Find the period of a simple pendulum of length $1$ m. (Take $g = 10$ m s$^{-2}$.)

$T = 2\pi\sqrt{\dfrac{l}{g}} = 2\pi\sqrt{0.1} = 2\pi \times 0.316 \approx 2.0$ s. Omitting the square root gives $2\pi \times 0.1 = 0.63$, and omitting the $2\pi$ gives about 0.32.