The volume of a cylinder is given by the formula $V = \pi r^2 h$, where $r$ is the radius of the cylinder and $h$ is the height (or length) of the cylinder.
Since the inside diameter of the larger pipe is 10'', the radius is 5''.
The volume of the larger pipe is then $V_1 = \pi (5^2) h$.
The inside diameter of the smaller pipe is 2'', so the radius is 1''.
The volume of the smaller pipe is $V_2 = \pi (1^2) h$.
To carry the same volume of water, we need to find the ratio of $V_1$ to $V_2$.
\[\frac{V_1}{V_2} = \frac{\pi (5^2) h}{\pi (1^2) h} = \frac{25}{1} = 25.\]
Therefore, we need $\boxed{25}$ 2''-inside-diameter pipes of the same length to carry the same volume of water as the larger pipe.
The answer is: 25
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
A pipe with inside diameter 10'' is to carry water from a reservoir to a small town in an arid land. Neglecting the friction and turbulence of the water against the inside of the pipes, what is the minimum number of 2''-inside-diameter pipes of the same length needed to carry the same volume of water to the arid town?

### Answer:
