Answer
$$0.805\;m$$
Work Step by Step
For a physical pendulum, time period is given by
$T=2\pi \sqrt {\frac{I}{mgh}}$
where, $m$ is the mass of the pendulum. $I$ is the rotational inertia of the pendulum about the pivot point, and $h$ is the distance of center of mass from the pivot point.
According to the given figure,
The rotational inertia for an axis through A is $I_A = I_{cm} + mh^2_A$
and for an axis through B is $I_B = I_{cm} + mh^2_B$
Now,
$T=T_A=T_B$
or, $2\pi \sqrt {\frac{I_{cm} + mh^2_A}{mgh_A}}=2\pi \sqrt {\frac{I_{cm} + mh^2_B}{mgh_B}}$
or, $\frac{I_{cm} + mh^2_A}{h_A}=\frac{I_{cm} + mh^2_B}{h_B}$
or, $I_{cm}(h_B-h_A)=mh_Ah_B(h_B-h_A)$
or, $I_{cm}=mh_Ah_B$
Thus,
$T=2\pi \sqrt {\frac{mh_Ah_B + mh^2_A}{mgh_A}}$
or, $T=2\pi \sqrt {\frac{h_B + h_A}{g}}$
or, $T=2\pi \sqrt {\frac{L}{g}}$
or, $L=\frac{T^2g}{4\pi^2}$
Substituting the given values
$L=\frac{(1.80)^2\times9.81}{4\times\pi^2}\;m$
or, $\boxed{L=0.805\;m}$
Therefore, the distance $L$ between A and B is $0.805\;m$
