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Q. For a simple pendulum, the graph between $ {{T}^{2}} $ and $ L $ is

KEAMKEAM 2007Oscillations

Solution:

$ T=2\pi \sqrt{\frac{L}{g}} $
Or $ T\propto \sqrt{L} $
or $ {{T}^{2}}\propto L $
It is linear relation between $ {{T}^{2}} $ and $ l $ hence the graph between $ {{T}^{2}} $ and L is a straight line passing through the origin.

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