Since the base of cylinder is a circle, the perimeter of the base of the right circular cylinder i.e. 22 feet is same as $\(2 \pi \mathrm{r}\)$, where r is the radius of the circular base of the cylinder. So, we get $\(2 \pi r=22 \Rightarrow r=\frac{7}{2}\)$

Now, the volume of the cylinder is $\(\pi r^2 h=\frac{22}{7} \times\left(\frac{7}{2}\right)^2 \times 2=77\)$

We know that the volume of the cylinder is to be raised to at least $\(90 \%\)$ from $\(80 \%\)$ i.e. it is to be increased by at least $\(10 \%\)$ of $\(77=7.7\)$

Let the number of lead balls of radius 0.10 feet each which to be dropped in the cylindrical tank so that the volume increases by at least 7.7 cubic feet be ' $\(n\)$ '

So, we get $\(n \times \frac{4}{3} \pi r^3=n \times \frac{4}{3} \pi(0.10)^3 \geq 7.7 \Rightarrow n \geq \frac{7.7 \times 3}{\pi \times(0.10)^3}=1837.5\)$ (Lead balls must be spherical in shape and the volume of Sphere is $\(\frac{4}{3} \pi r^3\)$ )

Thus, the minimum number of balls which need to be dropped in the cylinder is 1838 (Number of balls cannot be in decimal).

Hence the answer is (D).