Question

In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillations with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy, the accuracy in the determination of g is found to be x %. The value of x is _________ .

Answer 1

Time period is given as $T=2π\sqrt\frac{l}{g}$
$\frac{dg}{g}×100=\frac{2dT}{T}×100+\frac{dl}{l}×100$

$=2×\frac{1}{50}×100+\frac{1}{100}×100= 5$ %