Question

The time period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g}}$ . The following is known from experiments, $l=\left(\right.20.0\pm0.1\left.\right)cm$ and $T=\left(\right.0.90\pm0.01\left.\right)s$ . 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 $\Delta l=1 \, mm=10^{- 3} \, m, \, l=20\times 10^{- 2}m$
$\Delta t=1s,t=90s$
$\% \, error \, in \, g is$
$\frac{\Delta g}{g}\times 100=\left(\frac{\Delta l}{l} + \frac{2 \Delta t}{t}\right)\times 100$
$=\left(\frac{\left(10\right)^{- 3}}{20 \times \left(10\right)^{- 2}} + \frac{2 \times 1}{90}\right)\times 100$ $=\frac{1}{2}+\frac{20}{9}$ $=0.5+2.22$ $=2.72\%$
$\cong3\%$