If the area of the base of a coneis increased then it becomes 1.96 times oforiginal area. Its volume is increased by:

Let R be the Radius.

Area of the base of cone $$= \pi \times R^2$$

According to question,

Area of the base of a cone is increased then it becomes 1.96 times of original area,

$$\frac{New Area}{Old Area} = \frac{196}{100} $$

$$\frac{R_{New}}{R_{Old}}=\frac{14}{10}$$

Volume of a cone $$= \pi \times R^2 \times H$$ (Height of cone=H)

$$\frac{New Volume}{Old Volume} = \frac{\pi \times R_{New}^2 \times H}{\pi \times R_{Old}^2 \times H} = \frac{196}{100} = 96\%$$

Option C is correct.