Answer
Period as a Simple pendulum $= 1.87\space s$
Period as a physical pendulum $= 1.88\space s$
Error $=0.53\%$
Work Step by Step
Here we use the equation $T=2\pi \sqrt {\frac{L}{g}}$, where T - oscillation period, L - distance from the pivot to center of mass of the ball, g - gravitational acceleration ($9.8\space m/s^{2}$)
$T=2\pi \sqrt {\frac{L}{g}}$ Let's plug known values into this equation.
$T=2\pi \sqrt {\frac{(7.5+80)\times10^{-2}m}{9.8\space m/s^{2}}}=1.87s$
For a physical pendulum $T=2\pi \sqrt {\frac{I}{mgL}}$ , where T - oscillation period, I - Rotational inertia of the system through the pivot, L - distance from the pivot to center of mass of the ball, g - gravitational acceleration ($9.8\space m/s^{2}$)
$T=2\pi \sqrt {\frac{I}{mgL}}-(1)$
According to the parallel axis theorem we can write,
$I=I_{cm}+mL^{2}$ (Please see the attached image)
$I=\frac{2}{5}mR^{2}+mL^{2}= m(\frac{2R^{2}}{5}+L^{2})-(2)$
(2)=>(1)
$T=2\pi\sqrt {\frac{m(\frac{2}{5}R^{2}+L^{2})}{mgL}}=2\pi\sqrt {\frac{(\frac{2}{5}R^{2}+L^{2})}{gL}}$
$T=2\pi \sqrt {\frac{\frac{2}{5}\times(0.075m)^{2}+(0.8m)^{2}}{9.8m/s^{2}\times(0.8+0.075)m}}=1.88\space s$
$Error = \frac{(1.88-1.87)}{1.88}\times100\%=0.53\%$
