Question:
In a simple pendulum experiment for determination of acceleration due to gravity $(g)$, time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be $30 \mathrm{~s}$. The length of pendulum is measured by using a meter scale of least count $1 \mathrm{~mm}$ and the value obtained is $55.0 \mathrm{~cm}$. The percentage error in the determination of $g$ is close to :
Correct Option: 4
Solution:
(4) We have
$T=2 \pi \sqrt{\frac{\ell}{g}}$
or $g=4 \pi^{2} \frac{\ell}{T^{2}}$
$\frac{\Delta g}{g} \times 100=\frac{\Delta R}{Q} \times 100+2 \frac{\Delta T}{T} \times 100$
$=\frac{0.1}{55} \times 100+2\left(\frac{1}{30}\right) \times 100$
$=0.18+6.67=68 \%$