Question

In an experiment to determine the value of acceleration due to gravity, $g$, using a simple pendulum, the measured value of length of the pendulum is $31.4cm$ known to $1mm$ accuracy and the time period for $100$ oscillations of pendulum is $112.0s$ known to $0.01s$ accuracy. The accuracy in determining the value of $g$ is :

• A. $\Delta g=(987.2\pm 3.2$ %) $cm/s^2$
• B. $\Delta g=(98.72\pm 3.2$ %) $cm/s^2$
• C. $\Delta g=(9.872\pm 3.2$ %) $cm/s^2$
• D. $\Delta g=(987.2\pm 32$ %) $cm/s^2$

Answer 1

The correct option is A $\Delta g=(987.2\pm 3.2$ %) $cm/s^2$

$g=\frac{4{\pi }^{2}l}{T^2}=\frac{4{\pi }^2\times 31.4}{(\frac{112}{100}{)}^2}=987.2cm/s^2$

Now, $\frac{\Delta g}{g}\times 100\%=(\frac{\Delta l}{l}+\frac{2\Delta T}{T})\times 100\%$

$=(\frac{0.1}{31.4}+\frac{2\times 0.01}{112})\times 100\%$

$=3.2\%$

$\therefore$ Accuracy$=g\pm \Delta g=(987.2\pm 3.2$ %) $cm/s^2$