Question

To estimate $‘g’$(from $g = 4\pi^{2}\frac{L}{T^{2}})$ error in measurement of $L$ is $\pm2\%$ and error in measurement of $T$ is $\pm3\%$. The error in estimated $‘g'$ will be-

• A. $\pm8\%$
• B. $\pm6\%$
• C. $\pm3\%$
• D. $\pm5\%$

Answer 1

Answer: (A)

$g=4\pi^{2}\frac{L}{T^{2}}$
$\frac{\Delta L}{L}=\pm2\%=\pm2\times10^{-2}$
$\frac{\Delta T}{T}=\pm3\%=\pm3\times10^{-2}$
$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta L}{L}+\frac{2\Delta T}{T}$
$=2\times10^{-2}+2\times3\times10^{-2}$
$=8\times10^{-2}=\pm8\%$