Answer: Option B
Let the radius of a right circular cine be R cm and height be H cm
Volume of right circular cone $$ = \frac{1}{3}\pi {R^2}H{\text{ cu}}{\text{.cm}}$$
When height of right circular cone is increased by 200% and radius of the base is reduce by 50%
New volume :$$\eqalign{
& {\text{ = }}\frac{1}{3}\pi {\left( {\frac{R}{2}} \right)^2}.3H \cr
& = \frac{1}{3}\pi \frac{{{R^2}4}}{4}.3H \cr
& = \frac{{\pi {R^2}H}}{4} \cr} $$
Difference :
$$\eqalign{
& = \pi {R^2}H\left( {\frac{1}{3} - \frac{1}{4}} \right) \cr
& = \frac{1}{{12}}\pi {R^2}H \cr} $$
Decrease percentage :
$$\eqalign{
& = \frac{{\frac{1}{{12}}\pi {R^2}H}}{{\frac{1}{3}\pi {R^2}H}} \times 100 \cr
& = 25\% \cr} $$