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}\)

Answer A

Not sure this question is sub-600. SA of a hemisphere with no formula given?

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"