The diameter of a right circular cylinder is decreased to one third of its initial value. If the volume of the cylinder remains the same, then the height becomes how many times of the initial height?

Let for the initial cylinder 

Height = H1

Volume =V1

Diameter =D1 

Volume of the cylinder  = $$\pi \times R^2 \times H$$

Let for the final cylinder

Height = H2

Volume =V2

Diameter =D2

ATQ

D2 = $$\frac{1}{3}\times D1$$

V1= V2 =V

We know that, D = 2R

$$ \pi \times (\frac{D1}{2})^2 \times H1$$ =$$\pi \times (\frac{D1}{6})^2 \times H2$$

$$\pi \times D1 \times H1$$ =$$\pi \times (\frac{D1}{3})^2 \times H2$$

H2 = $$9\times H1$$