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Choose the correct length $$(L)$$ versus square of time period $$(T_2)$$ graph for a simple pendulum executing simple harmonic motion
From simple pendulum relation
$$T=2\pi\sqrt{\frac{L}{g}}$$
Squaring,
$$T^2=\frac{4\pi^2L}{g}$$
Rearrange:
$$\frac{1}{T^2}=\frac{g}{4\pi^2L}$$
So
$$T^2\propto L$$
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