Question

For a simple pendulum values are given as length of pendulum $l=(50\pm 0.1)$cm
Period of oscillation $T=(2\pm 0.1)$s
The maximum permissible error in the measurement of acceleration due to gravity g will be nearly.

• A. $\pm 1m/s^2$
• B. $\pm 2m/s^2$
• C. $\pm 3.5m/s^2$
• D. $\pm 0.4m/s^2$

Answer 1

Answer: (A)

$\pm 1m/s^2$

The period of oscillation of simple pendulum is given as
$T=2\pi \sqrt{\frac{l}{g}}$
So, $g=\frac{4{\pi }^{2}l}{T^2}$
$\therefore \frac{\Delta g}{g}=\pm \left(\frac{\Delta l}{l}+\frac{2\Delta T}{T}\right)$
$=\pm \left(\frac{0.1}{50}+2\times \frac{0.1}{2}\right)$
$=\pm (0.0002+0.1)=\pm 0.1002$
$\therefore \Delta g=\pm 0.1002\times 9.8$
$=\pm 0.98196=\pm 1m/s^2$.