If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one ?

Solution (By Examveda Team)

Let original radius = R
Then, new radius = $$\frac{{\text{R}}}{2}$$
$$\eqalign{ & \therefore \frac{{{\text{Volume of reduced cylinder }}}}{{{\text{Volume of original cylinder}}}} \cr & = \frac{{\pi \times {{\left( {\frac{R}{2}} \right)}^2} \times h}}{{\pi \times {R^2} \times h}} \cr & = \frac{1}{4}\,Or\,1:4 \cr} $$