Question

A spherical clay model of diameter 12cm is moulded into a frustum by a professional engineer with diameters 15cm and 10cm. What will be the slant height of the frustum obtained (in cm)?

• A. 18.187

Answer 1

The correct option is A 18.187

Given that the diameter of the sphere is 12 cm.
$⟹Radiusofthesphere=\frac{12}{2}=6cm$

Volume of a sphere of radius $r$ = $\frac{4}{3}\pi r^3$
$\therefore$ Volume of the original sphere
= $\frac{4}{3}\pi r^3$
$=\frac{4}{3}\pi (6{)}^3$
$=288\pi c{m}^3$

Now, Volume of a frustum of lower base radius $r$, upper base radius $R$ and height $h$ is given by
$\frac{1}{3}\pi (R^2+r^2+Rr)h$.
Thus, the volume of the frustum obtained from the sphere
= $\frac{1}{3}\pi (R^2+r^2+R\times r)h$

$=\frac{1}{3}\pi \times \left(\right(\frac{15}{2}{)}^2+5^2+\left(\frac{15}{2}\right)\times 5)hc{m}^3$

Note that the Volume of sphere = the volume of frustum

Thus,
$288\pi =\frac{1}{3}\pi \times \left(\right(\frac{15}{2}{)}^2+5^2+\left(\frac{15}{2}\right)\times 5)hc{m}^3$

$\Rightarrow h=7.275cm$

Also, Slant height $l=\sqrt{(R-r{)}^2+(h{)}^2}$

$l=\sqrt{(7.5-5{)}^2+(7.275{)}^2}$

$l=18.187cm$