The period of oscillation of a simple pendulum is given by T=2$\pi√(\frac{l}{g})$ here l is about 100 cm and is known to have 1mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is
Answer & explanation
Correct answer: option 3
T=2$\pi√(\frac{l}{g})$
$T^2=\frac{4\pi^2l}{g}$
$\Rightarrow g = \frac{4\pi^2 l}{T^2}$
$\Rightarrow \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$
Error in l = $\frac{0.1cm}{100cm}\times100$%=0.1%
Error in T = $\frac{0.1cm}{2\times100cm}\times100$%=0.05%
% Error in g = % Error in l + 2$\times$ % Error in T = 0.1% + 2$\times$ 0.05% = 0.2%