A hemispherical cup of radius 4 cm is filled to the brim with coffee. The coffee is then poured into a vertical cone of radius 8 cm and height 16 cm. The percentage of the volume of the cone that remains empty is:

Solution (By Examveda Team)

$$\eqalign{ & {\text{Volume of coffee}} = \frac{2}{3}\pi {r^3} \cr & = \frac{2}{3} \times \frac{{22}}{7} \times {\left( 4 \right)^3} \cr & = \frac{{128}}{3}\pi {\text{ c}}{{\text{m}}^3} \cr & {\text{Volume of cone}} = \frac{1}{3}\pi {r^2} \times h \cr & = \frac{1}{3}\pi {\left( 8 \right)^2} \times 16 \cr & = \frac{{1024}}{3}\pi \cr & \therefore {\text{Required percentage}} = \frac{{\frac{{1024}}{3} - \frac{{128}}{3}}}{{\frac{{1024}}{3}}} \times 100 \cr & = \frac{{896}}{{1024}} \times 100 \cr & = 87.5\% \cr} $$