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The period of oscillation of a simple pendulum is given by $T = 2\pi \sqrt {\frac{l}{g}} $ where $l$ is about $100 \,cm$ and is known to have $1\,mm$ accuracy. The period is about $2\,s$. The time of $100$ oscillations is measured by a stop watch of least count $0.1\, s$. The percentage error in $g$ is ......... $\%$
A
$0.1$
B
$1$
C
$0.2$
D
$0.8$
Solution
(c) $T = 2\pi \sqrt {l/g} $ $ \Rightarrow {T^2} = 4{\pi ^2}l/g$
$ \Rightarrow g = \frac{{4{\pi ^2}l}}{{{T^2}}}$
$Here \,\% \,error \,in \,l = \frac{{1mm}}{{100cm}} \times 100 = \frac{{0.1}}{{100}} \times 100 = 0.1\%$ and $\%\, error\, in\, T =\frac{{0.1}}{{2 \times 100}} \times 100 = 0.05\% $
$\% \,error\, in\, g = \% \,error\, in \,l + 2(\%\, error\, in \,T)$
$ = 0.1 + 2 \times 0.05= 0.2 \%$
Standard 11
Physics
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