Question

The altitude of a circualr cylinder is increased six times and the base area is decreases to one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is

(a) $\frac{2}{3}$ (b) $\frac{1}{2}$

(c) $\frac{3}{2}$ (d) 2

Answer 1

In first case,

Let r be the radius and h be the height of the cylinder. Then,

$\therefore$ Lateral surface area = $2\pi$ rh

In second case,

Base area $=\frac{1}{9}\pi r_1^2$

$\therefore Radius\left(R\right)\sqrt{\frac{\frac{1}{9}\pi r^2}{\pi }}=\sqrt{\frac{1}{9}{r}^2}=\frac{1}{3}r$

and height (H) = 6 h

$\therefore$ Lateral surface area

= $2\pi$ RH =$2\times \pi \times \frac{1}{3}r\times$ 6h

= $4\pi$ rh

= $2\left(2\pi rh\right)$

$\therefore$ Increase = 2 times