Question

The period of oscillation of a simple pendulum is $T =2 \pi \sqrt{\frac{ L }{ g }}$. Measured value of 'L' is $1.0 m$ from meter scale having a minimum division of $1 mm$ and time of one complete oscillation is 1.95 s measured from stopwatch of $0.01 s$ resolution. The percentage error in the determination of 'g' will be :

• A. $1.13 \%$
• B. $1.03 \%$
• C. $1.33 \%$
• D. $1.30 \%$

Answer 1

Answer: (A)

$T =2 \pi \sqrt{\frac{\ell}{ g }}$
$g =\frac{4 \pi^{2} \ell}{ T ^{2}}$
$\frac{\Delta g }{ g }=\frac{\Delta \ell}{\ell}+\frac{2 \Delta T }{ T }$
$\frac{\Delta g }{ g }=\frac{1 \times 10^{-3}}{1}+2 \times \frac{0.01}{1.95}$
$\frac{\Delta g }{ g}=0.0113$ or $1.13 \%$