The percentage error in the measurement of $g$ is $\left(\right.$ Given that $g=\frac{4 \pi^2 \mathrm{~L}}{\mathrm{~T}^2}, \mathrm{~L}=(10 \pm 0.1) \mathrm{cm}$, $\mathrm{T}=(100 \pm 1) \mathrm{s}):$

Question

Aicharya · Accepted Answer

Given:
$\begin{aligned}
g & =\frac{4 \pi^2 \mathrm{~L}}{\mathrm{~T}^2} \\
\mathrm{~L} & =(10 \pm 0.1) \mathrm{cm}, \text { and } \\
\mathrm{T} & =(100 \pm 1) \mathrm{s}
\end{aligned}$
Hence, error in measurement of $g$ is
$\begin{aligned}
\frac{\Delta g}{g} & =\frac{\Delta \mathrm{L}}{\mathrm{L}}+2 \frac{\Delta \mathrm{T}}{\mathrm{T}} \\
& =\frac{0.1}{10} \times 100+2 \times \frac{1}{100} \times 100 \\
& =3 \%
\end{aligned}$

The percentage error in the measurement of $g$ is $\left(\right.$ Given that $g=\frac{4 \pi^2…

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