Answer
0.53
Work Step by Step
Time Period of Physical Pendulum $$(T) =2π\sqrt \frac{I}{mgx}$$
Where $I=$ Moment of Inertia of Physical Pendulum, $m=$ Mass of Body, $g=$ Gravitational Acceleration, $x=$ Distance Between Point of Suspension and Center of Mass of Body
At Point of Suspension $I=\frac{ml^{2}}{12}+mx^{2}$ , ($l=$ Length of Rod)
By Putting Value of $I$
$$(T) =2π\sqrt \frac{\frac{ml^{2}}{12}+mx^{2}}{mgx}$$
For getting minimum Time Period,
$\frac{dT}{dx}=0$
$\frac{dT}{dx}=\frac{π}{\sqrt g}\sqrt \frac{x}{x^{2}+\frac{l^{2}}{12}}\frac{(2x^{2}-x^{2}-\frac{l^{2}}{12})}{x^{2}}=0$
Therefore, $x=\frac{l}{\sqrt 12}$
$x=\frac{1.85}{\sqrt 12}$
$x=0.534\approx0.53$
