A solid spherical ball remolded into a hollow spherical ball | CE Board Problem in Mathematics, Surveying and Transportation Engineering

Date of Exam

May 2010

Subject

Mathematics, Surveying and Transportation Engineering

Problem
A 523.6 cm³ solid spherical steel ball was melted and remolded into a hollow steel ball so that the hollow diameter is equal to the diameter of the original steel ball. Find the thickness of the hollow steel ball.

A. 1.3 cm	C. 1.2 cm
B. 1.5 cm	D. 1.6 cm

Answer Key

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[ A ]

Solution

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Radius of the original steel ball

$V = \frac{4}{3}\pi r^3$

$523.6 = \frac{4}{3}\pi r^3$

$r = 5 ~ \text{cm}$

The amount of material in the hollow steel ball is equal to the amount of material in the the original solid steel ball.

Let R = outer radius of the hollow steel ball.

$\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi r^3$

$\frac{4}{3}\pi R^3 = \frac{8}{3}\pi r^3$

$R^3 = 2r^3$

$R^3 = 2(5^3)$

$R = 6.3 ~ \text{cm}$

Thickness of the hollow steel ball

$t = R - r = 6.3 - 5$

$t = 1.3 ~ \text{cm}$ ← Answer: [ A ]

Category

Solid Geometry

Solid Mensuration

Sphere

Hollow Sphere

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