Determine the time period of small oscillations for a metre stick...

Length of the metre stick, $L = 1 \text{ m}$

Distance of the center of mass from the point of suspension, $d = 50 \text{ cm} - 20 \text{ cm} = 30 \text{ cm} = 0.3 \text{ m}$

Moment of inertia of the metre stick about the point of suspension is given by the parallel axis theorem:

$I = I_{CM} + M d^2 = \dfrac{M L^2}{12} + M d^2$

$I = M \left( \dfrac{1^2}{12} + (0.3)^2 \right) = M \left( \dfrac{1}{12} + \dfrac{9}{100} \right) = M \left( \dfrac{25 + 27}{300} \right) = \dfrac{52 M}{300} = \dfrac{13 M}{75}$

The time period of a physical pendulum is:

$T = 2\pi \sqrt{\dfrac{I}{M g d}}$

Substituting the values and taking $g = 10 \text{ m/s}^2$:

$T = 2\pi \sqrt{\dfrac{\dfrac{13 M}{75}}{M \times 10 \times 0.3}} = 2\pi \sqrt{\dfrac{13}{75 \times 3}} = 2\pi \sqrt{\dfrac{13}{225}} = \dfrac{2\pi \sqrt{13}}{15}$

$T \approx \dfrac{2 \times 3.14 \times 3.605}{15} \approx 1.51 \text{ s}$