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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