Question

In order to determine acceleration due to gravity $'g'$ , a simple pendulum experiment is performed, the time taken for $20$ oscillations is measured by using a watch with least count $1$ second. The length of pendulum is measured by meter scale with least count 1 mm. After the experiment, the mean value of time taken comes out to be $30s$ and the length of pendulum obtained is $55cm$ . The percentage error in the determination of 'g' is approximately equal to:

• A. $0.2\%$
• B. $0.7\%$
• C. $6.8\%$
• D. $3.5\%$

Answer 1

Answer: (C)

$g=\frac{4 \pi ^{2} \ell }{T^{2}}$
$\frac{\Delta g}{g}=\frac{\Delta \ell }{\ell }+2\frac{\Delta T}{T}$ $\frac{\Delta g}{g}=\frac{0 . 1}{55}+2\frac{1}{30}$
$\frac{\Delta g}{g}\times 100=\left(\right.\frac{10}{55}+\frac{20}{3}\left.\right)\times 100=6.8\%$