Solution:

The time period of a simple pendulum is \(T = 2\pi \sqrt{\frac{l}{g}}\). For a second's pendulum, \(T = 2\text{ s}\). Thus, \(2 = 2\pi \sqrt{\frac{l}{10}} ⇒ 1 = \pi^2 \frac{l}{10}\). Since \(\pi^2 \approx 10\), we find \(l \approx 1 \text{ m}\).