Answer
6.78 s
Work Step by Step
For a physical pendulum $T=2\pi\sqrt {\frac{I}{mgL}}$ ; Where, T - period, I -Rotational inertia of the system through the pivot, m - mass of the system, L - distance from the pivot to the center of gravity.
$T=2\pi\sqrt {\frac{I}{mgL}}-(1)$
According to the parallel axis theorem we can write,
$I=I_{cm}+mL^{2}$ (Please see the attached image)
$I=\frac{2}{5}mR^{2}+mL^{2}=m(\frac{2}{5}R^{2}+L^{2})-(2)$
$(2)=>(1)$
$T=2\pi\sqrt {\frac{m(\frac{2}{5}R^{2}+L^{2})}{mgL}}=2\pi\sqrt {\frac{2\frac{R^{2}}{5}+L^{2}}{gL}}$
Let's plug known values into this equation.
$T=2\pi\sqrt {\frac{2\frac{(2.745m)^{2}}{5}+(8.4m+2.745m)^{2}}{9.8m/s^{2}\times(8.4m+2.745m)}}= 6.78\space s$
