Find the length of a seconds' pendulum at a place where g = 10 ms^-2 (Take π = 3.14).

As we know,

$T = 2 π \sqrt{\dfrac{l}{g}}$

Given,

g = 10 ms^-2

π = 3.14

T = 2 s

Substituting the values in the formula above we get,

$2 = 2 \times 3.14 \sqrt{\dfrac{l}{10}} \\[0.5em] \Rightarrow (\dfrac{2}{2 \times 3.14}) = \sqrt{\dfrac{l}{10}} \\[0.5em] \Rightarrow (\dfrac{2}{2 \times 3.14})^2 = \dfrac{l}{10} \\[0.5em] \Rightarrow (\dfrac{1}{3.14})^2 = \dfrac{l}{10} \\[0.5em] \Rightarrow (0.3184)^2 = \dfrac{l}{10} \\[0.5em] \Rightarrow 0.10142 = \dfrac{l}{10} \\[0.5em] \Rightarrow l = 1.0142 \\[0.5em]$

Hence, length of a seconds’ pendulum = 1.0142 m