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Volumes Practice Test 4

Question : 6 of 16

Marks: +1, -0

2\frac{1}{3}

Solution:

Concept:

When a solid is melted and recast into another shape, its volume remains unchanged.

Explanation:

Radius of the iron ball

= \frac{14}{2} = 7\ \mathrm{cm}

Volume of the ball

= \frac{4}{3}\pi (7)^3

Height of the cylinder

= 2\frac{1}{3} = \frac{7}{3}\ \mathrm{cm}

Let the base radius of the cylinder be

r

Using volume conservation:

\pi r^2 \left(\frac{7}{3}\right) = \frac{4}{3}\pi (7)^3

Cancel

\pi

and multiply by

3

7r^2 = 4(7)^3

r^2 = 4(7)^2 = 196

r = 14\ \mathrm{cm}

Diameter of the cylinder base

= 2r = 2(14) = 28\ \mathrm{cm}

Answer:

\boxed{28\ \mathrm{cm}}

Correct option: D.

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