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?

• A. 5.212 cm
• B. 6.524 cm
• C. 8.4 cm
• D. 18.187 cm

Answer 1

The correct option is D

18.187 cm

Volume of a sphere of radius $r$ = $\frac{4}{3}\pi r^3$
Given that the diameter of the sphere is 12 cm.
$⟹Radiusofthesphere=\frac{12}{2}=6cm$
$\therefore$ 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$l$
$=\sqrt{(R-r{)}^2+(h{)}^2}$

$l=\sqrt{(7.5-5{)}^2+(7.275{)}^2}\Rightarrow l=18.187cm$