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05 Jan 2018, 10:38
Kudos
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A
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Difficulty:
45%
(medium)
Question Stats:
54% (01:31) correct
46%
(01:20)
wrong
based on 68
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
A) 2r/9
B) r/3
C) 4r/9
D) 2r/3
E) 4r/3
05 Jan 2018, 11:35
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mandeey
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:
\(\frac{\frac{2}{3} \pi r^3}{3\pi r^2}\\
= \frac{2r}{9}\)
Answer A
Not sure this question is sub-600. SA of a hemisphere with no formula given?
16 Apr 2022, 12:14
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