Question

The time period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$ . The following is known from experiments, $l=(20.0\pm 0.1)cm$ and $T=(0.90\pm 0.01)s$ . The percentage error in the measurement of the acceleration due to gravity is

• A. $5\%$
• B. $4\%$
• C. $3\%$
• D. $1\%$

Answer 1

Answer: (C)

$\therefore T=2\pi \sqrt{\frac{l}{g}}\Rightarrow g=4{\pi }^2\frac{l}{T^2}$
$\therefore$ Error in g can be calculated as
$\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}$
$\therefore$ Total time for n oscillation is $t=nT$ where T= time for oscillation.
$\Rightarrow \frac{\Delta t}{t}=\frac{\Delta T}{T}$
$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta t}{t}$
Given that $\Delta l=1mm=10^{-3}m,l=20\times 10^{-2}m$
$\Delta t=1s,t=90s$
$\%erroringis$
$\frac{\Delta g}{g}\times 100=\left(\frac{\Delta l}{l}+\frac{2\Delta t}{t}\right)\times 100$
$=\left(\frac{{\left(10\right)}^{-3}}{20\times {\left(10\right)}^{-2}}+\frac{2\times 1}{90}\right)\times 100$ $=\frac{1}{2}+\frac{20}{9}$ $=0.5+2.22$ $=2.72\%$
$\cong 3\%$