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The percentage error in the measurement of g is: (Given that $g=\frac{4\pi^2L}{T^2}$, $L=(10\pm0.1)$ cm, $T=(100\pm1)$ s)
Detailed Solution
$\frac{\Delta g}{g}\times100=\frac{\Delta L}{L}\times100+2\frac{\Delta T}{T}\times100=\left(\frac{0.1}{10}\times100\right)+2\left(\frac{1}{100}\times100\right)=1+2=3\%$.