Answer:
Explanation:
[tex]T = 2\pi \sqrt{(\frac{I}{mgh})}[/tex]
centre of mass will be in middle of meter stick
so we take the length h = 0.5m
Taking moment of inertia of rod about the top point
[tex]I = \frac{mL^2}{3}\\\\0.85 = 2\pi \sqrt {(\frac{mL^2}{3mgh})
}\\\\0.85 = 2\pi \sqrt{(\frac{1}{3g}(0.5))}
\\\\g = 36.43 m/s^2[/tex]