The height of a circular cylinder is increased six times and the base area is decreased to one ninth of its value. The factor by which the lateral surface of the cylinder increases is

Decrease in base radius = (Decrease in base area)$$^{\frac{1}{2}} = {\left( {\frac{1}{9}} \right)^{\frac{1}{2}}} = \frac{1}{3}$$
Let initial radius and height be 3r and h
∴ New radius and height are r and 6h
Old lateral surface area = 2 × π × 3r × h = 6πrh
New lateral surface area = 2 × π × r × 6h = 12πrh
Required factor $$ = \frac{{12\pi rh}}{{6\pi rh}} = 2$$