Question

An experiment is performed to obtain the value of acceleration due to gravity $g$ by using a simple pendulum of length $L$. In this experiment time for $100$ oscillations is measured by using a watch of $1$ second least count and the value is $90.0$ seconds. The length $L$ is measured by using a meter scale of least count $1\, mm$ and the value is $20.0\, cm$. The error in the determination of $g$ would be :

• A. 1.7 %
• B. 2.7 %
• C. 4.4 %
• D. 2.27 %

Answer 1

Answer: (B)

$T^{2}=\frac{4\pi^{2}\ell}{g}$
$g=4\pi^{2} \frac{\ell}{T^{2}}$
$\frac{\Delta g}{g}\times100=\left(\frac{\Delta\ell}{\ell}\times 100\right)+2\left(\frac{\Delta T}{T}\times 100\right)$
$=\left(\frac{0.1}{20}\times 100\right)+2\left(\frac{0.01}{.9}\times 100\right)$
$=0.5+2\times \frac{10}{9}=0.5+2.2=2.7\%$