Question

A sphere and a right circular cylinder of the same radius have equal volumes. By what percentage does the diameter of the cylinder exceed its height ?

Answer 1

Given volume of the sphere = volume of the cylinder.

              4/3 pi r^3 = pi r^2 h

                   h = 4/3 r.

                  3h = 4r.

Divide both sides by 2.

3/2 h = 4r/2

3/2 h = 2r

3/2 h = D  (We know that Diameter = 2r)

Diameter - Height = 3/2 h - h

                               = h/2.

% difference = h/2/h * 100

                     = 50%.

The diameter of the cylinder exceeds its height by 50%.

Hope this helps! :)

Answer 2

✬ 50 % ✬

Step-by-step explanation:

Given:

• A sphere and a right circular cylinder have same radius and have equal volumes.

To Find:

• By what percentage does the diameter of the Cylinder exceed it's height?

Solution: Let the given sphere and cylinder have same radius r and let the height of the cylinder be h . Then,

• Volume of sphere = Volume of cylinder

$\implies \: \frac{4}{3} \pi \: r ^{3} = \pi \: r ^{2} h \\ \\ \implies \: \frac{4}{3} r ^{3} = r ^{2} h \\ \\ \implies \: \frac{4}{3} (r ^{3} - r ^{2} ) = h \\ \\ \implies \frac{4}{3} r = h \\ \\ \implies \: r = \frac{3h}{4} \\ \\ \implies \: 2r = \frac{3h}{2} \\ \\ \implies \: d \: = \frac{3h}{2}$

$where \: d \: is \: the \: \: diameter \: of \: the \: Cylinder \\ \\ Thus, \: the \: diameter \: of \: the \: Cylinder \: is \frac{3h}{2} \: \\ \\ Excess \: of \: diameter \: of \: cylinder \: over \: its \: height \: = ( \frac{3h }{2} - h) \\ \\ \implies \: ( \frac{3h - 2h}{2} ) \\ \\ \implies \: \frac{h}{2} \\ \\ Excess \: \% \: = ( \frac{h}{2} \times \frac{1}{h} \times 100 \: \%) = 50 \: \%$