Find the length of a seconds pendulum at a place where g = 10 m s $^{- 2}$ (Take $π$ =3.14).

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Solution

$T = 2 π \sqrt{\frac{l}{g}}$
$2 = 2 π \sqrt{\frac{l}{g}}$
$l = \frac{g}{π^{2}} = \frac{10}{{3.14}^{2}} = 1.01 m$

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Find the length of a seconds pendulum at a place where g = 10 m s−2 (Take π =3.14).

T=2π√lg2=2π√lgl=gπ2=103.142=1.01m

Find the length of a seconds pendulum at a place where g = 10 m s $^{- 2}$ (Take $π$ =3.14).

$T = 2 π \sqrt{\frac{l}{g}}$
$2 = 2 π \sqrt{\frac{l}{g}}$
$l = \frac{g}{π^{2}} = \frac{10}{{3.14}^{2}} = 1.01 m$