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 ?
Options:
A . 1 : 2
B . 1 : 4
C . 1 : 8
D . 8 : 1
Answer: Option B Let original radius = R Then, new radius = $$\frac{{\text{R}}}{2}$$ $$\eqalign{ & \therefore \frac{{{\text{Volume of reduced cylinder }}}}{{{\text{Volume of original cylinder}}}} \cr & = \frac{{\pi \times {{\left( {\frac{R}{2}} \right)}^2} \times h}}{{\pi \times {R^2} \times h}} \cr & = \frac{1}{4}\,Or\,1:4 \cr} $$
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