In a simple pendulum experiment, the maximum | vedclass

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Question

Time period $T\, = \,\,2\pi \sqrt {\frac{1}{g}} \,\,\, = \,2\pi {\left( {\frac{1}{g}} \right)^{\frac{1}{2}}}\,\, = 2\pi \frac{{{{({1})}^{\frac{1}

Vedclass · Accepted Answer

$\pm 3 \%$

Vedclass · Answer

Time period $T\, = \,\,2\pi \sqrt {\frac{1}{g}} \,\,\, = \,2\pi {\left( {\frac{1}{g}} ight)^{\frac{1}{2}}}\,\, = 2\pi \frac{{{{({1})}^{\frac{1}{2}}}}}{{{{(g)}^{\frac{1}{2}}}}}$

$\left( {\frac{{\Delta T}}{t}\, 	imes \,100} ight)\,\% \, =  \pm \left[ {\frac{1}{2} 	imes \frac{{\Delta {l}}}{{l}}\, 	imes \,100 + \,\frac{1}{{2\,}} 	imes \,\frac{{\Delta g}}{g}\, 	imes \,100} ight]$

$ =  \pm \left[ {1\,\% \, + \,2\,\% } ight]\,$

$	herefore T\,\, =  \pm \,\,3\,\% $

In a simple pendulum experiment, the maximum percentage error in the measurement of length is $2\%$ and that in acceleration due to gravity $g$ is $4\%$. Then the maximum percentage error in determination of the time-period is

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