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• (4² - 2²)× 24 = 4/3 × r³

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• (16 - 4) × 24 = 4/3 × r³

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• 12 × 24 = 4/3 × r³

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• r³ = (12 × 24 × 3)/4

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• r³ = 3 × 24 × 3

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• r = ³√3 × 2 × 2 × 2× 3 × 3

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• r = ³√(3× 3 × 3 )× (2 × 2 × 2)

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$\large\underline{\red{\sf \pink{\bigstar} .}}$r = 3 × 2 = 6 cm$\large\underline{\red{\sf \purple{\bigstar} .}}$

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$\pink{\sf{\star\;r \;= \;6 cm}}$

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•Hence, the radius of a solid sphere is 6 cm${\boxed{\green{\checkmark{}}}}$

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