A metallic hemispherical bowl is made up of steel. The total steel used in making the bowl is $$486\pi cm^3$$. The bowl can hold $$144\pi cm^3$$ water. What is the thickness (in $$cm$$) of bowl and the curved surface area (in $$cm^2$$) of outer side?

Volume of hemisphere = $$\left(\frac{2}{3}\pi\times r^3\right)\ .$$

So, $$\left(\frac{2}{3}\pi\times r_1^3\right)=\ 486\pi\ \ .$$

or, $$r_1^3=\ 486\times\ \frac{3}{2}\ \ =729.$$

or, $$r_1=9.$$

And, $$\frac{2}{3}\times\ \pi\ \times\ r_2^3=144\pi\ .$$

or, $$r_2^3=144\ \times\frac{3}{2}=216\ .$$

or, $$r_2=6\ .$$

So, Thickness of bowl = $$r_1-r_2=9-6=3\ .$$

So, Curved surface area = $$2\times\ \pi\ \times\ r_1^2=2\times\ \pi\ \times\ 9^2=162\pi\ .$$

So, B is correct choice.