A solid hemisphere has radius 14 cm. It is melted to form a cylinder such that the ratio of its curved surface area and total surface area is 2 : 3. What is the radius (in cm) of its base?

Let say, radius of base of cylinder is r and height is h .

So, $$\frac{2\pi rh}{2\pi r\left(r+h\right)}=\frac{2}{3}\ .$$

or, $$\frac{h}{\left(r+h\right)}=\frac{2}{3}\ .$$

or, $$3h=2r+2h\ .$$

or, $$r=\frac{h}{2}\ .$$

According to question,

$$\pi r^2h=\frac{2}{3}\pi\left(14^3\right)\ .$$

or, $$\left(\frac{h}{2}\right)^2h=\frac{2}{3}\left(14^3\right)\ .$$

or,$$h^3=\frac{8}{3}\ 14^3\ .$$

or, $$h=28\sqrt[\ 3]{\frac{1}{3}}\ .$$

So, $$r=\frac{h}{2}=\frac{14}{\sqrt[\ 3]{3}}\ .$$

B is correct choice.