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If a solid spherical ball of iron with radius r centimetres

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Joined: 23 Feb 2017
Posts: 58
If a solid spherical ball of iron with radius r centimetres [#permalink]

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05 Jan 2018, 09:38
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45% (medium)

Question Stats:

50% (01:30) correct 50% (00:55) wrong based on 20 sessions

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If a solid spherical ball of iron with radius r centimetres is cut into two equal hemispheres, what is the ratio, in terms of r, of the volume and the surface area of each hemisphere?

A) 2r/9
B) r/3
C) 4r/9
D) 2r/3
E) 4r/3
[Reveal] Spoiler: OA

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Joined: 22 May 2016
Posts: 1260
If a solid spherical ball of iron with radius r centimetres [#permalink]

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05 Jan 2018, 10:35
mandeey wrote:
If a solid spherical ball of iron with radius r centimetres is cut into two equal hemispheres, what is the ratio, in terms of r, of the volume and the surface area of each hemisphere?

A) 2r/9
B) r/3
C) 4r/9
D) 2r/3
E) 4r/3

Careful here - don't forget the surface area of the circular flat base of the hemisphere

Volume of a sphere:
$$\frac{4}{3} \pi r^3$$
Volume of a hemisphere is half of that:
$$\frac{4}{3} \pi r^3 * \frac{1}{2} =\frac{2}{3} \pi r^3$$

Surface area of a sphere:$$4\pi r^2$$

Surface area of hemisphere =
$$\frac{1}{2}$$ the round part of sphere + surface area of circular base

Half of the round part of the sphere's SA = $$2\pi r^2$$
The hemisphere's circular base SA = $$\pi r^2$$
Add them. $$2\pi r^2 + \pi r ^2 = 3\pi r^2$$

Ratio of hemisphere's volume to hemisphere's SA:

$$\frac{\frac{2}{3} \pi r^3}{3\pi r^2} = \frac{2r}{9}$$

Not sure this question is sub-600. SA of a hemisphere with no formula given?
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If a solid spherical ball of iron with radius r centimetres   [#permalink] 05 Jan 2018, 10:35
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