Given radius r = 14 cm and height h = 5 cm.

Given radius $r = 14$ cm and height $h = 5$ cm.

$$V_{initial} = \pi (14)^2 (5) = \pi (196)(5) = 980\pi \text{ cm}^3$$

New radius $r' = (14+x)$ cm, height remains $h = 5$ cm.
New volume $V_r = \pi (14+x)^2 (5)$
Increase in volume $\Delta V_r = V_r - V_{initial}$

$$\Delta V_r = 5\pi (14+x)^2 - 980\pi$$ $$\Delta V_r = 5\pi [(14+x)^2 - 196]$$ $$\Delta V_r = 5\pi [196 + 28x + x^2 - 196]$$ $$\Delta V_r = 5\pi (x^2 + 28x)$$

Radius remains $r = 14$ cm, new height $h' = (5+x)$ cm.
New volume $V_h = \pi (14)^2 (5+x)$
Increase in volume $\Delta V_h = V_h - V_{initial}$

$$\Delta V_h = 196\pi (5+x) - 980\pi$$ $$\Delta V_h = 980\pi + 196\pi x - 980\pi$$ $$\Delta V_h = 196\pi x$$

The problem states that the increase in volume is the same in either case.

$$\Delta V_r = \Delta V_h$$ $$5\pi (x^2 + 28x) = 196\pi x$$

Divide both sides by $\pi$ (since $\pi \neq 0$):

$$5(x^2 + 28x) = 196x$$ $$5x^2 + 140x = 196x$$ $$5x^2 + 140x - 196x = 0$$ $$5x^2 - 56x = 0$$

Factor out $x$:

$$x(5x - 56) = 0$$

This gives two possible solutions: $x=0$ or $5x-56=0$.
Since the increase in volume is "not zero", $x$ cannot be 0.
Therefore, we take the second solution:

$$5x - 56 = 0$$ $$5x = 56$$ $$x = \frac{56}{5}$$ $$x = 11.2 \text{ cm}$$

Comparing this with the given options, the correct option is (D).

The final answer is $\boxed{\text{D) 11.2 cm}}$.