In a simple pendulum experiment, the maximum percentage error in the measurement of length is 2% and that in the observation of the time period is 3%. Then the maximum percentage error in determination of the acceleration due to gravity g is

Answer (4)

Given,

$\displaystyle \frac {\Delta L} {L}\times 100=2\%,\frac {\Delta T} {T}\times 100=3\%$

Time period, $\displaystyle T=2\pi\sqrt{\frac {L} {g} }$ ,

$\displaystyle \Rightarrow g=\frac {4\pi^2L} {T^2}$

The maximum percentage error in g,

$\displaystyle \begin{aligned}\frac {\Delta g} {g}\times 100 &= \frac {\Delta L} {L}\times 100+2\times \left(\frac {\Delta T} {T}\times 100\right)\\&=2\%+2(3\%)=8\%\end{aligned}$