Gujarati
13.Oscillations
easy

The acceleration due to gravity at a place is ${\pi ^2}\,m/se{c^2}$. Then the time period of a simple pendulum of length one metre is

A

$\frac{2}{\pi }\,sec$

B

$2\pi \,sec$

C

$2\,sec$

D

$\pi \,sec$

Solution

$T = 2\pi \sqrt {\frac{l}{g}}  = 2\pi \sqrt {\frac{1}{{{\pi ^2}}}} = 2 \,sec$

Standard 11
Physics

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