Answer
3.065
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,
$I=\frac{ml^{2}}{12}+mx^{2}$
Therefore,
$T=2π\sqrt \frac{\frac{ml^{2}}{12}+mx^{2}}{mgx}$
For getting minimum Time Period,
$\frac{dT}{dx}=0$
Therefore,
$x=\frac{l}{\sqrt 12}$
$x=\frac{2.20}{\sqrt 12}$
$x=0.1833$
Therefore,
$I=\frac{ml^{2}}{12}+mx^{2}$
$I=\frac{m(2.20)^{2}}{12}+m(0.1833)^{2}$
$I=m(0.4368)$
Therefore,
$(T)=2π\sqrt \frac{(0.4368)}{g(0.1833)}$
$(T)=2π\sqrt(0.2382)$
$(T)=2π(0.4880)$
$(T)=3.06464$
$$(T)\approx3.065$$
