A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. The radius of the base of the cone is :

Solution (By Examveda Team)

Let the radius of the cone be R cm
Then,
$$\eqalign{ & \frac{1}{3}\pi \times {R^2} \times 75 = \frac{2}{3}\pi \times 6 \times 6 \times 6 \cr & \Rightarrow {R^2} = \left( {\frac{{2 \times 6 \times 6 \times 6}}{{75}}} \right) \cr & \Rightarrow {R^2} = \frac{{144}}{{25}} \cr & \Rightarrow {R^2} = \frac{{{{\left( {12} \right)}^2}}}{{{{\left( 5 \right)}^2}}} \cr & \Rightarrow R = \frac{{12}}{5} \cr & \Rightarrow R = 2.4\,cm \cr} $$