$g=\frac{1}{4 \pi^{2}} \frac{T^{2}}{\ell}$
$\frac{\Delta g}{g}=\frac{2 \Delta T }{ T }+\frac{\Delta \ell}{\ell}$
$\frac{\Delta g}{g}=2 \cdot \frac{1}{100 \times 0.5}+\frac{1\,mm }{10\,cm }$
$\frac{\Delta g}{g}=\frac{5}{100}$
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[Take $\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}, c =3 \times 10^{8}$ $ms ^{-1}$ ]

A block is projected from the Point $A$ of inclined plane $A B$ along its surface with a velocity just sufficient to carry it to the top Point $B$ at a height $10 m$. After reaching the Point $B$ the block slides down on inclined plane $BC$. Time it takes to reach to the point $C$ from point $A$ is $t (\sqrt{2}+1) s$. The value of $t$ is........(use $g =10 m / s ^{2}$ )