Answer
.147 percent
Work Step by Step
For a simple pendulum, we find:
$T = 2\pi \sqrt{\frac{L}{g}}$
$T = 2\pi \sqrt{\frac{.8}{9.81}}=1.79428\ s$
For a physical pendulum $\underline{with\ a\ solid \ sphere\ attached}$, we know the following equation for period:
$T = 2\pi \sqrt{\frac{\frac{2}{5}r^2+l^2}{gl}}$
$T = 2\pi \sqrt{\frac{\frac{2}{5}(.075)^2+(.8)^2}{9.81(.8)}}=1.80$
Using the equation for percent error, we find that the percent error is $\fbox{.147 percent}$
