To estimate $g$ from $g=4 \pi^2 \frac{L}{T^2}$, error in measurement of $L$ is $\pm 2 \%$ and error in measurement of $T$ is $\pm 3 \%$. The error in estimated $g$ will be

Question

To estimate $g$ from $g=4 \pi^2 \frac{L}{T^2}$, error in measurement of $L$ is $\pm 2 \%$ and error in measurement of $T$ is $\pm 3 \%$. The error in estimated $g$ will be, $\pm 8 \%$, $\pm 5 \%$, $\pm 3 \%$, $\pm 6 \%$

MARKS App · Accepted Answer

$\because g=4 \pi^2 \frac{L}{T^2}$ Percentage error in the estimation of $g$ is given as, $\begin{aligned} \frac{\Delta g}{g} \times 100 & = \pm\left(\frac{\Delta L}{L} \times 100+2 \frac{\Delta T}{T} \times 100\right) \ & = \pm(2 \%+2 \times 3 \%) \ & = \pm(2 \%+6 \%)= \pm 8 \% \end{aligned}$

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