Question

In a simple pendulum experiment for determination of acceleration due to gravity $\left(\right.g\left.\right)$ , time taken for $20$ oscillations is measures by using a watch of $1$ second least count. The mean value of time taken comes out to be $30 \, s$ . The length of the pendulum is measured by using a meter scale of least count $1 \, mm$ and the value obtained is $55.0 \, cm$ . The percentage error in the determination of $g$ is close to

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

Answer 1

Answer: (B)

$g=\frac{4 \pi ^{2} l}{T^{2}}$
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2 \Delta T}{T}$
$=\frac{0.1}{55}+2\times \left(\frac{1}{30}\right)$
$=0.06848$
Percentage error $\approx6.8\%$