Question

Calculate the time period of a simple pendulum of length $1.12m$, when acceleration due to gravity is $9.8m{s}^{-2}$.

• A. 0
• B. 1.2
• C. 2.12
• D. 4.24

Answer 1

The correct option is C 2.12

$l=1.12m$; $g=9.8m{s}^{-2}$; $T=?$
$T=2\pi \sqrt{\frac{l}{g}}=\frac{2\times 22}{7}\sqrt{\frac{1.12}{9.8}}=\frac{44}{7}\sqrt{0.114}$$=\frac{44}{7}\times 0.338=2.12s$